Solutions Manual for Basic Engineering MathematicsDr. C. K. Chan, Dr. C. W. Chan and Dr. K. F. Hung August 17, 2007 Contents 1 Complex Numbers 1 2 Linear Algebra 13 3 Infinite series, Power series and Fourier series 41 4 Partial Differentiation 53 5 Multiple Integrals 83 6 Vector Calculus 125 7 Differential Equations 157 8 Laplace and Fourier Transformations 177 9 Partial Differential Equations 191 i Chapter 1 Complex Numbers (1) ¯ z = 2 −3i, [z[ = √ 2 2 + 3 2 = √ 13. (2) By definition, 1 z = _ a a 2 +b 2 _ +i _ −b a 2 +b 2 _ . Therefore, z 1 z = (a +ib) _ a a 2 +b 2 +i −b a 2 +b 2 _ = 1 + 0i = 1. (3) (a) 3 +i 2 −i = 3 +i 2 −i 2 +i 2 +i = (6 −1) +i(3 + 2) 2 2 + 1 2 = 5 + 5i 5 = 1 +i. (b) 2 + √ 3i 1 −4i = 2 + √ 3i 1 −4i 1 + 4i 1 + 4i == _ 2 −4 √ 3 _ + _ 8 + √ 3 _ i 17 (c) 1 −i 1 + √ 3i = 1 −i 1 + √ 3i 1 − √ 3i 1 − √ 3i = _ 1 − √ 3 _ − _ 1 + √ 3 _ i 4 (4) Let z = a + ib and w = c + id. We have (¯ z) = (a −ib) = a + ib = z. Since (z +w) = (a +c) +i(b +d) = a+c−i(b+d) and ¯ z+ ¯ w = (a−ib)+(c−id), we have (z +w) = ¯ z+ ¯ w. Moreover, observe that z w = (ac −bd) +i(ad +bc) = (ac −bd) −i(ad +bc) and ¯ z ¯ w = (a −ib)(c −id) = (ac −bd) +i(−bc −ad). We thus conclude that z w = ¯ z ¯ w. Finally, the last result follows from the identity [z w[ 2 = (z w)(z w) = z w ¯ z ¯ w = z¯ z w ¯ w = [z[ 2 [w[ 2 . (5) If z = a+ib, then z +¯ z = (a+ib) +(a−ib) = 2a = 2 Re z and z −¯ z = (a+ib) −(a−ib) = 1 CHAPTER 1. COMPLEX NUMBERS 2ib = 2i Imz. Since [Re z[ 2 = a 2 ≤ a 2 + b 2 = [z[ 2 , we may take square root to obtain [Re z[ ≤ [z[. (6) [z +w[ 2 = (z+w)(z +w) = (z+w)(¯ z+ ¯ w) = z ¯ z+w ¯ w+w ¯ z+z ¯ w = [z[ 2 +[w[ 2 +w ¯ z+w ¯ z = [z[ 2 + [w[ 2 + 2 Re(w ¯ z). Since the earlier results together imply Re(w ¯ z) ≤ [Re(w ¯ z)[ ≤ [w ¯ z[ = [w[ [¯ z[ = [w[ [z[, we obtain [z +w[ 2 ≤ [z[ 2 +[w[ 2 + 2 [z[ [w[ = ([z[ +[w[) 2 . The triangle inequality thus follows from taking square root of the above inequality. (7) (a) Distance between _ 1 2 , 1 _ and the center = _ _ 1 2 −0 _ 2 + (1 −1) 2 = 1 2 Thus, 1 2 +i lies inside the circle. (b) Distance between _ 1, i 2 _ and the center = _ (1 −0) 2 + _ 1 2 −1 _ 2 = √ 5 2 > 1 Thus, 1 + i 2 lies outside the circle. (c) Distance between _ 1 2 , √ 2 2 i _ and the center = _ _ 1 2 −0 _ 2 + _ √ 2 2 −1 _ 2 = 0.57947 Thus, 1 2 + √ 2 2 i lies inside the circle. (d) Distance between _ − 1 2 , √ 3i _ and the center = _ _ − 1 2 −0 _ 2 + _√ 3 −1 _ 2 = 0.88651 Thus, − 1 2 + √ 3i lies inside the circle. (8) For z = cos θ +i sinθ, we have z ¯ z = cos 2 θ + sin 2 θ = 1. Therefore, 1 z = ¯ z = cos θ − i sinθ. Therefore, z + 1 z = z + ¯ z = 2 cos θ. Furthermore, DeMoivre’s Theorem implies z 2 = cos 2θ +i sin2θ and therefore z 2 + 1 z 2 = z 2 + (¯ z) 2 = 2 cos 2θ. (9) (a) (3+4i) 2 −2(x−iy) = x+iy ⇒−7+24i = 3x−iy. Therefore, x = − 7 3 and y = −24. (b) Note that _ 1 +i 1 −i _ 2 = −1 and 1 x +iy = x −iy x 2 +y 2 . Therefore, _ 1 +i 1 −i _ 2 + 1 x +iy = 1 +i ⇒−1 + _ x x 2 +y 2 _ −i _ y x 2 +y 2 _ = 1 +i. Therefore, we have x x 2 +y 2 = 2 and y x 2 +y 2 = −1. Solving these equations and noting that x and y can not be both equal to zero, we conclude that x = 2 5 and y = − 1 5 . (c) (3 − 2i)(x + iy) = 2(x − 2iy) + 2i − 1 ⇒ x + 2y = −1 and −2x + 7y = 2. Solving these equations, one obtains x = −1 and y = 0. 2 CHAPTER 1. COMPLEX NUMBERS (10) Since √ 3 +i = 2 _ cos π 6 +i sin π 6 _ and 1 −i = √ 2 _ cos 7π 4 +i sin 7π 4 _ , we conclude that ( √ 3 +i) 4 (1 −i) 3 = 2 4 _ cos 4π 6 +i sin 4π 6 _ _√ 2 _ 3 _ cos 21π 4 +i sin 21π 4 _ = 4 √ 2 _ cos 55π 12 −i sin 55π 12 _ = 4 √ 2 _ cos 7π 12 −i sin 7π 12 _ . (11) (a) Since −1 −i √ 3 2 = cos 4π 3 +i sin 4π 3 , one concludes from DeMoivre’s Theorem that _ −1 −i √ 3 2 _ 3 = _ cos 4π 3 +i sin 4π 3 _ 3 = cos 12π 3 +i sin 12π 3 = 1. (b) √ 3 −i = 2 _ cos 11π 6 +i sin 11π 6 _ 1 +i = √ 2 _ cos π 4 +i sin π 4 _ _ √ 3 −i _ 4 (1 +i) 10 = _ 2 _ cos 11π 6 +i sin 11π 6 __ 4 _ √ 2 _ cos π 4 +i sin π 4 __ 10 = 2 9 _ cos _ 4 11π 6 + 10 π 4 _ +i sin _ 4 11π 6 + 10 π 4 __ = 2 9 _ cos 59π 6 +i sin 59π 6 _ = 2 9 _ cos 11π 6 +i sin 11π 6 _ = 256 √ 3 −256i (c) 1 − √ 3i = 2 _ cos 5π 3 +i sin 5π 3 _ 1 + √ 3i = 2 _ cos π 3 +i sin π 3 _ _ 1 − √ 3i _ 4 _ 1 + √ 3i _ 8 = _ 2 _ cos 5π 3 +i sin 5π 3 __ 4 _ 2 _ cos π 3 +i sin π 3 __ 8 = 2 12 _ cos _ 4 5π 3 + 8 π 3 _ +i sin _ 4 5π 3 + 8 π 3 __ = 2 12 _ cos 28π 3 +i sin 28π 3 _ = 2 12 _ cos 4π 3 +i sin 4π 3 _ = −2048 −2048 √ 3i (12) Since −32 = 32 (cos π +i sinπ) and 8i = 8 _ cos π 2 +i sin π 2 _ , the 5th roots of −32 are given 3 CHAPTER 1. COMPLEX NUMBERS by z k = 2 _ cos( 2k + 1 5 π) +i sin( 2k + 1 5 π) _ , k = 0, 1, 2, 3, 4; while the cubic roots of 8i are given by z k = 2 _ cos( 4k + 1 6 π) +i sin( 4k + 1 6 π) _ , k = 0, 1, 2. (13) (a) _ −8 + 8 √ 3i _ 1/4 = _ 16 _ − 1 2 + √ 3 2 i __ 1/4 = _ 16 _ cos _ 2kπ + 2π 3 _ +i sin _ 2kπ + 2π 3 ___ 1/4 = 2 _ cos _ kπ 2 + π 6 _ +i sin _ kπ 2 + π 6 __ , where k = 0, 1, 2 and 3 z 0 = 2 _ cos π 6 +i sin π 6 _ = √ 3 +i z 1 = 2 _ cos 2π 3 +i sin 2π 3 _ = −1 + √ 3i z 2 = 2 _ cos 7π 6 +i sin 7π 6 _ = − √ 3 −i z 3 = 2 _ cos 5π 3 +i sin 5π 3 _ = 1 − √ 3i (b) _ −32 + 32 √ 3i _ 1/6 = _ 64 _ − 1 2 + √ 3 2 i __ 1/6 = _ 64 _ cos _ 2kπ + 2π 3 _ +i sin _ 2kπ + 2π 3 ___ 1/6 = 2 _ cos _ kπ 3 + π 9 _ +i sin _ kπ 3 + π 9 __ , where k = 0, 1, ..., 5 z 0 = 2 _ cos π 9 +i sin π 9 _ = 1.8794 + 0.68404i z 1 = 2 _ cos 4π 9 +i sin 4π 9 _ = 0.34730 + 1.9696i z 2 = 2 _ cos 7π 9 +i sin 7π 9 _ = −1.5321 + 1.2856i z 3 = 2 _ cos 10π 9 +i sin 10π 9 _ = −1.8794 −0.68404i z 4 = 2 _ cos 13π 9 +i sin 13π 9 _ = −0.34730 −1.9696i 4 CHAPTER 1. COMPLEX NUMBERS z 5 = 2 _ cos 16π 9 +i sin 16π 9 _ = 1.5321 −1.2856i (14) If z = x +iy, then ¯ z z = ¯ z ¯ z z ¯ z = (¯ z) 2 [z[ 2 . If z →0 along the real axis, one has z = x +i0 and therefore ¯ z z →1. On the other hand, if z → 0 along the imaginary axis, then z = 0 + iy and ¯ z z → −1. We thus conclude that lim z→0 ¯ z z does not exist. When z 0 ,= 0, lim z→z 0 ¯ z z = z 0 z 0 . (15) (a) From DeMoivre’s theorem, we have (cos x +i sinx) 5 = cos 5x +i sin5x. On the other hand, by binomial theorem, we also have (cos x +i sinx) 5 = 5 k=0 _ 5 k _ cos k x(i sinx) 5−k Therefore, comparing the real and imaginary parts yields cos 5x = cos 5 x −10 cos 3 xsin 2 x + 5 cos xsin 4 x, sin5x = sin 5 x −10 cos 2 xsin 3 x + 5 cos 4 xsinx (b) cos xcos 5x = cos 6 x −10 cos 4 xsin 2 x + 5 cos 2 xsin 4 x sinxsin5x = sin 6 x −10 cos 2 xsin 4 x + 5 cos 4 xsin 2 x Adding these equations yields cos xcos 5x + sinxsin5x = sin 6 x + cos 6 x −5(cos 2 xsin 4 x + cos 4 xsin 2 x) cos 4x = sin 6 x + cos 6 x −5 cos 2 xsin 2 x Therefore, sin 6 x + cos 6 x = cos 4x + 5 sin 2 xcos 2 x = cos 4x + 5 4 sin 2 2x 5 CHAPTER 1. COMPLEX NUMBERS Since cos 4x = 1 −2 sin 2 2x, we have sin 6 x + cos 6 x = cos 4x + 5 8 (1 −cos 4x) = 3 8 cos 4x + 5 8 = 3 cos 4x + 5 8 (16) Using the identity z z = z 2 [z[ 2 for any non-zero complex no. z, we have 1 + sinx +i cos x 1 + sinx −i cos x = (1 + sinx +i cos x) 2 [1 + sinx +i cos x[ 2 = (1 + sinx) 2 −cos 2 x + 2i cos x(1 + sinx) (1 + sinx) 2 + cos 2 x = _ 1 + 2 sinx + sin 2 x −cos 2 x _ + 2i cos x(1 + sinx) 2 (1 + sinx) = sinx +i cos x = cos( π 2 −x) +i sin( π 2 −x) = e i( π 2 −x) . DeMoivre’s Theorem now gives _ 1 + sinx +i cos x 1 + sinx −i cos x _ n = _ e i( π 2 −x) _ n = e in( π 2 −x) = cos n( π 2 −x) +i sinn( π 2 −x). (17) If the given number is a +bi, we are looking for a number x +yi such that (x +yi) 2 = a +bi. This is equivalent to the system of equations x 2 −y 2 = a and 2xy = b. From these equations we obtain (x 2 +y 2 ) 2 = (x 2 −y 2 ) 2 + 4x 2 y 2 = a 2 +b 2 . Hence we must have x 2 +y 2 = _ a 2 +b 2 , 6 CHAPTER 1. COMPLEX NUMBERS where the square root is positive or zero. Together with the equation x 2 −y 2 = a we find x 2 = a + √ a 2 +b 2 2 and y 2 = −a + √ a 2 +b 2 2 . These equation yield, in general, two opposite value for x and two for y. But these values cannot be combined arbitrarily, for the equation 2xy = b is not a consequence of these equation. We must therefore be careful to select x and y so that their product has the sign of b. This leads to the general equation √ a +bi = ± _ _ r +a 2 +i(sgn b) _ r −a 2 _ where r = √ a 2 +b 2 . (18) (a) Let w = z 2 . Then the given equation becomes w 2 − 2w + 4 = 0. The quadratic formula implies z 2 = w = 2 ± √ 4 −16 2 = 1 ± √ 3i. Case z 2 = 1 + √ 3i: In polar form, 1 + √ 3i = 2 _ cos π 3 +i sin π 3 _ . Therefore, z = √ 2 _ cos _ π 3 + 2kπ 2 _ +i sin _ π 3 + 2kπ 2 __ , k = 0, 1. Equivalently, the square root formula with r = 2 and sgnb = 1 implies z = ± _ _ 2 + 1 2 +i _ 2 −1 2 _ = ± _ _ 3 2 +i _ 1 2 _ . Case z 2 = 1 − √ 3i: In polar form, 1 − √ 3i = 2 _ cos −π 3 +i sin −π 3 _ . Therefore, z = √ 2 _ cos _ − π 3 + 2kπ 2 _ +i sin _ − π 3 + 2kπ 2 __ , k = 0, 1. Equivalently, the square root formula with r = 2 and sgnb = −1 implies z = ± _ _ 2 + 1 2 −i _ 2 −1 2 _ = ± _ _ 3 2 −i _ 1 2 _ . (b) Let w = z 3 . Then the equation becomes w 2 + 2w + 2 = 0. Solving the equation yields z 3 = w = −2 ± √ 4 −8 2 = −1 ±i. 7 CHAPTER 1. COMPLEX NUMBERS Case z 3 = −1 +i: In polar form, −1 +i = √ 2 _ cos 3π 4 +i sin 3π 4 _ . Therefore, z = 2 1/6 _ cos _ 3π 4 + 2kπ 3 _ +i sin _ 3π 4 + 2kπ 3 __ , k = 0, 1, 2. Case z 3 = −1 −i: In polar form, −1 −i = √ 2 _ cos −3π 4 +i sin −3π 4 _ . Therefore, z = 2 1/6 _ cos _ − 3π 4 + 2kπ 3 _ +i sin _ − 3π 4 + 2kπ 3 __ , k = 0, 1, 2. (c) Let w = z 2 . Then the given equation becomes w 2 + 4w + 16 = 0. The quadratic formula implies z 2 = w = −4 ± √ 16 −64 2 = −2 ±2 √ 3i. Case z 2 = −2 + 2 √ 3i: In polar form, 1 + √ 3i = 4 _ cos 2π 3 +i sin 2π 3 _ . Therefore, z = 2 _ cos _ 2π 3 + 2kπ 2 _ +i sin _ 2π 3 + 2kπ 2 __ , k = 0, 1. Equivalently, the square root formula with r = 4 and sgnb = 1 implies z = ± _ _ 4 −2 2 +i _ 4 + 2 2 _ = ± _ 1 +i √ 3 _ . Case z 2 = −2 −2 √ 3i: In polar form, −2 −2 √ 3i = 4 _ cos −2π 3 +i sin −2π 3 _ . There- fore, z = 2 _ cos _ − 2π 3 + 2kπ 2 _ +i sin _ − 2π 3 + 2kπ 2 __ , k = 0, 1. Equivalently, the square root formula with r = 4 and sgnb = −1 implies z = ± _ _ 4 −2 2 −i _ 4 + 2 2 _ = ± _ 1 −i √ 3 _ . (d) z 4 = −1 = cos π +i sinπ. Therefore, z = cos _ π + 2kπ 4 _ +i sin _ π + 2kπ 4 _ , k = 0, 1, 2, 3. (19) (a) e i π 2 = cos π 2 +i sin π 2 = i 8 CHAPTER 1. COMPLEX NUMBERS (b) 4e −i π 2 = 4 _ cos _ − π 2 _ +i sin _ − π 2 _¸ = −4i (c) 8e i 7π 3 = 8 _ cos 7π 3 +i sin 7π 3 _ = 4 + 4 √ 3i (d) 2e −i 3π 4 = 2 _ cos _ − 3π 4 _ +i sin _ − 3π 4 _¸ = − √ 2 − √ 2i (e) 6e i 2π 3 e iπ = 6e i 5π 3 = 6 _ cos 5π 3 +i sin 5π 3 _ = 3 −3 √ 3i (f) e i π 4 e −iπ = e −i 3π 4 = _ cos _ − 3π 4 _ +i sin _ − 3π 4 _¸ = − √ 2 2 − √ 2 2 i (20) Let z = x +iy and define e z = e x (cos y +i siny). (a) If z = x + 0i, e z = e x (cos 0 +i sin0) = e x (b) [e z [ = _ (e x cos y) 2 + (e x sin y) 2 = e x . In particular ¸ ¸ e iy ¸ ¸ = e 0 = 1. (c) e z 1 +z 2 = e (x 1 +x 2 )+i(y 1 +y 2 ) = e x 1 +x 2 (cos (y 1 +y 2 ) +i sin(y 1 +y 2 )) = e z 1 e z 2 . (d) [e z [ = e x ,= 0 for any z ∈ C. Therefore, e z ,= 0. Furthermore, 1 e z = 1 e x (cos y +i siny) = e −x (cos y −i siny) = e −x+i(−y) = e −z . (e) Repeat application of (c) and (d), or by induction, yields e nz = (e z ) n for any integer n. (f) d dt e iwt = d dt cos wt +i d dt sinwt = −wsinwt +iwcos wt = iw(cos wt +i sinwt) = iwe iwt . (21) Since e i2πkt = cos (2πkt) + i sin(2πkt), the cosine function is an even function and the sine function is an odd function, we have n k=−n e i2πkt = n k=−n cos (2πkt) +i n k=−n sin (2πkt) = 1 + 2 n k=1 cos (2πkt) Recall that 1 +z +z 2 + +z n = 1 −z n+1 1 −z . 9 CHAPTER 1. COMPLEX NUMBERS Then 1 + n k=1 _ e iθ _ k = 1 − _ e iθ _ n+1 1 −(e iθ ) = 1 − _ e i(n+1)θ _ 1 −(e iθ ) = 1 −cos(n + 1)θ −i sin(n + 1)θ 1 −cos θ −i sinθ = 1 −cos(n + 1)θ −i sin(n + 1)θ (1 −cos θ) 2 + sin 2 θ = [1 −cos(n + 1)θ −i sin(n + 1)θ] [1 −cos θ −i sinθ] 2(1 −cos θ) Comparing the real part of this equation yields 1 + cos θ + cos 2θ + + cos nθ = (1 −cos(n + 1)θ)(1 −cos θ) + sin(n + 1)θ sinθ 4 sin 2 (θ/2) . Simplifying the right hand side of the equation, we have (1 −cos(n + 1)θ)(1 −cos θ) + sin(n + 1)θ sinθ 4 sin 2 (θ/2) = (1 −cos(n + 1)θ)(2 sin 2 (θ/2)) + sin(n + 1)θ(2 sin(θ/2) cos(θ/2)) 4 sin 2 (θ/2) = 2 sin 2 (θ/2) −2 sin(θ/2)[cos(n + 1)θ sin(θ/2) −sin(n + 1)θ cos(θ/2)] 4 sin 2 (θ/2) = 1 2 + sin[(2n + 1)θ/2] 2 sin(θ/2) Hence n k=1 cos (kθ) = − 1 2 + sin[(2n + 1)θ/2] 2 sin(θ/2) . Finally, by letting θ = 2πt, we have n k=−n e i2πkt = 1 + 2 n k=1 cos (2πkt) = 1 + 2 _ − 1 2 + sin[(2n + 1)πt] 2 sinπt _ = sin[(2n + 1)πt] sinπt . Integrating this result yields _ 1 0 sin[(2n + 1)πt] sinπt dt = _ 1 0 _ 1 + 2 n k=1 cos (2πkt) _ dt = 1 + 2 n k=1 _ 1 0 cos (2πkt) dt = 1. (22) Suppose w is a root of the polynomial P(z) = a 0 +a 1 z +a 2 z 2 + +a n z n , i.e., P(w) = 0. 10 CHAPTER 1. COMPLEX NUMBERS Then P (w) = a 0 +a 1 w +a 2 w 2 + +a n w n = a 0 +a 1 w +a 2 w 2 + +a n w n = a 0 +a 1 w +a 2 w 2 + +a n w n = P(w) = 0. (23) Let z = x +yi, then ¸ ¸ ¸ ¸ z −3 z + 3 ¸ ¸ ¸ ¸ = 2 =⇒ [z −3[ = 2 [z + 3[ =⇒ [(x −3) +yi[ = 2 [(x + 3) +yi[ =⇒ _ (x −3) 2 +y 2 = 2 _ (x + 3) 2 +y 2 =⇒ (x −3) 2 +y 2 = 4 _ (x + 3) 2 +y 2 _ =⇒ x 2 + 10x +y 2 + 9 = 0 =⇒ (x + 5) 2 +y 2 = 16 (24) Let z = x +yi, then [z + 3[ +[z −3[ = 10 =⇒ [(x + 3) +yi[ +[(x −3) +yi[ = 10 =⇒ _ (x + 3) 2 +y 2 + _ (x −3) 2 +y 2 = 10 =⇒ (x + 3) 2 +y 2 + 2 _ (x + 3) 2 +y 2 _ (x −3) 2 +y 2 + (x −3) 2 +y 2 = 100 =⇒ _ (x + 3) 2 +y 2 _ _ (x −3) 2 +y 2 _ = _ 41 −x 2 −y 2 _ 2 =⇒ 64x 2 + 100y 2 = 1600 =⇒ x 2 25 + y 2 16 = 1 (25) If z = a +bi and [z[ = 1, then a 2 +b 2 = 1. w = z −1 z + 1 = (a +bi) −1 (a +bi) + 1 = (a −1) +bi (a + 1) +bi = (a −1) +bi (a + 1) +bi (a + 1) −bi (a + 1) −bi = (a −1) (a + 1) +b 2 + 2bi (a + 1) 2 +b 2 = _ a 2 +b 2 _ −1 + 2bi (a + 1) 2 +b 2 = (1) −1 + 2bi (a + 1) 2 +b 2 = 2bi (a + 1) 2 +b 2 Therefore, w is a purely imaginary number. 11 Chapter 2 Linear Algebra (1) (a) A = _ ¸ ¸ _ 0 −1 −2 1 0 −1 2 1 0 _ ¸ ¸ _ (b) A = _ ¸ ¸ _ 1 1 2 1 3 2 1 2 3 3 3 2 1 _ ¸ ¸ _ (c) A = _ ¸ ¸ _ 0 −1 −2 1 2 0 − 1 2 2 3 1 3 0 _ ¸ ¸ _ (d) A = _ ¸ ¸ _ e e 2 e 3 e 2 e 4 e 6 e 3 e 6 e 9 _ ¸ ¸ _ (2) The entry at the i-th row and the j-th column of B T A T is given by the product of i-th row of B T = _ b 1i b 2i b ni _ and the j-th column of A T = _ a j1 a j2 a jn _ T which is equal to b ki a jk . On the other hand, the entry at the i-th row and the j-th column of (AB) T is the same as the the entry at the j-th row and i-th column of AB, and is therefore given by a jk b ki . If A −1 exists, the above result ⇒ A T _ A −1 _ T = _ A −1 A _ T = I T = I. (3) (a) B T = _ A _ A T A _ −1 A T _ T = _ A T _ T _ _ A T A _ −1 _ T A T = A _ _ A T A _ T _ −1 A T = A _ A T A _ −1 A T = B 13 CHAPTER 2. LINEAR ALGEBRA (b) B 2 = _ A _ A T A _ −1 A T _ 2 = _ A _ A T A _ −1 A T _ _ A _ A T A _ −1 A T _ = A _ A T A _ −1 _ A T A _ _ A T A _ −1 A T = A _ A T A _ −1 A T = B (4) (a) Let A = _ 1 0 1 1 _ and B = _ 1 1 0 1 _ . (A+B) 2 = __ 1 0 1 1 _ + _ 1 1 0 1 __ 2 = _ 5 4 4 5 _ A 2 + 2AB+B 2 = _ 1 0 1 1 _ 2 + 2 _ 1 0 1 1 __ 1 1 0 1 _ + _ 1 1 0 1 _ 2 = _ 4 4 4 6 _ (b) Let A = _ 1 0 1 1 _ and B = _ 1 1 0 1 _ . (A+B) (A−B) = __ 1 0 1 1 _ + _ 1 1 0 1 ____ 1 0 1 1 _ − _ 1 1 0 1 __ = _ 1 −2 2 −1 _ A 2 −B 2 = _ 1 0 1 1 _ 2 − _ 1 1 0 1 _ 2 = _ 0 −2 2 0 _ (c) Let C = _ 0 1 0 0 _ ,= 0. C 2 = _ 0 1 0 0 __ 0 1 0 0 _ = _ 0 0 0 0 _ (5) (a) _ ¸ ¸ _ 2 6 −4 10 4 1 3 −2 1 −3 4 −7 _ ¸ ¸ _ r 1 ←→r 3 −−−−−−→ _ ¸ ¸ _ 1 −3 4 −7 4 1 3 −2 2 6 −4 10 _ ¸ ¸ _ (b) _ ¸ ¸ _ 1 −3 4 −7 4 1 3 −2 2 6 −4 10 _ ¸ ¸ _ −4r 1 +r 2 −→r 2 −−−−−−−−−−−−→ _ ¸ ¸ _ 1 −3 4 −7 0 13 −13 26 2 6 −4 10 _ ¸ ¸ _ (c) _ ¸ ¸ _ 1 −3 4 −7 0 13 −13 26 2 6 −4 10 _ ¸ ¸ _ 1 2 r 3 −→r 3 −−−−−−−→ _ ¸ ¸ _ 1 −3 4 −7 0 13 −13 26 1 3 −2 5 _ ¸ ¸ _ (6) _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 2x 1 −x 2 + 3x 3 −2x 4 = 1 (1) x 2 −2x 3 + 3x 4 = 2 (2) 4x 3 + 3x 4 = 3 (3) 4x 4 = 4 (4) 14 CHAPTER 2. LINEAR ALGEBRA From (4), x 4 = 1. From (3), 4x 3 + 3 (1) = 3 =⇒x 3 = 0. From (2), x 2 −2 (0) + 3 (1) = 2 =⇒x 2 = −1. From (1), 2x 1 −(−1) + 3 (0) −2 (1) = 1 =⇒x 1 = 1. (7) _ ¸ ¸ _ 1 2 1 3 3 −1 −3 −1 2 3 1 4 _ ¸ ¸ _ −3r 1 +r 2 −→r 2 −−−−−−−−−−−−→ _ ¸ ¸ _ 1 2 1 3 0 −7 −6 −10 2 3 1 4 _ ¸ ¸ _ −2r 1 +r 3 −→r 3 −−−−−−−−−−−−→ _ ¸ ¸ _ 1 2 1 3 0 −7 −6 −10 0 −1 −1 −2 _ ¸ ¸ _ r 2 ←→r 3 −−−−−−→ _ ¸ ¸ _ 1 2 1 3 0 −1 −1 −2 0 −7 −6 −10 _ ¸ ¸ _ −7r 2 +r 3 −→r 3 −−−−−−−−−−−−→ _ ¸ ¸ _ 1 2 1 3 0 −1 −1 −2 0 0 1 4 _ ¸ ¸ _ −r 2 −→r 2 −−−−−−−→ _ ¸ ¸ _ 1 2 1 3 0 1 1 2 0 0 1 4 _ ¸ ¸ _ −r 3 +r 2 −→r 2 −−−−−−−−−−−→ _ ¸ ¸ _ 1 2 1 3 0 1 0 −2 0 0 1 4 _ ¸ ¸ _ −r 3 +r 1 −→r 1 −−−−−−−−−−−→ _ ¸ ¸ _ 1 2 0 −1 0 1 0 −2 0 0 1 4 _ ¸ ¸ _ −2r 2 +r 1 −→r 1 −−−−−−−−−−−−→ _ ¸ ¸ _ 1 0 0 3 0 1 0 −2 0 0 1 4 _ ¸ ¸ _ =⇒x = _ ¸ ¸ _ x 1 x 2 x 3 _ ¸ ¸ _ = _ ¸ ¸ _ 3 −2 4 _ ¸ ¸ _ (8) A → _ ¸ ¸ ¸ ¸ _ 0 1 2 0 −2 −4 1 1 1 0 2 4 _ ¸ ¸ ¸ ¸ _ → _ ¸ ¸ ¸ ¸ _ 0 1 2 0 0 0 1 1 1 0 0 0 _ ¸ ¸ ¸ ¸ _ → _ ¸ ¸ ¸ ¸ _ 1 0 −1 0 1 2 0 0 0 0 0 0 _ ¸ ¸ ¸ ¸ _ = R. Therefore, the augmented matrix [A[0] of Ax = 0 is reduced to [R[0]. Rx = 0 is of the form _ x 1 −x 3 = 0 x 2 +2x 3 = 0 , giving x 1 = x 3 and x 2 = −2x 3 . Therefore, we have x = _ t −2t t _ T where t is an arbitrary scalar as the solution. (9) (a) With A = _ ¸ ¸ _ 1 −2 1 2 1 8 1 −12 −11 _ ¸ ¸ _ and b = _ ¸ ¸ _ 1 3 −1 _ ¸ ¸ _ , we may reduce the augmented matrix 15 CHAPTER 2. LINEAR ALGEBRA of Ax = b as follows: [A[b] = _ ¸ ¸ _ 1 −2 1 2 1 8 1 −12 −11 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 3 −1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 −2 1 0 5 6 0 −10 −12 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 −2 _ ¸ ¸ _ → _ ¸ ¸ _ 1 −2 1 0 1 6 5 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 5 0 _ ¸ ¸ _ . We thus have _ ¸ ¸ _ ¸ ¸ _ x 1 −2x 2 +x 3 = 1 x 2 + 6 5 x 3 = 1 5 , giving x 2 = 1 5 − 6 5 x 3 and x 1 = 1 +2x 2 −x 3 . Putting x 3 = t, one obtains x 2 = 1 5 − 6 5 t and x 1 = 1+2 _ 1 5 − 6 5 t _ −t = 7 5 − 17 5 t. (b) With A = _ ¸ ¸ _ 3 −2 2 2 1 −3 1 −10 18 _ ¸ ¸ _ and b = _ ¸ ¸ _ 4 5 −8 _ ¸ ¸ _ , [A[b] = _ ¸ ¸ _ 3 −2 2 2 1 −3 1 −10 18 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 4 5 −8 _ ¸ ¸ _ → _ ¸ ¸ _ 0 28 −52 0 21 −39 1 −10 18 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 28 21 −8 _ ¸ ¸ _ → _ ¸ ¸ _ 0 28 −52 0 1 − 13 7 1 −10 18 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 28 1 −8 _ ¸ ¸ _ → _ ¸ ¸ _ 1 −10 18 0 1 − 13 7 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ −8 1 0 _ ¸ ¸ _ → _ ¸ ¸ _ 1 0 − 4 7 0 1 − 13 7 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 2 1 0 _ ¸ ¸ _ . We thus obtain x = _ 2 + 4 7 t 1 + 13 7 t t _ T , where t is an arbitrary scalar. (c) In this problem, A = _ ¸ ¸ _ 1 1 p 1 p 1 p 1 1 _ ¸ ¸ _ , b = _ ¸ ¸ _ 1 p p 2 _ ¸ ¸ _ . [A[b] = _ ¸ ¸ _ 1 1 p 1 p 1 p 1 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 p p 2 _ ¸ ¸ _ → _ ¸ ¸ _ 1 1 p 0 p −1 1 −p 0 1 −p 1 −p 2 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 p −1 p 2 −p _ ¸ ¸ _ → _ ¸ ¸ _ 1 1 p 0 p −1 1 −p 0 0 (1 −p)(p + 2) ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 p −1 p 2 −1 _ ¸ ¸ _ . There are 3 cases: 16 CHAPTER 2. LINEAR ALGEBRA (i) If p = 1, the above augmented matrix is reduced to _ ¸ ¸ _ 1 1 1 0 0 0 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 _ ¸ ¸ _ . Hence the sys- tem of linear equations is reduced to one single equation x 1 + x 2 + x 3 = 1. Thus x = _ 1 −s −t s t _ T , where t and s are arbitrary scalars. (ii) If p = −2, the augmented matrix becomes _ ¸ ¸ _ 1 1 −2 0 −3 3 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 −3 3 _ ¸ ¸ _ . Hence the correspond- ing system of linear equations is inconsistent. (iii) For p ,= 1, p ,= −2, the augmented matrix may be reduced to _ ¸ ¸ _ 1 1 p 0 1 −1 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 −1−p 2+p _ ¸ ¸ _ . We conclude that the corresponding linear system has one and only one solution given by x = _ (1+p) 2 2+p 1 2+p − 1+p 2+p _ T . (10) (a) _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 1 1 1 1 −1 −1 0 0 1 −1 −2 −2 0 0 3 1 0 0 1 1 3 3 1 1 2 2 4 4 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ −r 1 +r 5 −→r 5 2r 1 +r 3 −→r 3 r 1 +r 2 −→r 2 −−−−−−−−−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 1 1 1 1 0 0 0 0 2 0 0 0 2 2 5 3 0 0 1 1 3 3 0 0 1 1 3 3 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ r 2 ←→r 5 −−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 1 1 1 1 0 0 1 1 3 3 0 0 2 2 5 3 0 0 1 1 3 3 0 0 0 0 2 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ −r 2 +r 4 −→r 4 −2r 2 +r 3 −→r 3 −−−−−−−−−−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 1 1 1 1 0 0 1 1 3 3 0 0 0 0 −1 −3 0 0 0 0 0 0 0 0 0 0 2 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ r 4 ←→r 5 −r 3 −→r 3 −−−−−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 1 1 1 1 0 0 1 1 3 3 0 0 0 0 1 3 0 0 0 0 2 0 0 0 0 0 0 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ −2r 3 +r 4 −→r 4 −−−−−−−−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 1 1 1 1 0 0 1 1 3 3 0 0 0 0 1 3 0 0 0 0 0 −6 0 0 0 0 0 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ (b) The system of linear equations is inconsistent. If it is consistent, then there are three non-zero equations with five variables. Therefore, the system has infinitely many solutions. 17 CHAPTER 2. LINEAR ALGEBRA (11) _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 3 0 1 −2 0 0 −2 3 −1 0 −1 2 0 0 2 −3 1 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ −2r 1 +r 5 −→r 5 r 1 +r 4 −→r 4 2r 1 +r 3 −→r 3 −r 1 +r 2 −→r 2 −−−−−−−−−−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 3 0 0 −3 −3 0 0 5 5 0 0 3 3 0 0 −5 −5 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ − 1 5 r 5 −→r 5 1 3 r 4 −→r 4 1 5 r 3 −→r 3 − 1 3 r 2 −→r 2 −−−−−−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 3 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ −r 2 +r 5 −→r 5 −r 2 +r 4 −→r 4 −r 2 +r 3 −→r 3 −−−−−−−−−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 3 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ −r 2 +r 1 −→r 1 −−−−−−−−−−−−−→ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 0 2 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ The reduced system of equations is : _ x 1 + 2x 3 = 0 (1) x 2 +x 3 = 0 (2) Let x 3 = t, then from (2), x 2 = −x 3 = −t. From (1), x 1 = −2x 3 = −2t. Therefore, x = _ ¸ ¸ _ x 1 x 2 x 3 _ ¸ ¸ _ = _ ¸ ¸ _ −2t −t t _ ¸ ¸ _ . (12) _ ¸ ¸ _ 1 1 −1 1 2 3 a 3 1 a 3 2 _ ¸ ¸ _ −r 1 +r 3 −→r 3 −2r 1 +r 2 −→r 2 −−−−−−−−−−−−−−→ _ ¸ ¸ _ 1 1 −1 1 0 1 a + 2 1 0 a −1 4 1 _ ¸ ¸ _ −(a −1) r 2 +r 3 −→r 3 −−−−−−−−−−−−−−−−−→ _ ¸ ¸ _ 1 1 −1 1 0 1 a + 2 1 0 0 −(a + 3) (a −2) −(a −2) _ ¸ ¸ _ (a) If _ −(a + 3) (a −2) = 0 −(a −2) = 0 , then the system is consistent with infinitely many solu- tion. Therefore, a = 2. (b) If −(a + 3) (a −2) ,= 0, then the system is consistent with one and only one solution. Therefore, a ,= 2 or a ,= −3. (c) If _ −(a + 3) (a −2) = 0 −(a −2) ,= 0 , then the system is inconsistent. Therefore, a = −3. (13) _ ¸ ¸ _ 1 −1 1 2 3 1 4a −1 2 2 a a + 1 2 _ ¸ ¸ _ −2r 1 +r 3 −→r 3 −3r 1 +r 2 −→r 2 −−−−−−−−−−−−−−→ _ ¸ ¸ _ 1 −1 1 2 0 4 4a −4 −4 0 a + 2 a −1 −2 _ ¸ ¸ _ 1 4 r 2 −→r 2 −−−−−−−→ 18 CHAPTER 2. LINEAR ALGEBRA _ ¸ ¸ _ 1 −1 1 2 0 1 a −1 −1 0 a + 2 a −1 −2 _ ¸ ¸ _ −(a + 2) r 2 +r 3 −→r 3 −−−−−−−−−−−−−−−−−→ _ ¸ ¸ _ 1 −1 1 2 0 1 a −1 −1 0 0 −(a + 1) (a −1) a _ ¸ ¸ _ (a) If _ −(a + 1) (a −1) = 0 a = 0 , then the system is consistent with infinitely many so- lutions. However, the above system has no solution. Therefore, the system is not possible to have infinitely mant solutions. (b) If −(a + 1) (a −1) ,= 0, then the system is consistent with one and only one solution. Therefore, a ,= 1 or a ,= −1. (c) If _ −(a + 1) (a −1) = 0 a ,= 0 , then the system is inconsistent. Therefore, a = 1 or a = −1. (14) The augmented matrix _ ¸ ¸ _ 1 1 −1 −a −1 a a 2 1 −a ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ b 1 b 2 b 3 _ ¸ ¸ _ may be reduced to _ ¸ ¸ _ 1 1 −1 0 a −1 0 0 1 −a 2 a(a −1) ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ b 1 b 2 +ab 1 b 3 −a 2 b 1 _ ¸ ¸ _ ———(*) (a) If a ,= 0, 1, (*) is reduced to _ ¸ ¸ _ 1 1 −1 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ b 1 b 2 +ab 1 a−1 b 3 +(a+1)b 2 +ab 1 a(a−1) _ ¸ ¸ _ , which implies consis- tency of the linear system. On the other hand, if the system is consistent for any b 1 , b 2 and b 3 , then it follows from (*) that a ,= 0, 1. (b) If a = 0, (*) becomes _ ¸ ¸ _ 1 1 −1 0 1 0 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ b 1 b 3 b 2 +b 3 _ ¸ ¸ _ . It thus follows that the system is consistent if and only if b 2 +b 3 = 0. (c) If a = 1 and b = 0, (*) becomes _ ¸ ¸ _ 1 1 −1 0 0 0 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 0 0 0 _ ¸ ¸ _ . As such, we have one single equation x 1 +x 2 −x 3 = 0, and x = _ s −t t s _ T . 19 CHAPTER 2. LINEAR ALGEBRA (15) [A[b] = _ ¸ ¸ _ 1 3 −2 5 5 8 −5 12 3 16 −11 28 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ b 1 b 2 b 3 _ ¸ ¸ _ → _ ¸ ¸ _ 1 3 −2 5 0 −7 5 −13 0 7 −5 13 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ b 1 b 2 −5b 1 b 3 −3b 1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 3 −2 5 0 1 − 5 7 13 7 0 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ b 1 5b 1 −b 2 7 b 3 +b 2 −8b 1 _ ¸ ¸ _ . Therefore, the system is consistent if and only if b 3 +b 2 −8b 1 = 0. When b = [1 4 4] T , the system is reduced to one with 2 linear equations in 4 unknowns: _ ¸ ¸ _ ¸ ¸ _ x 1 +3x 2 −2x 3 +5x 4 = 1 x 2 − 5 7 x 3 + 13 7 x 4 = 1 7 . Therefore, solutions are x = _ − t 7 + 4s 7 + 4 7 5t 7 − 13s 7 + 1 7 t s _ T . (16) _ ¸ ¸ ¸ ¸ _ 1 1 1 α β γ β γ α γ α β ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 1 1 1 _ ¸ ¸ ¸ ¸ _ → _ ¸ ¸ ¸ ¸ _ 1 1 1 α β γ β γ α α +β +γ α +β +γ α +β +γ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 1 1 3 _ ¸ ¸ ¸ ¸ _ → _ ¸ ¸ ¸ ¸ _ 1 1 1 α β γ β γ α 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 1 1 3 −3(α +β +γ) _ ¸ ¸ ¸ ¸ _ → _ ¸ ¸ ¸ ¸ _ 1 1 1 0 β −α γ −α 0 γ −β α −β 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 1 −3α 1 −3β 3 −3(α +β +γ) _ ¸ ¸ ¸ ¸ _ . Therefore, if α+β+γ ,= 1, then the system is inconsistent . Assume now that α+β+γ = 1. 20 CHAPTER 2. LINEAR ALGEBRA The augmented matrix now takes the form _ ¸ ¸ ¸ ¸ _ 1 1 1 0 β −α γ −α 0 γ −β α −β 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 1 −3α 1 −3β 0 _ ¸ ¸ ¸ ¸ _ . Case (i) If α = β = γ = 1 3 , then the system is reduced to a single equation x 1 +x 2 +x 3 = 3. Therefore, it has infinitely many solutions. Case (ii) If α = β ,= γ, then the above matrix may be reduced to _ ¸ ¸ ¸ ¸ _ 1 1 1 0 1 0 0 0 1 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 1−3β γ−β 1−3α γ−α 0 _ ¸ ¸ ¸ ¸ _ . In this case, the system has one and only one solution. Similar, we have the same conclusion if β = γ ,= α or if α = γ ,= β. Case (iii) If α, β and γ are all distinct, then _ ¸ ¸ ¸ ¸ _ 1 1 1 0 β −α γ −α 0 γ −β α −β 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 1 −3α 1 −3β 0 _ ¸ ¸ ¸ ¸ _ → _ ¸ ¸ ¸ ¸ _ 1 1 1 0 1 γ−α β−α 0 0 δ 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 1−3α β−α η 0 _ ¸ ¸ ¸ ¸ _ , where δ = α 2 +β 2 +γ 2 −αβ −βγ −γα α −β and η = 1 −3β + (1 −3α)(β −γ) β −α . Since δ = (α −β) 2 + (β −γ) 2 + (γ −α) 2 2(α −β) ,= 0, the system of linear equations has a unique solution. 21 CHAPTER 2. LINEAR ALGEBRA (17) (a) _ ¸ ¸ _ 1 a b 0 2 c 0 0 −1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 0 1 0 0 0 1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 a b 0 1 c 2 0 0 −1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 0 1 2 0 0 0 1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 a b 0 1 c 2 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 0 1 2 0 0 0 −1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 a b 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 0 1 2 c 2 0 0 −1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 a 0 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 b 0 1 2 c 2 0 0 −1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 0 0 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 − a 2 b − ac 2 0 1 2 c 2 0 0 −1 _ ¸ ¸ _ Therefore, _ ¸ ¸ _ 1 a b 0 2 c 0 0 −1 _ ¸ ¸ _ −1 = _ ¸ ¸ _ 1 − a 2 b − ac 2 0 1 2 c 2 0 0 −1 _ ¸ ¸ _ . (b) _ ¸ ¸ _ 3 1 2 −1 3 −4 4 0 5 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 0 1 0 0 0 1 _ ¸ ¸ _ → _ ¸ ¸ _ 0 10 −10 −1 3 −4 0 12 −11 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 3 0 0 1 0 0 4 1 _ ¸ ¸ _ → _ ¸ ¸ _ 0 1 −1 −1 3 −4 0 12 −11 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 10 3 10 0 0 1 0 0 4 1 _ ¸ ¸ _ → _ ¸ ¸ _ 0 1 −1 −1 0 −1 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 10 3 10 0 − 3 10 1 10 0 − 12 10 4 10 1 _ ¸ ¸ _ → _ ¸ ¸ _ 0 1 0 −1 0 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ − 11 10 7 10 1 − 15 10 5 10 1 − 12 10 4 10 1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 0 0 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 15 10 − 5 10 −1 − 11 10 7 10 1 − 12 10 4 10 1 _ ¸ ¸ _ . 22 CHAPTER 2. LINEAR ALGEBRA (c) _ ¸ ¸ _ −2 3 −1 −1 2 −1 −6 9 −4 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 0 1 0 0 0 1 _ ¸ ¸ _ → _ ¸ ¸ _ 0 −1 1 1 −2 1 0 −3 2 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 −2 0 0 −1 0 0 −6 1 _ ¸ ¸ _ → _ ¸ ¸ _ 0 1 −1 1 0 −1 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ −1 2 0 −2 3 0 3 0 −1 _ ¸ ¸ _ → _ ¸ ¸ _ 0 1 0 1 0 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 2 2 −1 1 3 −1 3 0 −1 _ ¸ ¸ _ → _ ¸ ¸ _ 1 0 0 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 3 −1 2 2 −1 3 0 −1 _ ¸ ¸ _ . (18) Let ∆ = αδ −βγ. Case (1) If α ,= 0, a simple calcultion shows that _ α β γ δ _ → _ 1 β α γ δ _ → _ 1 β α 0 ∆ α _ . If ∆ ,= 0, then the last matrix may be reduced to I by two more elementary row operations, which implies that A is nonsingular. If ∆ = 0, then the reduced row echelon form of A is of the form _ 1 β α 0 0 _ , meaning that A is singular. Case (2) Suppose α = 0. When ∆ ,= 0, then both γ and β are non-zero so that A = _ 0 β γ δ _ → _ γ δ 0 β _ → _ 1 δ γ 0 β _ → _ 1 δ γ 0 1 _ →I. If ∆ = 0, then either γ = 0 or β = 0. Hence either A = _ 0 β 0 δ _ or A = _ 0 0 γ δ _ . Therefore, A is singular in both cases. 23 CHAPTER 2. LINEAR ALGEBRA (19) _ ¸ ¸ _ 1 1 1 1 2 4 1 3 9 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 1 1 0 0 0 1 _ ¸ ¸ _ σ 1 → _ ¸ ¸ _ 1 1 1 0 1 3 1 3 9 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 −1 1 0 0 0 1 _ ¸ ¸ _ σ 2 → _ ¸ ¸ _ 1 1 1 0 1 3 0 2 8 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 −1 1 0 −1 0 1 _ ¸ ¸ _ σ 3 → _ ¸ ¸ _ 1 1 1 0 1 3 0 0 2 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 −1 1 0 1 −2 1 _ ¸ ¸ _ σ 4 → _ ¸ ¸ _ 1 1 1 0 1 3 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 −1 1 0 1 2 −1 1 2 _ ¸ ¸ _ σ 5 → _ ¸ ¸ _ 1 1 1 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 0 − 5 2 4 − 3 2 1 2 −1 1 2 _ ¸ ¸ _ σ 6 → _ ¸ ¸ _ 1 1 0 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 2 1 − 1 2 − 5 2 4 − 3 2 1 2 −1 1 2 _ ¸ ¸ _ σ 7 → _ ¸ ¸ _ 1 0 0 0 1 0 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 3 −3 1 − 5 2 4 − 3 2 1 2 −1 1 2 _ ¸ ¸ _ . For 1 ≤ j ≤ 7, let E j be the elementary matrix obtained by applying σ j to I. Let F j = E −1 j . Since E 7 E 6 E 5 E 4 E 3 E 2 E 1 A = I, we conclude that A −1 = E 7 E 6 E 5 E 4 E 3 E 2 E 1 . Therefore, A = (E 7 E 6 E 5 E 4 E 3 E 2 E 1 ) −1 = F 1 F 2 F 3 F 4 F 5 F 6 F 7 . All F j can be written down easily. For example, F 1 = _ ¸ ¸ _ 1 0 0 −1 1 0 0 0 1 _ ¸ ¸ _ −1 = _ ¸ ¸ _ 1 0 0 1 1 0 0 0 1 _ ¸ ¸ _ . (20) (a) −2r 1 +r 2 −→r 2 : E 1 = _ ¸ ¸ _ 1 0 0 −2 1 0 0 0 1 _ ¸ ¸ _ r 2 +r 3 −→r 3 : E 2 = _ ¸ ¸ _ 1 0 0 0 1 0 0 1 1 _ ¸ ¸ _ −r 3 −→r 3 : E 3 = _ ¸ ¸ _ 1 0 0 0 1 0 0 0 −1 _ ¸ ¸ _ 24 CHAPTER 2. LINEAR ALGEBRA (b) _ ¸ ¸ _ 1 2 1 0 1 2 0 0 1 _ ¸ ¸ _ = E 3 E 2 E 1 A =⇒ _ ¸ ¸ _ 1 2 1 0 1 2 0 0 1 _ ¸ ¸ _ −1 = (E 3 E 2 E 1 A) −1 = A −1 E −1 1 E −1 2 E −1 3 =⇒ _ ¸ ¸ _ 1 2 1 0 1 2 0 0 1 _ ¸ ¸ _ −1 E 3 E 2 E 1 = A −1 E −1 1 E −1 2 E −1 3 E 3 E 2 E 1 = A −1 A −1 = _ ¸ ¸ _ 1 −2 3 0 1 −2 0 0 1 _ ¸ ¸ _ _ ¸ ¸ _ 1 0 0 0 1 0 0 0 −1 _ ¸ ¸ _ _ ¸ ¸ _ 1 0 0 0 1 0 0 1 1 _ ¸ ¸ _ _ ¸ ¸ _ 1 0 0 −2 1 0 0 0 1 _ ¸ ¸ _ = _ ¸ ¸ _ 11 −5 −3 −6 3 2 2 −1 −1 _ ¸ ¸ _ (21) det(A) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 2 3 4 0 −4 −8 −12 0 −8 −16 −24 13 14 15 16 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 2 3 4 0 −4 −8 −12 0 0 0 0 13 14 15 16 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 0. (22) (a) i. x = _ ¸ ¸ _ 1 2 3 3 8 11 1 5 7 _ ¸ ¸ _ −1 _ ¸ ¸ _ −1 1 1 _ ¸ ¸ _ = _ ¸ ¸ _ 1 2 1 2 −1 −5 2 −1 7 2 − 3 2 1 _ ¸ ¸ _ _ ¸ ¸ _ −1 1 1 _ ¸ ¸ _ = _ ¸ ¸ _ −1 6 −4 _ ¸ ¸ _ . ii. ∆ = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 2 3 3 8 11 1 5 7 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 2, ∆ 1 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ −1 2 3 1 8 11 1 5 7 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −2, ∆ 2 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 −1 3 3 1 11 1 1 7 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 12, ∆ 3 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 2 −1 3 8 1 1 5 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −8. Therefore, x 1 = ∆ 1 ∆ = −1, x 2 = ∆ 2 ∆ = 6, x 3 = ∆ 3 ∆ = −4. 25 CHAPTER 2. LINEAR ALGEBRA (b) i. A −1 = _ ¸ ¸ ¸ ¸ _ 1 0 1 0 1 1 0 0 0 0 1 1 0 1 1 1 _ ¸ ¸ ¸ ¸ _ −1 = _ ¸ ¸ ¸ ¸ _ 0 1 1 −1 0 0 −1 1 1 −1 −1 1 −1 1 2 −1 _ ¸ ¸ ¸ ¸ _ x = _ ¸ ¸ ¸ ¸ _ x 1 x 2 x 3 x 4 _ ¸ ¸ ¸ ¸ _ = A −1 b = _ ¸ ¸ ¸ ¸ _ 0 1 1 −1 0 0 −1 1 1 −1 −1 1 −1 1 2 −1 _ ¸ ¸ ¸ ¸ _ _ ¸ ¸ ¸ ¸ _ 1 1 1 0 _ ¸ ¸ ¸ ¸ _ = _ ¸ ¸ ¸ ¸ _ 2 −1 −1 2 _ ¸ ¸ ¸ ¸ _ ii. det A = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 1 0 1 1 0 0 0 0 1 1 0 1 1 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −1, det A 1 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −2, det A 2 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 1 0 1 1 0 0 0 1 1 1 0 0 1 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 1, det A 3 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 1 0 1 1 1 0 0 0 1 1 0 1 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 1, det A 4 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 0 1 1 1 1 0 1 0 0 1 1 0 1 1 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −2 Therefore, x 1 = ∆A 1 ∆A = −2 −1 = 2, x 2 = ∆A 2 ∆A = −1, x 3 = ∆A 3 ∆A = −1 and x 4 = ∆A 4 ∆A = 2. (c) i. A −1 = _ ¸ ¸ ¸ ¸ _ 1 1 1 1 1 1 0 0 1 0 1 1 0 1 0 1 _ ¸ ¸ ¸ ¸ _ −1 = _ ¸ ¸ ¸ ¸ _ −1 1 1 0 1 0 −1 0 2 −1 −1 −1 −1 0 1 1 _ ¸ ¸ ¸ ¸ _ x = _ ¸ ¸ ¸ ¸ _ x 1 x 2 x 3 x 4 _ ¸ ¸ ¸ ¸ _ = A −1 b = _ ¸ ¸ ¸ ¸ _ −1 1 1 0 1 0 −1 0 2 −1 −1 −1 −1 0 1 1 _ ¸ ¸ ¸ ¸ _ _ ¸ ¸ ¸ ¸ _ 1 1 1 1 _ ¸ ¸ ¸ ¸ _ = _ ¸ ¸ ¸ ¸ _ 1 0 −1 1 _ ¸ ¸ ¸ ¸ _ ii. ∆A = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 1 1 1 1 0 0 1 0 1 1 0 1 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −1, ∆A 1 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 1 1 1 1 0 0 1 0 1 1 1 1 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −1 26 CHAPTER 2. LINEAR ALGEBRA ∆A 2 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 1 1 1 1 0 0 1 1 1 1 0 1 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 0, ∆A 3 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 1 1 1 1 1 0 1 0 1 1 0 1 1 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 1 ∆A 4 = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 1 1 1 1 0 1 1 0 1 1 0 1 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −1 Therefore, x 1 = ∆A 1 ∆A = −1 −1 = 1, x 2 = ∆A 2 ∆A = 0, x 3 = ∆A 3 ∆A = −1 and x 4 = ∆A 4 ∆A = 1. (23) (a) By Kirchoff’s Law, one obtains the circuit equations _ ¸ ¸ _ ¸ ¸ _ i 1 −i 2 −i 3 = 0 (R 1 +R 2 +R 3 ) i 1 +R 4 i 2 = E 1 +E 2 R 4 i 2 −R 5 i 3 = E 2 . Cramer’s rule then yields i 1 = −E 1 (R 4 +R 5 ) −E 2 R 5 ∆ , i 2 = −E 1 R 5 −E 2 (R 1 +R 2 +R 3 +R 5 ) ∆ and i 3 = −E 1 R 4 +E 2 (R 1 +R 2 +R 3 ) ∆ , where ∆ = −R 1 R 4 −R 1 R 5 −R 2 R 4 −R 2 R 5 −R 3 R 4 −R 3 R 5 −R 4 R 5 . (b) Using Kirchoff’s Law, currents in the circuit satisfy _ ¸ ¸ _ ¸ ¸ _ i 1 −i 2 −i 3 = 0 R 2 i 2 −R 3 i 3 = 0 (R 1 +R 4 ) i 1 +R 3 i 3 = −E 0 . Using Cramer’s rule, we obtain i 3 = − E 0 R 2 R , i 2 = − E 0 R 3 R and i 1 = − E 0 (R 2 +R 3 ) R , where R = R 1 R 2 +R 1 R 3 +R 2 R 3 +R 2 R 4 +R 3 R 4 . (24) Let x 1 , x 2 and x 3 be the numbers of batches of Milky, Extra Milky and Supreme respec- 27 CHAPTER 2. LINEAR ALGEBRA tively. We therefore obtain _ ¸ ¸ _ ¸ ¸ _ x 1 +x 2 +x 3 = 6 x 1 +2x 2 +3x 3 = 14 x 1 +4x 2 +9x 3 = 36 . Therefore x 1 = 1, x 2 = 2, x 3 = 3. (25) x 1 , x 2 , x 3 , x 4 satisfy _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ −x 1 +x 4 = 100 x 1 −x 2 = 300 −x 3 +x 4 = 500 x 2 −x 3 = 100 . Solving the system, we obtain x 1 = x 4 −100, x 2 = x 4 −400, x 3 = x 4 −500, where x 4 is any integer ≥ 500. (26) (a) The matrices are _ 0.8 0.37 0.2 0.63 _ and _ 0.75 0.15 0.25 0.85 _ . The combined network is repre- sented by _ 0.75 0.15 0.25 0.85 __ 0.8 0.37 0.2 0.63 _ = _ 0.63 0.372 0.37 0.628 _ . (b) _ v w _ = _ 0.75 0.15 0.25 0.85 __ 0.8 0.37 0.2 0.63 __ x y _ = _ 0.63x + 0.372y 0.37x + 0.628y _ . (27) The system may be re-written as _ ¸ ¸ _ R 3 +R 4 +R 6 −R 3 −R 4 −R 3 R 2 +R 3 +R 5 −R 5 −R 4 −R 5 R 1 +R 4 +R 5 _ ¸ ¸ _ _ ¸ ¸ _ i 1 i 2 i 3 _ ¸ ¸ _ = _ ¸ ¸ _ E 0 0 0 _ ¸ ¸ _ . Using Cramer’s rule, one has i 2 = ∆ 2 ∆ and i 3 = ∆ 3 ∆ , where ∆ is the determinant of the coefficient matrix, ∆ 2 = E 0 (R 1 R 3 +R 3 R 4 +R 3 R 5 +R 4 R 5 ) , ∆ 3 = E 0 (R 2 R 4 +R 3 R 4 +R 3 R 5 +R 4 R 5 ) . Therefore, i 2 = i 3 ⇒∆ 2 = ∆ 3 ⇒R 1 R 3 = R 2 R 4 . 28 CHAPTER 2. LINEAR ALGEBRA (28) Consider the homogeneous systems in 3 unknowns _ ¸ ¸ _ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 _ ¸ ¸ _ _ ¸ ¸ _ x 1 x 2 x 3 _ ¸ ¸ _ = _ ¸ ¸ _ 0 0 0 _ ¸ ¸ _ and the non-homogeneous system in 2 unknowns _ ¸ ¸ _ a 1 b 1 a 2 b 2 a 3 b 3 _ ¸ ¸ _ _ y 1 y 2 _ = _ ¸ ¸ _ c 1 c 2 c 3 _ ¸ ¸ _ If y = _ α β _ T is a solution of the nonhomogeneous system, then x = _ α β −1 _ T is a solution of the homogeneous system. Therefore, consistency of the nonhomogeneous system implies existence of nontrivial solutions of the homogeneous system, which also implies the coefficient matrix of the homogeneous system has zero determinant, i.e., det _ _ _ _ _ ¸ ¸ _ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 _ ¸ ¸ _ _ _ _ _ = 0. (29) Elementary row operations give _ ¸ ¸ _ 1 3 −2 5 4 1 3 −2 −1 3 −4 7 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 0 0 0 _ ¸ ¸ _ → _ ¸ ¸ _ 1 0 1 −1 0 1 −1 2 0 0 0 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 0 0 0 _ ¸ ¸ _ . Solutions of the homogeneous system are _ −s +t s −2t s t _ T , where s and t are arbi- trary scalars. Taking s = 3, t = 1, we have −2 _ ¸ ¸ _ 1 4 −1 _ ¸ ¸ _ + _ ¸ ¸ _ 3 1 3 _ ¸ ¸ _ + 3 _ ¸ ¸ _ −2 3 −4 _ ¸ ¸ _ + _ ¸ ¸ _ 5 −2 7 _ ¸ ¸ _ = 0. Hence the given vectors are linearly dependent. (30) Since |u ±v| 2 = ¸u ±v, u ±v) = ¸u, u) + ¸v, v) ± 2 ¸u, v), we deduce that |u +v| 2 + |u −v| 2 = 2 ¸u, u) + 2 ¸v, v) = 2 |u| 2 + 2 |v| 2 . 29 CHAPTER 2. LINEAR ALGEBRA (31) w = _ ¸v, u) |u| 2 _ u = (−4) 1 + 5 (−2) + 6 3 1 2 + (−2) 2 + 3 2 _ ¸ ¸ _ 1 −2 3 _ ¸ ¸ _ = _ ¸ ¸ _ 2 7 − 4 7 6 7 _ ¸ ¸ _ . Furthermore, ¸v −w, u) = _ _ −4 5 6 _ T − _ 2 7 − 4 7 6 7 _ T , _ 1 −2 3 _ T _ = _ _ − 30 7 39 7 36 7 _ T , _ 1 −2 3 _ T _ = − 30 7 − 78 7 + 108 7 = 0. (32) The jth row of Q T is the same as the transpose of the jth column of Q. Therefore, the entry at the jth row and the kth column of the product Q T Q is just n i=1 q ij q ik . This implies (a) ⇔(b). The equivalence of (a) and (c) may be similarily proved by considering the product QQ T . (33) det (A−λI) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ −2 −λ 2 −3 2 1 −λ −6 −1 0 −λ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −λ 3 −λ 2 + 9λ + 9 = −(λ + 3) (λ −3) (λ + 1) Therefore, the eigenvalues of A are λ = 3, λ = −1 and λ = −3. (34) det (A−λI) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 −λ 2 1 0 3 −λ 1 0 5 −1 −λ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −λ 3 + 3λ 2 + 6λ −8 = −(λ −1) (λ + 2) (λ −4) Therefore, the eigenvalues of A are λ = 1, λ = 4 and λ = −2. (35) When λ = 0, (A−λI) v = 0 =⇒ _ ¸ ¸ _ 3 −1 −2 0 2 0 −2 0 2 −1 −1 0 _ ¸ ¸ _ → _ ¸ ¸ _ 1 0 −1 0 0 1 −1 0 0 0 0 0 _ ¸ ¸ _ and v = _ ¸ ¸ _ 1 1 1 _ ¸ ¸ _ r, where r ∈ R. 30 CHAPTER 2. LINEAR ALGEBRA When λ = 1, (A−λI) v = 0 =⇒ _ ¸ ¸ _ 2 −1 −2 0 2 −1 −2 0 2 −1 −2 0 _ ¸ ¸ _ → _ ¸ ¸ _ 1 − 1 2 −1 0 0 0 0 0 0 0 0 0 _ ¸ ¸ _ and v = _ ¸ ¸ _ 1 2 s +t s t _ ¸ ¸ _ = _ ¸ ¸ _ 1 2 1 0 _ ¸ ¸ _ s + _ ¸ ¸ _ 1 0 1 _ ¸ ¸ _ t, where s, t ∈ R. Thus, the eigenvectors are _ ¸ ¸ _ 1 1 1 _ ¸ ¸ _ , _ ¸ ¸ _ 1 2 1 0 _ ¸ ¸ _ and _ ¸ ¸ _ 1 0 1 _ ¸ ¸ _ . Therefore, D = _ ¸ ¸ _ 0 0 0 0 1 0 0 0 1 _ ¸ ¸ _ and P = _ ¸ ¸ _ 1 1 2 1 1 1 0 1 0 1 _ ¸ ¸ _ . AP = _ ¸ ¸ _ 3 −1 −2 2 0 −2 2 −1 −1 _ ¸ ¸ _ _ ¸ ¸ _ 1 1 2 1 1 1 0 1 0 1 _ ¸ ¸ _ = _ ¸ ¸ _ 0 1 2 1 0 1 0 0 0 1 _ ¸ ¸ _ PD = _ ¸ ¸ _ 1 1 2 1 1 1 0 1 0 1 _ ¸ ¸ _ _ ¸ ¸ _ 0 0 0 0 1 0 0 0 1 _ ¸ ¸ _ = _ ¸ ¸ _ 0 1 2 1 0 1 0 0 0 1 _ ¸ ¸ _ = AP (36) When λ = 2, (A−λI) v = 0 =⇒ _ ¸ ¸ _ 0 1 −1 0 1 0 −1 0 1 1 −2 0 _ ¸ ¸ _ → _ ¸ ¸ _ 1 0 −1 0 0 1 −1 0 0 0 0 0 _ ¸ ¸ _ and v = _ ¸ ¸ _ 1 1 1 _ ¸ ¸ _ r, where r ∈ R. When λ = 1, (A−λI) v = 0 =⇒ _ ¸ ¸ _ 1 1 −1 0 1 1 −1 0 1 1 −1 0 _ ¸ ¸ _ → _ ¸ ¸ _ 1 1 −1 0 0 0 0 0 0 0 0 0 _ ¸ ¸ _ and v = _ ¸ ¸ _ −s +t s t _ ¸ ¸ _ = _ ¸ ¸ _ −1 1 0 _ ¸ ¸ _ s + _ ¸ ¸ _ 1 0 1 _ ¸ ¸ _ t, where s, t ∈ R. Thus, the eigenvectors are _ ¸ ¸ _ 1 1 1 _ ¸ ¸ _ , _ ¸ ¸ _ −1 1 0 _ ¸ ¸ _ and _ ¸ ¸ _ 1 0 1 _ ¸ ¸ _ . 31 CHAPTER 2. LINEAR ALGEBRA Therefore, D = _ ¸ ¸ _ 2 0 0 0 1 0 0 0 1 _ ¸ ¸ _ and P = _ ¸ ¸ _ 1 −1 1 1 1 0 1 0 1 _ ¸ ¸ _ . AP = _ ¸ ¸ _ 2 1 −1 1 2 −1 1 1 0 _ ¸ ¸ _ _ ¸ ¸ _ 1 −1 1 1 1 0 1 0 1 _ ¸ ¸ _ = _ ¸ ¸ _ 2 −1 1 2 1 0 2 0 1 _ ¸ ¸ _ PD = _ ¸ ¸ _ 1 −1 1 1 1 0 1 0 1 _ ¸ ¸ _ _ ¸ ¸ _ 2 0 0 0 1 0 0 0 1 _ ¸ ¸ _ = _ ¸ ¸ _ 2 −1 1 2 1 0 2 0 1 _ ¸ ¸ _ = A (37) (a) The characteristic polynomial f(λ) = det _ ¸ ¸ _ −1 −λ 2 −3 2 2 −λ −6 −1 −2 1 −λ _ ¸ ¸ _ = −λ 3 + 2λ 2 + 20λ + 24 = (6 −λ) (λ + 2) 2 . Eigenvalues are λ = 6 and λ = −2(double root). (i) For λ = −2, (A − λI) = _ ¸ ¸ _ 1 2 −3 2 4 −6 −1 −2 3 _ ¸ ¸ _ . We use Gaussian elimination to solve the homogeneous system _ ¸ ¸ _ 1 2 −3 2 4 −6 −1 −2 3 _ ¸ ¸ _ v = 0 and conclude that v 1 = [3 0 1] T and v 2 = [−2 1 0] T are corresponding eigenvectors. (ii) For λ = 6, (A−λI) = _ ¸ ¸ _ −7 2 −3 2 −4 −6 −1 −2 −5 _ ¸ ¸ _ . Use Gaussian elimination to solve the homo- geneous system _ ¸ ¸ _ −7 2 −3 2 −4 −6 −1 −2 −5 _ ¸ ¸ _ v = 0 to obtain v 3 = [−1 −2 1] T as an eigenvector. Finally, P = _ ¸ ¸ _ 3 −2 −1 0 1 −2 1 0 1 _ ¸ ¸ _ diagonalizes A, i.e., P −1 AP = _ ¸ ¸ _ −2 0 0 0 −2 0 0 0 6 _ ¸ ¸ _ . 32 CHAPTER 2. LINEAR ALGEBRA (b) f(λ) = det _ ¸ ¸ _ 0 −λ −2 0 −2 1 −λ −2 0 −2 2 −λ _ ¸ ¸ _ = −λ 3 + 3λ 2 + 6λ −8 = (1 −λ)(λ + 2)(λ −4). Eigenvalues are λ = 1, λ = −2, λ = 4. (i) For λ = 1, (A − λI) = _ ¸ ¸ _ −1 −2 0 −2 0 −2 0 −2 1 _ ¸ ¸ _ . Solving (A − λI)v = 0, we obtain v 1 = [−2 1 2] T as a corresponding eigenvector. (ii) For λ = −2, (A− λI) = _ ¸ ¸ _ 2 −2 0 −2 3 −2 0 −2 4 _ ¸ ¸ _ . Gaussian elimination when applied to the homogeneous system (A−λI)v = 0 gives v 2 = [2 2 1] T as a corresponding eigenvector. (iii) For λ = 4, (A − λI) = _ ¸ ¸ _ −4 −2 0 −2 −3 −2 0 −2 −2 _ ¸ ¸ _ . We now solve the homogeneous system (A − λI)v = 0 and obtain the third eigenvector v 3 = [1 −2 2] T . Therefore, P = _ ¸ ¸ _ −2 2 1 1 2 −2 2 1 2 _ ¸ ¸ _ diagonalizes A, i.e., P −1 AP = _ ¸ ¸ _ 1 0 0 0 −2 0 0 0 4 _ ¸ ¸ _ . (c) f(λ) = det _ ¸ ¸ _ 4 −λ 2 2 2 1 −λ −4 2 −4 1 −λ _ ¸ ¸ _ = −λ 3 + 6λ 2 + 15λ −100 = −(λ + 4) (λ −5) 2 . Eigenvalues of A are λ = 5 (double root), −4. For λ = 5, (A − λI) = _ ¸ ¸ _ −1 2 2 2 −4 −4 2 −4 −4 _ ¸ ¸ _ . We thus obtain two linearly independent eigenvectors v 1 = [2 0 1] T and v 2 = [2 1 0] T by solving the homogeneous system (A−λI)v = 0. For λ = −4, (A−λI) = _ ¸ ¸ _ 8 2 2 2 5 −4 2 −4 5 _ ¸ ¸ _ . Therefore, v 3 = [1 −2 −2] T is a correspond- 33 CHAPTER 2. LINEAR ALGEBRA ing eigenvector. Furthermore, putting P = _ ¸ ¸ _ 2 2 1 0 1 −2 1 0 −2 _ ¸ ¸ _ , we have P −1 AP = _ ¸ ¸ _ 5 0 0 0 5 0 0 0 −4 _ ¸ ¸ _ . (38) (i) (a) det (A−λI) = ¸ ¸ ¸ ¸ ¸ 2 −λ 3 −1 6 −λ ¸ ¸ ¸ ¸ ¸ = λ 2 −8λ + 15 = (λ −3) (λ −5) = 0 Thus, the eigenvalues of A are λ = 3 and λ = 5. When λ = 3, (A−λI) v = 0 =⇒ _ −1 3 0 −1 3 0 _ → _ 1 −3 0 0 0 0 _ and v = _ 3 1 _ s, where s ∈ R. When λ = 5, (A−λI) v = 0 =⇒ _ −3 3 0 −1 1 0 _ → _ 1 −1 0 0 0 0 _ and v = _ 1 1 _ t, where t ∈ R. (b) Using the eigenvalues and eigenvectors, Therefore, D = _ 3 0 0 5 _ and P = _ 3 1 1 1 _ AP = _ 2 3 −1 6 __ 3 1 1 1 _ = _ 9 5 3 5 _ PD = _ 3 1 1 1 __ 3 0 0 5 _ = _ 9 5 3 5 _ = AP (c) D n = _ λ n 1 0 0 λ n 2 _ P −1 = _ 3 1 1 1 _ −1 = _ 1 2 − 1 2 − 1 2 3 2 _ A 6 −6A 4 + 11A 2 = PD 6 P −1 −6PD 4 P −1 + 11PD 2 P −1 = P _ D 6 −6D 4 + 11D 2 _ P −1 = _ 3 1 1 1 ___ 3 6 0 0 5 6 _ −6 _ 3 4 0 0 5 4 _ + 11 _ 3 2 0 0 5 2 ___ 1 2 − 1 2 − 1 2 3 2 _ = _ −5562 17712 −5904 18054 _ 34 CHAPTER 2. LINEAR ALGEBRA (ii) (a) det (A−λI) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ −1 −λ 2 0 1 1 −λ 0 0 0 1 −λ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −λ 3 +λ 2 + 3λ −3 = −(λ −1) _ λ − √ 3 __ λ + √ 3 _ = 0 Thus, the eigenvalues of A are λ = 1, λ = √ 3, and λ = − √ 3. When λ = 1, (A−λI) v = 0 =⇒ v = _ ¸ ¸ _ 0 0 1 _ ¸ ¸ _ r, where r ∈ R. When λ = √ 3, (A−λI) v = 0 =⇒ v = _ ¸ ¸ _ √ 3 −1 1 0 _ ¸ ¸ _ s, where s ∈ R. When λ = − √ 3, (A−λI) v = 0 =⇒ v = _ ¸ ¸ _ − √ 3 −1 1 0 _ ¸ ¸ _ t, where t ∈ R. (b) Using the eigenvalues and eigenvectors, Therefore, D = _ ¸ ¸ _ 1 0 0 0 √ 3 0 0 0 − √ 3 _ ¸ ¸ _ and P = _ ¸ ¸ _ 0 √ 3 −1 − √ 3 −1 0 1 1 1 0 0 _ ¸ ¸ _ AP = _ ¸ ¸ _ −1 2 0 1 1 0 0 0 1 _ ¸ ¸ _ _ ¸ ¸ _ 0 √ 3 −1 − √ 3 −1 0 1 1 1 0 0 _ ¸ ¸ _ = _ ¸ ¸ _ 0 3 − √ 3 3 + √ 3 0 √ 3 − √ 3 1 0 0 _ ¸ ¸ _ PD = _ ¸ ¸ _ 0 √ 3 −1 − √ 3 −1 0 1 1 1 0 0 _ ¸ ¸ _ _ ¸ ¸ _ 1 0 0 0 √ 3 0 0 0 − √ 3 _ ¸ ¸ _ = _ ¸ ¸ _ 0 3 − √ 3 3 + √ 3 0 √ 3 − √ 3 1 0 0 _ ¸ ¸ _ 35 CHAPTER 2. LINEAR ALGEBRA (c) P −1 = _ ¸ ¸ _ 0 √ 3 −1 − √ 3 −1 0 1 1 1 0 0 _ ¸ ¸ _ −1 = _ ¸ ¸ _ 0 0 1 √ 3 6 3+ √ 3 6 0 − √ 3 6 3− √ 3 6 0 _ ¸ ¸ _ A 6 −6A 4 + 11A 2 = P _ D 6 −6D 4 + 11D 2 _ P −1 = _ ¸ ¸ _ 0 √ 3 −1 − √ 3 −1 0 1 1 1 0 0 _ ¸ ¸ _ _ ¸ ¸ _ 6 0 0 0 6 0 0 0 6 _ ¸ ¸ _ _ ¸ ¸ _ 0 0 1 √ 3 6 3+ √ 3 6 0 − √ 3 6 3− √ 3 6 0 _ ¸ ¸ _ = _ ¸ ¸ _ 6 0 0 0 6 0 0 0 6 _ ¸ ¸ _ (39) det (A−λI) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 4 −λ 2 0 4 0 2 −λ −1 0 0 0 3 −λ 3 0 4 0 7 −λ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = λ 4 −16λ 3 + 89λ 2 −194λ + 120 = (λ −1) (λ −4) (λ −5) (λ −6) = 0. Thus, the fourth eigenvalue of A is λ = 5. When λ = 5, (A−λI) v = 0 =⇒ _ ¸ ¸ ¸ ¸ _ −1 2 0 4 0 0 −3 −1 0 0 0 0 −2 3 0 0 4 0 2 0 _ ¸ ¸ ¸ ¸ _ → _ ¸ ¸ ¸ ¸ _ 1 0 0 −3 0 0 1 0 1 2 0 0 0 1 − 3 2 0 0 0 0 0 0 _ ¸ ¸ ¸ ¸ _ . There- fore, v = _ ¸ ¸ ¸ ¸ _ 3 − 1 2 3 2 1 _ ¸ ¸ ¸ ¸ _ s, where s ∈ R. and, the eigenvector associated with λ = 5 is _ ¸ ¸ ¸ ¸ _ 6 −1 3 2 _ ¸ ¸ ¸ ¸ _ . 36 CHAPTER 2. LINEAR ALGEBRA (40) det (A−λI) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 −λ 2 0 2 0 −2 −λ −1 0 0 0 3 −λ 3 0 0 0 −1 −λ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = λ 4 −λ 3 −7λ 2 +λ + 6 = (λ −1) (λ −3) (λ + 2) (λ + 1) = 0 Thus, the eigenvalues of A are λ = 1, λ = 3, λ = −1, and λ = −2. When λ = 1, v = _ ¸ ¸ ¸ ¸ _ 1 0 0 0 _ ¸ ¸ ¸ ¸ _ . When λ = 3, v = _ ¸ ¸ ¸ ¸ _ 1 1 −5 0 _ ¸ ¸ ¸ ¸ _ . When λ = −1, v = _ ¸ ¸ ¸ ¸ _ 7 −3 3 −4 _ ¸ ¸ ¸ ¸ _ . When λ = −2, v = _ ¸ ¸ ¸ ¸ _ 2 −3 0 0 _ ¸ ¸ ¸ ¸ _ . (41) Assume A is a matrix with eigenvalue λ and eigenvector x, i.e., Ax = λx. A 2 x = A(Ax) = A(λx) = λ(Ax) = λ(λx) = λ 2 x (42) (a) For A = _ 0.7 0.4 0.3 0.6 _ , f(λ) = (λ −0.3)(λ −1). Therefore, eigenvalues are λ 1 = 0.3 and λ 2 = 1, with eigenvectors v 1 = _ 1 −1 _ T and v 2 = _ 4 3 _ T respectively. Taking P = _ 1 4 −1 3 _ , we have A = P _ 0.3 0 0 1 _ P −1 . Therefore, A k = P _ 0.3 k 0 0 1 _ P −1 = _ 1 4 −1 3 __ 0.3 k 0 0 1 __ 3 7 − 4 7 1 7 1 7 _ = 1 7 _ 4 + 3 0.3 k 4 −4 0.3 k 3 −3 0.3 k 3 + 4 0.3 k _ . (b) If the Red Party is in power now, the probability that it is to be in power after k elections is _ 1 0 _ A k _ 1 0 _ = 4 + 3 0.3 k 7 . If the Red Party is not in power now, the probability that it is to be in power after k elections is _ 0 1 _ A k _ 0 1 _ 3 −3 0.3 k 7 . 37 CHAPTER 2. LINEAR ALGEBRA When k →∞, the respective probabilities are 4 7 and 3 7 . (43) (a) _ 1 2 1 0 __ a n−1 a n−2 _ = _ a n−1 + 2a n−2 a n−1 _ = _ a n a n−1 _ for n ≥ 3. (b) A = _ 1 2 1 0 _ has eigenvalues λ 1 = 2 and λ 2 = −1, with eigenvectors v 1 = _ 2 1 _ and v 2 = _ −1 1 _ respectively. Therefore, one obtains A = _ 2 −1 1 1 __ 2 0 0 −1 __ 2 −1 1 1 _ −1 and A k = _ 2 −1 1 1 __ 2 k 0 0 (−1) k __ 1 3 1 3 − 1 3 2 3 _ = 1 3 _ 2 k+1 + (−1) k 2 k+1 −2 (−1) k 2 k −(−1) k 2 k + 2 (−1) k _ . (c) _ a n a n−1 _ = _ 1 2 1 0 __ a n−1 a n−2 _ = _ 1 2 1 0 _ 2 _ a n−2 a n−3 _ = . . . = _ 1 2 1 0 _ n−2 _ a 2 a 1 _ Therefore, a n = 2 n −(−1) n 3 for n ≥ 3. (44) (a) Eigenvalues Eigenvectors Orthonormal eigenvectors λ = 3 _ 1 1 1 _ T _ 1 √ 3 1 √ 3 1 √ 3 _ T λ = 6 _ 1 1 −2 _ T _ 1 √ 6 1 √ 6 − 2 √ 6 _ T λ = 8 _ 1 −1 0 _ T _ 1 √ 2 − 1 √ 2 0 _ T Therefore, Q = _ ¸ ¸ _ 1 √ 3 1 √ 6 1 √ 2 1 √ 3 1 √ 6 −1 √ 2 1 √ 3 −2 √ 6 0 _ ¸ ¸ _ and D = _ ¸ ¸ _ 3 0 0 0 6 0 0 0 8 _ ¸ ¸ _ (b) Eigenvalues Eigenvectors λ = 7 (double root) _ 1 −2 0 _ T and _ 0 2 1 _ T λ = −2 _ 2 1 −2 _ T 38 CHAPTER 2. LINEAR ALGEBRA _ 1 −2 0 _ T and _ 0 2 1 _ T are not orthogonal to each other, although each of them is orthogonal to _ 2 1 −2 _ T . Using Gram Schmidt Process, one obtains 3 orthonor- mal eigenvectors w 1 = _ 1 √ 5 − 2 √ 5 0 _ T , w 2 = _ 4 3 √ 5 2 3 √ 5 5 3 √ 5 _ T , w 3 = _ 2 3 1 3 − 2 3 _ T . Therefore, Q = _ ¸ ¸ _ 1 √ 5 4 3 √ 5 2 3 − 2 √ 5 2 3 √ 5 1 3 0 5 3 √ 5 − 2 3 _ ¸ ¸ _ and D = _ ¸ ¸ _ 7 0 0 0 7 0 0 0 −2 _ ¸ ¸ _ 39 Chapter 3 Infinite series, Power series and Fourier series (1) (a) Convergent, Comparison with 1 n 3/2 . 1 _ n(n 2 + 1) = 1 √ n 3 +n ≤ 1 √ n 3 . (b) Divergent, Comparison with 1 n . π/4 n ≤ tan −1 n n (c) Convergent, Comparison with 1 n 2 . lnn n 3 ≤ n n 3 = 1 n 2 (d) Convergent, Comparison with 1 n 3 . 1 (n + 1)(n + 2)(n + 3) ≤ 1 n 3 (e) Convergent, ratio test ((n+1)!) 2 (2(n+1))! (n!) 2 (2n)! = (2n)!((n + 1)!) 2 (2(n + 1))!(n!) 2 = (n + 1) 2 (2n + 1)(2n + 2) = n + 1 4n + 2 → 1 4 41 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES (f) Convergent, ratio test (n+1)! (n+1) n+1 n! n n = (n + 1)!n n n!(n + 1) n+1 = n n (n + 1) n = 1 _ n+1 n _ n = 1 _ 1 + 1 n _ n → 1 e (g) Convergent, ratio test x n+1 (n+1) n+1 x n n n = x n+1 n n x n (n + 1) n+1 = xn n (n + 1) n+1 = n n (n + 1) n x n + 1 →0 (h) Divergent, integral test _ ∞ a dx xlnxln(lnx) = [ln(ln(lnx))] ∞ a (i) Convergent, integral test _ ∞ a e n 9 +e 2n dx = _ 1 3 tan −1 _ e x 3 __ ∞ a (j) Divergent, integral test _ ∞ a 2 ln(ln n) nlnn dx = _ 2 ln(ln x) ln2 _ ∞ a (2) (a) Conditionally. The series is not absolutely convergent because x < sin −1 x for 0 < x ≤ 1 , i.e. 1 n ≤ sin −1 _ 1 n _ . Since sin −1 x is an increasing continuous function, we have sin −1 _ 1 n _ > sin −1 _ 1 n + 1 _ n→∞ −−−→0. By Leibniz test, the series converges conditionally. (b) Absolutely ¸ ¸ ¸ ¸ (−1) n n 3 e n ¸ ¸ ¸ ¸ = n 3 e n , which converges by ratio test. 42 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES (c) Absolutely ¸ ¸ ¸ ¸ (−1) n 2 + 2n +n 2 ¸ ¸ ¸ ¸ = 1 2 + 2n +n 2 ≤ 1 n 2 , which converges by Comparison test. (d) Conditionally ¸ ¸ ¸ ¸ (−1) n nlnn ¸ ¸ ¸ ¸ = 1 nlnn , which diverges by Integral test. Since xlnx is an increasing function for x > 1, 1 nlnn > 1 (n + 1) ln(n + 1) n→∞ −−−→0. By Leibniz test, the series converges conditionally. (3) Consider the k-th partial sum: s k = k n=1 1 (4n −1)(4n + 3) = k n=1 _ 1 4(4n −1) − 1 4(4n + 3) _ = _ 1 12 − 1 28 _ + _ 1 28 − 1 44 _ + + _ 1 4(4n −1) − 1 4(4n + 3) _ = 1 12 − 1 4(4k + 3) → 1 12 (4) (a) [−1, 1), 0 < s ≤ 1; [−1, 1], s > 1 R = lim n→∞ ¸ ¸ ¸ ¸ a n a n+1 ¸ ¸ ¸ ¸ = lim n→∞ ¸ ¸ ¸ ¸ ¸ 1 n s 1 (n+1) s ¸ ¸ ¸ ¸ ¸ = lim n→∞ ¸ ¸ ¸ ¸ _ 1 + 1 n _ s ¸ ¸ ¸ ¸ = 1 At x = 1, the series becomes 1 n s which converges if s > 1 and diverges if s ≤ 1. At x = −1, the series becomes the alternating series (−1) n n s , which converges absolutely if s > 1 and conditionally if s ≤ 1. (b) (−e, e) R = lim n→∞ ¸ ¸ ¸ ¸ a n a n+1 ¸ ¸ ¸ ¸ = lim n→∞ ¸ ¸ ¸ ¸ ¸ ¸ n! n n (n+1)! (n+1) n+1 ¸ ¸ ¸ ¸ ¸ ¸ = lim n→∞ _ n + 1 n _ n = e 43 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES At x = e, the series becomes n!e n n n , which diverges because a n = n!e n n n = n! n n _ 1 +n + + n n n! + _ > 1 and therefore a n does not converges to 0. Similarly, at x = −e, the series becomes the alternating series (−1) n n!e n n n , which diverges because b n = (−1) n n!e n n n does not converge to 0. (c) (−1, 1) R = lim n→∞ ¸ ¸ ¸ ¸ a n a n+1 ¸ ¸ ¸ ¸ = lim n→∞ ¸ ¸ ¸ ¸ n n + 1 ¸ ¸ ¸ ¸ = 1 At x = −1 or x = 1, the terms of the series are (−1) n n and n respectively, and do not converge to 0. (d) (−2, 0] R = lim n→∞ ¸ ¸ ¸ ¸ a n a n+1 ¸ ¸ ¸ ¸ = lim n→∞ ¸ ¸ ¸ ¸ ¸ (−1) n−1 n (−1) n n+1 ¸ ¸ ¸ ¸ ¸ = lim n→∞ n + 1 n = 1 Translating to the left by 1 units, we are interested in the interval (−2, 0) At x = −2, the series becomes (−1) 2n−1 n = − 1 n , which is the negative harmonic series and diverges. At x = 0, the series becomes (−1) n−1 n , which is the alternating harmonic series and converges. (e) (−4, 0) R = lim n→∞ ¸ ¸ ¸ ¸ a n a n+1 ¸ ¸ ¸ ¸ = lim n→∞ ¸ ¸ ¸ ¸ ¸ n 2 2 n (n+1) 2 2 n+1 ¸ ¸ ¸ ¸ ¸ = lim n→∞ 2n 2 (n + 1) 2 = 2 Translating to the left by 2 units, we are interested in the interval (−4, 0). At x = −4 or x = 1, the terms of the series are (−1) 2 n 2 and n 2 respectively, and do not converge to 0. (5) The Maclaurin Series of a function f(x) is given by f(x) = f(0) +f ′ (0)x + f ′′ (0) 2! x 2 + f ′′′ (0) 3! x 3 + f (4) (0) 4! x 4 + (a) 1 −2x + 2x 2 − 4 3 x 3 + 2 3 x 4 Note: e −2x = ∞ n=0 (−2x) n n! = ∞ n=0 (−2) n n! x n 44 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES (b) 2x − 4 3 x 3 Note sin(2x) = ∞ n=0 (−1) n (2x) 2n+1 (2n + 1)! (c) x 2 +x 3 + x 4 2! + x 5 3! Note x 2 e x = x 2 ∞ n=0 x n n! = ∞ n=0 x n+2 n! (d) ln3 + 2 3 x − 2 9 x 2 + 8 81 x 3 − 4 81 x 4 Note ln(3 + 2x) = ln(1 + 2(1 +x)) = ∞ n=1 (−1) n−1 n [2(1 +x)] n (6) (a) Let z = ax. Then e z = ∞ n=0 z n n! = ∞ n=0 (ax) n n! (b) Let z = x 2 . Then cos z = ∞ n=0 (−1) n z 2n (2n)! = ∞ n=0 (−1) n x 2n 4 n (2n)! (c) e x = 1 +x + x 2 2! + x 3 3! + + x n n! + e −x = 1 −x + x 2 2! − x 3 3! + + x n n! + cosh x = e x +e −x 2 = ∞ n=0 x 2n (2n)! (d) ln[1 +x[ = x − 1 2 x 2 + 1 3 x 3 − 1 4 x 4 + + (−1) n−1 n x n + ln[1 −x[ = −x + 1 2 x 2 − 1 3 x 3 + 1 4 x 4 + + (−1) n n x n + 1 2 ln ¸ ¸ ¸ ¸ 1 +x 1 −x ¸ ¸ ¸ ¸ = 1 2 (ln[1 +x[ −ln[1 −x[) = ∞ n=1 x 2n−1 2n −1 45 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES (7) (a) ∞ n=1 (−1) n+1 n (x −1) n lnx = ln(1 + (x −1)) = ∞ n=1 (−1) n+1 n (x −1) n (b) ∞ n=0 (−1)(x + 1) n 1 x = − 1 1 −(x + 1) = − ∞ n=0 (x + 1) n = ∞ n=0 (−1)(x + 1) n (c) ∞ n=0 (−1) n π 2n (2n)! _ x − 1 2 _ 2n sinπx = cos _ π _ x − 1 2 __ = ∞ n=0 (−1) n (2n)! _ π _ x − 1 2 __ 2n (8) (a) Since [x[ is an even function, all b n = 0. a 0 = 1 π _ π −π [x[ dx = 1 π _ π 0 xdx + 1 π _ 0 −π −xdx = π a n = 1 π _ π −π [x[ cos nxdx = 2 1 π _ π 0 xcos nxdx = 2 π −1 + cos πn n 2 = 2 π (−1) n −1 n 2 Hence, f (x) = π 2 + 2 π ∞ n=1 (−1) n −1 n 2 cos nx Note: Choose f = x, g ′ = cos nx and use integration by parts, we get _ xcos nxdx = x sinnx n − _ sinnx n dx = x sinnx n + cos nx n 2 (b) f (x) = sin 2 x −2 cos 3 x = 1 2 (1 −cos 2x) −2 _ cos 2 x _ cos x = 1 2 (1 −cos 2x) −(1 + cos 2x) cos x = 1 2 (1 −cos 2x) −cos x −cos 2xcos x = 1 2 (1 −cos 2x) −cos x − 1 2 (cos 3x + cos x) = 1 2 − 3 2 cos x − 1 2 cos 2x − 1 2 cos 3x 46 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES Note: cos 2x = 1 −2 sin 2 x = 2 cos 2 x −1 cos mxcos nx = 1 2 [cos (m+n) x + cos (m−n) x] (c) a 0 = 1 π _ π 0 sin tdt = 2 π For n ,= 1, b n = 1 π _ π 0 sin t sinntdt = 1 π _ π 0 − 1 2 [cos (1 +n) t −cos (1 −n) t] dt = _ − 1 2 sin(1 +n) t 1 +n + 1 2 sin (1 −n) t 1 −n _ π 0 = 0 For n = 1, b 1 = 1 π _ π 0 sint sintdt = 1 2π _ π 0 (1 −cos 2t) dt = _ 1 2π _ t − 1 2 sin2t __ π 0 = 1 2 a n = 1 π _ π 0 sint cos ntdt = 1 π _ π 0 1 2 [sin(1 +n) t + sin(1 −n) t] dt = _ 1 π _ − 1 2 cos (1 +n) t 1 +n − 1 2 cos (1 −n) t 1 −n __ π 0 = − 1 π cos πn + 1 n 2 −1 = − 1 π (−1) n + 1 n 2 −1 Hence, if n is even, i.e. n = 2m, a 2m = − 1 π 2 4m 2 −1 if n is odd a n = 0 f (x) = 1 π + 1 2 sint − 1 π ∞ m=1 2 4m 2 −1 cos 2mt Note: sin mxsinnx = − 1 2 [cos (m+n) x −cos (m−n) x] sinmxcos nx = 1 2 [sin(m+n) x + sin (m−n) x] 47 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES (9) (a) a 0 = 2 π _ π 0 x(π −x) dx = π 2 3 a n = 2 π _ π 0 x(π −x) cos nxdx = −2 cos πn + 1 n 2 = −2 (−1) n + 1 n 2 b n = 2 π _ π 0 x(π −x) sinnxdx = − 4 π cos πn −1 n 3 = 4 π 1 −(−1) n n 3 The half range cosine series of f (x) is f (x) = π 2 6 −2 ∞ n=1 (−1) n + 1 n 2 cos nx = π 2 3 −2 ∞ m=1 (−1) 2m + 1 4m 2 cos 2mx = π 2 6 − ∞ m=1 1 m 2 cos 2mx The half range sine series of f (x) is f (x) = 4 π ∞ n=1 1 −(−1) n n 3 sinnx = 8 π ∞ m=1 1 (2m−1) 3 sin(2m−1) x (b) a 0 = 4 π _ π/2 0 cos xdx = 4 π a n = 4 π _ π/2 0 cos xcos 2nxdx = − 4 π cos πn 4n 2 −1 = 4 π (−1) n−1 4n 2 −1 f (x) = 2 π + 4 π ∞ m=1 (−1) n−1 4n 2 −1 cos 2nx (10) (a) Since [sinx[ is an even function, all b n = 0. a 0 = 1 π _ π −π [sinx[ dx = 1 π _ π 0 sinxdx + 1 π _ 0 −π −sinxdx = 4 π a n = 1 π _ π −π [sinx[ cos nxdx = 2 1 π _ π 0 sinxcos nxdx = − 2 π cos πn + 1 −1 +n 2 = 2 π 1 + (−1) n n 2 −1 48 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES Hence, f (x) = 2 π − 4 π ∞ m=1 1 4m 2 −1 cos 2mx (b) a 0 = 1 π _ π −π e x dx = 1 π _ e π −e −π _ a n = 1 π _ π −π e x cos nxdx = 1 π __ n 2 n 2 + 1 __ e x sinnx n −e x cos nx n 2 __ π −π = 1 π _ n 2 n 2 + 1 _ e π cos nπ n 2 −e −π cos nπ n 2 _ _ = 1 π _ (−1) n n 2 + 1 _ e π −e −π _ _ b n = 1 π _ π −π e x sinnxdx = _ e x sinnx − n 2 n 2 −1 _ e x sinnx n −e x cos nx n 2 __ π 0 = − n π _ (−1) n n 2 + 1 _ e π −e −π _ _ f (x) = 1 2π _ e π −e −π _ + ∞ n=1 1 π _ (−1) n n 2 + 1 _ e π −e −π _ cos nx _ + ∞ n=1 n π _ (−1) n n 2 + 1 _ e π −e −π _ sinnx _ = 1 π _ e π −e −π _ _ 1 2 + ∞ n=1 (−1) n n 2 + 1 (cos nx −nsinnx) _ = 2 sinhπ π _ 1 2 + ∞ n=1 (−1) n n 2 + 1 (cos nx −nsinnx) _ _ e x cos nxdx = e x sinnx n − _ e x sinnx n dx = e x sinnx n +e x cos nx n 2 + _ _ −e x cos nx n 2 _ dx 49 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES Therefore, _ e x cos nxdx = _ n 2 n 2 −1 __ e x sinnx n −e x cos nx n 2 _ _ e x sinnx n dx = e x sinnx n − _ e x cos nxdx = e x sinnx − _ n n 2 −1 __ e x sin nx n −e x cos nx n 2 _ Note: sinhx = e x −e −x 2 (11) a 0 = 1 π _ π 0 (x −π) 2 dx + 1 π _ 2π π π 2 dx = 1 3 π 2 +π 2 = 4 3 π 2 a n = 1 π _ π 0 (x −π) 2 cos nxdx + 1 π _ 2π π π 2 cos nxdx = 2 π −sinπn +πn n 3 + _ π sin nx n _ 2π π = 2 n 2 b n = 1 π _ π 0 (x −π) 2 sinnxdx + 1 π _ 2π π π 2 sinnxdx = 1 π 2 cos πn −2 +n 2 π 2 n 3 − _ π cos nx n _ 2π π = 2 π (−1) n −1 n 3 + π n − π n + (−1) n π n = 2 π (−1) n −1 n 3 + (−1) n π n f (x) = 2 3 π 2 + ∞ n=1 2 n 2 cos nx + ∞ n=1 _ 2 π (−1) n −1 n 3 + (−1) n π n _ sinnx 50 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES Put x = 0 π 2 = 2 3 π 2 + 2 ∞ n=1 1 n 2 π 2 3 = 2 ∞ n=1 1 n 2 ∞ n=1 1 n 2 = π 2 6 Put x = π, π 2 2 = 1 2 _ f _ π − _ +f _ π + __ = 2π 2 3 + 2 ∞ n=1 cos nπ n 2 − π 2 6 = 2 ∞ n=1 (−1) n n 2 ∞ n=1 (−1) n−1 n 2 = π 2 12 Note: _ (x −π) 2 cos nxdx = (x −π) 2 sinnx n − _ (2x −2π) sinnx n dx = (x −π) 2 sinnx n + (2x −2π) cos nx n 2 − _ _ −2 cos nx n 2 _ dx = (x −π) 2 sinnx n + (2x −2π) cos nx n 2 + 2 sin nx n 3 _ (x −π) 2 sinnxdx = −(x −π) 2 cos nx n + _ _ (2x −2π) cos nx n _ dx = −(x −π) 2 cos nx n + (2x −2π) sin nx n 2 − _ 2 sinnx n 2 dx = −(x −π) 2 cos nx n + (2x −2π) sin nx n 2 −2 cos nx n 3 (12) (a) a 0 = 1 π _ 0 −π −1dx = −1 a n = 1 π _ 0 −π −cos nxdx = − sinπn πn = 0 b n = 1 π _ 0 −π −sinnxdx = 1 −cos πn πn = 1 −(−1) n πn 51 CHAPTER 3. INFINITE SERIES, POWER SERIES AND FOURIER SERIES f (x) = − 1 2 + ∞ n=1 1 −(−1) n πn sinnx (b) Since x 2 is an even function, all b n = 0. a 0 = _ 1 −1 x 2 dx = 2 3 a n = _ 1 −1 x 2 cos nπxdx = _ x 2 sinnπx nπ + 2x cos nπx n 2 π 2 + 2 sinnπx n 3 π 3 _ 1 −1 = (−1) n 4 n 2 π 2 Therefore, f (x) = 1 3 + ∞ n=1 4 n 2 π 2 (−1) n cos nπx Note: Choose f = x 2 and g ′ = cos nπx, _ x 2 cos nπxdx = x 2 sinnπx nπ − _ 2x sinnπx nπ dx. Choose f = 2x and g ′ = sinnπx, _ 2x sinnπx nπ dx = −2x cos nπx n 2 π 2 − _ _ −2 cos nπx n 2 π 2 _ dx = −2x cos nπx n 2 π 2 −2 sinnπx n 3 π 3 Hence, _ x 2 cos nπxdx = x 2 sin nπx nπ + 2x cos nπx n 2 π 2 + 2 sin nπx n 3 π 3 52 Chapter 4 Partial Differentiation (1) f (x, y) = 2x−3y x+y Suppose that (x, y) approaches (0, 0) along the x-axis. Then lim x→0 y=0 f (x, y) = lim x→0 2x x = 2. Suppose that (x, y) approaches (0, 0) along the y-axis. Then lim y→0 x=0 f (x, y) = lim x→0 −3y y = −3. Since the limits are not equal, then the limit does not exist. (2) w = _ x 2 +y 2 ∂w ∂x = ∂ _ √ x 2 +y 2 _ ∂x = _ 1 2 _ _ x 2 +y 2 _ − 1 2 (2x) = x √ x 2 +y 2 ∂w ∂y = y √ x 2 +y 2 Using polar coordinates of the point (x, y), i.e., x = r cos θ and y = r sin θ, lim r→0 r= √ x 2 +y 2 x _ x 2 +y 2 = lim r→0 r cos θ _ (r cos θ) 2 + (r sin θ) 2 = cos θ and lim r→0 r= √ x 2 +y 2 y _ x 2 +y 2 = lim r→0 r sinθ _ (r cos θ) 2 + (r sin θ) 2 = sinθ. 53 CHAPTER 4. PARTIAL DIFFERENTIATION Since both of the limits are dependent on θ, then there is no unique value for the derivatives for w at (0, 0). Therefore, the derivatives do not exist. (3) By the product formula, we have ∂w ∂t = e − x 2 4t _ − 1 2 t − 3 2 + x 2 4 t − 5 2 _ = ∂ 2 w ∂x 2 . (4) (a) ∂w ∂x = e x cos y + 2x + y, ∂w ∂y = −e x siny + x − 2y, ∂ 2 w ∂x 2 = e x cos y + 2, ∂ 2 w ∂y 2 = −e x cos y −2y, ∂ 2 w ∂y∂x = −e x sin y + 1 = ∂ 2 w ∂x∂y . (b) ∂w ∂x = 2x x 2 +y 2 −z 2 , ∂w ∂y = 2y x 2 +y 2 −z 2 , ∂w ∂z = −2z x 2 +y 2 −z 2 , ∂ 2 w ∂x 2 = 2(y 2 −x 2 −z 2 ) (x 2 +y 2 −z 2 ) 2 , ∂ 2 w ∂y 2 = 2(x 2 −y 2 −z 2 ) (x 2 +y 2 −z 2 ) 2 , ∂ 2 w ∂z 2 = − 2(x 2 +y 2 +z 2 ) (x 2 +y 2 −z 2 ) 2 , ∂ 2 w ∂x∂y = −4xy (x 2 +y 2 −z 2 ) 2 , ∂ 2 w ∂z∂y = 4yz (x 2 +y 2 −z 2 ) 2 , ∂ 2 w ∂z∂x = 4zx (x 2 +y 2 −z 2 ) 2 . (c) ∂w ∂x = e 2x+y cos (x −y) + 2e 2x+y sin(x −y) = e 2x+y [cos (x −y) + 2 sin(x −y)] ∂w ∂y = −e 2x+y cos (x −y) +e 2x+y sin(x −y) = e 2x+y [sin(x −y) −cos (x −y)] ∂ 2 w ∂x 2 = 2e 2x+y [cos (x −y) + 2 sin(x −y)] +e 2x+y [−sin(x −y) + 2 cos (x −y)] = e 2x+y [4 cos (x −y) + 3 sin(x −y)] ∂ 2 w ∂y 2 = e 2x+y [sin(x −y) −cos (x −y)] +e 2x+y [−cos (x −y) −sin(x −y)] = −2e 2x+y cos (x −y) ∂ 2 w ∂x∂y = 2e 2x+y [sin(x −y) −cos (x −y)] +e 2x+y [cos (x −y) + sin(x −y)] = e 2x+y [3 sin(x −y) −cos (x −y)] ∂ 2 w ∂y∂x = e 2x+y [cos (x −y) + 2 sin (x −y)] +e 2x+y [sin(x −y) −2 cos (x −y)] = e 2x+y [3 sin(x −y) −cos (x −y)] 54 CHAPTER 4. PARTIAL DIFFERENTIATION (d) ∂w ∂x = 1 √ x 2 +2y 2 _ 1 2 _ _ x 2 + 2y 2 _ −1/2 (2x) = x x 2 +2y 2 ∂w ∂y = 1 √ x 2 +2y 2 _ 1 2 _ _ x 2 + 2y 2 _ −1/2 (4y) = 2y x 2 +2y 2 ∂ 2 w ∂x 2 = (x 2 +2y 2 )−x(2x) (x 2 +2y 2 ) 2 = 2y 2 −x 2 (x 2 +2y 2 ) 2 ∂ 2 w ∂y 2 = 2(x 2 +2y 2 )−2y(4y) (x 2 +2y 2 ) 2 = 2x 2 −4y 2 (x 2 +2y 2 ) 2 ∂ 2 w ∂x∂y = −2y(2x) (x 2 +2y 2 ) 2 = −4xy (x 2 +2y 2 ) 2 ∂ 2 w ∂y∂x = −x(4y) (x 2 +2y 2 ) 2 = −4xy (x 2 +2y 2 ) 2 (e) ∂w ∂x = 2e 2x−y 2 sin(x +y) +e 2x−y 2 cos(x +y) = e 2x−y 2 [2 sin(x +y) + cos(x +y)] ∂w ∂y = −2ye 2x−y 2 sin(x +y) +e 2x−y 2 cos(x +y) = e 2x−y 2 [cos(x +y) −2y sin(x +y)] ∂ 2 w ∂x 2 = 2e 2x−y 2 [2 sin(x +y) + cos(x +y)] +e 2x−y 2 [2 cos(x +y) −sin(x +y)] = e 2x−y 2 [4 cos (x +y) + 3 sin (x +y)] ∂ 2 w ∂y 2 = −2ye 2x−y 2 [cos(x +y) −2y sin(x +y)] +e 2x−y 2 [−sin(x +y) −2 sin(x +y) −2y cos (x +y)] = e 2x−y 2 _ −4y cos (x +y) + _ 4y 2 −3 _ sin (x +y) ¸ ∂ 2 w ∂x∂y = 2e 2x−y 2 [cos(x +y) −2y sin(x +y)] +e 2x−y 2 [−sin(x +y) −2y cos(x +y)] = e 2x−y 2 [(2 −2y) cos (x +y) −(4y + 1) sin (x +y)] ∂ 2 w ∂y∂x = −2ye 2x−y 2 [2 sin(x +y) + cos(x +y)] +e 2x−y 2 [2 cos(x +y) −sin(x +y)] = e 2x−y 2 [(2 −2y) cos (x +y) −(4y + 1) sin (x +y)] = ∂ 2 w ∂x∂y (f) ∂w ∂x = y x lny +ye xy lny 55 CHAPTER 4. PARTIAL DIFFERENTIATION ∂w ∂y = xy x−1 +xe xy lny + 1 y e xy ∂ 2 w ∂x 2 = (lny) 2 y x +y 2 e xy lny ∂ 2 w ∂y 2 = x(x −1) y x−2 +x 2 e xy lny + 2 x y e xy − 1 y 2 e xy ∂ 2 w ∂x∂y = y x−1 +xy x−1 lny +e xy lny +xye xy lny +e xy ∂ 2 w ∂y∂x = xy x−1 lny +y x−1 +e xy (lny + 1) +xye xy lny = ∂ 2 w ∂x∂y (5) (a) ∂w ∂x = sin(2y −3z 2 ), ∂w ∂y = 2xcos(2y −3z 2 ), ∂w ∂z = −6xz cos(2y −3z 2 ). (b) ∂w ∂x = 2Ax +Dy +Fz, ∂w ∂y = 2By +Dx +Ez, ∂w ∂z = 2Cz +Ey +Fx. (c) ∂w ∂x = −(2y)−sin x) (y+cos x) 2 = 2y sin x (y+cos x) 2 , ∂w ∂y = 2(y+cos x)−(2y)(1) (y+cos x) 2 = 2 cos x (y+cos x) 2 . (d) ∂w ∂x = ye x 2 +y 2 + 2x 2 ye x 2 +y 2 , ∂w ∂y = xe x 2 +y 2 + 2xy 2 e x 2 +y 2 . (e) ∂w ∂x = (x 2 +2y 2 )−(x−y)(2x) (x 2 +2y 2 ) 2 = x 2 +2y 2 −2x 2 +2xy (x 2 +2y 2 ) 2 = 2y 2 −x 2 +2xy (x 2 +2y 2 ) 2 , ∂w ∂y = (x 2 +2y 2 )(−1)−(x−y)(4y) (x 2 +2y 2 ) 2 = −x 2 −2y 2 −4xy+4y 2 (x 2 +2y 2 ) 2 = 2y 2 −x 2 −4xy (x 2 +2y 2 ) 2 . (6) 1 R = 1 R 1 + 1 R 2 + 1 R 3 = R 2 R 3 +R 1 R 3 +R 1 R 2 R 1 R 2 R 3 R = R 1 R 2 R 3 R 2 R 3 +R 1 R 3 +R 1 R 2 ∂R ∂R 1 = (R 2 R 3 +R 1 R 3 +R 1 R 2 )(R 2 R 3 )−(R 1 R 2 R 3 )(R 3 +R 2 ) (R 2 R 3 +R 1 R 3 +R 1 R 2 ) 2 = (R 2 R 3 ) 2 (R 2 R 3 +R 1 R 3 +R 1 R 2 ) 2 , ∂R ∂R 2 = (R 1 R 3 ) 2 (R 2 R 3 +R 1 R 3 +R 1 R 2 ) 2 , ∂R ∂R 3 = (R 1 R 2 ) 2 (R 2 R 3 +R 1 R 3 +R 1 R 2 ) 2 . (7) (a) a 2 = c 2 +b 2 −2bc cos A =⇒cos A = c 2 +b 2 −a 2 2bc or A = cos −1 _ c 2 +b 2 −a 2 2bc _ ∂ ∂a (cos A) = ∂ ∂a _ c 2 +b 2 −a 2 2bc _ =⇒−sinA ∂A ∂a = − a bc =⇒ ∂A ∂a = a bc sin A , ∂A ∂b = c 2 −a 2 −b 2 2b 2 c sin A , ∂A ∂c = b 2 −a 2 −c 2 2bc 2 sin A . (b) sin A a = sin B b = sin C c =⇒ a = b sin A sin B ∂a ∂A = b cos A sin B , ∂a ∂B = −b sin Acos B (sin B) 2 . (8) (a) ∂w ∂x = 2xy + 2y 2 sinxy +y cos x, ∂w ∂y = x 2 −2 cos xy + 2xy sin xy + sinx, ∂ 2 w ∂x 2 = 2y + 2y 3 cos xy −y sinx, ∂ 2 w ∂y 2 = 4xsin xy −2x 2 y cos xy, ∂ 2 w ∂x∂y = ∂ ∂x _ ∂w ∂y _ = 2x + 4y sin xy + 2xy 2 cos xy + cos x, ∂ 2 w ∂y∂x = ∂ ∂y _ ∂w ∂x _ = 2x + 4y sinxy + 2xy 2 cos xy + cos x. (b) ∂w ∂x = x x 2 +y 2 , ∂w ∂y = y x 2 +y 2 , ∂ 2 w ∂x 2 = y 2 −x 2 (x 2 +y 2 ) 2 , ∂ 2 w ∂y 2 = x 2 −y 2 (x 2 +y 2 ) 2 , ∂ 2 w ∂x∂y = ∂ ∂x _ ∂w ∂y _ = −2xy (x 2 +y 2 ) 2 , ∂ 2 w ∂y∂x = ∂ ∂y _ ∂w ∂x _ = −2xy (x 2 +y 2 ) 2 . (c) ∂w ∂x = yx y−1 , ∂w ∂y = x y lnx, ∂ 2 w ∂x 2 = y (y −1) x y−2 , ∂ 2 w ∂y 2 = (lnx) 2 x y ∂ 2 w ∂x∂y = ∂ ∂x _ ∂w ∂y _ = x y−1 (1 +y lnx), ∂ 2 w ∂y∂x = ∂ ∂y _ ∂w ∂x _ = x y−1 (1 +y lnx). 56 CHAPTER 4. PARTIAL DIFFERENTIATION (d) w = xsin 2 y +e xy = 1 2 x(1 −cos 2y) +e xy ∂w ∂x = sin 2 y +ye xy = 1 2 (1 −cos 2y) +ye xy , ∂w ∂y = xsin2y +xe xy , ∂ 2 w ∂x 2 = y 2 e xy , ∂ 2 w ∂y 2 = 2xcos 2y +x 2 e xy , ∂ 2 w ∂x∂y = ∂ ∂x _ ∂w ∂y _ = sin2y +e xy +xye xy , ∂ 2 w ∂y∂x = ∂ ∂y _ ∂w ∂x _ = sin2y +e xy +xye xy . (e) ∂w ∂x = (y 2 +sin x+1)(2y−2x)−(2xy−x 2 )(cos x) (y 2 +sin x+1) 2 , ∂w ∂y = (y 2 +sin x+1)(2x)−(2xy−x 2 )(2y) (y 2 +sin x+1) 2 , ∂ 2 w ∂x 2 = 2 cos 2 x(2xy−x 2 ) (y 2 +sin x+1) 3 + sin x(2xy−x 2 )+4(x−y) cos x (y 2 +sin x+1) 2 − 2 y 2 +sin x+1 , ∂ 2 w ∂y 2 = 8y 2 (2xy−x 2 ) (y 2 +sin x+1) 3 + 2x 2 −12xy (y 2 +sin x+1) 2 , ∂ 2 w ∂x∂y = ∂ ∂x _ ∂w ∂y _ = 4y cos x(2xy−x 2 ) (y 2 +sin x+1) 3 − (4y 2 −4xy)+2xcos x (y 2 +sin x+1) 2 + 2 y 2 +sin x+1 , ∂ 2 w ∂y∂x = ∂ ∂y _ ∂w ∂x _ = 4y cos x(2xy−x 2 ) (y 2 +sin x+1) 3 − (4y 2 −4xy)+2xcos x (y 2 +sin x+1) 2 + 2 y 2 +sin x+1 . (9) f (x, y, z) = cos _ 2xy +z 2 _ f x = −2y sin(2xy +z 2 ) f xy = −2 _ sin(2xy +z 2 ) + 2xy cos(2xy +z 2 ) ¸ = −2 sin(2xy +z 2 ) −4xy cos(2xy +z 2 ) f xyz = −4z cos(2xy +z 2 ) + 8xyz sin(2xy +z 2 ) f z = −2z sin(2xy +z 2 ) f zz = −4z 2 cos(2xy +z 2 ) −2 sin(2xy +z 2 ) f zzx = 8yz 2 sin(2xy +z 2 ) −4y cos(2xy +z 2 ) (10) ∂w ∂x = −x (x 2 +y 2 +z 2 ) 3 2 , ∂ 2 w ∂x 2 = 2x 2 −y 2 −z 2 (x 2 +y 2 +z 2 ) 5 2 , ∂w ∂y = −y (x 2 +y 2 +z 2 ) 3 2 , ∂ 2 w ∂y 2 = 2y 2 −x 2 −z 2 (x 2 +y 2 +z 2 ) 5 2 , ∂w ∂z = −z (x 2 +y 2 +z 2 ) 3 2 , ∂ 2 w ∂z 2 = 2z 2 −x 2 −y 2 (x 2 +y 2 +z 2 ) 5 2 . ∂ 2 w ∂x 2 + ∂ 2 w ∂y 2 + ∂ 2 w ∂z 2 = 2x 2 −y 2 −z 2 (x 2 +y 2 +z 2 ) 5 2 + 2y 2 −x 2 −z 2 (x 2 +y 2 +z 2 ) 5 2 + 2z 2 −x 2 −y 2 (x 2 +y 2 +z 2 ) 5 2 = 0 (x 2 +y 2 +z 2 ) 5 2 = 0 (11) (a) ∂w ∂x = A(Ct +D), ∂ 2 w ∂x 2 = 0, ∂w ∂t = C (Ax +B), ∂ 2 w ∂t 2 = 0 ∂ 2 w ∂t 2 −c 2 ∂ 2 w ∂x 2 = (0) −c 2 (0) = 0 57 CHAPTER 4. PARTIAL DIFFERENTIATION (b) ∂w ∂x = _ Ake kx −Bke −kx __ Ce ckt +De −ckt _ , ∂ 2 w ∂x 2 = _ Ak 2 e kx +Bk 2 e −kx __ Ce ckt +De −ckt _ = k 2 _ Ae kx +Be −kx __ Ce ckt +De −ckt _ , ∂w ∂t = _ Ae kx +Be −kx __ Ccke ckt −Dcke −ckt _ , ∂ 2 w ∂t 2 = _ Ae kx +Be −kx __ Cc 2 k 2 e ckt +Dc 2 k 2 e −ckt _ = c 2 k 2 _ Ae kx +Be −kx __ Ce ckt +De −ckt _ , ∂ 2 w ∂t 2 −c 2 ∂ 2 w ∂x 2 = c 2 k 2 _ Ae kx +Be −kx __ Ce ckt +De −ckt _ − _ c 2 _ k 2 _ Ae kx +Be −kx __ Ce ckt +De −ckt _ = 0 (c) ∂w ∂x = g ′ (x +ct), ∂ 2 w ∂x 2 = g ′′ (x +ct), ∂w ∂t = cg ′ (x +ct), ∂ 2 w ∂t 2 = c 2 g ′′ (x +ct) ∂ 2 w ∂t 2 −c 2 ∂ 2 w ∂x 2 = _ c 2 g ′′ (x +ct) ¸ −c 2 [g ′′ (x +ct)] = 0 (12) ∂w ∂x = 2x x 2 +y 2 , ∂w ∂y = 2y x 2 +y 2 , ∂ 2 w ∂x 2 = 2(y 2 −x 2 ) (x 2 +y 2 ) 2 , ∂ 2 w ∂y 2 = 2(x 2 −y 2 ) (x 2 +y 2 ) 2 ∂ 2 w ∂x 2 + ∂ 2 w ∂y 2 = 2(y 2 −x 2 ) (x 2 +y 2 ) 2 + 2(x 2 −y 2 ) (x 2 +y 2 ) 2 = 0 (13) (a) f x (x, y) = (x+3y)(2)−(2x−y)(1) (x+3y) 2 = 7y (x+3y) 2 , f y (x, y) = −7x (x+3y) 2 L.H.S. = xf x (x, y) +yf y (x, y) = (x) _ 7y (x+3y) 2 _ + (y) _ −7x (x+3y) 2 _ = 0 = R.H.S. (b) f xx (x, y) = −14y (x+3y) 3 , f yy (x, y) = 42x (x+3y) 3 , f xy (x, y) = 7x−21y (x+3y) 3 L.H.S. = x 2 −14y (x+3y) 3 + 2xy 7x−21y (x+3y) 3 +y 2 42x (x+3y) 3 = 0 = R.H.S. or ∂ ∂x [xf x (x, y) +yf y (x, y)] = 0 =⇒f x (x, y) +xf xx (x, y) +yf yx (x, y) = 0 ... (1) ∂ ∂y [xf x (x, y) +yf y (x, y)] = 0 =⇒xf xy (x, y) +f y (x, y) +yf yy (x, y) = 0 ... (2) x (1) +y (2) : xf x (x, y) +x 2 f xx (x, y) +xyf yx (x, y) +xyf xy (x, y) +yf y (x, y) +y 2 f yy (x, y) = 0 [xf x (x, y) +yf y (x, y)] +x 2 f xx (x, y) +xyf yx (x, y) +xyf xy (x, y) +y 2 f yy (x, y) = 0 x 2 f xx (x, y) +xyf yx (x, y) +xyf xy (x, y) +y 2 f yy (x, y) = 0 58 CHAPTER 4. PARTIAL DIFFERENTIATION (14) ∂F ∂u = ∂f ∂x ∂x ∂u + ∂f ∂y ∂y ∂u = ∂f ∂x (2u) + ∂f ∂y (2v) = 2u ∂f ∂x + 2v ∂f ∂y ∂F ∂v = ∂f ∂x ∂x ∂v + ∂f ∂y ∂y ∂v = ∂f ∂x (−2v) + ∂f ∂y (2u) = −2v ∂f ∂x + 2u ∂f ∂y _ ∂F ∂u _ 2 + _ ∂F ∂v _ 2 = _ 2u ∂f ∂x + 2v ∂f ∂y _ 2 + _ −2v ∂f ∂x + 2u ∂f ∂y _ 2 = 4u 2 _ ∂f ∂x _ 2 + 4v 2 _ ∂f ∂y _ 2 + 4v 2 _ ∂f ∂x _ 2 + 4u 2 _ ∂f ∂y _ 2 = 4 _ u 2 +v 2 _ _ _ ∂f ∂x _ 2 + _ ∂f ∂y _ 2 _ If u ∂F ∂u −v ∂F ∂v = 0, then u _ 2u ∂f ∂x + 2v ∂f ∂y _ −v _ −2v ∂f ∂x + 2u ∂f ∂y _ = 0 =⇒2 _ u 2 +v 2 _ ∂f ∂x = 0 =⇒ ∂f ∂x = 0 or u = 0 or v = 0. f (x, y) = g (y), which is dependent on y only. Therefore, f (x, y) is independent of x. (15) Since ∂w ∂x = y(y 2 −x 2 ) (x 2 +y 2 ) 2 and ∂w ∂y = x(x 2 −y 2 ) (x 2 +y 2 ) 2 , therefore x ∂w ∂x +y ∂w ∂y = 0. Since ∂ 2 w ∂x 2 = 2xy (x 2 −3y 2 ) (x 2 +y 2 ) 3 , ∂ 2 w ∂y 2 = 2xy (y 2 −3x 2 ) (x 2 +y 2 ) 3 and ∂ 2 w ∂x∂y = −x 4 −y 4 + 6x 2 y 2 (x 2 +y 2 ) 3 , therefore, x 2 ∂ 2 w ∂x 2 + 2xy ∂ 2 w ∂x∂y +y 2 ∂ 2 w ∂y 2 = 0. (16) Let w = g (u, v) with u = x y and v = z y . ∂w ∂x = ∂w ∂u ∂u ∂x + ∂w ∂v ∂v ∂x = g u (u, v) _ 1 y _ +g v (u, v) (0) = 1 y g u (u, v) ∂w ∂y = ∂w ∂u ∂u ∂y + ∂w ∂v ∂v ∂y = g u (u, v) _ − x y 2 _ +g v (u, v) _ − z y 2 _ = − 1 y 2 [xg u (u, v) +zg v (u, v)] ∂w ∂z = ∂w ∂u ∂u ∂z + ∂w ∂v ∂v ∂z = g u (u, v) (0) +g v (u, v) _ 1 y _ = 1 y g v (u, v) x ∂w ∂x +y ∂w ∂y +z ∂w ∂z = x y g u (u, v) − 1 y [xg u (u, v) +zg v (u, v)] + z y g v (u, v) = 0 (17) ∂w ∂x = ∂w ∂u ∂u ∂x + ∂w ∂v ∂v ∂x = ∂w ∂u (1) + ∂w ∂v (1) = ∂w ∂u + ∂w ∂v ∂w ∂t = ∂w ∂u ∂u ∂t + ∂w ∂v ∂v ∂t = ∂w ∂u (c) + ∂w ∂v (−c) = c ∂w ∂u −c ∂w ∂v ∂ 2 w ∂x 2 = ∂ ∂x ∂w ∂u + ∂ ∂x ∂w ∂v = _ ∂ 2 w ∂u 2 ∂u ∂x + ∂w ∂v∂u ∂v ∂x _ + _ ∂ 2 w ∂u∂v ∂u ∂x + ∂ 2 w ∂v 2 ∂v ∂x _ = _ ∂ 2 w ∂u 2 + ∂ 2 w ∂v∂u _ + _ ∂ 2 w ∂v 2 + ∂ 2 w ∂u∂v _ = ∂ 2 w ∂u 2 + 2 ∂ 2 w ∂u∂v + ∂ 2 w ∂v 2 59 CHAPTER 4. PARTIAL DIFFERENTIATION ∂ 2 w ∂t 2 = c ∂ ∂t ∂w ∂u −c ∂ ∂t ∂w ∂v = c _ ∂ 2 w ∂u 2 ∂u ∂t + ∂w ∂v∂u ∂v ∂t _ −c _ ∂ 2 w ∂u∂v ∂u ∂t + ∂ 2 w ∂v 2 ∂v ∂t _ = c _ c ∂ 2 w ∂u 2 −c ∂ 2 w ∂v∂u _ −c _ c ∂ 2 w ∂u∂v −c ∂ 2 w ∂v 2 _ = c 2 ∂ 2 w ∂u 2 −2c 2 ∂ 2 w ∂u∂v +c 2 ∂ 2 w ∂v 2 If c 2 ∂ 2 w ∂x 2 = ∂ 2 w ∂t 2 , then c 2 _ ∂ 2 w ∂u 2 + 2 ∂ 2 w ∂u∂v + ∂ 2 w ∂v 2 _ = c 2 ∂ 2 w ∂u 2 −2c 2 ∂ 2 w ∂u∂v +c 2 ∂ 2 w ∂v 2 =⇒4c 2 ∂ 2 w ∂u∂v = 0 =⇒ ∂ 2 w ∂u∂v = 0. (18) ∂w ∂x = dw du ∂u ∂x = f ′ (u) (2x) ∂w ∂y = dw du ∂u ∂y = f ′ (u) (2y) y ∂w ∂x = 2xyf ′ (u) x ∂w ∂y = 2xyf ′ (u) = y ∂w ∂x (19) ∂w ∂x = ∂w ∂u ∂u ∂x + ∂w ∂v ∂v ∂x = ∂w ∂u (1) + ∂w ∂v (1) = ∂w ∂u + ∂w ∂v ∂w ∂y = ∂w ∂u ∂u ∂y + ∂w ∂v ∂v ∂y = ∂w ∂u (−1) + ∂w ∂v (1) = − ∂w ∂u + ∂w ∂v Thus, _ ∂w ∂x _ _ ∂w ∂y _ = _ ∂w ∂u + ∂w ∂v _ _ − ∂w ∂u + ∂w ∂v _ = − _ ∂w ∂u _ 2 + _ ∂w ∂v _ 2 . (20) ∂f ∂x = ∂f ∂u ∂u ∂x + ∂f ∂v ∂v ∂x + ∂f ∂w ∂w ∂x = ∂f ∂u − ∂f ∂w ∂f ∂y = ∂f ∂u ∂u ∂y + ∂f ∂v ∂v ∂y + ∂f ∂w ∂w ∂y = − ∂f ∂u + ∂f ∂v ∂f ∂z = ∂f ∂u ∂u ∂z + ∂f ∂v ∂v ∂z + ∂f ∂w ∂w ∂z = − ∂f ∂v + ∂f ∂w Thus, ∂f ∂x + ∂f ∂y + ∂f ∂z = _ ∂f ∂u − ∂f ∂w _ + _ − ∂f ∂u + ∂f ∂v _ + _ − ∂f ∂v + ∂f ∂w _ = 0. (21) ∂w ∂x = g _ x y _ ∂ ∂x √ xy + √ xy ∂ ∂x g _ x y _ = 1 2 _ y x g _ x y _ + _ x y g ′ _ x y _ ∂w ∂y = g _ x y _ ∂ ∂y √ xy + √ xy ∂ ∂y g _ x y _ = 1 2 _ x y g _ x y _ − _ x 3 y 3 g ′ _ x y _ Thus, x ∂w ∂x +y ∂w ∂y −w = x _ 1 2 _ y x g _ x y _ + _ x y g ′ _ x y __ +y _ 1 2 _ x y g _ x y _ − ¸ x 3 y 3 g ′ _ x y _ _ − _ √ xyg _ x y __ = 0. (22) With w = g(u) and u = x −ct, one has ∂w ∂x = dw du ∂u ∂x = g ′ (u), 60 CHAPTER 4. PARTIAL DIFFERENTIATION ∂w ∂t = dw du ∂u ∂t = −c g ′ (u), ∂ 2 w ∂x 2 = ∂ ∂x _ ∂w ∂x _ = ∂ ∂x _ g ′ (u) _ = d du _ g ′ (u) _ ∂u ∂x = g ′′ (u) and ∂ 2 w ∂t 2 = ∂ ∂t _ ∂w ∂t _ = ∂ ∂t _ −cg ′ (u) _ = (−c) d du _ g ′ (u) _ ∂u ∂t = (−c) 2 g ′′ (u). We therefore conclude that ∂ 2 w ∂t 2 = c 2 ∂ 2 w ∂x 2 . (23) With w = g(u) and u = x 2 −t, one has ∂w ∂x = dw du ∂u ∂x = g ′ (u)2x ∂w ∂t = dw du ∂u ∂t = g ′ (u)(−1) ∂ 2 w ∂x 2 ∂ ∂x _ ∂w ∂x _ = 2 ∂ ∂x _ xg ′ (u) _ = 2g ′ (u) + 2x ∂ ∂x _ g ′ (u) _ = 2g ′ (u) + 4x 2 g ′′ (u) and ∂ 2 w ∂t 2 = ∂ ∂t _ ∂w ∂t _ = ∂ ∂t _ −g ′ (u) _ = (−1) 2 g ′′ (u). Hence the result. (24) ∂w ∂s = ∂w ∂x ∂x ∂s + ∂w ∂y ∂y ∂s = 4xy 3 s + 6x 2 y 2 s, ∂w ∂t = ∂w ∂x ∂x ∂t + ∂w ∂y ∂y ∂t = 4xy 3 t −6x 2 y 2 t. (25) (a) f(x, y) = _ 9x 2 +y 2 △f ≃ ∂f ∂x(x 0 ,y 0 ) △x + ∂f ∂y (x 0 ,y 0 ) △y ∂f ∂x = _ 1 2 _ _ 9x 2 +y 2 _ − 1 2 (9) (2x) = 9x √ 9x 2 +y 2 ∂f ∂y = _ 1 2 _ _ 9x 2 +y 2 _ − 1 2 (2y) = y √ 9x 2 +y 2 f(1.95, 8.1) ≃ f(2, 8) +△f = f(2, 8) + ∂f ∂x(x 0 ,y 0 ) △x + ∂f ∂y (x 0 ,y 0 ) △y = 10 + (1.8)(−0.05) + (0.8)(0.1) = 9.99 Note: _ 9 (1.95) 2 + (8.1) 2 = 9.9916. (b) Let z = f (x, y) = 1 3 √ x 2 +y 2 . z x = −2x 3(x 2 +y 2 ) 4/3 and z y = −2y 3(x 2 +y 2 ) 4/3 . dz = −2(x∆x+y∆y) 3(x 2 +y 2 ) 4/3 = −2[(2)(−0.1)+(2)(0.04)] 3[(2) 2 +(2) 2 ] 4/3 = 0.005 61 CHAPTER 4. PARTIAL DIFFERENTIATION f (1.9, 2.04) ≃ f (2, 2) +dz = 1 2 + 0.005 = 0.505 Note: 1 3 √ 1.9 2 +2.04 2 = 0.50485. (c) Let z = f (x, y) = 1 (x 2 +y 2 ) 3/2 , then ∂z ∂x = −3x (x 2 +y 2 ) 5/2 and ∂z ∂y = −3y (x 2 +y 2 ) 5/2 . 1 (4.10 2 + 2.95 2 ) 3/2 = f (4.10, 2.95) = f (4 + 0.1, 3 −0.05) ≃ f (4, 3) + ∂z ∂x(4,3) ∆x + ∂z ∂y (4,3) ∆y = 1 125 + _ −12 3125 _ (0.1) + _ −9 3125 _ (−0.05) = 0.00776 Note: 1 (4.10 2 +2.95 2 ) 3/2 = 0.0077602. (d) f (x, y, z) = 3 _ x 2 +y 2 +z 2 f x = 2x 3(x 2 +y 2 +z 2 ) 2/3 , f y = 2y 3(x 2 +y 2 +z 2 ) 2/3 , f z = 2z 3(x 2 +y 2 +z 2 ) 2/3 3 _ 4.98 2 + 1.05 2 + 0.96 2 = f (4.98, 1.05, 0.96) = f (5 −0.02, 1 + 0.05, 1 −0.04) ≃ f (5, 1, 1) + ∂f ∂x(5,1,1) ∆x + ∂f ∂y (5,1,1) ∆y + ∂f ∂z (5,1,1) ∆z = 3 + 2 [(5) (−0.02) + (1) (0.05) + (1) (−0.04)] 27 = 2.9933 Note: 3 √ 4.98 2 + 1.05 2 + 0.96 2 = 2.9935. (26) f x (x, y) = 2x + 2y −4 f y (x, y) = 2x f(0.99, 2.04) ≃ f(1, 2) +△f = f(1, 2) +f x (x 0 , y 0 ) △x +f y (x 0 , y 0 ) △y = (1) + (2) (−0.01) + (2) (0.04) = 1.06 Note: f(0.99, 2.04) = 1. 0593. (27) (a) ∂w ∂ρ = ∂w ∂x ∂x ∂ρ + ∂w ∂y ∂y ∂ρ + ∂w ∂z ∂z ∂ρ = (2x) (sinφcos θ) + (2y) (sinφsinθ) + (−2z) (cos φ) = 2 (ρ sinφcos θ) (sinφcos θ) + 2 (ρ sinφsinθ) (sinφsin θ) −2 (ρ cos φ) (cos φ) = 2ρ sin 2 φcos 2 θ + 2ρ sin 2 φsin 2 θ −2ρ cos 2 φ = −2ρ _ cos 2 φ −sin 2 φ _ = −2ρ cos 2φ 62 CHAPTER 4. PARTIAL DIFFERENTIATION (b) ∂w ∂θ = ∂w ∂x ∂x ∂θ + ∂w ∂y ∂y ∂θ + ∂w ∂z ∂z ∂θ = (2x) (−ρ sinφsin θ) + (2y) (ρ sinφcos θ) + (−2z) (0) = 2 (ρ sinφcos θ) (−ρ sinφsin θ) + 2 (ρ sinφsinθ) (ρ sinφcos θ) = −2ρ 2 sin 2 φsin θ cos θ + 2ρ 2 sin 2 φsinθ cos θ = 0 (28) Let f (x, y) = 1 1+x−y , then f x = −1 (1+x−y) 2 and f x = 1 (1+x−y) 2 . f (0, 0) = 1, f x (0, 0) = −1 and f y (0, 0) = 1 Using the Taylor’s formula, f (x, y) ≃ f (x 0 , y 0 ) + [(x −x 0 ) f x (x 0 , y 0 ) + (y −y 0 ) f y (x 0 , y 0 )] = 1 + [(x −0) (−1) + (y −0) (1)] = 1 −x +y. (29) ∆w ≃ ∂w ∂x ∆x + ∂w ∂y ∆y + ∂w ∂z ∆z. With w = x 2 ye 3z , we have ∂w ∂x = −2, ∂w ∂y = 1 and ∂w ∂z = −3 at the point (1, −1, 0). When ∆x = 0.01, ∆y = −0.03 and ∆z = 0.02, the linear approximation formula gives ∆w ≃ −0.11. (30) Let f (x, y) = _ x 2 +y 2 , then f x (x, y) = x √ x 2 +y 2 and f y (x, y) = y √ x 2 +y 2 . △f ≈ f x (x, y) △x +f y (x, y) △y = x(△x)+y(△y) √ x 2 +y 2 ¸ ¸ ¸ △f f ¸ ¸ ¸ ≈ ¸ ¸ ¸ ¸ x(△x)+y(△y) √ x 2 +y 2 1 √ x 2 +y 2 ¸ ¸ ¸ ¸ = ¸ ¸ ¸ x(△x)+y(△y) x 2 +y 2 ¸ ¸ ¸ = x 2 |△x/x|+y 2 |△y/y| x 2 +y 2 = (0.005)x 2 +(0.005)y 2 x 2 +y 2 = 0.005 = 0.5% (31) △T ≃ dT = ¸ ¸ ¸ ∂T ∂l (l 0 ,g 0 ) ¸ ¸ ¸ [△l[ + ¸ ¸ ¸ ∂T ∂g (l 0 ,g 0 ) ¸ ¸ ¸ [△g[ ∂T ∂l = 2π _ 1 g _ 1 2 √ l _ = π √ gl , ∂T ∂g = 2π √ l _ − 1 2g 3/2 _ = −π _ l g 3 △T ≃ dT = π √ (3.5)(9.79) (0.05) + ¸ ¸ ¸−π _ (3.5) (9.79) 3 ¸ ¸ ¸ (0.05) = 0.036428 (32) 1 R = 1 R 1 + 1 R 2 ⇒R = R 1 R 2 R 1 +R 2 and ∆R ≃ ∂R ∂R 1 ∆R 1 + ∂R ∂R 2 ∆R 2 = R 2 2 (R 1 +R 2 ) 2 ∆R 1 + R 2 1 (R 1 +R 2 ) 2 ∆R 2 . When R 1 = 3, ∆R 1 = 0.2, R 2 = 8 and ∆R 2 = −0.5, we obtain ∆R ≃ 64 121 0.2 + 9 121 (−0.5) = 0.0686. 63 CHAPTER 4. PARTIAL DIFFERENTIATION (33) ∆A ∼ = ∂A ∂a ∆a + ∂A ∂b ∆b + ∂A ∂θ ∆θ = b sin θ 2 ∆a + a sin θ 2 ∆b + ab cos θ 2 ∆θ. Therefore, one has [∆A[ ≤ 200 2 sin π 3 [∆a[ + 150 2 sin π 3 [∆b[ + 150 200 2 cos π 3 [∆θ[ ≤ 43.75 √ 3 + 7500 1.5 π 180 ≃ 272.1 (34) Let S = f (x, y) = x x−y , then ∂S ∂x = −y (x−y) 2 and ∂S ∂y = x (x−y) 2 . [∆S[ ≃ [dS[ = ¸ ¸ f x(9,5) ¸ ¸ [∆x[ + ¸ ¸ f y(9,5) ¸ ¸ [∆y[ = ¸ ¸ ¸ ¸ − 5 (9 −5) 2 ¸ ¸ ¸ ¸ [0.05[ + ¸ ¸ ¸ ¸ 9 (9 −5) 2 ¸ ¸ ¸ ¸ [0.1[ = 0.071875 ≤ 0.072 (35) Let Q = f (P, R, V, L) = πPR 4 5V L , then ∂Q ∂P = πR 4 5V L = Q P , ∂Q ∂R = 4 πPR 3 5V L = 4 Q R , ∂Q ∂V = − 1 V πPR 4 5V L = − Q V , and ∂Q ∂L = − 1 L πPR 4 5V L = − Q L ¸ ¸ ¸ ¸ ∆Q Q ¸ ¸ ¸ ¸ ≃ ¸ ¸ ¸ ¸ dQ Q ¸ ¸ ¸ ¸ = 1 Q __ Q P _ ∆P + _ 4 Q R _ ∆R + _ − Q V _ ∆V + _ − Q L _ ∆L _ = ¸ ¸ ¸ ¸ ∆P P ¸ ¸ ¸ ¸ + 4 ¸ ¸ ¸ ¸ ∆R R ¸ ¸ ¸ ¸ + ¸ ¸ ¸ ¸ − ∆V V ¸ ¸ ¸ ¸ + ¸ ¸ ¸ ¸ − ∆L L ¸ ¸ ¸ ¸ = (0.5%) + 4 (0.25%) + (0.15%) + (0.3%) = 1.95% (36) (a) f x (x, y) = −2xcos _ y 2 −x 2 _ f x (1, 1) = −2 f y (x, y) = 2y cos _ y 2 −x 2 _ f y (1, 1) = 2 (b) f (x, y) ≈ f (1, 1) + [(x −1) f x (1, 1) + (y −1) f y (1, 1)] = (0) + (x −1) (−2) + (y −1) (2) = −2x + 2y f _ 0, 1 2 _ ≈ −2 (0) + 2 _ 1 2 _ = 1 (c) f xx (x, y) = −2 cos _ y 2 −x 2 _ −4x 2 sin _ y 2 −x 2 _ f xx (1, 1) = −2 f yy (x, y) = 2 cos _ y 2 −x 2 _ −4y 2 sin _ y 2 −x 2 _ f yy (1, 1) = 2 f xy (x, y) = 4xy sin _ y 2 −x 2 _ f yx (x, y) = 4xy sin _ y 2 −x 2 _ = f xy (x, y) 64 CHAPTER 4. PARTIAL DIFFERENTIATION f xy (1, 1) = 0 f (x, y) ≈ f (1, 1) + [(x −1) f x (1, 1) + (y −1) f y (1, 1)] + 1 2! _ (x −1) 2 f xx (1, 1) + 2 (x −1) (y −1) f xy (1, 1) + (y −1) 2 f yy (1, 1) _ = (0) + (x −1) (−2) + (y −1) (2) + 1 2 _ (x −1) 2 (−2) + 0 + (y −1) 2 (2) _ = −2x + 2y −(x −1) 2 + (y −1) 2 f _ 0, 1 2 _ ≈ −2 (0) + 2 _ 1 2 _ −(0 −1) 2 + _ 1 2 −1 _ 2 = 1 4 Note: f _ 0, 1 2 _ = sin _ 1 4 _ = 0.2474. (37) We calculate the partial derivatives of f(x, y) at (0, 1) as follows: f x (0, 1) = 1 4 , f y (0, 1) = − 1 4 , f xx (0, 1) = 1 4 , f yy (0, 1) = 1 4 , f xy (0, 1) = − 1 4 . Since f(0, 1) = 1 2 , Taylor’s Formula then yields f(x, y) ∼ = 1 2 + 1 4 x − 1 4 (y −1) + 1 2 _ 1 4 x 2 − 1 2 x(y −1) + 1 4 (y −1) 2 _ . (38) (a) Solving the equations f x = 4x 3 − 4y = 0 and f y = 4y 3 − 4x = 0, one obtains (0, 0), (1, 1) and (−1, −1) as the critical points of the function. Using the second order derivative test with f xx = 12x 2 , f xy = −4 and f yy = 12y 2 , we conclude that H = f xx f yy −f 2 xy = 144x 2 y 2 −16. At (0, 0), H = −16 < 0. We therefore conclude that (0, 0) is a saddle point. At (1, 1) and (−1, −1), H = 128 > 0 and A = 12 > 0. As such, both are relative minimum points. (b) Solving the equations f x = 12x + 6y −6x 2 = 0 and f y = 6x + 6y = 0, one obtains (0, 0) and (1, −1) as the critical points. Since f xx = 12 −12x, f xy = 6 and f yy = 6, we conclude that H = f xx f yy − f 2 xy = 72(1 − x) − 36. At (0, 0), H = 36 > 0 and f xx = 12 > 0. Thus (0, 0) is a relative minimum. At (1, −1), H = −36 < 0 and therefore (1, −1) is a saddle point. (c) f (x, y) = 9x 3 −4xy + 1 3 y 3 f x = 27x 2 −4y f y = −4x +y 2 f xx = 54x f yy = 2y f xy = f yx = −4 _ f x = 0 =⇒y = 27 4 x 2 f y = 0 =⇒x = 1 4 y 2 =⇒(x, y) = (0, 0) or _ 4 9 , 4 3 _ At (0, 0), H = (0) (0) −(−4) 2 = −16 < 0 A = 0. At _ 4 9 , 4 3 _ , H = (24) _ 8 3 _ −(−4) 2 = 48 > 0 A = 24 > 0. Thus, (0, 0) is a saddle point and _ 4 9 , 4 3 _ is a local minimum. 65 CHAPTER 4. PARTIAL DIFFERENTIATION (d) Solve _ ¸ ¸ _ ¸ ¸ _ ∂w ∂x = 3x 2 −96y = 0 ∂w ∂y = 192y 2 −96x = 0 to obtain two critical points (0, 0) and (8, 2) for the given function. As ∂ 2 w ∂x 2 = 6x, ∂ 2 w ∂y 2 = 384y and ∂ 2 w ∂x∂y = −96, one obtains A B C H Conclusion (0, 0) 0 −96 0 −9216 saddle point (8, 2) 48 −96 768 27648 relative minimum (e) Solve _ ¸ ¸ _ ¸ ¸ _ ∂w ∂x = 2x + 4 = 0 ∂w ∂y = −6y + 6y 2 −12 = 0 to obtain critical points (−2, 2) and (−2, −1) for the given function. As ∂ 2 w ∂x 2 = 2, ∂ 2 w ∂y 2 = 12y −6 and ∂ 2 w ∂x∂y = 0, one obtains A B C H Conclusion (−2, 2) 2 0 18 36 relative minimum (−2, −1) 2 0 −18 −36 saddle point (f) f x = 3x 2 −12, f y = 3y 2 −3, f xx = 6x, f yy = 6y and f xy = f yx = 0. _ f x = 0 =⇒x = ±2 f y = 0 =⇒y = ±1 _ =⇒(x, y) = (2, 1) or (2, −1) or (−2, 1) or (−2, −1) At (2, 1), H = (12) (6) −(0) 2 = 96 > 0 A = 12 > 0. At (2, −1), H = (12) (−6) −(0) 2 = −96 < 0 A = 12 > 0. At (−2, 1), H = (−12) (6) −(0) 2 = −96 < 0 A = −12 < 0. At (−2, −1), H = (−12) (−6) −(0) 2 = 96 > 0 A = −12 < 0. Thus, (2, −1) and (−2, 1) are saddle points, (2, 1) is a local minimum, and (−2, −1) is a local maximum. (g) Solve _ ¸ ¸ _ ¸ ¸ _ f x = 6xy −6x = 0 f y = 3x 2 + 3y 2 −6y = 0 to obtain four critical points (0, 0), (0, 2) and (±1, 1) for the given function. As f xx = 6y − 6, f yy = 6y − 6 and f xy = f yx = 6x, then f H A Nature (0, 0) 2 36 −6 relative maximum (0, 2) −2 36 6 relative minimum (±1, 1) 0 −36 saddle point 66 CHAPTER 4. PARTIAL DIFFERENTIATION (h) f x = y − 64 x 2 , f y = x − 27 y 2 , f xx = 128 x 3 , f yy = 54 y 3 and f xy = f yx = 1 _ f x = 0 =⇒y = 64 x 2 f y = 0 =⇒x = 27 y 2 =⇒(x, y) = _ 16 3 , 9 4 _ At _ 16 3 , 9 4 _ , H = _ 27 32 _ _ 128 27 _ −(1) 2 = 3 > 0 A = 27 32 > 0. Thus, _ 16 3 , 9 4 _ is a local minimum. (i) Since _ f x = cos x −cos(x +y) = 0 f y = cos y −cos(x +y) = 0 , we conclude that cos x−cos y = 0, or sin _ x−y 2 _ sin _ x+y 2 _ = 0. Therefore, x−y = 2nπ or x+y = 2mπ for some integers m, n. As x, y are required to satisfy 0 < x+y < 2π, the second possibility is ruled out, and we have x = y + 2nπ for some integer n. Substituting this into the equation cos x−cos(x+y) = 0, one has cos y −cos 2y = 0, or 2 cos 2 y −cos y −1 = 0. Solving this quadratic equation, one obtains cos y = − 1 2 or cos y = 1. If cos y = 1, then y = 2kπ and x = (2k+2n)π where n, k are integers. This however implies that x +y = (4k + 2n)π, which is impossible because 0 < x +y < 2π. If cos y = − 1 2 , then y = 2π 3 + 2kπ or y = 4π 3 + 2lπ for some integers k and l. Let us consider the following two cases: i. When y = 2π 3 + 2kπ, one has x = 2π 3 + (2k + 2n)π and therefore x + y = 4π 3 +(4k +2n)π. The condition 0 < x+y < 2π forces 4k +2n = 0, or n = −2k. As such, the critical points are given by ( 2π 3 −2kπ, 2π 3 + 2kπ) for any integer k. ii. When y = 4π 3 + 2kπ, one has x = 4π 3 + (2k + 2n)π and therefore x + y = 8π 3 + (4k + 2n)π. Once again, the condition 0 < x +y < 2π ⇒−2k − 4 3 < n < −2k − 1 3 , or n = −2k − 1. We therefore conclude that the critical points are ( 4π 3 −2kπ −2π, 4π 3 + 2kπ) for any integer k. A simple calculation shows that f xx = −sinx+sin(x+y), f yy = −siny +sin(x+y) and f xy = sin(x + y). At ( 2π 3 − 2kπ, 2π 3 + 2kπ), f xx = f yy = − √ 3 and f xy = − √ 3 2 . Therefore, those critical points are relative maximum. On the other hand, at ( 4π 3 −2kπ−2π, 4π 3 +2kπ), one has f xx = f yy = √ 3 and f xy = √ 3 2 . We thus conclude that those critical points are relative minimum. (39) The cost C is given by C = 9xy +16z(x +y), with xyz = 36. With z = 36 xy , we conclude that the cost is given by C = 9xy + 576 y + 576 x . By solving equations ∂C ∂x = 0 and ∂C ∂y = 0, one obtains x = 4 and y = 4 as the critical point of C (as a function of two variables x, y), and second order derivative test confirms that (4, 4) is a relative minimum. As such, the optimal dimensions of the box are 4m4m9m. 67 CHAPTER 4. PARTIAL DIFFERENTIATION (40) Solving the equations f x = y +2 − 2 x = 0 and f y = x − 1 y = 0, we obtain ( 1 2 , 2) is the only critical point of the function f(x, y) = xy + 2x − lnx 2 y. Since f xx = 2 x 2 , f yy = 1 y 2 and f xy = 1, we have f xx ( 1 2 , 2) f yy ( 1 2 , 2) − _ f xy ( 1 2 , 2) ¸ 2 = 8 1 4 −1 = 3 > 0. The point is thus a relative minimum. (41) (a) R(2) = _ 8, 2, 2 √ 6 ¸ R ′ (t) = _ 3t 2 , 1, √ 6t ¸ R ′ (2) = _ 12, 1, 2 √ 6 ¸ Equation of the line tangent: x−8 12 = y−2 1 = z−2 √ 6 2 √ 6 . (b) Arc length = _ 3 0 ¸ ¸ R ′ (t) ¸ ¸ dt = _ 3 0 _ (3t 2 ) 2 + (1) 2 + _ √ 6t _ 2 dt = _ 3 0 _ 9t 4 + 6t 2 + 1 dt = _ 3 0 _ 3t 2 + 1 _ dt = _ t 3 +t ¸ 3 0 = 30 (42) (a) 2 (x −1) + 3 (y −2) −1 (z −1) = 0 =⇒2x + 3y −z = 7 (b) 3 (x + 3) −2 (y + 1) + 2 (z −1) = 0 =⇒3x −2y + 2z = −5 (c) 3 (x −2) −(y + 2) −2 (z −3) = 0 =⇒3x −y −2z = 2 (43) (a) Since 2 2 + 2 1 2 = 6, P 0 = (2, 1, 6) is a point on the surface. Putting f(x, y, z) = x 2 + 2y 2 −z, one has ∇f = 2xi + 4yj − k. Thus ∇f(2, 1, 6) = 4i + 4j − k, and this is a vector normal to the surface at P 0 . Finally, equation of the tangent plane is given by 4(x −2) + 4(y −1) −(z −6) = 0, or 4x + 4y −z = 6. (b) Since z(0, 0, 1) = (0) cos 0 − (0) e 0 + 1 = 1, P 0 = (0, 0, 1) is a point on the surface. Putting f(x, y, z) = xcos y −ye x + 1 −z, one has ∇f = (cos y −ye x ) i + (−xsiny −e x ) j − k. Thus ∇f(0, 0, 1) = i −j − k, and this is a vector normal to the surface at P 0 . 68 CHAPTER 4. PARTIAL DIFFERENTIATION Finally, equation of the tangent plane is given by (x −0) −(y −0) −(z −1) = 0, or x −y −z = 1. (44) (a) x = 1 + 3t, y = 2 −2t. (b) x = 1 +t, y = 2 + 3t, z = 3 −2t. (c) x = −3 +t, y = 2 −2t, z = 2 +t. (45) (a) Let f (x, y, z) = xy +yz +zx −11, then ∇f (x, y, z) = ¸y +z, x +z, x +y) and ∇f (1, 2, 3) = ¸5, 4, 3) . Equation of the tangent plane 5 (x −1)+4 (y −2)+3 (z −3) = 0 i.e., 5x+4y+3z = 22 (b) Let f (x, y, z) = x 3 +y 3 +z 3 −18, then ∇f (x, y, z) = ¸ 3x 2 , 3y 2 , 3z 2 _ and ∇f (−2, 3, −1) = ¸12, 27, 3) . Equation of the tangent plane 12 (x + 2)+27 (y −3)+3 (z + 1) = 0 i.e., 4x+9y+z = 18 (c) Let f (x, y, z) = ze y sinx −1, then ∇f (x, y, z) = ¸ze y cos x, ze y sin x, e y sinx) and ∇f _ π 2 , 0, 1 _ = ¸0, 1, 1) . Equation of the tangent plane 0 _ x − π 2 _ + (y −0) + (z −1) = 0 i.e., y +z = 1 (d) Let f (x, y, z) = x 2 +y 2 −xyz −7, then ∇f (x, y, z) = ¸2x −yz, 2y −xz, −xy) and ∇f (−2, 3, −1) = ¸−1, 4, 6) . Equation of the tangent plane (−1) (x + 2) +4 (y −3) +6 (z + 1) = 0 i.e., −x+4y + 6z = 8 (46) Equation of the cone is x 2 + y 2 −z 2 = 0. At P = (x 0 , y 0 , z 0 ), the normal vector to the cone is given by n = ∇ _ x 2 +y 2 −z 2 _ at P = 2x 0 i + 2y 0 j −2z 0 k. (a) Equation of the tangent plane at P is n • (r −r 0 ) = 0, or x 0 (x −x 0 ) +y 0 (y −y 0 ) + z 0 (z −z 0 ) = 0. Since x 2 0 +y 2 0 −z 2 0 = 0, the equation is satisfied when x = y = z = 0. Therefore, the tangent plane passes through the origin. 69 CHAPTER 4. PARTIAL DIFFERENTIATION (b) Equation of the normal line is r −r 0 = tn, where t is any real parameter. This vector equation also takes the form x = x 0 +x 0 t, y = y 0 +y 0 t, z = z 0 +z 0 t. When x = y = 0, we have t = −1. This implies z = 2z 0 . Hence the normal line intersects the z−axis at (0, 0, 2z 0 ). (47) The normal vector to the ellipsoid at (1, 1, 3) is given by n = ∇ _ x 2 + 2y 2 + 3z 2 _ at (1,1,3) = 2i + 4 j+18 k. As such, equation of the tangent plane is 2(x−1)+4(y−1)+18(z−3) = 0, or x+2y+9z = 30. (48) Tangent vector = _ dr dt _ at t= 1 4 = _ i + 1 2 √ t j _ at t= 1 4 = i + j. When t = 1 4 , x = 1 4 and y = 1 2 . Vector equation of the normal line is given by T• (r −r 0 ) = 0, where T = i + j is the tangent vector, r =xi +yj and r 0 = 1 4 i + 1 2 j. (49) (a) By (45) b., ∇f (−2, 3, −1) = ¸12, 27, 3), then the normal line to the surface at (−2, 3, −1) is x+2 12 = y−3 27 = z+1 3 or x = −2 + 12t, y = 3 + 27t and z = −1 + 3t. (b) By (45) c., ∇f _ π 2 , 0, 1 _ = ¸0, 1, 1), then the normal line to the surface at _ π 2 , 0, 1 _ is x−π/2 0 = y 1 = z−1 1 or x = π 2 , y = t and z = 1 +t. (c) By (45) d., ∇f (−2, 3, −1) = ¸−1, 4, 6), then the normal line to the surface at (−2, 3, −1) is x+2 −1 = y−3 4 = z+1 6 or x = −2 −t, y = 3 + 4t and z = −1 + 6t. (50) (a) r = ¸ t 2 , 3t, 0 _ , ∇r (x, y, z) = ¸2t, 3, 0) and ∇r (1, 3, 0) = ¸2, 3, 0). x = 1 + 2s, y = 3 + 3s and z = 0. (b) r = ¸ 2t 2 , 3t, t 3 _ , ∇r (x, y, z) = ¸ 4t, 3, 3t 2 _ and ∇r (2, 3, 3) = ¸4, 3, 3). x = 4 + 4s, y = 3 + 3s and z = 3 + 3s. (c) r = ¸cos t, sint, t), ∇r (x, y, z) = ¸−sint, cos t, 1) and ∇r _ 1 2 , √ 3 2 , π 3 _ = _ − √ 3 2 , 1 2 , 1 _ . x = 1 2 − √ 3 2 s, y = √ 3 2 + 1 2 s and z = π 3 +s. (51) (a) r = ¸ t, −1, e 2t _ , ∇r (x, y, z) = ¸ 1, 0, 2e 2t _ and |v| = |∇r (x, y, z)| = √ 1 + 4e 4t . (b) r = __ 9 2 t 2 , 3t, t 3 _ , ∇r (x, y, z) = ¸√ 18t, 3, 3t 2 _ and |v| = √ 18t 2 + 9 + 9t 4 = 3 _ t 2 + 1 _ . (52) Arc length of a curve is given by the integral _ β α ¸ _ dx dt _ 2 + _ dy dt _ 2 + _ dz dt _ 2 dt. 70 CHAPTER 4. PARTIAL DIFFERENTIATION (a) arc length = _ π 0 _ (−sint) 2 + (1 + cos t) 2 dt = _ π 0 √ 2 + 2 cos tdt = 4. (b) arc length = _ 4 0 _ (t) 2 + _√ 2t + 1 _ 2 dt = _ 4 0 √ t 2 + 2t + 1dt = _ 4 0 (t + 1)dt = 12. (c) arc length = _ 2π 0 _ (−a sint) 2 + (a cos t) 2 +k 2 dt = 2π √ a 2 +k 2 . (53) (a) ∇f (x, y) = ¸ 2x, 6y 2 _ , ∇f (2, −1) = ¸4, 6) and b b = 1 √ 5 ¸1, −2) = _ 1 √ 5 , − 2 √ 5 _ . ∇f (2, −1) • b b = ¸4, 6) • _ 1 √ 5 , − 2 √ 5 _ = −8 √ 5 (b) ∇f (x, y) = ¸e y cos x, e y sinx), ∇f _ π 3 , 0 _ = _ 1 2 , √ 3 2 _ and b b = ¸ 4 5 , − 3 5 _ . ∇f _ π 3 , 0 _ • b b = _ 1 2 , √ 3 2 _ • ¸ 4 5 , − 3 5 _ = 4−3 √ 3 10 (c) ∇f (x, y, z) = ¸yze xy , xze xy , e xy ), ∇f (2, 1, −1) = ¸ −e 2 , −2e 2 , e 2 _ and b b = _ 1 √ 3 , 1 √ 3 , 1 √ 3 _ . ∇f (2, 1, −1) • b b = ¸ −e 2 , −2e 2 , e 2 _ • _ 1 √ 3 , 1 √ 3 , 1 √ 3 _ = − 2 √ 3 e 2 (54) (a) ∇T (x, y) = ¸e x cos y −e y sinx, −e x siny +e y cos x) and ∇T (0, 0) = i + j, then |∇T (x, y)| = _ (1) 2 + (1) 2 = √ 2. (b) −∇T (0, 0) = −i −j (55) ∂T ∂x = 2x and ∂T ∂y = −2y. _ dx dt = ∂T ∂x = 2x dy dt = ∂T ∂y = −2y =⇒ _ x(t) = C 1 e 2t y (t) = C 2 e −2t With initial condition (−2, 1), x(t) = −2e 2t and y (t) = e −2t . (56) ∇f(1, −1, 0) = 2i + 2j −k. Therefore D u f(1, −1, 0) = ∇f(1, −1, 0) • u |u| = (2i + 2j −k) • 2i −3j + 6k _ 2 2 + (−3) 2 + 6 2 = − 8 7 . The direction along which f increases most rapidly at the point (1, −1, 0) is 2i + 2j −k, and the direction along which f decreases most rapidly is −2i − 2j + k. The rate of increase/decrease is given by ∇f(1, −1, 0) • ±(2i + 2j −k) |±(2i + 2j −k)| = ±3. (57) To minimize f(x, y, z) = (x −1) 2 +y 2 + (z + 1) 2 subject to g = 2x −y + 5z −3 = 0, we 71 CHAPTER 4. PARTIAL DIFFERENTIATION use Lagrange Multiplier Method: _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ f x = λg x f y = λg y f z = λg z g = 0 ⇒ _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 2(x −1) = 2λ 2y = −λ 2(z + 1) = 5λ 2x −y + 5z = 3 From the first 3 equations, we conclude that x = λ + 1, y = − λ 2 and z = −1 + 5λ 2 . Substitution into the last equation yields λ = 2 5 , which implies that the minimum point is ( 7 5 , − 1 5 , 0). Hence the minimum distance is equal to ¸ ( 7 5 −1) 2 + _ − 1 5 _ 2 + (0 + 1) 2 = _ 6 5 . (58) Label the pentagon as ABCDE where △ABE is an isoceles triangle with AB = AE = x, BE = 2y and BC = ED = z. We wish to minimize P = 2x + 2y + 2z, subject to the constraint g = y _ x 2 −y 2 + 2yz − 1000 = 0. Lagrange Multiplier Method yields the equations _ ¸ ¸ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ ¸ ¸ _ 2 = λxy √ x 2 −y 2 2 = λ _ _ x 2 −y 2 − y 2 √ x 2 −y 2 + 2z _ 2 = λ2y y _ x 2 −y 2 + 2yz = 1000 . The third equation implies that λ = 1 y . Substitution into the first and the second equation yields y = √ 3x 2 , z = ( √ 3+1)x 2 and _ x 2 −y 2 = x 2 . These, together with the fourth equation, give x = ¸ 4000 6 + 3 √ 3 , y = ¸ 1000 2 + √ 3 , z = _ 1 + √ 3 _ ¸ 1000 6 + 3 √ 3 . (59) We minimize the function f(x, y) = (x − a) 2 + y 2 , subject to the constraint g(x, y) = y 2 −4bx = 0, where a > 0, b > 0. Lagrange Multiplier method leads to _ ¸ ¸ _ ¸ ¸ _ 2(x −a) = λ(−4b) 2y = λ2y y 2 −4bx = 0 . There are two cases: y = 0: The third equation yields x = 0 as well. Therefore, (0, 0) is the point on the 72 CHAPTER 4. PARTIAL DIFFERENTIATION parabola which minimizes f(x, y). As such, the shortest distance from (a, 0) to the parabola is given by _ (0 −a) 2 + 0 2 = a. y ,= 0: The second equation implies λ = 1. Substituting λ = 1 into the first equation, we obtain x = a − 2b. Therefore y = ± _ 4b(a −2b). To summarize, there are two points on the parabola which minimize f(x, y), i.e., (a − 2b, _ 4b(a −2b)) and (a − 2b, − _ 4b(a −2b)). The shortest distance from (a, 0) to the parabola is given by 4b(a −b). Note: You may prove that case (b) occurs if and only if a > 2b by considering the intersection of the circle (x −a) 2 +y 2 = a 2 and the parabola y 2 = 4bx. (60) We apply Lagrange Multiplier Method to the function T = 2x 2 + 5y 2 + 11z 2 + 20xy − 4xz + 16yz, subject to g = x 2 +y 2 +z 2 −1 = 0, to obtain the equations _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 4x + 20y −4z = λ2x 20x + 10y + 16z = λ2y −4x + 16y + 22z = λ2z x 2 +y 2 +z 2 = 1 The first three equations are equivalent to an eigenvalue problem _ ¸ ¸ _ 4 20 −4 20 10 16 −4 16 22 _ ¸ ¸ _ _ ¸ ¸ _ x y z _ ¸ ¸ _ = λ _ ¸ ¸ _ x y z _ ¸ ¸ _ . Solving the characteristic equation ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 4 −λ 20 −4 20 10 −λ 16 −4 16 22 −λ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 0, one obtains eigenvalues λ = 9, −9 and 18. Corresponding eigenvectors are obtained as follows: when λ = −9, _ ¸ ¸ _ x y z _ ¸ ¸ _ = _ ¸ ¸ _ 2t −2t t _ ¸ ¸ _ ; when λ = 9, _ ¸ ¸ _ x y z _ ¸ ¸ _ = _ ¸ ¸ _ 2s s −2s _ ¸ ¸ _ ; when λ = 18, _ ¸ ¸ _ x y z _ ¸ ¸ _ = _ ¸ ¸ _ r 2r 2r _ ¸ ¸ _ . We then substitute these solutions into the constraint and obtain t = ± 1 3 , s = ± 1 3 , r = ± 1 3 . Therefore, the critical points of the function T on the sphere are ( 2 3 , − 2 3 , 1 3 ), (− 2 3 , 2 3 , − 1 3 ), ( 2 3 , 1 3 , − 2 3 ), (− 2 3 , − 1 3 , 2 3 ), ( 1 3 , 2 3 , 2 3 ) and (− 1 3 , − 2 3 , − 2 3 ). Substituting these points into T, we conclude that ( 1 3 , 2 3 , 2 3 ) and (− 1 3 , − 2 3 , − 2 3 ) are the hottest points on the sphere, with maximum temperature = 18. (61) We minimize H = R 1 i 2 1 +R 2 i 2 2 +R 3 i 2 3 +R 4 i 2 4 , subject to the constraint g = i 1 +i 2 +i 3 + i 4 −I = 0. Lagrange Multiplier Method gives 2R k i k = λ for k = 1, 2, 3, 4, from which it 73 CHAPTER 4. PARTIAL DIFFERENTIATION follows that i k = λ 2R k for each k. Substitution into the constraint gives λ 2 = I 1 R 1 + 1 R 2 + 1 R 3 + 1 R 4 , from which it follows that i k = R R k I for k = 1, 2, 3, 4, where 1 R = 1 R 1 + 1 R 2 + 1 R 3 + 1 R 4 . (62) We maximize A = 4xy, subject to the constraint g = 9x 2 + 4y 2 − 36 = 0. Lagrange Multiplier Method yields _ ¸ ¸ _ ¸ ¸ _ 4y = λ18x 4x = λ8y 9x 2 + 4y 2 = 36 . We may eliminate λ from the first two equations to obtain 4y 2 = 9x 2 . Substitution into the third equation ⇒ x = √ 2 and y = 3 √ 2 . As a result, the largest rectangle that can be inscribed in the ellipse has vertices at ( √ 2, 3 √ 2 ), (− √ 2, 3 √ 2 ), (− √ 2, − 3 √ 2 ) and ( √ 2, − 3 √ 2 ). Its area is given by A = 4 √ 2 3 √ 2 = 12. (63) We wish to minimize S = 2πrh + 2πr 2 , subject to g = πr 2 h − 1000 = 0. Lagrange Multiplier Method gives _ ¸ ¸ _ ¸ ¸ _ 2πh + 4πr = λ2πrh 2πr = λπr 2 πr 2 h = 1000 . The first two equations together yield h = 2r. Substituting this into the last equation, we conclude that r = 3 _ 500 π ≃ 5.42 and h = 2 3 _ 500 π ≃ 10.84. (64) T x = 2x −1, T y = 4y, T xx = 2, T yy = 4 and T xy = T yx = 0. T x = 0 =⇒x = 1 2 T y = 0 =⇒y = 0 H = f xx _ 1 2 , 0 _ f yy _ 1 2 , 0 _ − _ f xy _ 1 2 , 0 _¸ 2 = 8 > 0 A = f xx = 2 > 0 Thus, f (x, y) has a local minimum (coldest point) at _ 1 2 , 0 _ and f _ 1 2 , 0 _ = − 1 4 . It is evident that the maximum occurs at the boundary. Then we have to use the Lagrange Multipliers method to find the maximum point. Let f (x, y) = x 2 −x + 2y 2 and g (x, y) = x 2 +y 2 −1. _ ¸ ¸ _ ¸ ¸ _ 2x −1 = 2λx 4y = 2λy x 2 +y 2 = 1 =⇒(x, y, λ) = _ ±1, 0, 1 2 _ or _ − 1 2 , ± √ 3 2 , 2 _ For (±1, 0), f (1, 0) = 0 and f (−1, 0) = 2. For _ − 1 2 , ± √ 3 2 _ , f _ − 1 2 , ± √ 3 2 _ = 2.25. Thus, the maximum (hottest point) is at _ − 1 2 , ± √ 3 2 _ and f _ − 1 2 , ± √ 3 2 _ = 2.25. 74 CHAPTER 4. PARTIAL DIFFERENTIATION (65) Let f (A, B, C) = 4 i=1 (Ax i +By i +C −z i ) 2 , then f A = 2 4 i=1 (Ax i +By i +C −z i ) x i , f B = 2 4 i=1 (Ax i +By i +C −z i ) y i and f C = 2 4 i=1 (Ax i +By i +C −z i ). f A = 0 =⇒2 4 i=1 (Ax i +By i +C −z i ) x i = 0 =⇒2A+B + 2C = 0 f B = 0 =⇒2 4 i=1 (Ax i +By i +C −z i ) y i = 0 =⇒A+ 2B + 2C + 2 = 0 f C = 0 =⇒2 4 i=1 (Ax i +By i +C −z i ) = 0 =⇒2A+ 2B + 4C + 1 = 0 Solving the system of linear equations, A = 1 2 , B = − 3 2 and C = 1 4 . Thus, the ”fitted” plane is z = 1 2 x − 3 2 y − 1 4 . (66) f x = −2x + 2, f y = −2y + 2, f xx = −2, f yy = −2 and f xy = f yx = 0. _ f x = 0 f x = 0 =⇒ _ x = 1 y = 1 H = f xx (1, 1) f yy (1, 1) −[f xy (1, 1)] 2 = 4 > 0 A = f xx = −2 < 0 Thus, f (x, y) has a maxima at (1, 1) and f (1, 1) = 4. (67) Let f (x, y, z) = (x −1) 2 + (y −1) 2 + (z + 2) 2 and g (x, y, z) = x − 2y + 5z − 4, then f x = 2 (x −1), f y = 2 (y −1), f z = 2 (z + 2), g x = 1, g y = −2 and g z = 5. _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ f x = λg x f y = λg y f z = λg z x −2y + 5z = 4 =⇒ _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 2 (x −1) = λ 2 (y −1) = −2λ 2 (z + 2) = 5λ x −2y + 5z = 4 =⇒ _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ x = 1 2 λ + 1 y = −λ + 1 z = 5 2 λ −2 x −2y + 5z = 4 =⇒ _ 1 2 λ + 1 _ −2 (−λ + 1) + 5 _ 5 2 λ −2 _ = 15λ −11 = 4 =⇒λ = 1 Thus, x = 3 2 , y = 0, z = 1 2 and f (x, y, z) = 15 2 . The minimum distance is √ 30 2 . (68) Let f (x, y, z) = x 2 + y 2 + z 2 and g (x, y, z) = x 2 + y 2 − z 2 − 1, then f x = 2x, f y = 2y, f z = 2z, g x = 2x, g y = 2y and g z = −2z. _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ f x = λg x f y = λg y f z = λg z x 2 +y 2 −z 2 = 1 =⇒ _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 2x = 2λx 2y = 2λy 2z = −2λz x 2 +y 2 −z 2 = 1 =⇒ _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ x(λ −1) = 0 y (λ −1) = 0 z (λ + 1) = 0 x 2 +y 2 −z 2 = 1 While λ ,= 0, z = 0 and x 2 +y 2 = 1. If x = 0, then y = ±1. If y = 0, then x = ±1. Thus, f (0, ±1, 0) = 1 and f (±1, 0, 0) = 1. The minimum distance is 1. (69) Let f (x, y, z) = xyz and g (x, y, z) = x + y + z 2 − 16, then f x = yz, f y = xz, f z = xy, g x = 1, g y = 1 and g z = 2z. 75 CHAPTER 4. PARTIAL DIFFERENTIATION _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ f x = λg x f y = λg y f z = λg z x +y +z 2 = 16 =⇒ _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ yz = λ ... (1) xz = λ ... (2) xy = 2λz ... (3) x +y +z 2 = 16 ... (4) From (1) and (2), yz = xz =⇒z (y −x) = 0 =⇒y = x as z ,= 0. Again, from (1) and (2), x = y = λ z ... (5). From (3) and (5), _ λ z _ _ λ z _ = 2λz =⇒λ = 2z 3 ... (6) as λ ,= 0. For (4) , _ λ z _ + _ λ z _ +z 2 = 16 =⇒ 2λ z +z 2 = 16 =⇒4z 2 +z 2 = 16 =⇒z = 4 √ 5 . So, λ = 128 5 √ 5 and x = y = 128 5 √ 5 / 4 √ 5 = 32 5 . Thus, f (x, y, z) = _ 32 5 _ _ 32 5 _ _ 4 √ 5 _ = 4096 125 √ 5. (70) Let f (x, y, z) = x − 2y + 5z and g (x, y, z) = x 2 + y 2 + z 2 − 36, then f x = 1, f y = −2, f z = 5, g x = 2x, g y = 2y and g z = 2z. _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 1 = 2λx −2 = 2yλ 5 = 2λz x 2 +y 2 +z 2 = 36 =⇒(x, y, z, λ) = __ 6 5 , −2 _ 6 5 , √ 30, _ 5 24 _ _ − _ 6 5 , 2 _ 6 5 , − √ 30, _ 5 24 _ f __ 6 5 , −2 _ 6 5 , √ 30 _ = __ 6 5 _ −2 _ −2 _ 6 5 _ + 5 _√ 30 _ = 6 √ 30 = 32.863 f _ − _ 6 5 , 2 _ 6 5 , − √ 30 _ = −6 √ 30 = −32.863 Thus, the maximum value of f is 32.863 and the minimum value of f is −32.863. (71) Let x, y and z be respectively the length, the width and the height of the box. We wish to minimize f(x, y, z) = 2xy + 8[xy + 2xz + 2yz], subject to the constraint g(x, y, z) = xyz −36 = 0. Using the Lagrange Multiplier method, we solve the equations _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 10y + 16z = λyz 10x + 16z = λxz 16x + 16y = λxy xyz = 36 to obtain x = y = 3 _ 288 5 ≃ 3.862 and z = 10 16 x = 5 8 3 _ 288 5 ≃ 2.414. Physical consideration indicates that this must be a minimum. Thus the cost for building such a box is approximately $447.48. 76 CHAPTER 4. PARTIAL DIFFERENTIATION (72) We wish to minimize f(x, y) = x 2 +(y +2) 2 , subject to the constraint g(x, y) = x 2 −2x+ y 2 = 0. Using the Lagrange Multiplier method, we simplify the equations _ ¸ ¸ _ ¸ ¸ _ 2x = λ(2x −2) 2(y + 2) = λ2y x 2 −2x +y 2 = 0 to conclude that y = 2x −2 and 5x 2 −10x + 4 = 0. Thus x = 1 − 1 √ 5 , y = − 2 √ 5 or x = 1 + 1 √ 5 , y = 2 √ 5 . For mininum, we have x = 1 − 1 √ 5 , y = − 2 √ 5 . The shortest distance is given by _ _ 1 − 1 √ 5 _ 2 + _ − 2 √ 5 + 2 _ 2 = √ 5 −1 ≃ 1.236. (73) Using the Lagrange Multiplier method, we solve the equations _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 1 = 2λx 3 = 2λy 5 = 2λz x 2 +y 2 +z 2 = 1 to obtain two critical points (x, y, z, λ) = ± _ 1 √ 35 , 3 √ 35 , 5 √ 35 , 2 √ 35 _ . Thus the maximum value is f _ 1 √ 35 , 3 √ 35 , 5 √ 35 _ = √ 35 and the minimum value is f _ − 1 √ 35 , − 3 √ 35 , − 5 √ 35 _ = − √ 35. (74) We minimize f(x, y, z) = (x + 1) 2 + (y −2) 2 + (z −3) 2 , subject to _ g(x, y, z) = x + 2y −3z −4 = 0 h(x, y, z) = 2x −y + 2z −5 = 0 . 77 CHAPTER 4. PARTIAL DIFFERENTIATION By the Lagrange Multiplier method, we solve the equations _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 2(x + 1) = λ + 2µ 2(y −2) = 2λ −µ 2(z −3) = −3λ + 2µ x + 2y −3z = 4 2x −y + 2z = 5 to obtain x = 37 15 , y = 49 15 , z = 25 15 . Physical consideration indicates that this must be a minimum. Therefore, the shortest distance is given by ¸ _ 37 15 + 1 _ 2 + _ 49 15 −2 _ 2 + _ 25 15 −3 _ 2 ≃ 3.9243. (75) Let f (x, y, z) = x 2 + y 2 + z 2 and g (x, y, z) = xyz − 1, then f x = 2x, f y = 2y, f z = 2z, g x = yz, g y = xz and g z = xy. _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 2x = yz 2y = xz 2z = xy xyz = 1 =⇒(x, y, z) = _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ (1, 1, 1) (−1, −1, 1) (−1, 1, −1) (1, −1, −1) f (1, 1, 1) = f (−1, −1, 1) = f (−1, 1, −1) = f (1, −1, −1) = 3 Therefore, the minimum distance is √ 3 and the maximum distance is ∞. (76) We differentiate the equation x 2 −2yw +w 3 + 1 = 0 with respect to x to obtain 2x −2y ∂w ∂x + 3w 2 ∂w ∂x = 0, (4.0.1) which implies ∂w ∂x = −2x 3w 2 −2y . If we now differentiate (4.0.1) with respect to x, we obtain 2 −2y ∂ 2 w ∂x 2 + 6w _ ∂w ∂x _ 2 + 3w 2 ∂ 2 w ∂x 2 = 0. From this equation, one concludes that ∂ 2 w ∂x 2 = − 2 + 6w _ ∂w ∂x _ 2 3w 2 −2y . (77) Let F (x, y, z) = x 2 + y 2 − z 2 − 2x(y +z) = 0. Then F x = 2x − 2y − 2z, F y = 2y − 2x and F z = −2z −2x. 78 CHAPTER 4. PARTIAL DIFFERENTIATION Thus, ∂z ∂x = − F x F z = x−y−z x+z and ∂z ∂y = − F y F z = y−x x+z . ∂ 2 z ∂y 2 = (x +z) (1) −(y −x) ∂z ∂y (x +z) 2 = (x +z) −(y −x) y−x x+z (x +z) 2 = (x +z) 2 −(y −x) 2 (x +z) 3 = (y +z) (2x −y +z) (x +z) 3 . ∂ 2 z ∂y∂x = ∂ ∂y _ ∂z ∂x _ = (x +z) _ −1 − ∂z ∂y _ −(x −y −z) ∂z ∂y (x +z) 2 = (x +z) _ −1 − y−x x+z _ −(x −y −z) y−x x+z (x +z) 2 = x 2 −3xy −2xz +y 2 −z 2 (x +z) 3 . ∂ 2 z ∂x∂y = ∂ ∂x _ ∂z ∂y _ = (x +z) (−1) −(y −x) _ 1 + ∂z ∂x _ (x +z) 2 = (x +z) (−1) −(y −x) _ 1 + x−y−z x+z _ (x +z) 2 = (x +z) 2 (−1) −(y −x) (2x −y) (x +z) 3 = x 2 −3xy −2xz +y 2 −z 2 (x +z) 3 . (78) Let F (x, y, w) = x +y +xyw −w 3 , then F x = 1 +yw, F y = 1 +xw and F w = xy −3w 2 . ∂w ∂x = − F x F w = − 1+yw xy−3w 2 and ∂w ∂y = − F y F w = − 1+xw xy−3w 2 ∂ 2 w ∂x 2 = ∂ ∂x _ 1 +yw 3w 2 −xy _ = _ 3w 2 −xy _ ∂ ∂x (1 +yw) −(1 +yw) ∂ ∂x _ 3w 2 −xy _ (3w 2 −xy) 2 = _ 3w 2 −xy _ _ y ∂w ∂x _ −(1 +yw) _ 6w ∂w ∂x −y _ (3w 2 −xy) 2 = _ 3w 2 −xy _ _ y 1+yw 3w 2 −xy _ −(1 +yw) _ 6w 1+yw 3w 2 −xy −y _ (3w 2 −xy) 2 = −2 _ xy 2 + 3w _ (wy + 1) (3w 2 −xy) 3 79 CHAPTER 4. PARTIAL DIFFERENTIATION ∂ 2 w ∂y 2 = ∂ ∂y _ 1 +xw 3w 2 −xy _ = _ 3w 2 −xy _ ∂ ∂y (1 +xw) −(1 +xw) ∂ ∂y _ 3w 2 −xy _ (3w 2 −xy) 2 = _ 3w 2 −xy _ _ x ∂w ∂y _ −(1 +xw) _ 6w ∂w ∂y −x _ (3w 2 −xy) 2 = _ 3w 2 −xy _ _ x 1+xw 3w 2 −xy _ −(1 +xw) _ 6w 1+xw 3w 2 −xy −x _ (3w 2 −xy) 2 = −2 _ x 2 y + 3w _ (wx + 1) (3w 2 −xy) 3 ∂ 2 w ∂x∂y = ∂ ∂x _ 1 +xw 3w 2 −xy _ = _ 3w 2 −xy _ ∂ ∂x (1 +xw) −(1 +xw) ∂ ∂x _ 3w 2 −xy _ (3w 2 −xy) 2 = _ 3w 2 −xy _ _ w +x ∂w ∂x _ −(1 +xw) _ 6w ∂w ∂x −y _ (3w 2 −xy) 2 = _ 3w 2 −xy _ _ w +x 1+yw 3w 2 −xy _ −(1 +xw) _ 6w 1+yw 3w 2 −xy −y _ (3w 2 −xy) 2 = w _ 3w 2 −xy _ 2 +x _ 3w 2 −xy _ (1 +yw) −6w(1 +yw) 2 −y _ 3w 2 −xy _ (1 +xw) (3w 2 −xy) 3 ∂ 2 w ∂y∂x = ∂ ∂y _ 1 +yw 3w 2 −xy _ = _ 3w 2 −xy _ ∂ ∂y (1 +yw) −(1 +yw) ∂ ∂y _ 3w 2 −xy _ (3w 2 −xy) 2 = _ 3w 2 −xy _ _ w +y ∂w ∂y _ −(1 +yw) _ 6w ∂w ∂y −x _ (3w 2 −xy) 2 = _ 3w 2 −xy _ _ w +y 1+xw 3w 2 −xy _ −(1 +yw) _ 6w 1+xw 3w 2 −xy −x _ (3w 2 −xy) 2 = w _ 3w 2 −xy _ 2 +x _ 3w 2 −xy _ (1 +yw) −6w(1 +yw) 2 −y _ 3w 2 −xy _ (1 +xw) (3w 2 −xy) 3 (79) Differentiating the equation with respect to x and y, we obtain 3w 2 ∂w ∂x + 4 ∂w ∂x −2x +y 2 = 0 and 3w 2 ∂w ∂y + 4 ∂w ∂y + 2xy + 8 = 0. Therefore, ∂w ∂x = 2x−y 2 3w 2 +4 and ∂w ∂y = −2xy−8 3w 2 +4 . Solving _ 2x −y 2 = 0 −2xy −8 = 0 , one obtains 80 CHAPTER 4. PARTIAL DIFFERENTIATION (2, −2) as the only critical point of the function. At (2, −2), further differentiations yield ∂ 2 w ∂x 2 = 2 _ 3w 2 + 4 _ − _ 2x −y 2 _ (6w ∂w ∂x ) (3w 2 + 4) 2 , ∂ 2 w ∂y 2 = (−2x) _ 3w 2 + 4 _ + (2xy + 8) (6w ∂w ∂x ) (3w 2 + 4) 2 , ∂ 2 w ∂x∂y = −2y _ 3w 2 + 4 _ − _ 2x −y 2 _ (6w ∂w ∂y ) (3w 2 + 4) 2 . Since ∂w ∂x = 0 at (2, −2), conclude that ∂ 2 w ∂x 2 > 0 and ∂ 2 w ∂y 2 < 0 at that point. Therefore, ∂ 2 w ∂x 2 ∂ 2 w ∂y 2 − _ ∂ 2 w ∂x∂y _ 2 < 0 at (2, −2) and therefore the point (2, −2) is a saddle point. (80) Differentiating the equation with respect to x and y, we obtain 1 +yw +xy ∂w ∂x −e w ∂w ∂x = 0 and 1 +xw +xy ∂w ∂y −e w ∂w ∂y = 0. This imply ∂w ∂x = 1+yw e w −xy and ∂w ∂y = 1+xw e w −xy . Further differentiation gives ∂ 2 w ∂x 2 = (e w −xy) y ∂w ∂x −(1 +yw) _ e w ∂w ∂x −y _ (e w −xy) 2 , ∂ 2 w ∂y 2 = x(e w −xy) ∂w ∂y −(1 +xw) _ e w ∂w ∂y −x _ (e w −xy) 2 , ∂ 2 w ∂x∂y = (e w −xy) _ w +y ∂w ∂y _ −(1 +yw) _ e w ∂w ∂y −x _ (e w −xy) 2 . (81) Differentiating the equations with respective to x and y, we obtain _ 2u ∂u ∂x −4v ∂v ∂x = −2x −5y (v cos uv + 1 −6v) ∂u ∂x +(ucos uv −3 −6u) ∂v ∂x = −e x cos y _ 2u ∂u ∂y −4v ∂v ∂y = 6y −5x (v cos uv + 1 −6v) ∂u ∂y +(ucos uv −3 −6u) ∂v ∂y = e x siny . We may now use Cramer’s rule to obtain ∂u ∂x , ∂v ∂x , ∂u ∂y and ∂v ∂y . For instance, we have ∂u ∂x = 1 D ¸ ¸ ¸ ¸ ¸ −2x −5y −4v −e x cos y ucos uv −3 −6u ¸ ¸ ¸ ¸ ¸ 81 CHAPTER 4. PARTIAL DIFFERENTIATION and ∂v ∂x = 1 D ¸ ¸ ¸ ¸ ¸ 2u −2x −5y v cos uv + 1 −6v −e x cos y ¸ ¸ ¸ ¸ ¸ , where D = ¸ ¸ ¸ ¸ ¸ 2u −4v v cos uv + 1 −6v ucos uv −3 −6u ¸ ¸ ¸ ¸ ¸ . (82) (a) If T and Φ are defined as functions of P and V by the equations, we differentiate the equations to obtain ∂T ∂P = V R , ∂Φ ∂P = C V P , ∂T ∂V = P R and ∂Φ ∂V = C P V . Therefore, ∂T ∂P ∂Φ ∂V − ∂T ∂V ∂Φ ∂P = C P −C V R = 1. (b) If P and V are defined as functions of T and Φ by the equations, we differentiate the equations with respect to T and Φ to obtain _ V ∂P ∂T +P ∂V ∂T = R C V P ∂P ∂T + C P V ∂V ∂T = 0 and _ V ∂P ∂Φ +P ∂V ∂Φ = 0 C V P ∂P ∂Φ + C P V ∂V ∂Φ = 1 . Cramer’s rule then yields ∂P ∂T = C P V , ∂V ∂T = − C V P , ∂P ∂Φ = − P R and ∂V ∂Φ = V R . Therefore, ∂P ∂T ∂V ∂Φ − ∂V ∂T ∂P ∂Φ = 1. 82 Chapter 5 Multiple Integrals (1) (a) _ 3 0 _ 2 0 4xydxdy = _ 3 0 4y _ x 2 2 _ 2 0 dy = _ 3 0 8ydy = _ 4y 2 ¸ 3 0 = 36 (b) _ 4 2 _ 3 1 4xydxdy = _ 4 2 4y _ x 2 2 _ 3 1 dy = _ 4 2 16ydy = _ 8y 2 ¸ 4 2 = 96 (c) _ 1 0 _ 1 y 4xydxdy = _ 1 0 4y _ x 2 2 _ 1 y dy = _ 1 0 _ 2y −2y 3 _ dy = _ y 2 − 1 2 y 4 _ 1 0 = 1 2 (d) _ 2 0 _ x+1 x/2 4xydydx = _ 2 0 4x _ y 2 2 _ x+1 x/2 dx = _ 2 0 4x _ (x + 1) 2 2 − x 2 8 _ dx = 1 2 _ 2 0 _ 3x 3 + 8x 2 + 4x _ dx = 1 2 _ 3 4 x 4 + 8 3 x 3 + 2x 2 _ 2 0 = 62 3 (e) _ 1 0 _ 3 √ y y 2 4xydxdy = _ 1 0 4y _ x 2 2 _ 3 √ y y 2 dy = 2 _ 1 0 _ y 5/3 −y 5 _ dy = 2 _ 3 8 y 8/3 − 1 6 y 6 _ 1 0 = 5 12 (f) _ 1 0 _ √ y 0 4xydxdy = _ 1 0 4y _ x 2 2 _ √ y 0 dy = _ 1 0 2y 2 dy = _ 2 3 y 3 _ 1 0 = 2 3 83 CHAPTER 5. MULTIPLE INTEGRALS (2) _ y = x 2 y = x 3 =⇒(x, y) = (0, 0) or (1, 1) _ 1 0 _ x 2 x 3 _ x 4 +y 2 _ dydx = _ 1 0 _ x 4 y + 1 3 y 3 _ x 2 x 3 dx = _ 1 0 _ 4 3 x 6 −x 7 − 1 3 x 9 _ dx = _ 4 21 x 7 − 1 8 x 8 − 1 30 x 10 _ 1 0 = 9 280 = 0.032143 (3) _ x = y 2 y = x =⇒(x, y) = (0, 0) or (1, 1) _ 1 0 _ y y 2 √ xydxdy = _ 1 0 √ y _ 2 3 x 3/2 _ y y 2 dy = _ 1 0 2 3 _ y 2 −y 7/2 _ dx = 2 3 _ 1 3 y 3 − 2 9 y 9/2 _ 1 0 = 2 27 (4) The parabola and the line intersect each other at (−1, −3) and (4, 12). Hence, _ R xdA = _ 4 −1 __ 3x x 2 −4 xdy _ dx = _ 4 −1 x _ 3x −x 2 + 4 ¸ dx = _ 4 −1 _ −x 3 + 3x 2 + 4x ¸ dx = 125 4 . (5) (a) __ R 2xydxdy = X +Y X = _ 1 0 _ x+2 0 2xydydx = _ 1 0 x _ y 2 ¸ x+2 0 dx = _ 1 0 _ x 3 + 4x 2 + 4x _ dx = 43 12 Y = _ 3 1 _ 4−x 0 2xydydx = _ 3 1 x _ y 2 ¸ 4−x 0 dx = _ 3 1 _ x 3 −8x 2 + 16x _ dx = 44 3 Thus, _ _ R 2xydxdy = 43 12 + 44 3 = 73 4 (b) __ R 2xydxdy = X +Y 84 CHAPTER 5. MULTIPLE INTEGRALS X = _ 2 1 _ 6−2y y 2xydxdy = _ 2 1 y _ x 2 ¸ 6−2y y dy = _ 2 1 _ 3y 3 −24y 2 + 36y _ dy = 37 4 Y = _ 1 0 _ 4y y 2xydxdy = _ 1 0 y _ x 2 ¸ 4y y dy = _ 1 0 15y 3 dy = 15 4 Thus, _ _ R 2xydxdy = 37 4 + 15 4 = 13 (c) __ R 2xydxdy = X +Y X = _ 3 1 _ (9−y)/2 3(y−1)/2 2xydxdy = _ 3 1 y _ x 2 ¸ (9−y)/2 3(y−1)/2 dy = 1 4 _ 3 1 _ 72y −8y 3 _ dy = 32 Y = _ 1 0 _ 3y+1 1−y 2xydxdy = _ 1 0 y _ x 2 ¸ 3y+1 1−y dy = _ 1 0 _ 8y 3 + 8y 2 _ dy = 14 3 Thus, _ _ R 2xydxdy = 32 + 14 3 = 110 3 (d) __ R 2xydxdy = X +Y X = _ 1 0 _ x+2 2−2x 2xydydx = _ 1 0 x _ y 2 ¸ x+2 2−2x dx = _ 1 0 _ 12x 2 −3x 3 _ dx = 13 4 Y = _ 3 1 _ (7−x)/2 (x−1)/2 2xydydx = _ 3 1 x _ y 2 ¸ (7−x)/2 (x−1)/2 dx = 1 4 _ 3 1 _ 48x −12x 2 _ dx = 22 Thus, __ R 2xydxdy = 13 4 + 22 = 101 4 (6) y x 1 3 85 CHAPTER 5. MULTIPLE INTEGRALS (a) _ 3 0 _ 1 0 x 2 dxdy = _ 3 0 _ 1 3 x 3 _ 1 0 dy = _ 1 3 y _ 1 0 = 1 (b) _ 3 0 _ 1 0 e x+y dxdy = _ 3 0 _ e x+y ¸ 1 0 dy = _ 3 0 _ e 1+y −e y _ dy = _ e 1+y −e y ¸ 3 0 = e 4 −e 3 −e + 1 (c) _ 3 0 _ 1 0 xy 2 dxdy = _ 3 0 _ 1 2 y 2 x 2 _ 1 0 dy = _ 3 0 1 2 y 2 dy = _ 1 6 y 3 _ 3 0 = 9 2 (7) y = 1 (a) _ 1 0 _ 1 y x 3 ydxdy = _ 1 0 _ 1 4 x 4 y _ 1 y dy = _ 1 0 _ 1 4 y − 1 4 y 5 _ dy = _ 1 8 y 2 − 1 24 y 6 _ 1 0 = 1 12 (b) _ 1 0 _ 1 y xy 3 dxdy = _ 1 0 _ 1 2 x 2 y 3 _ 1 y dy = _ 1 0 _ 1 2 y 3 − 1 2 y 5 _ dy = _ 1 8 y 4 − 1 12 y 6 _ 1 0 = 1 24 (c) _ 1 0 _ 1 y x 2 y 2 dxdy = _ 1 0 _ 1 3 x 3 y 2 _ 1 y dy = _ 1 0 _ 1 3 y 2 − 1 3 y 5 _ dy = _ 1 9 y 3 − 1 18 y 6 _ 1 0 = 1 18 86 CHAPTER 5. MULTIPLE INTEGRALS (8) y x 2 2 (a) _ π/2 0 _ π/2 0 sin(x +y) dxdy = _ π/2 0 [−cos (x +y)] π/2 0 dy = _ π/2 0 _ cos y −cos _ π 2 +y __ dy = _ siny −sin _ π 2 +y __ π/2 0 = 2 (b) _ π/2 0 _ π/2 0 cos (x +y) dxdy = _ π/2 0 [sin(x +y)] π/2 0 dy = _ π/2 0 _ sin _ π 2 +y _ −sin (y) _ dy = _ cos _ π 2 +y _ −cos (y) _ π/2 0 = 0 (c) _ π/2 0 _ π/2 0 (1 +xy) dxdy = _ π/2 0 _ x + 1 2 x 2 y _ π/2 0 dy = _ π/2 0 _ 1 2 π + 1 8 π 2 y _ dy = _ 1 2 πy + 1 16 π 2 y 2 _ π/2 0 = 1 4 π 2 + 1 64 π 4 (9) R = _ (x, y) [ 3 √ y ≤ x ≤ 2, 0 ≤ y ≤ 8 _ = _ (x, y) [0 ≤ x ≤ 2, 0 ≤ y ≤ x 3 _ _ 8 0 _ _ 2 3 √ y e x 4 dx _ dy = _ 2 0 _ _ x 3 0 e x 4 dy _ dx = _ 2 0 e x 4 x 3 dx = 1 4 _ e 16 −1 _ (10) R = _ (x, y) [0 ≤ x ≤ 1, 1 ≤ y ≤ x 2 _ = _ (x, y) [0 ≤ x ≤ √ y, 0 ≤ y ≤ 1 _ _ 1 0 __ 1 x 2 siny √ y dy _ dx = _ 1 0 _ _ + √ y 0 sin y √ y dx _ dy = _ 1 0 siny √ y √ ydy = 1 −cos 1 = 0.4597 87 CHAPTER 5. MULTIPLE INTEGRALS (11) (a) _ 1 −1 _ √ 1−y 2 − √ 1−y 2 _ x 2 + 3y 3 _ dxdy = _ 1 −1 _ 1 3 x 3 + 3y 3 x _ √ 1−y 2 − √ 1−y 2 dy = 2 3 _ 1 −1 _ (1 −y 2 )dy − 2 3 _ 1 −1 _ (1 −y 2 )y 2 dy + 6 _ 1 −1 y 3 _ (1 −y 2 )dy = 2 3 _ π/2 −π/2 cos 2 θdθ − 2 3 _ π/2 −π/2 sin 2 θ cos 2 θdθ + 6 _ π/2 −π/2 sin 3 θ cos 2 θdθ = 2 3 _ π/2 −π/2 1 + cos 2θ 2 dθ − 2 3 _ π/2 −π/2 sin 2 2θ 4 dθ −6 _ π/2 −π/2 _ cos 2 θ −cos 4 θ _ d (cos θ) = 2 3 _ 1 2 π _ − 2 3 _ π/2 −π/2 1 −cos 4θ 8 dθ −6 __ cos 3 θ 3 − cos 5 θ 5 __ π/2 −π/2 = 1 3 π − 2 3 _ 1 8 π _ + 0 = 1 4 π (b) _ 1 0 _ y y 2 √ xydxdy = _ 1 0 _ 2 3 x 3/2 y 1/2 _ y y 2 dy = _ 1 0 2 3 _ y 2 −y 7/2 _ dy = _ 2 9 y 3 − 4 27 y 9/2 _ 1 0 = 2 27 (c) _ 1 0 _ y 2 0 ye x dxdy = _ 1 0 [ye x ] y 2 0 dy = _ 1 0 _ ye y 2 −y _ dy = _ 1 2 e y 2 − 1 2 y 2 _ 1 0 = 1 2 e −1 (d) 88 CHAPTER 5. MULTIPLE INTEGRALS 1 4 -3 -2 -1 1 2 3 3 2 y 2 = 2x y 2 = 8 - 2x y x _ 2 −2 _ (8−y 2 )/2 y 2 /2 _ 4 −y 2 _ dxdy = _ 2 −2 __ 4 −y 2 _ x ¸ (8−y) 2 /2 y 2 /2 dy = _ 2 −2 _ 16 −8y 2 +y 4 _ dy = _ 16y − 8 3 y 3 + 1 5 y 5 _ 2 −2 = 512 15 (e) _ 1 0 _ 3 √ y √ y _ x 4 +y 2 _ dxdy = _ 1 0 _ 1 5 x 5 +y 2 x _ 3 √ y √ y dy = _ 1 0 _ 1 5 y 5/3 +y 7/3 − 6 5 y 5/2 _ dy = _ 3 40 y 8/3 + 3 10 y 10/3 − 12 35 y 7/2 _ 1 0 = 9 280 = 0.03214 (f) y = x y = -x y x 1.0 1.0 -1.0 -1.0 0.5 0.5 -0.5 -0.5 _ 1 0 _ x −x _ 3xy 2 −y _ dydx + _ 0 −1 _ −x x _ 3xy 2 −y _ dydx = _ 1 0 _ 3 2 x 2 y 2 −xy _ x −x dx = _ 1 0 2x 4 dx − _ 0 −1 2x 4 dx = 0 (g) 89 CHAPTER 5. MULTIPLE INTEGRALS y x 2y = x 2 0 _ 2 0 _ x/2 0 e x 2 dydx = _ 2 0 _ e x 2 y _ x/2 0 dx = _ 2 0 1 2 xe x 2 dx = _ 1 4 e x 2 _ 2 0 = 1 4 e 4 − 1 4 (h) 1 0.5 -0.5 -1 0 0.5 1 -0.5 -1 y x _ 1 0 _ x 3 x 4 (x +y) dydx + _ 0 −1 _ x 4 x 3 (x +y) dydx = _ 1 0 _ 1 2 x 2 +yx _ x 3 x 4 dx + _ 0 −1 _ 1 2 x 2 +yx _ x 4 x 3 dx = _ 1 0 _ x 4 + 1 2 x 6 −x 5 − 1 2 x 8 _ dx + _ 0 −1 _ x 5 + 1 2 x 8 −x 4 − 1 2 x 6 _ dx = _ 1 5 x 5 + 1 14 x 7 − 1 6 x 6 − 1 18 x 9 _ 1 0 + _ 1 6 x 6 + 1 18 x 9 − 1 5 x 5 − 1 14 x 7 _ 0 −1 = 31 630 − 241 630 = − 1 3 (i) y x 2y = x 1 0 90 CHAPTER 5. MULTIPLE INTEGRALS _ 1 0 _ 2y 0 e −y 2 /2 dxdy = _ 1 0 _ xe −y 2 /2 _ 2y 0 dy = _ 1 0 2ye −y 2 /2 dy = _ −2e − 1 2 y 2 _ 1 0 = 2 −2e − 1 2 (12) Boundaries: u = 0, u = 1, v = 0, v = 1 Jacobian: ∂ (x, y) ∂ (u, v) = ¸ ¸ ¸ ¸ ¸ ∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ ∂(u−uv) ∂u ∂(uv) ∂u ∂(u−uv) ∂v ∂(uv) ∂v ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ 1 −v v −u u ¸ ¸ ¸ ¸ ¸ = u _ 1 0 _ 1−x 0 e y/(x+y) dydx = _ 1 0 _ 1 0 e uv u ∂ (x, y) ∂ (u, v) dudv = _ 1 0 _ 1 0 ue v dvdu = _ 1 0 [ue v ] 1 0 du = _ 1 0 (ue −u) du = __ u 2 e 2 − u 2 2 __ 1 0 = e −1 2 (13) _ 1 0 f (x) dx = A =⇒ _ 1 0 f (y) dy = A _ 1 0 _ 1 0 f (x) f (y) dxdy = _ 1 0 f (y) _ _ 1 0 f (x) dx _ dy = _ 1 0 Af (y) dy = A _ 1 0 f (y) dy = A 2 _ 1 0 _ x 0 f (x) f (y) dydx + _ 1 0 _ 1 x f (x) f (y) dydx = _ 1 0 _ 1 0 f (x) f (y) dxdy = A 2 Let A = ¦(x, y) [0 ≤ x ≤ 1, 0 ≤ y ≤ x¦ and B = ¦(x, y) [0 ≤ x ≤ 1, x ≤ y ≤ 1¦. Consider the following transformation that transforms the space in Ato B, _ v (x, y) = x u(x, y) = y . Then J = ¸ ¸ ¸ ¸ ¸ ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ 0 1 1 0 ¸ ¸ ¸ ¸ ¸ = −1. 91 CHAPTER 5. MULTIPLE INTEGRALS _ 1 0 _ x 0 f (x) f (y) dydx = _ 1 0 _ 1 u f (u) f (v) dvdu = _ 1 0 _ 1 x f (x) f (y) dydx Thus, _ 1 0 _ x 0 f (x) f (y) dydx + _ 1 0 _ 1 x f (x) f (y) dydx = 2 _ 1 0 _ 1 x f (x) f (y) dydx = A 2 . Therefore, _ 1 0 _ 1 x f (x) f (y) dydx = A 2 2 . (14) Let u = xy and v = xy 3 , where 4 ≤ u ≤ 8 and 5 ≤ v ≤ 15. Jacobian: ∂ (u, v) ∂ (x, y) = ¸ ¸ ¸ ¸ ¸ ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ ¸ ∂(xy) ∂x ∂(xy 3 ) ∂x ∂(xy) ∂y ∂(xy 3 ) ∂y ¸ ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ y y 3 x 3xy 2 ¸ ¸ ¸ ¸ ¸ = 2xy 3 = 2v __ D dxdy = __ D 1 ∂ (x, y) ∂ (u, v) dudv = __ D 1 1 ∂(u,v) ∂(x,y) dudv = _ 8 4 _ 15 5 1 2v dvdu = _ 8 4 _ lnv 2 _ 15 5 du = _ 8 4 _ ln 15 2 − ln5 2 _ du = _ 8 4 _ 1 2 ln3 _ du = 2 ln 3 (15) (a) _ _ Ω (x +y) dxdy = _ π 0 _ 2 0 (r cos θ +r sinθ) rdrdθ = _ π 0 (cos θ + sinθ) dθ _ 2 0 r 2 dr = [sinθ −cos θ] π 0 _ 1 3 r 3 _ 2 0 = (2) _ 8 3 _ = 16 3 (b) _ _ Ω (x +y) dxdy = _ π/2 −π/2 _ 3 2 (r cos θ +r sinθ) rdrdθ = _ π/2 −π/2 (cos θ + sinθ) dθ _ 3 2 r 2 dr = [sinθ −cos θ] π/2 −π/2 _ 1 3 r 3 _ 3 2 = (2) _ 27 −8 3 _ = 38 3 92 CHAPTER 5. MULTIPLE INTEGRALS (c) _ _ Ω (x +y) dxdy = _ π/2 −π/2 _ 2 cos θ 0 (r cos θ +r sinθ) rdrdθ = _ π/2 −π/2 (cos θ + sinθ) _ 2 cos θ 0 r 2 drdθ = _ π/2 −π/2 (cos θ + sinθ) _ 1 3 r 3 _ 2 cos θ 0 dθ = 8 3 _ π/2 −π/2 _ cos 4 θ + cos 3 θ sinθ _ dθ = 8 3 _ π/2 −π/2 cos 4 θdθ + 8 3 _ π/2 −π/2 cos 3 θ sinθdθ = 8 3 (2) _ π/2 0 cos 4 θdθ + 8 3 _ π/2 −π/2 cos 3 θ sinθdθ = 16 3 _ 3π 16 _ + 8 3 _ − 1 4 cos 4 θ _ π/2 −π/2 = π + 0 = π (d) _ _ Ω (x +y) dxdy = _ π/2 0 _ 3 √ θ 0 (r cos θ +r sin θ) rdrdθ = _ π/2 0 (cos θ + sinθ) _ 3 √ θ 0 r 2 drdθ = _ π/2 0 (cos θ + sinθ) _ 1 3 r 3 _ 3 √ θ 0 dθ = 1 3 _ π/2 0 θ (cos θ + sinθ) dθ3 = 1 3 [cos θ +θ sinθ + sinθ −θ cos θ] π/2 0 = π 6 (16) Distance from (0, 0) to _ 1, √ 3 _ = _ (1) 2 + _√ 3 _ 2 = 2 sec θ = 2 =⇒θ = π 3 _ _ R _ x 2 +y 2 dxdy = _ π/3 0 _ sec θ 0 (r) rdrdθ = _ π/3 0 _ 1 3 r 3 _ sec θ 0 dθ = 1 3 _ π/3 0 sec 3 θdθ = √ 3 3 + ln _ 2 + √ 3 _ 6 = 0.79684 (17) R = _ (x, y) [ −2 ≤ x ≤ 4, x 2 −6 2 ≤ y ≤ x + 1 _ _ _ R xydA = _ 4 −2 _ x+1 x 2 −6 2 xydydx = 1 8 _ 4 −2 (−x 5 + 16x 3 + 8x 2 −32x)dx = 36 93 CHAPTER 5. MULTIPLE INTEGRALS (18) Let x = r cos θ and y = r sinθ. Jacobian J = ∂(x,y) ∂(r,θ) = r. R = ¦(r, θ) [0 ≤ r ≤ 4 sin θ, 0 ≤ θ ≤ π¦ _ π 0 _ 4 sin θ 0 r 2 sinθdrdθ = 64 3 _ π 0 sin 4 θdθ = 8π (19) Let x = r cos θ and y = r sinθ. Jacobian J = ∂(x,y) ∂(r,θ) = r. R = _ (r, θ) [2 cos θ ≤ r ≤ 4 cos θ, − π 2 ≤ θ ≤ π 2 _ _ π 2 − π 2 _ 4 cos θ 2 cos θ r 2 cos θdrdθ = 56 3 _ π 2 − π 2 cos 4 θdθ = 7π (20) A sketch will convince you that _ 2 0 _ _ √ 4−y 2 0 _ x 2 +y 2 dx _ dy = _ R _ x 2 +y 2 dxdy, where R is the region given by ¦(x, y) : x ≥ 0, y ≥ 0 and x 2 +y 2 ≤ 4. The above integral is therefore equal to _ π 2 0 __ 2 0 r r dr _ dθ = 4π 3 . (21) (a) __ x 2 +y 2 ≤1 cos _ x 2 +y 2 _ dxdy = _ 2π 0 _ 1 0 _ cos r 2 _ rdrdθ = _ 2π 0 _ 1 2 sinr 2 _ 1 0 dθ = _ 2π 0 _ 1 2 sin1 _ dθ = π sin1 (b) __ 1≤x 2 +y 2 ≤4 cos _ x 2 +y 2 _ dxdy = _ 2π 0 _ 2 1 _ cos r 2 _ rdrdθ = _ 2π 0 _ 1 2 sinr 2 _ 2 1 dθ = _ 2π 0 _ 1 2 sin4 − 1 2 sin 1 _ dθ = π sin4 −π sin 1 94 CHAPTER 5. MULTIPLE INTEGRALS (22) (a) __ x 2 +y 2 ≤1 sin _ _ x 2 +y 2 _ dxdy = _ 2π 0 _ 1 0 (sinr) rdrdθ = _ 2π 0 [sinr −r cos r] 1 0 dθ = _ 2π 0 (sin1 −cos 1) dθ = 2π sin 1 −2π cos 1 (b) __ 1≤x 2 +y 2 ≤4 sin _ x 2 +y 2 _ dxdy = _ 2π 0 _ 2 1 (sinr) rdrdθ = _ 2π 0 [sinr −r cos r] 2 1 dθ = _ 2π 0 (sin2 −2 cos 2 −sin1 + cos 1) dθ = 2π sin2 −4π cos 2 −2π sin1 + 2π cos 1 (23) (a) x y 1 1 __ 0≤x 2 +y 2 ≤1 (x +y) dxdy = _ π/2 0 _ 1 0 (r cos θ +r sin θ) rdrdθ = _ π/2 0 _ 1 0 r 2 (cos θ + sinθ) drdθ = _ π/2 0 _ 1 3 r 3 (cos θ + sinθ) _ 1 0 dθ = _ π/2 0 1 3 (cos θ + sinθ) dθ = 1 3 [sinθ −cos θ] π/2 0 = 2 3 (b) 2 2 1 1 y x 95 CHAPTER 5. MULTIPLE INTEGRALS __ 1≤x 2 +y 2 ≤4 (x +y) dxdy = _ π/2 0 _ 2 1 (r cos θ +r sin θ) rdrdθ = _ π/2 0 _ 2 1 r 2 (cos θ + sinθ) drdθ = _ π/2 0 _ 1 3 r 3 (cos θ + sinθ) _ 2 1 dθ = _ π/2 0 _ 8 3 − 1 3 _ (cos θ + sinθ) dθ = 7 3 [sinθ −cos θ] π/2 0 = 14 3 (24) y x 1 y x = 3 __ Ω _ x 2 +y 2 dxdy = _ π/3 0 _ sec θ 0 _ (r cos θ) 2 + (r sin θ) 2 rdrdθ = _ π/3 0 _ sec θ 0 r 2 drdθ = _ π/3 0 _ 1 3 r 3 _ sec θ 0 dθ = _ π/3 0 1 3 sec 3 θdθ = 1 3 _ π/3 0 sec θ _ 1 + tan 2 θ _ dθ = 1 3 _ π/3 0 sec θdθ + 1 3 _ π/3 0 sec θ tan 2 θdθ = 1 3 [ln(sec θ + tanθ)] π/3 0 + 1 3 _ π/3 0 tanθd (sec θ) = 1 3 ln _ 2 + √ 3 _ + 1 3 [tanθ sec θ] π/3 0 − _ π/3 0 sec 3 θdθ = 1 6 ln _ 2 + √ 3 _ + 1 3 √ 3 (25) (a) 1 1 1 − x 96 CHAPTER 5. MULTIPLE INTEGRALS _ 1 −1 _ √ 1−y 2 0 _ x 2 +y 2 dxdy = _ π/2 −π/2 _ 1 0 r 2 drdθ = _ π/2 −π/2 _ 1 3 r 3 _ 1 0 dθ = _ π/2 −π/2 1 3 dθ = _ 1 3 θ _ π/2 −π/2 = 1 3 π (b) x y 2 2 _ 2 0 _ √ 4−x 2 0 _ x 2 +y 2 dxdy = _ π/2 0 _ 2 0 r 2 drdθ = _ π/2 0 _ 1 3 r 3 _ 2 0 dθ = _ π/2 0 8 3 dθ = _ 8 3 θ _ π/2 0 = 4 3 π (c) x y Ω 3 π 1 1 1 2 _ 2 1/2 _ √ 1−x 2 0 3 _ x 2 +y 2 dxdy = _ π/3 0 _ 1 1 2 sec θ r 4 drdθ = _ π/3 0 _ 1 5 r 5 _ 1 1 2 sec θ dθ = _ π/3 0 _ 1 5 − sec 5 θ 160 _ dθ = 1 15 π − 11 640 √ 3 − 3 1280 ln _ 2 + √ 3 _ Note: For n ≥ 2, _ sec n θdθ = 1 n −1 sec n−2 θ tanθ + n −2 n −1 _ sec n−2 θdθ. 97 CHAPTER 5. MULTIPLE INTEGRALS (d) x y Ω 3 π 1 1 1 2 _ 1/2 0 _ √ 1−x 2 0 xy _ x 2 +y 2 dxdy = _ π/2 π/3 _ 1 0 r 4 cos θ sinθdrdθ + _ π/3 0 _ 1 2 sec θ 0 r 4 cos θ sinθdrdθ = _ π/2 π/3 _ 1 5 r 5 cos θ sinθ _ 1 0 dθ + _ π/3 0 _ 1 5 r 5 cos θ sinθ _1 2 sec θ 0 dθ = 1 5 _ π/2 π/3 cos θ sin θdθ + _ π/3 0 1 160 sec 5 θ cos θ sinθdθ = 1 5 _ π/2 π/3 sin2θ 2 dθ − _ π/3 0 1 160 sec 4 θd (cos θ) = 1 5 _ − 1 4 cos 2θ _ π/2 π/3 − 1 160 _ − 1 3 cos −3 θ _ π/3 0 = 1 40 + 7 480 = 19 480 (26) __ x 2 +y 2 ≤b 2 (y +b) dxdy = _ 2π 0 _ b 0 (r sinθ +b) rdrdθ = _ 2π 0 _ 1 3 (sinθ) r 3 + 1 2 br 2 _ b 0 dθ = _ 2π 0 _ 1 3 (sinθ) b 3 + 1 2 b 3 _ dθ = _ − 1 3 (cos θ) b 3 + 1 2 b 3 θ _ 2π 0 = πb 3 (27) __ x 2 +y 2 ≤1 _ 1 − _ x 2 +y 2 _¸ dxdy = _ 2π 0 _ 1 0 _ 1 −r 2 _ rdrdθ = _ 2π 0 _ − 1 4 r 4 + 1 2 r 2 _ 1 0 dθ = _ 2π 0 1 4 dθ = _ 1 4 θ _ 2π 0 = π 2 98 CHAPTER 5. MULTIPLE INTEGRALS (28) 2 __ x 2 +y 2 ≤4 ¸ 3 _ 1 − x 2 +y 2 4 _ dxdy = 2 _ 2π 0 _ 2 0 ¸ 3 _ 1 − r 2 4 _ rdrdθ = 2 _ 2π 0 _ − 1 18 _ 12 −3r 2 _ 3/2 _ 2 0 dθ = 2 _ 2π 0 _ 2 3 √ 12 _ dθ = 2 __ 2 3 √ 12 _ θ _ 2π 0 = 16 3 π √ 3 (29) __ x 2 +y 2 ≤5 _ 5 −x 2 −y 2 _ 1/6 dxdy = _ 2π 0 _ √ 5 0 _ 5 −r 2 _ 1/6 rdrdθ = _ 2π 0 _ − 3 7 _ 5 −r 2 _ 7/6 _ √ 5 0 dθ = _ 2π 0 _ 15 7 6 √ 5 _ dθ = _ 15 7 6 √ 5θ _ 2π 0 = 30 7 π 6 √ 5 = 17.6 (30) __ x 2 +y 2 ≤1 _ 4 −x 2 −y 2 dA = _ 2π 0 _ 1 0 _ 4 −r 2 rdrdθ = _ 2π 0 _ − 1 3 _ 4 −r 2 _ 3/2 _ 1 0 dθ = _ 2π 0 _ − √ 3 + 8 3 _ dθ = _ − √ 3 + 8 3 _ 2π 0 = 16π 3 −2π √ 3 (31) x Ω 99 CHAPTER 5. MULTIPLE INTEGRALS __ _ 1 −x 2 −y 2 _ dA = _ π/2 −π/2 _ cos θ 0 _ 1 −r 2 _ rdrdθ = _ π/2 −π/2 _ − 1 4 r 4 + 1 2 r 2 _ cos θ 0 dθ = _ π/2 −π/2 _ − 1 4 cos 4 θ + 1 2 cos 2 θ _ dθ = 5 32 π Note: _ cos 2 θdθ = _ 1 + cos 2θ 2 dθ = 1 2 θ + 1 4 sin2θ and _ cos n θdθ = 1 n −1 cos n−1 θ sinθ + n −1 n _ cos n−2 θdθ (32) 2 __ 2xdA = _ π/2 −π/2 _ 2 cos θ 0 2r 2 cos θdrdθ = _ π/2 −π/2 _ 2 3 r 3 cos θ _ 2 cos θ 0 dθ = _ π/2 −π/2 _ 16 3 cos 4 θ _ dθ = 2π (33) __ _ x 2 +y 2 dA = _ π/2 −π/2 _ 2a cos θ 0 r 2 drdθ = _ π/2 −π/2 _ 1 3 r 3 _ 2a cos θ 0 dθ = _ π/2 −π/2 _ 8 3 a 3 cos 3 θ _ dθ = _ π/2 −π/2 _ 8 3 a 3 _ 1 −sin 2 θ _ _ d (sinθ) = _ 8 3 a 3 _ sin θ − 1 3 sin 3 θ __ π/2 −π/2 = 32 9 a 3 (34) 100 CHAPTER 5. MULTIPLE INTEGRALS __ _ b 2 − b 2 a 2 (x 2 +y 2 )dA = b _ π 0 _ a sin θ 0 _ 1 − r 2 a 2 rdrdθ = b _ π 0 _ − 1 3 a 2 _ 1 − r 2 a 2 _ 3/2 _ a sin θ 0 dθ = 2a 2 b 3 _ π/2 0 _ 1 −cos 3 θ _ dθ = 2a 2 b 3 _ π/2 0 dθ − 2a 2 b 3 _ π/2 0 _ 1 −sin 2 θ _ d (sinθ) = 1 3 a 2 bπ − 4 9 a 2 b (35) x = - y I x = __ R y 2 ρ (x, y) dA = _ 2 0 _ y−y 2 −y y 2 (x +y) dxdy = _ 2 0 __ 1 2 x 2 y 2 +y 3 x __ y−y 2 −y dy = _ 2 0 _ −2y 5 + 1 2 y 6 + 2y 4 _ dy = _ − 1 3 y 6 + 1 14 y 7 + 2 5 y 5 _ 2 0 = 64 105 (36) x = y y = 2 - x 101 CHAPTER 5. MULTIPLE INTEGRALS M = __ R ρ (x, y) dA = _ 1 0 _ 2−y y (6x + 3y + 3) dxdy = _ 1 0 _ 3x 2 + 3xy + 3x ¸ 2−y y dy = _ 1 0 _ −6y 2 −12y + 18 _ dx = _ −2y 3 −6y 2 + 18y ¸ 1 0 = 10 M y = __ R xρ (x, y) dA = _ 1 0 _ 2−y x y (6x + 3y + 3) dxdy = _ 1 0 _ 22 −24y + 6y 2 −4y 3 _ dx = _ 22y −12y 2 + 2y 3 −y 4 ¸ 1 0 = 11 M x = __ R yρ (x, y) dA = _ 1 0 _ 2−y y y (6x + 3y + 3) dxdy = _ 1 0 _ 18y −12y 2 −6y 3 _ dx = _ 9y 2 −4y 3 − 3 2 y 4 _ 1 0 = 7 2 Hence, ¯ x = M y M = 11 10 and ¯ y = M x M = 7/2 10 = 7 20 (37) M = __ R ρ (x, y) dA = _ 1 0 _ 6 0 (x +y + 1) dxdy = _ 1 0 _ 1 2 x 2 +yx +x _ 6 0 dy = _ 1 0 (24 + 6y) dy = _ 24y + 3y 2 ¸ 1 0 = 27 M x = __ R yρ (x, y) dA = _ 1 0 _ 6 0 y (x +y + 1) dxdy = _ 1 0 _ 1 2 x 2 y +y 2 x +xy _ 6 0 dy = _ 1 0 _ 24y + 6y 2 _ dy = _ 12y 2 + 2y 3 ¸ 1 0 = 14 102 CHAPTER 5. MULTIPLE INTEGRALS M y = __ R xρ (x, y) dA = _ 1 0 _ 6 0 x(x +y + 1) dxdy = _ 1 0 _ 1 3 x 3 + 1 2 (y + 1) x 2 _ 6 0 dy = _ 1 0 (90 + 18y) dy = _ 90y + 9y 2 ¸ 1 0 = 99 Hence, ¯ x = M y M = 99 27 = 11 3 and ¯ y = M x M = 14 27 I y = __ R x 2 ρ (x, y) dA = _ 1 0 _ 6 0 x 2 (x +y + 1) dxdy = _ 1 0 _ 1 4 x 4 + 1 3 (y + 1) x 3 _ 6 0 dy = _ 1 0 (396 + 72y) dy = 432 (38) x = 4 - y Let k be the constant density, M = __ R ρ (x, y) dA = _ 2 0 _ 4−y y 2 /2 kdxdy = k _ 2 0 _ 4 −y − y 2 2 _ dy = k _ 4y − 1 2 y 2 − 1 6 y 3 _ 2 0 = 14k 3 M y = __ R xρ (x, y) dA = _ 2 0 _ 4−y y 2 /2 kxdxdy = k _ 2 0 _ x 2 2 _ 4−y y 2 /2 = − 1 8 k _ 2 0 _ −64 + 32y −4y 2 +y 4 _ dy = − 1 8 k _ −64y + 16y 2 − 4 3 y 3 + 1 5 y 5 _ 2 0 = 128k 15 103 CHAPTER 5. MULTIPLE INTEGRALS M x = __ R yρ (x, y) dA = _ 2 0 _ 4−y y 2 /2 ykdxdy = k _ 2 0 _ 4y −y 2 − y 3 2 _ dy = k _ 2y 2 − 1 3 y 3 − 1 8 y 4 _ 2 0 = 10k 3 Hence, ¯ x = M y M = 128k/15 14k/3 = 64 35 and ¯ y = M x M = 10k/3 14k/3 = 5 7 (39) a -a 0 a Density = kr M = __ R ρ (x, y) dA = _ π 0 _ a 0 kr rdrdθ = _ π 0 dθ _ a 0 kr 2 dr = π _ kr 3 3 _ a 0 = 1 3 πka 3 M x = __ R yρ (x, y) dA = _ π 0 _ a 0 kr (r sin θ) rdrdθ = _ π 0 sinθdθ _ a 0 kr 3 dr = [−cos θ] 2π 0 _ kr 4 4 _ a 0 = 1 2 ka 4 Hence, ¯ y = M x M = ka 4 2 1 3 πka 3 = 3a 2π and by symmetry, ¯ x = 0. (40) (a) 104 CHAPTER 5. MULTIPLE INTEGRALS M = __ R ρ (x, y) dA = _ 1 0 _ 1−y 0 dxdy = _ 1 0 (1 −y) dy = _ y − 1 2 y 2 _ 1 0 = 1 2 M x = __ R yρ (x, y) dA = _ 1 0 _ 1−y 0 ydxdy = _ 1 0 [yx] 1−y 0 dy = _ 1 0 _ y −y 2 _ dy = _ 1 2 y 2 − 1 3 y 3 _ 1 0 = 1 6 M y = __ R xρ (x, y) dA = _ 1 0 _ 1−y 0 xdxdy = _ 1 0 _ x 2 2 _ 1−y 0 dy = _ 1 0 1 2 (−1 +y) 2 dy = _ 1 6 (−1 +y) 3 _ 1 0 = 1 6 Hence, ¯ x = M y M = 1 6 1 2 = 1 3 and ¯ y = M x M = 1 6 1 2 = 1 3 (b) a a 0 a M = __ R ρ (x, y) dA = _ π 4 − π 4 _ a 0 rdrdθ = πa 2 4 M y = __ R xρ (x, y) dA = 2 _ π 4 0 _ a 0 (r cos θ) rdrdθ = 1 3 √ 2a 3 Hence, ¯ x = M y M = 1 3 √ 2a 3 πa 2 4 = 4 √ 2a 3π 105 CHAPTER 5. MULTIPLE INTEGRALS and by symmetry, ¯ y = 0. (41) (a) ___ D f (x, y, z) dxdy dz = _ 2 0 _ 1 0 _ 1 −1 yz dxdy dz = _ 1 −1 dx _ 1 0 y dy _ 2 0 z dz = [x] 1 −1 _ 1 2 y 2 _ 1 0 _ 1 2 z 2 _ 2 0 = (2) _ 1 2 _ (2) = 2 (b) ___ D f (x, y, z) dxdy dz = _ 2 0 _ 1 0 _ 1 −1 (x +y +z) dxdy dz = _ 2 0 _ 1 0 _ 1 2 x 2 +xy +xz _ 1 −1 dy dz = _ 2 0 _ 1 0 (2y + 2z) dy dz = 2 _ 2 0 __ 1 2 y 2 +yz __ 1 0 dz = 2 _ 2 0 _ 1 2 +z _ dz = 6 (42) (a) ___ D f (x, y, z) dxdy dz = _ 1 0 _ 1−x 0 _ 1−x−y 0 dz dy dx = _ 1 0 _ 1−x 0 (1 −x −y) dy dx = _ 1 0 _ y −xy − 1 2 y 2 _ 1−x 0 dx = _ 1 0 _ 1 2 −x + 1 2 x 2 _ dx = 1 6 (b) ___ D f (x, y, z) dxdy dz = _ 1 0 _ 1−x 0 _ 1−x−y 0 xy dz dy dx = _ 1 0 x _ 1−x 0 _ y −xy −y 2 _ dy dx = _ 1 0 x _ 1 2 y 2 − 1 2 xy 2 − 1 3 y 3 _ 1−x 0 dx = 1 6 _ 1 0 _ x −3x 2 + 3x 3 −x 4 _ dx = 1 120 106 CHAPTER 5. MULTIPLE INTEGRALS (43) (a) ___ D f (x, y, z) dxdy dz = _ 1 −1 _ 2 0 _ 1−x 2 0 (x +y) dz dy dx = _ 1 −1 _ 2 0 (x +y) _ 1 −x 2 _ dy dx = _ 1 −1 _ 2 0 _ x −x 3 +y −x 2 y _ dy dx = _ 1 −1 _ 2x −2x 3 + 2 −2x 2 _ dx = 8 3 (b) ___ D f (x, y, z) dxdy dz = _ 1 −1 _ 2 0 _ 1−x 2 0 y 2 dz dy dx = _ 2 0 y 2 dy _ 1 −1 _ 1−x 2 0 dz dx = _ 1 3 y 3 _ 2 0 _ 1 −1 _ 1 −x 2 _ dx = _ 8 3 __ x − 1 3 x 3 _ 1 −1 = 32 9 (44) (a) ___ R yz 2 dxdydz = _ 1 0 _ 1 0 _ 1 0 yz 2 dzdydx = _ 1 0 _ 1 0 _ 1 3 yz 3 _ 1 0 dydx = _ 1 0 _ 1 0 1 3 ydydx = _ 1 0 _ 1 6 y 2 _ 1 0 dx = _ 1 0 1 6 dx = 1 6 (b) 107 CHAPTER 5. MULTIPLE INTEGRALS ___ R ydxdydz = _ 1 0 _ 1−y 0 _ 1−x−y 0 ydzdxdy = _ 1 0 _ 1−y 0 [yz] 1−x−y 0 dxdy = _ 1 0 _ 1−y 0 (y (1 −y −x)) dxdy = _ 1 0 _ y _ x −yx − 1 2 x 2 __ 1−y 0 dy = _ 1 0 _ 1 2 y −y 2 + 1 2 y 3 _ dy = _ 1 4 y 2 − 1 3 y 3 + 1 8 y 4 _ 1 0 = 1 24 (c) 1 0.75 0.5 0.25 0 1 0.75 0.5 0.25 0 1 0.75 0.5 0.25 0 x y z x y z ___ R xydxdydz = _ 1 0 _ 1 0 _ 1−y 0 xydzdydx = _ 1 0 xdx _ 1 0 _ 1−y 0 ydzdy = _ x 2 2 _ 1 0 _ 1 0 [yz] 1−y 0 dy = 1 2 _ 1 0 _ y −y 2 _ dy = 1 2 _ 1 2 y 2 − 1 3 y 3 _ 1 0 = 1 12 (45) (a) ___ D _ x 2 +y 2 _ dxdy dz = _ 2π 0 _ 1 0 _ r cos θ+r sin θ+2 0 r 3 dz dr dθ = _ 2π 0 _ 1 0 _ r 4 cos θ +r 4 sinθ + 2r 3 _ dr dθ = _ 2π 0 _ 1 5 r 5 cos θ + 1 5 r 5 sinθ + 1 2 r 4 _ 1 0 dθ = _ 2π 0 _ 1 5 cos θ + 1 5 sin θ + 1 2 _ dθ = _ 1 5 sinθ − 1 5 cos θ + 1 2 θ _ 2π 0 = π 108 CHAPTER 5. MULTIPLE INTEGRALS (b) ___ D f (x, y, z) dxdy dz = _ π −π _ 1+cos θ 0 _ r cos θ+r sin θ+2 0 r 3 dz dr dθ = _ π −π _ 1+cos θ 0 _ r 4 cos θ +r 4 sinθ + 2r 3 _ dr dθ = _ π −π _ 1 5 r 5 cos θ + 1 5 r 5 sinθ + 1 2 r 4 _ 1+cos θ 0 dθ = _ π −π _ 1 5 (1 + cos θ) 5 cos θ + 1 5 (1 + cos θ) 5 sin θ + 1 2 (1 + cos θ) 4 _ dθ = 1 5 _ 105π 8 _ + 1 5 (0) + 1 2 _ 35π 4 _ = 7π (46) D = _ (x, y, z) [ 0 ≤ x 2 +y 2 +z 2 ≤ 1 _ R = ¦(ρ, φ, θ) [ 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π¦ ___ D ρ (x, y, z) dxdy dz = _ _ _ R ρ (ρ sinφcos θ, ρ sinφsin θ, ρ cos φ) ρ 2 sinφdρ dφdθ = _ 2π 0 _ π 0 _ 1 0 ρ 4 sinφdρ dφdθ = _ 2π 0 dθ _ π 0 sinφdφ _ 1 0 ρ 4 dρ = [θ] 2π 0 [−cos φ] π 0 _ 1 5 ρ 5 _ 1 0 = (2π) (2) _ 1 5 _ = 4 5 π (47) D = _ (x, y, z) [ 1 ≤ x 2 +y 2 +z 2 ≤ 4 _ R = ¦(ρ, φ, θ) [ 1 ≤ ρ ≤ 2, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π¦ ___ D ρ (x, y, z) dxdy dz = _ 2π 0 _ π 0 _ 2 1 ρ 3 sinφdρ dφdθ = _ 2π 0 dθ _ π 0 sinφdφ _ 2 1 ρ 3 dρ = [θ] 2π 0 [−cos φ] π 0 _ 1 4 ρ 4 _ 2 1 = (2π) (2) _ 15 4 _ = 15π (48) _ z = _ x 2 +y 2 x 2 +y 2 +z 2 = 1 =⇒x 2 +y 2 = 1 2 (The intersection of the surface is a cycle centred at the origin and radius is √ 2 2 .) 109 CHAPTER 5. MULTIPLE INTEGRALS D = _ (x, y, z) [ 0 ≤ x 2 +y 2 ≤ 1 2 , _ x 2 +y 2 ≤ z ≤ _ 1 −x 2 −y 2 _ R = _ (ρ, φ, θ) [ 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π 4 , 0 ≤ θ ≤ 2π _ ___ D ρ (x, y, z) dxdy dz = _ 2π 0 _ π/4 0 _ 1 0 ρ 2 sinφdρ dφdθ = _ 2π 0 _ π/4 0 sinφ _ 1 3 ρ 3 _ 1 0 dφdθ = 1 3 _ 2π 0 dθ _ π/4 0 sinφdφ = 1 3 [θ] 2π 0 [−cos φ] π/4 0 = 1 3 (2π) _ 1 − √ 2 2 _ = _ 2 − √ 2 _ π 3 (49) S = ¦(r, θ, z) [0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 4 −r¦, V = _ 2π 0 _ 4 0 _ 4−r 0 r dz dr dθ = _ 2π 0 _ 4 0 _ 4r −r 2 _ dr dθ = (2π) _ 32 − 64 3 _ = 64 3 π (50) R h y z x (a) Let σ = kz. Mass = M = ___ T σdV = _ 2π 0 _ R 0 _ h 0 rkzdzdrdθ = _ 2π 0 _ R 0 _ 1 2 rkz 2 _ h 0 drdθ = _ 2π 0 _ R 0 _ 1 2 rkh 2 drdθ = _ 2π 0 _ 1 4 r 2 kh 2 _ R 0 dθ = _ 2π 0 1 4 R 2 kh 2 dθ = _ 1 4 R 2 kh 2 θ _ 2π 0 = 1 2 kπR 2 h 2 110 CHAPTER 5. MULTIPLE INTEGRALS (b) By symmetry, ¯ x = ¯ y = 0 M¯ z = ___ T zσdV = _ 2π 0 _ R 0 _ h 0 rkz 2 dzdrdθ = _ 2π 0 dθ _ R 0 rdr _ h 0 kz 2 dz = (2π) _ 1 2 r 2 _ R 0 _ 1 3 kz 3 _ h 0 = (2π) 1 2 R 2 1 3 kh 3 = 1 3 πkR 2 h 3 ¯ z = 1 3 πkR 2 h 3 1 2 kπR 2 h 2 = 2 3 h (c) I z = ___ T σ _ x 2 +y 2 _ dV = _ 2π 0 dθ _ R 0 r 3 dr _ h 0 kzdz = (2π) _ 1 4 r 4 _ R 0 _ 1 2 kz 2 _ h 0 = (2π) 1 4 R 4 1 2 kh 2 = 1 4 πR 4 kh 2 = 1 2 MR 2 (51) (a) Since σ = constant M = σ ___ T dV = σ _ 2π 0 dθ _ R 0 rdr _ h 0 dz = σ (2π) _ 1 2 r 2 _ R 0 [z] h 0 = πσR 2 h and I z = σ ___ T _ x 2 +y 2 _ dV = σ _ 2π 0 dθ _ R 0 r 3 dr _ h 0 dz = σ (2π) _ 1 4 r 4 _ R 0 [z] h 0 = σ (2π) 1 4 R 4 h = 1 2 πσR 4 h = M 2 R 2 (b) I x = σ ___ T _ y 2 +z 2 _ dV = σ _ 2π 0 dθ _ R 0 rdr _ h 0 _ r 2 sin 2 θ +z 2 _ dz = σ _ 2π 0 sin 2 θdθ _ R 0 r 3 dr _ h 0 dz +σ _ 2π 0 dθ _ R 0 rdr _ h 0 z 2 dz = σ _ 2π 0 sin 2 θdθ _ R 0 r 3 dr _ h 0 dz +σ (2π) _ 1 2 r 2 _ R 0 _ 1 3 z 3 _ h 0 = σ _ 1 2 θ − sin2θ 4 _ 2π 0 _ 1 4 r 4 _ R 0 [z] h 0 +σ (2π) 1 4 R 4 h = 1 4 σπR 4 h + 1 3 σπR 2 h 3 = M _ R 2 4 + h 2 3 _ 111 CHAPTER 5. MULTIPLE INTEGRALS (c) By part (b), the moment of inertia is M _ R 2 4 + h 2 3 _ −M _ h 2 _ 2 = M _ R 2 4 + h 2 12 _ (52) h y x z R h r z = (a) V = ___ T dV = _ 2π 0 dθ _ R 0 rdr _ h hr R dz = _ 2π 0 dθ _ R 0 r _ h − hr R _ dr = (2π) _ 1 2 hr 2 − 1 3 hr 3 _ R 0 = πhR 2 − 2 3 πhR 2 = 1 3 πhR 2 (b) M = σV = 1 3 σπhR 2 ¯ zM = σ ___ T zdV = σ _ 2π 0 dθ _ R 0 rdr _ h hr R zdz = σ _ 2π 0 dθ _ R 0 r _ h 2 2 − h 2 r 2 2R 2 _ dr = σ _ 2π 0 1 8 h 2 R 2 dθ = 1 4 σπh 2 R 2 Hence, ¯ z = 3 4 h By symmetry, ¯ x = ¯ y = 0. 112 CHAPTER 5. MULTIPLE INTEGRALS (c) I z = σ ___ T _ x 2 +y 2 _ dV = σ _ 2π 0 dθ _ R 0 r 3 dr _ h hr R dz = σ _ 2π 0 dθ _ R 0 r 3 _ h − hr R _ dr = σ _ 2π 0 1 20 hR 4 dθ = 1 10 σπhR 4 = 3 10 MR 2 (d) I x = σ ___ T _ y 2 +z 2 _ dV = σ _ 2π 0 dθ _ R 0 rdr _ h hr R _ r 2 sin 2 θ +z 2 _ dz = σ _ 2π 0 sin 2 θdθ _ R 0 r 3 dr _ h hr R dz +σ _ 2π 0 dθ _ R 0 rdr _ h hr R z 2 dz = σ _ 2π 0 sin 2 θdθ _ R 0 r 3 _ h − hr R _ dr +σ _ 2π 0 dθ _ R 0 1 3 r _ h 3 − h 3 r 3 R 3 _ dr = σ _ 2π 0 1 20 hR 4 sin 2 θdθ +σ _ 2π 0 1 10 h 3 R 2 dθ = 1 20 σπhR 4 + 1 5 σπR 2 h 3 (53) (a) V = ___ T dV = _ 2π 0 dθ _ 1 0 rdr _ 1−r 2 0 dz = _ 2π 0 dθ _ 1 0 r _ 1 −r 2 _ dr = _ 2π 0 1 4 dθ = π 2 (b) Note that σ = kz M = ___ T σdV = k ___ T zdV = k _ 2π 0 dθ _ 1 0 rdr _ 1−r 2 0 zdz = k _ 2π 0 dθ _ 1 0 1 2 r _ 1 −r 2 _ 2 dr = k _ 2π 0 1 12 dθ = πk 6 113 CHAPTER 5. MULTIPLE INTEGRALS (c) Note that σ = k _ r 2 +z 2 _ M = ___ T σdV = k ___ T _ r 2 +z 2 _ dV = k _ 2π 0 dθ _ 1 0 rdr _ 1−r 2 0 _ r 2 +z 2 _ dz = k _ 2π 0 dθ _ 1 0 r _ r 2 _ 1 −r 2 _ + 1 3 _ 1 −r 2 _ 3 _ dr = k _ 2π 0 1 8 dθ = πk 4 (54) V = ___ dV = _ π/2 −π/2 dθ _ 2a cos θ 0 rdr _ r 2 /a 0 dz = _ π/2 −π/2 dθ _ 2a cos θ 0 r 3 a dr = _ π/2 −π/2 4a 3 cos 4 θdθ = 3 2 πa 3 (55) V = ___ dV = _ π/2 −π/2 dθ _ 2a cos θ 0 rdr _ r 0 dz = _ π/2 −π/2 dθ _ 2a cos θ 0 r 2 dr = _ π/2 −π/2 8 3 a 3 cos 3 θdθ = 32 9 a 3 (56) V = ___ dV = _ π/2 −π/2 dθ _ 2 cos θ 0 rdr _ 1 2 (4+r cos θ) 0 dz = _ π/2 −π/2 dθ _ 2 cos θ 0 1 2 r (4 +r cos θ) dr = _ π/2 −π/2 _ 4 3 cos 4 θ + 4 cos 2 θ _ dθ = 5 2 π (57) V = ___ dV = _ π/2 −π/2 dθ _ a cos θ 0 rdr _ a−r 0 dz = _ π/2 −π/2 dθ _ a cos θ 0 r (a −r) dr = _ π/2 −π/2 _ − 1 3 a 3 cos 3 θ + 1 2 a 3 cos 2 θ _ dθ = 1 4 πa 3 − 4 9 a 3 114 CHAPTER 5. MULTIPLE INTEGRALS (58) z = z = r + 1 5 2 - r 2 Where the sphere x 2 + y 2 + z 2 = 25 intersects the cone z = _ x 2 +y 2 + 1, we have r 2 + (r + 1) 2 = 25, thus, r = 3. V = _ 2π 0 dθ _ 3 0 rdr _ √ 25−r 2 r+1 dz = _ 2π 0 dθ _ 3 0 r _ _ 25 −r 2 −(r + 1) _ dr = _ 2π 0 _ − 1 3 _ 25 −r 2 _ 3/2 − 1 3 r 3 − 1 2 r 2 _ 3 0 dθ = _ 2π 0 41 6 dθ = 41 3 π (59) x x y 1 z = x 2 + y 2 z = x The surfaces intersect at x 2 +y 2 = x, thus r = cos θ. V = ___ dV = _ π/2 −π/2 dθ _ cos θ 0 rdr _ r cos θ r 2 dz = _ π/2 −π/2 dθ _ cos θ 0 r _ r cos θ −r 2 _ dr = _ π/2 −π/2 _ − 1 4 cos 4 θ + 1 3 cos 4 θ _ dθ = 1 32 π 115 CHAPTER 5. MULTIPLE INTEGRALS (60) cone hyperbola z Where the hyperboloid intersects the cone, we have r 2 +a 2 = 2r 2 , thus r = a. V = _ 2π 0 dθ _ a 0 rdr _ √ a 2 +r 2 √ 2r dz = _ 2π 0 dθ _ a 0 r _ _ a 2 +r 2 − √ 2r _ dr = _ 2π 0 _ 1 3 _ a 2 +r 2 _ 3/2 − √ 2 3 r 3 _ a 0 dθ = _ 2π 0 _ 1 3 a 3 √ 2 − 1 3 a 3 _ dθ = 2 3 πa 3 √ 2 − 2 3 πa 3 = 2 3 πa 3 _ √ 2 −1 _ (61) y x z 2 1 Sphere: x 2 + y 2 + z 2 = 1 Cone: z = 3(x 2 + y 2 ) 2 1 The cone z = _ 3 (x 2 +y 2 ) has equation z = √ 3r. It intersects the sphere z 2 +r 2 = 1 at 3r 2 +r 2 = 1 or r = 1 2 . V = _ 2π 0 dθ _ 1 2 0 rdr _ √ 1−r 2 √ 3r dz = _ 2π 0 dθ _ 1 2 0 r _ _ 1 −r 2 − √ 3r _ dr = _ 2π 0 _ − 1 3 _ 1 −r 2 _ 3/2 − 1 3 r 3 √ 3 _1 2 0 dθ = _ 2π 0 _ − 1 6 √ 3 + 1 3 _ dθ = 2 3 π − 1 3 π √ 3 116 CHAPTER 5. MULTIPLE INTEGRALS (62) V = _ 2π 0 dθ _ 2 1 rdr _ _ 9− r 2 4 0 dz = _ 2π 0 dθ _ 2 1 1 2 r _ (36 −r 2 )dr = _ 2π 0 _ − 1 6 _ 36 −r 2 _ 3/2 _ 2 1 dθ = _ 2π 0 _ − 16 3 √ 32 + 35 6 √ 35 _ dθ = − 128 3 π √ 2 + 35 3 π √ 35 = 27.273 (63) (a) r = 1. Cylinder of radius 1, with the z-axis as its axis. φ = π 2 . The xy-plane. φ = 3π 4 . The cone shown in diagram. = 3 4 y x z θ = π 4 . A plane prependicular to the xy-plane with the angle between this plane and the xz-plane is π 4 . z = 1. The plane parallel to the xy-plane through (0, 0, 1) . ρ = cos φ. The sphere with centre at (x, y, z) = _ 0, 0, 1 2 _ and radius 1 2 . (64) V = _ 2π 0 _ π 0 _ R 0 ρ 2 sin φdρdφdθ = _ 2π 0 _ π 0 1 3 R 3 sinφdφdθ = _ 2π 0 2 3 R 3 dθ = 4 3 πR 3 (65) r = ρ sinφ, θ = θ, z = ρ cos φ. (66) R = _ (x, y, z) [ −2 ≤ x ≤ 2, − _ 4 −x 2 ≤ y ≤ _ 4 −x 2 , 0 ≤ z ≤ _ 4 −(x 2 +y 2 ) _ = _ (ρ, φ, θ) [0 ≤ ρ ≤ 2, 0 ≤ φ ≤ π 2 , 0 ≤ θ ≤ 2π _ 117 CHAPTER 5. MULTIPLE INTEGRALS _ 2 −2 _ √ 4−x 2 − √ 4−x 2 _ √ 4−(x 2 +y 2 ) 0 z 2 _ x 2 +y 2 +z 2 dzdydx = _ 2π 0 _ π/2 0 _ 2 0 _ ρ 3 cos 2 φ _ _ ρ 2 sinφ _ dρdφdθ = _ 2π 0 _ π/2 0 _ 1 6 ρ 6 _ 2 0 cos 2 φsinφdφdθ = 32 3 _ 2π 0 _ π/2 0 cos 2 φsinφdφdθ = 32 3 _ 2π 0 _ − 1 3 cos 3 φ _ π/2 0 dθ = 32 9 _ 2π 0 dθ = 64π 9 (67) Substitute x = ρ sinφcos θ, y = ρ sinφsin θ, z = ρ cos φ into z = _ 3x 2 + 3y 2 , then ρ cos φ = _ 3 (ρ sinφcos θ) 2 + 3 (ρ sinφsinθ) 2 = √ 3ρ sinφ =⇒tanφ = 1 √ 3 =⇒φ = π 6 . S = _ (ρ, φ, θ) [0 ≤ ρ ≤ 4, 0 ≤ φ ≤ π 6 , 0 ≤ θ ≤ 2π _ V = _ 2π 0 _ π/6 0 _ 4 0 ρ 2 sinφdρ dφdθ = (2π) _ 1 − √ 3 2 _ _ 64 3 _ = 64 3 π _ 2 − √ 3 _ = 17.958 (68) _ 3 0 _ √ 9−z 2 0 _ x 0 xy dydxdz = _ 3 0 _ √ 9−z 2 0 x _ y 2 2 _ x 0 dxdz = 1 2 _ 3 0 _ √ 9−z 2 0 x 3 dxdz = 1 2 _ 3 0 _ x 4 4 _ √ 9−z 2 0 dz = 1 8 _ 3 0 _ 9 −z 2 _ 2 dz = 1 8 _ 3 0 _ 81 −18z 2 +z 4 _ dz = 81 5 (69) V = _ α 0 _ π 0 _ R 0 ρ 2 sinφdρdφdθ = _ α 0 _ π 0 1 3 R 3 sin φdφdθ = _ α 0 2 3 R 3 dθ = 2 3 αR 3 118 CHAPTER 5. MULTIPLE INTEGRALS (70) Now, σ = k (R −ρ) , the mass is M = _ _ _ σdV = _ 2π 0 _ π 0 _ R 0 k (R −ρ) ρ 2 sinφdρdφdθ = _ 2π 0 _ π 0 _ k (sinφ) _ − 1 4 ρ 4 + 1 3 Rρ 3 __ R 0 dφdθ = _ 2π 0 _ π 0 1 12 R 4 k sinφdφdθ = _ 2π 0 1 6 R 4 kdθ = 1 3 πR 4 k (71) h y R Base Now, σ = kρ, tanα = R h . On the base, ρ cos φ = h. The mass is M = _ _ _ kρdV = _ 2π 0 _ α 0 _ hsec φ 0 kρ 3 sin φdρdφdθ = _ 2π 0 _ α 0 _ 1 4 kρ 4 sinφ _ hsec φ 0 dφdθ = _ 2π 0 _ α 0 1 4 h 4 cos 4 φ k sin φdφdθ = _ 2π 0 1 12 h 4 k _ sec 3 α −1 _ dθ = 1 6 πh 4 k _ sec 3 α −1 _ = 1 6 πh 4 k _ _ R 2 +h 2 h 2 _ 3/2 −1 _ = 1 6 πhk _ _ R 2 +h 2 _ 3/2 −h 3 _ (72) V = ___ cone dV = _ 2π 0 _ α 0 _ hsec φ 0 ρ 2 sinφdρdφdθ = _ 2π 0 _ α 0 _ 1 3 ρ 3 sinφ _ hsec φ 0 dφdθ = _ 2π 0 _ α 0 1 3 h 3 cos 3 φ sinφdφdθ = _ 2π 0 1 6 h 3 _ sec 3 α −1 _ dθ = 1 3 πh 3 _ sec 2 α −1 _ = 1 3 πh 3 tan 2 α = 1 3 πR 2 h 119 CHAPTER 5. MULTIPLE INTEGRALS (73) (a) i. I z = ___ ball σ _ x 2 +y 2 _ dV = σ _ 2π 0 _ π 0 _ R 0 ρ 2 sin 2 φ ρ 2 sinφdρdφdθ = σ _ 2π 0 _ π 0 _ R 0 ρ 4 sin 3 φdρdφdθ = σ _ 2π 0 _ π 0 R 5 5 sin 3 φdφdθ = σ _ 2π 0 4 15 R 5 dθ = 8 15 σπR 5 = 2 5 MR 2 By parallel axis theorem, I = _ 2 5 MR 2 +MR 2 _ = 7 5 MR 2 (b) M xy = ___ T zσdV = σ _ 2π 0 _ π/2 0 _ R 0 ρ 2 sinφ ρ cos φdρdφdθ = σ _ 2π 0 _ π/2 0 _ R 0 ρ 3 sinφcos φdρdφdθ = σ _ 2π 0 _ π/2 0 R 4 4 sinφcos φdφdθ = σ _ 2π 0 _ π/2 0 R 4 4 sin2φ 2 dφdθ = σ _ 2π 0 1 8 R 4 dθ = 1 4 σπR 4 Thus ¯ z = M xy M/2 = 1 4 σπR 4 2 3 σπR 3 = 3 8 R and by symmetry ¯ x = ¯ y = 0. (74) (a) I z = ___ T σ _ x 2 +y 2 _ dV = σ _ 2π 0 _ π 0 _ R 2 R 1 ρ 2 sin 2 φ ρ 2 sinφdρdφdθ = σ _ 2π 0 _ π 0 _ R 2 R 1 ρ 4 sin 3 φdρdφdθ = σ _ 2π 0 _ π 0 _ R 5 2 5 − R 5 1 5 _ sin 3 φdφdθ = σ _ 2π 0 4 15 _ R 5 2 −R 5 1 _ dθ = 8 15 σπ _ R 5 2 −R 5 1 _ = 2 5 M R 5 2 −R 5 1 R 3 2 −R 3 1 (b) Setting R 2 = R and R 1 →R in part (a), we get I z = lim R 1 →R 2 5 M R 5 −R 5 1 R 3 −R 3 1 = 2 5 M lim R 1 →R R 5 −R 5 1 R 3 −R 3 1 = 2 5 M lim R 1 →R −5R 4 1 −3R 2 1 = 2 3 MR 2 120 CHAPTER 5. MULTIPLE INTEGRALS (c) By parallel axis theorem, I = 2 3 MR 2 +MR 2 = 5 3 R 2 M (d) M xy = ___ T zσdV = σ _ 2π 0 _ π/2 0 _ R 2 R 1 ρ 2 sinφ ρ cos φdρdφdθ = σ _ 2π 0 _ π/2 0 _ R 2 R 1 ρ 3 sinφcos φdρdφdθ = σ _ 2π 0 _ π/2 0 _ R 4 2 4 − R 4 1 4 _ sinφcos φdφdθ = σ _ 2π 0 _ π/2 0 _ R 4 2 4 − R 4 1 4 _ sin2φ 2 dφdθ = σ _ 2π 0 _ R 4 2 8 − R 4 1 8 _ dθ = 1 4 πσ _ R 4 2 −R 4 1 _ and M = ___ T σdV = σ _ 2π 0 _ π/2 0 _ R 2 R 1 ρ 2 sinφdρdφdθ = 2 3 πσ _ R 3 2 −R 3 1 _ Thus ¯ z = M xy M = 1 4 πσ _ R 4 2 −R 4 1 _ 2 3 πσ _ R 3 2 −R 3 1 _ = 3 8 R 4 2 −R 4 1 R 3 2 −R 3 1 and by symmetry ¯ x = ¯ y = 0. (e) Setting R 2 = R, R 1 →R in part (a), we get ¯ z = lim R 1 →R 3 8 R 4 −R 4 1 R 3 −R 3 1 = 1 2 R (75) (a) Rsec φ ≤ ρ ≤ 2Rcos φ, 0 ≤ φ ≤ π 4 , 0 ≤ θ ≤ 2π. (b) Sphere: ρ 2 sin 2 φ + (ρ cos φ −R) 2 = R 2 , therefore ρ = 2Rcos φ 121 CHAPTER 5. MULTIPLE INTEGRALS (76) (a) σ = kρ M = ___ σdV = _ 2π 0 _ π/2 0 _ 2Rcos φ 0 ρ 2 sinφ kρdρdφdθ = _ 2π 0 _ π/2 0 _ 1 4 kρ 4 sinφ _ 2Rcos φ 0 dφdθ = k _ 2π 0 _ π/2 0 4R 4 cos 4 φsin φdφdθ = k _ 2π 0 4 5 R 4 dθ = 8 5 kπR 4 (b) σ = kρ sinφ M = ___ σdV = _ 2π 0 _ π/2 0 _ 2Rcos φ 0 ρ 2 sin φ kρ sinφdρdφdθ = _ 2π 0 _ π/2 0 _ 1 4 kρ 4 sin 2 φ _ 2Rcos φ 0 dφdθ = k _ 2π 0 _ π/2 0 4R 4 cos 4 φsin 2 φdφdθ = k _ 2π 0 1 8 πR 4 dθ = 1 4 π 2 kR 4 (c) σ = kρ cos 2 θ sinφ M = ___ σdV = _ 2π 0 _ π/2 0 _ 2Rcos φ 0 ρ 2 sinφ kρ cos 2 θ sinφdρdφdθ = k _ 2π 0 _ π/2 0 _ 1 4 kρ 4 sin 2 φcos 2 θ _ 2Rcos φ 0 dφdθ = k _ 2π 0 _ π/2 0 4R 4 cos 4 φsin 2 φcos 2 θdφdθ = k _ 2π 0 1 8 πR 4 cos 2 θdθ = 1 8 π 2 kR 4 (77) x z 4 122 CHAPTER 5. MULTIPLE INTEGRALS Where the sphere ρ = 2 √ 2 cos φ intersects the sphere ρ = 2, we have 2 = 2 √ 2 cos φ giving φ = π 4 . V = _ 2π 0 _ π/4 0 _ 2 0 ρ 2 sin φdρdφdθ + _ 2π 0 _ π/2 π/4 _ 2 √ 2 cos φ 0 ρ 2 sinφdρdφdθ = _ 2π 0 _ π/4 0 8 3 sinφdφdθ + _ 2π 0 _ π/2 π/4 16 3 √ 2 cos 3 φsin φdφdθ = _ 2π 0 _ 8 3 − 4 3 √ 2 _ dθ + _ 2π 0 1 3 √ 2dθ = 16 3 π − 8 3 π √ 2 + 2 3 π √ 2 = 16 3 π −2π √ 2 (78) x z V = _ 2π 0 _ π 0 _ 1−cos φ 0 ρ 2 sinφdρdφdθ = _ 2π 0 _ π 0 1 3 (1 −cos φ) 3 sinφdφdθ = _ 2π 0 1 12 _ (1 −cos φ) 4 _ π 0 dθ = _ 2π 0 4 3 dθ = 8 3 π 123 Chapter 6 Vector Calculus (1) Let θ be the angle the two vectors a and b. Since (a • b) = |a| |b| cos θ, |a b| 2 = (|a| |b| sinθ) 2 = |a| 2 |b| 2 sin 2 θ = |a| 2 |b| 2 _ 1 −cos 2 θ _ = |a| 2 |b| 2 −|a| 2 |b| 2 cos 2 θ = |a| 2 |b| 2 −(a • b) 2 A B C (2) −−→ AB = b 1 i +b 2 j −(a 1 i +a 2 j) = (b 1 −a 1 ) i + (b 2 −a 2 ) j −→ AC = c 1 i +c 2 j −(a 1 i +a 2 j) = (c 1 −a 1 ) i + (c 2 −a 2 ) j The area of the triangle is 1 2 _ _ _ −−→ AB _ _ _ _ _ _ −→ AC _ _ _sinθ = 1 2 _ _ _ −−→ AB −→ AC _ _ _ = 1 2 _ _ _ _ _ _ _ _ det _ _ _ _ i j k (b 1 −a 1 ) (b 2 −a 2 ) 0 (c 1 −a 1 ) (c 2 −a 2 ) 0 _ _ _ _ _ _ _ _ _ _ _ _ = 1 2 _ _ _ _ _ det _ (b 1 −a 1 ) (b 2 −a 2 ) (c 1 −a 1 ) (c 2 −a 2 ) _ k _ _ _ _ _ = 1 2 ¸ ¸ ¸ ¸ ¸ det _ (b 1 −a 1 ) (b 2 −a 2 ) (c 1 −a 1 ) (c 2 −a 2 ) _¸ ¸ ¸ ¸ ¸ where θ is the angle between −−→ AB and −→ AC. (3) Let x be the position vector of the point (x, y, z) . 125 CHAPTER 6. VECTOR CALCULUS B 0 x (XB) 2 = (OX) 2 −(OB) 2 = |x| 2 − _ _ _ −−→ OB _ _ _ 2 = |x| 2 −(|x| cos θ) 2 = |x| 2 − _ |x| x • a |x| |a| _ 2 = |x| 2 − (x • a) 2 |a| 2 . (4) (a) ∇f[ (2,1) = ∂ (y −3x) ∂x i + ∂ (y −3x) ∂y j ¸ ¸ ¸ ¸ (2,1) = −3i +j (b) ∇f[ (2,1,−1) = 2x 2 +z x i + 2yj + (−4z + lnx) k ¸ ¸ ¸ ¸ (2,1,−1) = 7 2 i + 2j + (4 + ln2) k (5) (a) ∇• F = _ ∂ ∂x i + ∂ ∂y j + ∂ ∂z k _ • __ x 2 −y _ i + (4z) j + _ x 2 _ k _ = ∂ ∂x _ x 2 −y _ + ∂ ∂y (4z) + ∂ ∂z _ x 2 _ = 2x ∇F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z _ x 2 −y _ (4z) _ x 2 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −4i−2xj +k (b) ∇• F = _ ∂ ∂x i + ∂ ∂y j + ∂ ∂z k _ • ((yz) i + (xz) j + (xy) k) = 0 ∇F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z (yz) (xz) (xy) ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 0 126 CHAPTER 6. VECTOR CALCULUS (c) ∇• F = _ ∂ ∂x i + ∂ ∂z j _ • ((−y) i + (x) j) = ∂ ∂x (−y) + ∂ ∂y (x) = 0 ∇F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z (−y) (x) 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 2k (d) ∇• F = _ ∂ ∂x i + ∂ ∂y j + ∂ ∂z k _ • _ (2xz) i + (−xy) j + _ −z 2 _ k _ = −x ∇F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z (2xz) (−xy) 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 2xj −yk (6) (a) div F(x, y, z) = ∇• F(x, y, z) = 2x + 6y −4z div F(x 0 , y 0 , z 0 ) = 2x 0 + 6y 0 −4z 0 curl F(x, y, z) = ∇F(x, y, z) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z x 2 3y 2 −2z 2 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 0 (b) div F(x, y, z) = 0 curl F(x, y, z) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z e yz e zx e xy ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸xe xy −xe zx , ye yz −ye xy , ze zx −ze yz ) curl F(0, 1, 2) = ¸ 0, e 2 −1, 2 −2e 2 _ (c) div F(x, y, z) = 0 curl F(x, y, z) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z sinyz cos zx xy ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸x −xsinzx, y cos yz −y, −z sinzx −z cos yz) curl F _ 0, 1, π 2 _ = ¸0, −1, 0) (d) div F(x, y, z) = 2xy− 6xy 2 z 2 + 1 div F(2, 1, 1) = 4 −12 + 1 = −7 127 CHAPTER 6. VECTOR CALCULUS curl F(x, y, z) = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z x 2 y −z 2 −2xy 3 z 2 xy +z ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸ x + 4xy 3 z, −2z −y, −2y 3 z 2 −x 2 _ curl F(2, 1, 1) = ¸10, −3, −6) (7) (a) ∇ 2 f = ∇• ∇f = ∇• _ ∂ _ ln _ x 2 +y 2 __ ∂x i + ∂ _ ln _ x 2 +y 2 __ ∂y j _ = _ ∂ ∂x i + ∂ ∂y j _ • _ 2x x 2 +y 2 i + 2y x 2 +y 2 j _ = ∂ ∂x _ 2x x 2 +y 2 _ + ∂ ∂y _ 2y x 2 +y 2 _ = 2 −x 2 +y 2 (x 2 +y 2 ) 2 −2 −x 2 +y 2 (x 2 +y 2 ) 2 = 0 (b) ∇ 2 f = ∇• ∇f = _ ∂ ∂x i + ∂ ∂y j _ • __ 3Ax 2 + 2Bxy +Cy 2 _ i + _ Bx 2 + 2Cxy + 3Dy 2 _ j _ = ∂ ∂x _ 3Ax 2 + 2Bxy +Cy 2 _ + ∂ ∂y _ Bx 2 + 2Cxy + 3Dy 2 _ = 6Ax + 2By + 2Cx + 6Dy (c) ∇ 2 f = ∇• ∇f = ∇• _ ∂r −1 ∂x i + ∂r −1 ∂y j + ∂r −1 ∂z j _ = _ ∂ ∂x i + ∂ ∂y j + ∂ ∂z k _ • _ −r −2 x r i −r −2 y r j −−r −2 z r k _ = − ∂ ∂x _ r −3 x _ − ∂ ∂y _ r −3 y _ − ∂ ∂z _ r −3 z _ = −r −3 + 3r −4 x 2 r −r −3 + 3r −4 y 2 r −r −3 + 3r −4 z 2 r = −3r −3 + 3r −5 _ x 2 +y 2 +z 2 _ = −3r −3 + 3r −3 = 0 128 CHAPTER 6. VECTOR CALCULUS (8) ∇• (f (r) r) = _ ∂ ∂x i+ ∂ ∂y j+ ∂ ∂z k _ • (f (r) xi+f (r) yj+f (r) zk) = _ f (r) ∂x ∂x +x ∂f (r) ∂x _ + _ f (r) ∂y ∂y +y ∂f (r) ∂y _ + _ f (r) ∂z ∂z +z ∂f (r) ∂z _ = _ f (r) +xf ′ (r) ∂r ∂x _ + _ f (r) +yf ′ (r) ∂r ∂y _ + _ f (r) +zf ′ (r) ∂r ∂z _ = _ f (r) +f ′ (r) x 2 r _ + _ f (r) +f ′ (r) y 2 r _ + _ f (r) +f ′ (r) z 2 r _ = 3f (r) +f ′ (r) x 2 +y 2 +z 2 r = 3f (r) +rf ′ (r) (9) (a) To find a φ such that ∇φ = A, φ x = e x cos y +yz ⇒φ = _ (e x cos y +yz) dx = e x cos y +xyz +f (y, z) φ y = xz −e x siny ⇒φ = _ (xz −e x siny) dy = e x cos y +xyz +g (x, z) φ z = xy +z + 2 ⇒φ = _ (xy +z + 2) dz = xyz + 1 2 z 2 + 2z +h(x, y) Thus, φ = e x cos y +xyz + 1 2 z 2 + 2z +C. (b) To find a φ such that ∇φ = A, φ x = y +z ⇒φ = _ (y +z) dx = xy +xz +f (y, z) φ y = x +z ⇒φ = _ (x +z) dy = xy +yz +g (x, z) φ z = x +y ⇒φ = _ (x +y) dz = xz +yz +h(x, y) Thus, φ = xy +yz +xz +C. (c) To find a φ such that ∇φ = A, φ x = y sinz ⇒φ = _ y sinzdx = xy sinz +f (y, z) φ y = xsinz ⇒φ = _ xsinzdy = xy sinz +g (x, z) φ z = xy cos z ⇒φ = _ xy cos zdz = xy sinz +h(x, z) Thus, φ = xy sin z +C. 129 CHAPTER 6. VECTOR CALCULUS (10) ∇F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z _ 2x 2 −3 _ (−4z) cos z ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = _ ∂ ∂y (cos z) − ∂ ∂z (−4z) _ i − _ ∂ ∂x (cos z) − ∂ ∂z _ 2x 2 −3 _ _ j + _ ∂ ∂x (−4z) − ∂ ∂y _ 2x 2 −3 _ _ k = 4i ,= 0 Hence, F is not a conservative vector field. (11) Let r (t) = _ 4 −t 2 , t ¸ , −3 ≤ t ≤ 2, then r ′ (t) = [−2t, 1]. _ Γ y 2 dx +xdy = _ 2 −3 _ (t) 2 (−2t) + _ 4 −t 2 _ (1) _ dt = _ 2 −3 _ 4 −t 2 −2t 3 _ dt = 245 6 (12) Let r (t) = [cos t, sint, t], 0 ≤ t ≤ 2π, then r ′ (t) = [−sint, cos t, 1]. _ Γ y sinz ds = _ 2π 0 (sint) sin(t) _ (−sint) 2 + (cos t) 2 + (1) 2 dt = √ 2 _ 2π 0 sin 2 t dt = √ 2 2 _ 2π 0 (1 −cos 2t) dt = √ 2π (13) Let x(t) = 1 + 2t, y (t) = 2, z (t) = 3 −2t and 0 ≤ t ≤ 1. f (x(t) , y (t) , z (t)) = (1 + 2t) + 2 (3 −2t) = 7 −2t x ′ (t) = 2, y ′ (t) = 0, z ′ (t) = −2 and _ _ dx dt _ 2 + _ dy dt _ 2 + _ dz dt _ 2 = √ 8 _ C (x +yz) ds = _ 1 0 (7 −2t) _√ 8 _ dt = 12 √ 2 (14) F = _ x 4 , xy ¸ R = ¦(x, y) [0 ≤ x ≤ 1, 0 ≤ y ≤ x¦ _ Γ x 4 dx +xy dy = _ Γ F • dr = _ 1 0 _ x 0 _ ∂ ∂x (xy) − ∂ ∂y _ x 4 _ _ dydx = _ 1 0 _ x 0 y dydx = _ 1 0 _ y 2 2 _ x 0 dx = 1 2 _ 1 0 x 2 dx = 1 6 (15) F = _ y 2 , 3xy ¸ 130 CHAPTER 6. VECTOR CALCULUS R = ¦(r, θ) [1 ≤ r ≤ 2, 0 ≤ θ ≤ π¦ _ Γ y 2 dx + 3xy dy = _ Γ F • dr = _ π 0 _ 2 1 _ ∂ ∂x (3xy) − ∂ ∂y _ y 2 _ _ rdrdθ = _ π 0 _ 2 1 y rdrdθ = _ π 0 _ 2 1 r 2 sinθ drdθ = 7 3 _ π 0 sin θ dθ = 14 3 (16) (a) Let x = t, y = t, z = 0, 0 ≤ t ≤ 1, then dx dt = 1, dy dt = 1, dz dt = 0. _ AB f (x, y, z) ds = _ 1 0 (t) (t) 2 _ 1 2 + 1 2 + 0 2 dt = √ 2 _ 1 0 t 3 dt = √ 2 4 (b) Let x = 2t, y = 3t, z = 6t, 0 ≤ t ≤ 1, then dx dt = 2, dy dt = 3, dz dt = 6. _ AB f (x, y, z) ds = _ 1 0 _ (2t) (3t) + (6t) 2 _ √ 2 2 + 3 2 + 6 2 dt = 294 _ 1 0 t 2 dt = 98 (17) (a) Let x = cos t, y = sint, where 0 ≤ t ≤ π 2 . _ C xyds = _ π 2 0 cos t sint ¸ _ d (cos t) dt _ 2 + _ d (sint) dt _ 2 dt = _ π 2 0 cos t sintdt = 1 2 _ π 2 0 sin2tdt = _ − 1 4 cos 2t _ π/2 0 = 1 2 (b) Let x = t, y = 2t, z = −t where 0 ≤ t ≤ 1. _ C (xy +y +z) ds = _ 1 0 (t 2t + 2t −t) ¸ _ d (t) dt _ 2 + _ d (2t) dt _ 2 + _ d (−t) dt _ 2 dt = _ 1 0 _ 2t 2 +t _ √ 6dt = √ 6 _ 1 0 _ 2t 2 +t _ dt = 7 6 √ 6 (c) Let x = cos t, y = sint, z = t where 0 ≤ t ≤ 2π. _ C yds = _ 2π 0 sint ¸ _ d (cos t) dt _ 2 + _ d (sint) dt _ 2 + _ d (t) dt _ 2 dt = _ 2π 0 √ 2 sintdt = 0 131 CHAPTER 6. VECTOR CALCULUS (18) (a) Let x = cos θ, y = sinθ, 0 ≤ θ ≤ π, then dx dθ = −sinθ, dy dθ = cos θ. _ C f (x, y) ds = _ π 0 (cos θ + sinθ) _ (−sinθ) 2 + (cos θ) 2 dθ = _ π 0 (cos θ + sinθ) dθ = [sinθ −cos θ] π 0 = 2 (b) _ C f (x, y) ds = _ π 0 (cos θ) 2 sinθ _ (−sinθ) 2 + (cos θ) 2 dθ = _ π 0 cos 2 θ sinθ dθ = _ − 1 3 cos 3 θ _ π 0 = 2 3 (19) x = t cos t, y = t sint, z = t, 0 ≤ t ≤ 1, then dx dt = cos t −t sint, dy dt = sint +t cos t, dz dt = 1 and _ _ dx dt _ 2 + _ dy dt _ 2 + _ dz dt _ 2 = _ (cos t −t sint) 2 + (sint +t cos t) 2 + 1 2 = √ t 2 + 2. (a) _ C f (x, y, z) ds = _ 1 0 (t) √ t 2 + 2 dt = _ 1 3 _ t 2 + 2 _ 3/2 _ 1 0 = √ 3 − 2 3 √ 2 (b) _ C f (x, y, z) ds = _ 1 0 _ t 2 + 2 dt = _ t 2 _ t 2 + 2 + ln ¸ ¸ ¸t + _ t 2 + 2 ¸ ¸ ¸ _ 1 0 = ln _ √ 3 + 1 _ −ln √ 2 + 1 2 √ 3 = 1.5245 = 1 2 _ ln _ √ 3 + 2 _ + √ 3 _ (20) (a) x = cos t + t sint, y = sint − t cos t, 0 ≤ t ≤ 2π, then dx dt = t cos t, dy dt = t sint and _ _ dx dt _ 2 + _ dy dt _ 2 = t. Arc length = _ 2π 0 _ _ dx dt _ 2 + _ dy dt _ 2 dt = _ 2π 0 t dt = _ 1 2 t 2 ¸ 2π 0 = 2π 2 (b) x = θ − sinθ, y = 1 − cos θ, 0 ≤ θ ≤ π, then dx dθ = 1 − cos θ, dy dt = sinθ and _ _ dx dθ _ 2 + _ dy dθ _ 2 = √ 2 −2 cos θ. Arc length = _ π 0 √ 2 −2 cos θ dθ = _ π 0 2 sin θ 2 dθ = _ −4 cos θ 2 ¸ π 0 = 4 132 CHAPTER 6. VECTOR CALCULUS (21) (a) curl F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z 3 + 2xy x 2 −3y 2 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = _ 0 − ∂ ∂z _ x 2 −3y 2 _ , ∂ ∂z (3 + 2xy) −0, ∂ ∂x _ x 2 −3y 2 _ − ∂ ∂y (3 + 2xy) _ = [0, 0, 0] = 0 Since curl F = 0, the F is conservative. (b) Since F is conservative, then the potential function f exists. f x = 3 + 2xy =⇒f (x, y) = _ (3 + 2xy) dx = 3x +x 2 y +g (y) f y = ∂ ∂y _ 3x +x 2 y +g (y) ¸ = x 2 +g ′ (y) = x 2 −3y 2 So, g ′ (y) = −3y 2 =⇒g (y) = _ _ −3y 2 _ dy = −y 3 +C. Thus, f (x, y) = 3x +x 2 y −y 3 +C. r (π) = [e π sinπ, e π cos π] = [0, −e π ] r (0) = _ e 0 sin 0, e 0 cos 0 ¸ = [0, 1] By the Foundamental Theorem of Line Integrals, _ Γ F • dr =f (0, −e π ) −f (0, 1) = _ −(−e π ) 3 _ − _ −(1) 3 _ = e 3π + 1. (22) Let f (x, y, z) = 3x −6yz, g (x, y, z) = 2y + 3xz and h(x, y, z) = 1 −4xz 2 . (a) Let x = t, y = t 2 , z = t 3 , 0 ≤ t ≤ 1, then dx dt = 1, dy dt = 2t, dz dt = 3t 2 . f dx dt +g dy dt +h dz dt = (3x −6yz) + 2t (2y + 3xz) + 3t 2 _ 1 −4xz 2 _ = _ 3t −6t 5 _ + 2t _ 2t 2 + 3t 4 _ + 3t 2 _ 1 −4t 7 _ = −12t 9 + 4t 3 + 3t 2 + 3t _ AB F • dr = _ 1 0 _ −12t 9 + 4t 3 + 3t 2 + 3t _ dt = 23 10 (b) Let x = t, y = t, z = t, 0 ≤ t ≤ 1, then dx dt = dy dt = dz dt = 1. f dx dt +g dy dt +h dz dt = _ 3t −6t 2 _ + _ 2t + 3t 2 _ + _ 1 −4t 3 _ = −4t 3 −3t 2 + 5t + 1 _ AB F • dr = _ 1 0 _ −4t 3 −3t 2 + 5t + 1 _ dt = 3 2 (c) From A to C, let x = 0, y = 0, z = t, 0 ≤ t ≤ 1, then dx dt = dy dt = 0, dz dt = 1. f dx dt +g dy dt +h dz dt = 0 + 0 + (1) = 1 133 CHAPTER 6. VECTOR CALCULUS _ AC F • dr = _ 1 0 dt = 1 From C to B, let x = s, y = s, z = 0, 0 ≤ s ≤ 1, then dx ds = dy ds = 1, dz ds = 0. f dx ds +g dy ds +h dz ds = (3s −6s) + (2s + 3s) + 0 = 2s _ CB F • dr = _ 1 0 2s ds = 1 _ AB F • dr = _ AC F • dr + _ CB F • dr = 1 + 1 = 2 (23) Let x(t) = t and y (t) = t 3 , −2 ≤ t ≤ 3. Then r (t) = _ t, t 3 ¸ , F(r (t)) = _ (t) _ t 3 _ , t 3 −t ¸ = _ t 4 , t 3 −t ¸ and r ′ (t) = _ 1, 3t 2 ¸ . Work done = _ C F • dr = _ 3 −2 _ t 4 , t 3 −t ¸ • _ 1, 3t 2 ¸ dt = _ 3 −2 _ t 4 + 3t 5 −3t 3 _ dt = 1355 4 (24) r (t) = [cos t, sint], 0 ≤ t ≤ π, then r ′ (t) = [−sint, cos t]. F(r (t)) = _ −(sint) 2 , (cos t) (sint) _ = _ −sin 2 t, sint cos t ¸ Work done = _ π 0 _ −sin 2 t, sint cos t ¸ • [−sint, cos t] dt = _ π 0 _ sin 3 t + sint cos 2 t _ dt = _ π 0 sint dt = 2 (25) (a) Let x = 1 +t, y = 2 −t and 0 ≤ t ≤ 1, then dx dt = 1 and dy dt = −1. F(r (t)) = ¸2 (1 +t) −(2 −t) + 4, 3 (1 +t) + 5 (2 −t) −6) = ¸3t + 4, 7 −2t) r ′ (t) = ¸1, −1) _ Γ F • dr = F(r (t)) • r ′ (t) dt = _ 1 0 [(3t + 4) −(7 −2t)] dt = _ 1 0 (5t −3) dt = − 1 2 (b) F(r (t)) = _ _ t 2 _ _ t 3 _ , _ t 2 _ 2 , t 3 _ = ¸ t 5 , t 4 , t 3 _ r ′ (t) = ¸ 1, 2t, 3t 2 _ _ Γ F • dr = _ 1 0 __ t 5 _ (1) + _ t 4 _ (2t) + _ t 3 _ _ 3t 2 _¸ dt = _ 1 0 6t 5 dt = 1 (c) From (0, 0) to (1, 0), let x = t, y = 0 and 0 ≤ t ≤ 1, then dx dt = 1and dy dt = 0. Then F(r (t)) = ¸0, 0) and _ Γ 1 F• dr = 0. From (1, 0) to (1, 1), let x = 1, y = s and 0 ≤ s ≤ 1, then dx ds = 0 and dy ds = 1. Then F(r (s)) = ¸6s, 2s), r ′ (s) = ¸0, 1) and _ Γ 2 F • dr = _ 1 0 2s ds = 1. From (1, 1) to (0, 1), let x = 1 −u, y = 1 and 0 ≤ u ≤ 1, then dx du = −1, dy du = 0. F(r (u)) = _ 6 (1 −u) 2 , 2 _ , r ′ (u) = ¸−1, 0) and _ Γ 3 F • dr = _ 1 0 −6u 2 du = −2. From (0, 1) to (0, 0), let x = 0, y = 1 −v and 0 ≤ v ≤ 1, then dx dv = 0 and dy dv = −1. F(r (v)) = ¸0, 2 (1 −v)), r ′ (v) = ¸0, −1) and _ Γ 4 F • dr = _ 1 0 2 (v −1) dv = −1. Therefore, _ Γ F • dr = _ Γ 1 + _ Γ 2 + _ Γ 3 + _ Γ 4 = 0 + 1 −2 −1 = −2. 134 CHAPTER 6. VECTOR CALCULUS (26) (a) Let x = t, y = t and z = t, where 0 ≤ t ≤ 1. _ C F•dr = _ C __ x 2 −y _ i+ _ y 2 −z _ j+ _ z 2 −x _ k ¸ • [dxi+dyj+dzk] = _ C _ x 2 −y _ dx+ _ y 2 −z _ dy+ _ z 2 −x _ dz = _ 1 0 _ t 2 −t _ dt+ _ t 2 −t _ dt+ _ t 2 −t _ dt = 3 _ 1 0 _ t 2 −t _ dt = − 1 2 (b) Let x = t, y = t 2 and z = t 3 , where 0 ≤ t ≤ 1. _ C F•dr = _ C __ x 2 −y _ i+ _ y 2 −z _ j+ _ z 2 −x _ k ¸ • [dxi+dyj+dzk] = _ C _ x 2 −y _ dx+ _ y 2 −z _ dy+ _ z 2 −x _ dz = _ 1 0 _ t 2 −t 2 _ dt+ _ _ t 2 _ 2 −t 3 _ dt 2 + _ _ t 3 _ 2 −t _ dt 3 = _ 1 0 _ t 2 −t 2 _ dt+2t _ t 4 −t 3 _ dt+3t 2 _ t 6 −t _ dt = _ 1 0 _ 2t 5 −2t 4 + 3t 8 −3t 3 _ dt = − 29 60 (27) (a) Let x = cos t and y = sint, where 0 ≤ t ≤ 2π, _ C −ydx +xdy = _ 1 0 −sintd (cos t) + cos td (sint) = _ 2π 0 _ sin 2 t + cos 2 t _ dt = 2π. (b) Let x = 1 −t and y = t, where 0 ≤ t ≤ 1 _ C (x +y) dx + _ x 2 +y 2 _ dy = _ 1 0 (1) d (1 −t) + _ 1 0 _ (1 −t) 2 +t 2 _ dt = −1 + _ 1 0 _ 2t 2 + 1 −2t _ dt = − 1 3 . 135 CHAPTER 6. VECTOR CALCULUS (28) Since ∇F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z y sinz xsinz xy cos z ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = _ ∂ ∂y (xy cos z) − ∂ ∂z (xsinz) _ i+ _ ∂ ∂x (xy cos z) − ∂ ∂z (y sinz) _ j + _ ∂ ∂x (xsinz) − ∂ ∂y (y sinz) _ k = 0 Therefore, F is conservative and there exists a function φ such that ∇φ = F. In this case, φ = xy sinz+ constant and _ (1,2,3) (0,0,0) [(xy sin z) i + (xsinz) j + (xy cos z) k] • dr = xy sinz[ (1,2,3) (0,0,0) = (1) (2) sin(3) −(0) (0) sin0 = 2 sin3 (29) Since cos 2 t + sin 2 t = 1, so, let x −1 = cos t and y = sint, then x = cos t −1, y = sint. (30) (a) F = ¸3y, 5x), P = 3y, Q = 5x and R = ¦(r, θ) [0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π¦. _ ∂R F • dr = _ _ R _ ∂Q ∂x − ∂P ∂y _ dA = _ 2π 0 _ 1 0 (5 −3) r dr dθ = 2 [θ] 2π 0 _ 1 2 r 2 ¸ 1 0 = 2π (b) F = ¸ y 2 , x 2 _ , P = y 2 , Q = x 2 and R = ¦(x, y) [0 ≤ x ≤ 1, 0 ≤ y ≤ 1¦. _ ∂R F • dr = _ 1 0 _ 1 0 (2x −2y) dxdy = _ 1 0 _ x 2 −2xy ¸ 1 0 dy = _ 1 0 (1 −2y) dy = _ y −y 2 ¸ 1 0 = 0 (c) F = ¸xy, x −y), P = xy, Q = x −y, and R = ¦(x, y) [0 ≤ x ≤ 1, 1 −x ≤ y ≤ 1¦. _ ∂R F • dr = _ 1 0 _ 1 1−x (1 −x) dy dx = _ 1 0 _ x −x 2 _ dx = _ 1 2 x 2 − 1 3 x 3 ¸ 1 0 = 1 6 136 CHAPTER 6. VECTOR CALCULUS (31) (a) Let x = a cos t and y = a sint, where 0 ≤ t ≤ 2π. _ C _ −x 2 y _ dx + _ xy 2 _ dy = _ 2π 0 _ −a 2 cos 2 t a sint _ d (a cos t) + _ a 2 cos t a sin 2 t _ d (a sint) = _ 2π 0 _ a 4 cos 2 t sin 2 t +a 4 cos 2 t sin 2 t _ dt = 2a 4 _ 2π 0 _ cos 2 t sin 2 t _ dt = 2a 4 _ 2π 0 _ sin2t 2 _ 2 dt = 1 2 a 4 _ 2π 0 sin 2 2tdt = 1 2 a 4 _ 2π 0 _ 1 −cos 4t 2 _ dt = 1 2 πa 4 and __ x 2 +y 2 ≤a 2 _ ∂ _ xy 2 _ ∂x − ∂ _ −x 2 y _ ∂y _ dxdy = __ x 2 +y 2 ≤a 2 _ y 2 +x 2 _ dxdy Use polar coordinates, x = r cos θ and y = r sinθ, we have __ x 2 +y 2 ≤a 2 _ ∂ _ xy 2 _ ∂x − ∂ _ −x 2 y _ ∂y _ dxdy = _ 2π 0 _ a 0 _ r 2 cos 2 θ +r 2 sin 2 θ _ rdrdθ = _ 2π 0 _ a 0 r 3 drdθ = _ 2π 0 1 4 a 4 dθ = 1 2 πa 4 (b) x y (0,1) (1,0) (-1,0) (0,-1) 137 CHAPTER 6. VECTOR CALCULUS On C 1 – the line x +y = 1, x = 1 −t and y = t, where 0 ≤ t ≤ 1. _ C 1 (−y) dx + (x) dy = _ 1 0 −(t) d (1 −t) + (1 −t) dt = _ 1 0 tdt + (1 −t) dt = _ 1 0 dt = 1 On C 2 – the line −x +y = 1, x = −t and y = 1 −t, where 0 ≤ t ≤ 1. _ C 2 (−y) dx + (x) dy = _ 1 0 −(1 −t) d (−t) + (−t) d (1 −t) = _ 1 0 (1 −t) dt +tdt = _ 1 0 dt = 1 On C 3 – the line −x −y = 1, x = t −1 and y = −t, where 0 ≤ t ≤ 1. _ C 3 (−y) dx + (x) dy = _ 1 0 −(−t) d (t −1) + (t −1) d (−t) = _ 1 0 tdt −(t −1) dt = _ 1 0 dt = 1 On C 4 – the line x −y = 1, x = t and y = t −1, where 0 ≤ t ≤ 1. _ C 4 (−y) dx + (x) dy = _ 1 0 −(t −1) dt +td (t −1) = _ 1 0 −(t −1) dt +tdt = _ 1 0 dt = 1 Hence, _ C (−y) dx + (x) dy = _ C 1 + _ C 2 + _ C 3 + _ C 4 (−y) dx + (x) dy = 1 + 1 + 1 + 1 = 4 and __ square _ ∂ (x) ∂x − ∂ (−y) ∂y _ dxdy = __ square 2dxdy = 2 area of square = 2 √ 2 √ 2 = 4 Therefore, _ C (−y) dx + (x) dy = __ square _ ∂ (x) ∂x − ∂ (−y) ∂y _ dxdy. 138 CHAPTER 6. VECTOR CALCULUS and the Green’s Theorem is true in this case. (32) By Green’s Theorem, _ xy 2 dx + _ x 2 y + 2x _ dy = __ Region _ ∂ _ x 2 y + 2x _ ∂x − ∂ _ xy 2 _ ∂y _ dxdy = __ Region 2dxdy = 2 area of the region (33) (a) X = cos usinv, Y = sinusinv, Z = cos v X u = −sinusinv, Y u = cos usinv, Z u = 0 X v = cos ucos v, Y v = sinucos v, Z v = −sinv and E = (X u ) 2 + (Y u ) 2 + (Z u ) 2 = sin 2 v G = (X v ) 2 + (Y v ) 2 + (Z v ) 2 = 1 F = X u X v +Y u Y v +Z u Z v = 0 _ S f (x, y, z) dxdydz = _ π/2 0 _ π/2 0 cos usin v sinusinv cos v _ sin 2 vdudv = _ π/2 0 cos usinudu _ π/2 0 cos v sin 3 vdv = 1 8 (b) O x y z 139 CHAPTER 6. VECTOR CALCULUS On S 1 , the plane x = 0 _ S 1 f (x, y, z) dS = _ 1 0 _ 1−z 0 (y +z) dydz = 1 3 On S 2 , the plane y = 0 _ S 2 f (x, y, z) dS = _ 1 0 _ 2 0 (z) dxdz = 1 On S 3 , the plane x = 2 _ S 3 f (x, y, z) dS = _ 1 0 _ 1−z 0 (y +z) dydz = 1 3 On S 4 , the plane z = 0 _ S 4 f (x, y, z) dS = _ 1 0 _ 2 0 ydxdy = 1 On S 5 , the plane y +z = 1 _ S 5 f (x, y, z) dS = _ 1 0 _ 2 0 (1) _ 1 + (0) 2 + (−1) 2 dxdy = 2 √ 2 Hence, _ S f (x, y, z) dS = 1 + 1 3 + 1 + 1 3 + 2 √ 2 = 8 3 + 2 √ 2 (34) z = _ 2 −x 2 −y 2 , z x = −x _ 2 −x 2 −y 2 , z y = −y _ 2 −x 2 −y 2 . 140 CHAPTER 6. VECTOR CALCULUS Let x = r cos θ, y = r sin θ, 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. __ x 2 +y 2 ≤1 _ 1 +z 2 x +z 2 y dxdy = __ x 2 +y 2 ≤1 ¸ 1 + x 2 2 −x 2 −y 2 + y 2 2 −x 2 −y 2 dxdy = _ 2π 0 _ 1 0 r _ 1 + r 2 2 −r 2 drdθ = _ 2π 0 _ 1 0 r _ 2 2 −r 2 drdθ = − _ 2π 0 _ √ 2 _ 2 −r 2 _ 1 0 dθ = _ 2π 0 _ 2 − √ 2 _ dθ = 2π _ 2 − √ 2 _ (35) Let X = a cos u, Y = a sinu, Z = v, where 0 ≤ u ≤ 2π and 0 ≤ v ≤ 10 −a cos u −2a sinu. Then X u = −a sinu, Y u = a cos u, Z u = 0 and X v = 0, Y v = 0, Z v = 1. E = (−a sinu) 2 + (a cos u) 2 + (0) 2 = a 2 G = (0) 2 + (0) 2 + (1) 2 = 1 F = (−a sinu) (0) + (a cos u) (0) + (0) (1) = 0 and _ EG−F 2 = _ a 2 cos 2 u +a 2 sin 2 u = a A = _ 2π 0 _ 10−a cos u−2a sin u 0 advdu = a _ 2π 0 (10 −a cos u −2a sinu) du = 20πa 141 CHAPTER 6. VECTOR CALCULUS (36) z = 2x, z x = 2, z y = 0. __ x 2 +y 2 ≤1 _ 1 +z 2 x +z 2 y dxdy = __ x 2 +y 2 ≤1 _ 1 + (2) 2 + (0) 2 dxdy = √ 5 __ x 2 +y 2 ≤1 dxdy = √ 5π (37) z = 4−x 2 −y 2 , z x = −2x, z y = −2y. Let x = r cos θ, y = r sinθ, 1 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π __ 1≤x 2 +y 2 ≤4 _ 1 +z 2 x +z 2 y dxdy = __ 1≤x 2 +y 2 ≤4 _ 1 + 4x 2 + 4y 2 dxdy = _ 2π 0 _ 2 1 r _ 1 + 4r 2 drdθ = _ 2π 0 _ 1 12 _ 1 + 4r 2 _ 3/2 _ 2 1 dθ = _ 2π 0 _ 17 12 √ 17 − 5 12 √ 5 _ dθ = 1 6 π _ 17 √ 17 −5 √ 5 _ (38) (a) Let f (x, y, z) = 1, z = φ(x, y) = 1 −x −y, and ∂z ∂x = ∂z ∂y = −1. R is the region inside x 2 +y 2 = a 2 or R = ¦(r, θ) [ 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π¦. S = _ _ S f (x, y, z) dS = _ _ R f (x, y, φ(x, y)) ¸ _ ∂z ∂x _ 2 + _ ∂z ∂y _ 2 + 1 dxdy = _ 2π 0 _ a 0 _ √ 3 _ r dr dθ = √ 3πa 2 (b) Let f (x, y, z) = 1, z = φ(x, y) = 2 3 _ x 3/2 +y 3/2 _ , and ∂z ∂x = √ x, ∂z ∂y = √ y. 142 CHAPTER 6. VECTOR CALCULUS R = ¦(x, y) [ 0 ≤ x ≤ 1, 0 ≤ y ≤ 1¦ S = _ _ S f (x, y, z) dS = _ 1 0 _ 1 0 _ _√ x _ 2 + ( √ y) 2 + 1 dxdy = _ 1 0 _ 2 3 (x +y + 1) 3/2 _ 1 0 dy = 2 3 _ 1 0 _ (y + 2) 3/2 −(y + 1) 3/2 _ dy = 2 3 _ 2 5 (y + 2) 5/2 − 2 5 (y + 1) 5/2 _ 1 0 = 4 15 __ 3 5/2 −2 5/2 _ − _ 2 5/2 −1 5/2 __ = 4 15 _ 3 5/2 −2 7/2 + 1 _ = 1.4066 (39) (a) σ (x, y, z) = z, let z = φ(x, y) = _ 4 −x 2 −y 2 , and ∂z ∂x = − x z , ∂z ∂y = − y z . R = ¦(r, θ) [ 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π¦ S = _ _ S σ (x, y, z) dS = _ 2π 0 _ 2 0 (z) _ _ − x z _ 2 + _ − y z _ 2 + 1r dr dθ = _ 2π 0 _ 2 0 _ x 2 +y 2 +z 2 r dr dθ = _ 2π 0 dθ _ 2 0 2r dr = [θ] 2π 0 _ r 2 ¸ 2 0 = 8π (b) Find the equation for the area, a plane, within ∆ABC: −−→ AB = −−→ OB − −→ OA = ¸0, 1, 0) −¸1, 0, 0) = ¸−1, 1, 0) and −→ AC = ¸−1, 0, 1). normal vector = n = −−→ AB −→ AC = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k −1 1 0 −1 0 1 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸1, 1, 1) Equation for the plane with normal vector n and passing through A(1, 0, 0): (x −1) + (y −0) + (z −0) = 0 =⇒x +y +z = 1 σ (x, y, z) = xyz, let z = φ(x, y) = 1 −x −y, and ∂z ∂x = ∂z ∂y = −1. R = ¦(x, y) [ 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 −x¦ S = _ _ S σ (x, y, z) dS = _ 1 0 _ 1−x 0 (xyz) _ (−1) 2 + (−1) 2 + 1 dy dx = √ 3 _ 1 0 _ 1−x 0 xy (1 −x −y) dy dx = √ 3 _ 1 0 _ 1 2 xy 2 − 1 2 x 2 y 2 − 1 3 xy 3 _ 1−x 0 dx = √ 3 _ 1 0 _ − 1 6 x 4 + 1 2 x 3 − 1 2 x 2 + 1 6 x _ dx = √ 3 120 (c) σ (x, y, z) = z 2 On the surface S 1 , z = 0 and σ (x, y, z) = 0, thus _ _ S 1 σ (x, y, z) dS 1 = 0. On the surface S 2 , z = 1 and σ (x, y, z) = 1 and S 2 = ¦(x, y) [0 ≤ x ≤ 1, 0 ≤ y ≤ 1¦, thus _ _ S 2 σ (x, y, z) dS 2 = _ _ S 2 dS 2 = _ 1 0 _ 1 0 dxdy = 1. 143 CHAPTER 6. VECTOR CALCULUS On the surface S 3 and S 4 , x = 0 and 1, and S 3 = S 4 = ¦(y, z) [0 ≤ y ≤ 1, 0 ≤ z ≤ 1¦, thus _ _ S 3 σ (x, y, z) dS 3 = _ _ S 4 σ (x, y, z) dS 4 = _ 1 0 _ 1 0 z 2 dy dz = 1 3 . On the surface S 5 and S 6 , y = 0 and 1, and S 5 = S 6 = ¦(x, z) [0 ≤ x ≤ 1, 0 ≤ z ≤ 1¦, thus _ _ S 5 σ (x, y, z) dS 5 = _ _ S 6 σ (x, y, z) dS 6 = _ 1 0 _ 1 0 z 2 dxdz = 1 3 . Therefore, _ _ S σ (x, y, z) dS = _ _ S 1 + _ _ S 2 +... + _ _ S 6 = 7 3 . (40) (a) The equation for the plane that including ∆ABC: unit normal vector = 6,3,2 √ 6 2 +3 2 +2 2 = ¸ 6 7 , 3 7 , 2 7 _ 6 (x −1) + 3 (y −0) + 2 (z −0) = 0 =⇒6x + 3y + 2z = 6 Let z = φ(x, y) = 3 −3x − 3 2 y, then F = ¸xy, y, zx) = ¸ xy, y, 3x −3x 2 − 3 2 xy _ Using the formula _ _ S F • ndS = _ _ R _ −P ∂z ∂x −Q ∂z ∂y +R _ dxdy F • n = ¸ xy, y, 3x −3x 2 − 3 2 xy _ • ¸ 6 7 , 3 7 , 2 7 _ = 6 7 x + 3 7 y + 3 7 xy − 6 7 x 2 R = ¦(x, y) [ 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 −2x¦ dS = 1 |n • k| dxdy = 7 2 dxdy _ _ S F • ndS = _ 1 0 _ 2−2x 0 _ 6 7 x + 3 7 y + 3 7 xy − 6 7 x 2 _ 7 2 dy dx = 7 4 = _ 1 0 _ 3xy −3x 2 y − 3 4 xy 2 _ 2−2x 0 dx or using the formula __ S F • ndS = __ D F(r (u, v)) • (r u r v ) dudv Let r (x, y) = ¸ x, y, 3 −3x − 3 2 y _ and D = ¦(x, y) [0 ≤ x ≤ 1, 0 ≤ y ≤ 2 −2x¦, then r x = ¸1, 0, −3) and r y = ¸ 0, 1, − 3 2 _ . r x r y = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k 1 0 −3 0 1 − 3 2 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸ 3, 3 2 , 1 _ 144 CHAPTER 6. VECTOR CALCULUS F(r (x, y)) = ¸ xy, y, 3x −3x 2 − 3 2 xy _ __ D F(r (x, y)) • (r x r y ) dxdy = _ 1 0 _ 2−2x 0 _ xy, y, 3x −3x 2 − 3 2 xy _ • _ 3, 3 2 , 1 _ dy dx = _ 1 0 _ 2−2x 0 _ 3 2 xy + 3 2 y + 3x −3x 2 _ dy dx = _ 1 0 _ 9x 3 −15x 2 + 3x + 3 _ dx = 7 4 (b) Let f (x, y, z) = x 2 +y 2 +z 2 −a 2 and z = φ(x, y) = _ a 2 −x 2 −y 2 . F = ¸x, y, z) = _ x, y, _ a 2 −x 2 −y 2 _ Using the formula _ _ S F • ndS = _ _ R _ −P ∂z ∂x −Q ∂z ∂y +R _ dxdy n = ∇f ∇f = 2x,2y,2z 2 √ x 2 +y 2 +z 2 = _ 2x,2y,2 √ a 2 −x 2 −y 2 _ 2a = _ x a , y a , 1 a _ a 2 −x 2 −y 2 _ F • n = x 2 a + y 2 a + 1 a _ a 2 −x 2 −y 2 _ = a R = ¦(r, θ) [ 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π¦ dS = 1 |n • k| r dr dθ = a _ a 2 −x 2 −y 2 r dr dθ = ar √ a 2 −r 2 dr dθ _ _ S F • ndS = _ 2π 0 _ a 0 (a) ar √ a 2 −r 2 dr dθ = a 2 _ 2π 0 _ − √ a 2 −r 2 _ a 0 dθ = a 3 _ 2π 0 dθ = 2πa 3 or using the formula _ _ S F • ndS = _ _ D F(r (u, v)) • (r u r v ) dudv Let r (φ, θ) = ¸a sinφcos θ, a sinφsin θ, a cos φ) and D = _ (φ, θ) [0 ≤ φ ≤ π 2 , 0 ≤ θ ≤ 2π _ then r φ = ¸a cos φcos θ, a cos φsinθ, −a sinφ) and r θ = ¸−a sinφsin θ, a sinφcos θ, 0). 145 CHAPTER 6. VECTOR CALCULUS r φ r θ = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k a cos φcos θ a cos φsin θ −a sinφ −a sinφsin θ a sinφcos θ 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸ a 2 sin 2 φcos θ, a 2 sin 2 φsinθ, a 2 sinφcos φ _ F(r (φ, θ)) = ¸a sinφcos θ, a sinφsin θ, a cos φ) F(r (φ, θ)) • (r φ r θ ) = a 3 sinφ Therefore, __ D F(r (φ, θ)) • (r φ r θ ) dφdθ = _ 2π 0 _ π/2 0 a 3 sinφdφdθ = 2πa 3 (c) On the surface S 1 , z = 0, F = ¸x, 2y, 0) and n = ¸0, 0, −1), thus __ S 1 F • ndS = __ S 1 (0) dS = 0. On the surface S 2 , z = 1, F = ¸x, 2y, 3), n = ¸0, 0, 1), dS = 1 |n • k| dxdy = dxdy, and S 2 = ¦(x, y) [0 ≤ x ≤ 1, 0 ≤ y ≤ 1¦, thus __ S 2 F • ndS = __ S 2 (3) dS = 3 _ 1 0 _ 1 0 dxdy = 3. On the surface S 3 , x = 0, F = ¸0, 2y, 3z), n = ¸−1, 0, 0), thus __ S 3 F • ndS = __ S 3 (0) dS = 0. On the surface S 4 , x = 1, F = ¸1, 2y, 3z), n = ¸1, 0, 0), dS 4 = 1 |n • i| dxdy = dy dz, and S 4 = ¦(y, z) [0 ≤ y ≤ 1, 0 ≤ z ≤ 1¦, thus __ S 4 F • ndS = __ S 4 (1) dS = _ 1 0 _ 1 0 dy dz = 1. On the surface S 5 , y = 0, F = ¸x, 0, 3z), n = ¸0, −1, 0), thus __ S 5 F • ndS = __ S 5 (0) dS = 0. On the surface S 6 , y = 1, F = ¸x, 2, 3z), n = ¸0, 1, 0), dS 6 = 1 |n•j| dxdy = dxdy, and 146 CHAPTER 6. VECTOR CALCULUS S 6 = ¦(x, z) [0 ≤ x ≤ 1, 0 ≤ z ≤ 1¦, thus __ S 6 F • ndS = __ S 6 (2) dS = 2 _ 1 0 _ 1 0 dxdz = 2. Therefore, __ S F • ndS = __ S 1 + __ S 2 +... + __ S 6 = 6. (41) For the top end-face of the cylinder x 2 + y 2 = 4, 0 ≤ z ≤ 3, n = ¸0, 0, 1). Using the formula __ S F • ndS = __ R _ −P ∂z ∂x −Q ∂z ∂y +R _ dxdy F = ¸ 4x, −2y 2 , z 2 x 2 _ = ¸ 4x, −2y 2 , 9x 2 _ F • n = 9x 2 R = ¦(r, θ) [ 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π¦ dS = 1 |n • k| r dr dθ = r dr dθ __ S F • ndS = _ 2π 0 _ 2 0 9 (r cos θ) 2 r dr dθ = 9 _ 2π 0 _ 2 0 r 3 cos 2 θ dr dθ = 9 _ 2π 0 1 + cos 2θ 2 dθ _ 2 0 r 3 dr = 9 _ θ 2 + 1 4 sin2θ _ 2π 0 _ 1 4 r 4 _ 2 0 = 36π or using the formula __ S F • ndS = _ _ D F(r (u, v)) • (r u r v ) dudv Let r (r, θ) = ¸r cos θ, r sinθ, 3) and D = ¦(r, θ) [0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π¦ then r φ = ¸cos θ, sinθ, 0) and r θ = ¸−r sinθ, r cos θ, 0). r r r θ = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k cos θ sin θ 0 −r sinθ r cos θ 0 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸0, 0, r) F(r (r, θ)) = ¸ 4r cos θ, −2r 2 sin 2 θ, 9r 2 cos 2 θ _ F(r (r, θ)) • (r r r θ ) = 9r 3 cos 2 θ __ D F(r (r, θ)) • (r r r θ ) dr dθ = _ 2π 0 _ 2 0 9r 3 cos 2 θ dr dθ = 36π 147 CHAPTER 6. VECTOR CALCULUS (42) (a) O x y z Let f = x + 2y +z −3 = 0, then n = ∇f = i + 2j +k _ (1) 2 + (2) 2 + (1) 2 = 1 √ 6 (i + 2j +k) _ S F • ndS = _ S (−i + 2j + 3k) • _ 1 √ 6 (i + 2j +k) _ dS = _ S √ 6dS = √ 6 _ 3 0 _ 3−x 2 0 _ 1 + (−1) 2 + (−2) 2 dydx = 27 2 (b) O x y z Let f = e x −y = 0, then n = ∇f = e x i −j _ (e x ) 2 + (1) 2 = e x i −j √ e 2x + 1 _ S F • ndS = _ S (−2i + 2yj +zk) • _ e x i −j √ e 2x + 1 _ dS = _ S (−2e x −2y) √ e 2x + 1 dS Let Z = v, X = u, Y = e u , where 0 ≤ v ≤ 1 and 0 ≤ u ≤ ln2. Note that X u = 1, Y u = e u , Z u = 0 and X v = 0, Y v = 0, Z v = 1 E = (X u ) 2 + (Y u ) 2 + (Z u ) 2 = (1) 2 + (e u ) 2 + 0 2 = 1 +e 2u G = (X v ) 2 + (Y v ) 2 + (Z v ) 2 = (0) 2 + (0) 2 + (1) 2 = 1 F = X u X v +Y u Y v +Z u Z v = (1) (0) + (e u ) (0) + (0) (1) = 0 _ S F • ndS = _ 1 0 _ ln2 0 (−2e u −2e u ) √ e 2u + 1 _ (1 +e 2u ) (1) −0 2 dudv = −4 _ 1 0 _ ln 2 0 e u dudv = −4 _ 1 0 dv = −4 148 CHAPTER 6. VECTOR CALCULUS (43) (a) _ D ∇• Fdxdydz = _ D _ ∂ ∂x i+ ∂ ∂y j+ ∂ ∂z k _ • _ x 3 i+y 3 j+z 3 k _ dxdydz = _ D _ ∂x 3 ∂x + ∂y 3 ∂y + ∂z 3 ∂z _ dxdydz = 3 _ D _ x 2 +y 2 +z 2 _ dxdydz = 3 _ π/2 0 _ 2π 0 _ 1 0 r 4 sinφdrdθdφ = 6 5 π On S 1 – the surface of the hemi-sphere x 2 +y 2 +z 2 = 1 and n = r. _ S 1 F • ndS = _ S 1 F • rdS = _ S 1 _ x 3 i+y 3 j+z 3 k _ • (xi+yj+zk) dS = _ S 1 _ x 4 +y 4 +z 4 _ dS Let X = sinv cos u, Y = sinv sinu, Z = cos v, then √ EG−F 2 = √ sin 2 v = sinv. _ S 1 F • ndS = _ 2π 0 _ π/2 0 _ (sinv cos u) 4 +(sinv sinu) 4 +(cos v) 4 _ sinvdvdu = _ 2π 0 _ π/2 0 _ sin 4 v cos 4 u + sin 4 v sin 4 u + cos 4 v _ sinvdvdu = _ 2π 0 _ π/2 0 _ sin 4 v _ cos 4 u + sin 4 u _ + cos 4 v _ sinvdvdu = − _ 2π 0 _ _ cos 4 u + sin 4 u _ _ 1 5 cos 5 v − 2 3 cos 3 v + cos v _ + 1 5 cos 5 v _ π/2 0 du = _ 2π 0 _ 8 15 _ cos 4 u + sin 4 u _ + 1 5 _ du = 8 15 _ 2π 0 _ _ 1 + cos 2u 2 _ 2 + _ 1 −cos 2u 2 _ 2 _ du + 2 5 π = 8 15 _ 2π 0 _ 1 + 2 cos 2u + cos 2 2u 4 + 1 −2 cos 2u + cos 2 2u 4 _ du + 2 5 π = 4 15 _ 2π 0 _ 1 + cos 2 2u _ du + 2 5 π = 4 15 _ 2π 0 _ 1 + _ 1 + cos 4u 2 __ du + 2 5 π = 4 15 _ 3u 2 + sin4u 8 _ 2π 0 + 2 5 π = 6π 5 149 CHAPTER 6. VECTOR CALCULUS On S 2 – the bottom surface (z = 0) and n = −k _ S 2 F • ndS = _ S 2 _ x 3 i+y 3 j+z 3 k _ • (−k) dS = − _ S 2 z 3 dS = 0 (b) _ D ∇• Fdxdydz = _ D _ ∂ ∂x i+ ∂ ∂y j+ ∂ ∂z k _ • _ 2xzi+yj−z 2 k _ dxdydz = _ D _ ∂2xz ∂x + ∂y ∂y + ∂ _ −z 2 _ ∂z _ dxdydz = _ D (2z + 1 −2z) dxdydz = __ x 2 +4y 2 ≤16 _ y 0 dzdxdy = _ _ x 2 +4y 2 ≤16 ydxdy Let x = 4r cos θ, y = 2r sinθ and [J[ = 8r _ D ∇• Fdxdydz = _ π/2 0 _ 1 0 16r 2 sinθdrdθ = 16 3 On the plane z = y : n = −j +k √ 2 _ S 1 F • ndS = _ S 1 _ 2xzi+yj−z 2 k _ • −j +k √ 2 dS = 1 √ 2 _ S 1 _ −y −z 2 _ dS = − 1 √ 2 __ R 1 _ y +y 2 _ _ 1 +z 2 x +z 2 y dxdy = − 1 √ 2 __ R 1 _ y +y 2 _ √ 1 + 1dxdy = − __ R 1 _ y +y 2 _ dxdy = −8 _ π/2 0 _ 1 0 _ 2r 2 sinθ + 4r 3 sin 2 θ _ drdθ = −2π − 16 3 On the plane z = 0 : n = −k _ S 2 F • ndS = _ S 2 _ 2xzi+yj−z 2 k _ • (−k) dS = − _ S 2 z 2 dS = 0 150 CHAPTER 6. VECTOR CALCULUS On the plane x 2 + 4y 2 = 16 : n = xi + 4yj _ x 2 + 16y 2 _ S 3 F • ndS = _ S 3 _ 2xzi+yj−z 2 k _ • _ xi + 4yj _ x 2 + 16y 2 _ dS = _ S 4 2x 2 z + 4y 2 _ x 2 + 16y 2 dS Let X = 4 cos u, Y = 2 sinu and Z = v Then X u = −4 sinu, Y u = 2 cos u, Z u = 0, and X v = 0, Y v = 0, Z v = 1. E = X 2 u +Y 2 u +Z 2 u = 16 sin 2 u + 4 cos 2 u, G = 1, F = 0 _ S 3 F • ndS = _ π/2 0 _ 2 sin u 0 2 (4 cos u) 2 (v) + 4 (2 sinu) 2 _ (4 cos u) 2 + 16 (2 sinu) 2 _ 16 sin 2 u + 4 cos 2 udvdu = _ π/2 0 _ 2 sin u 0 _ 16v cos 2 u + 8 sin 2 u _ dvdu = _ π/2 0 _ 32 sin 2 ucos 2 u + 16 sin 3 u _ du = 2π + 32 3 On the plane x = 0 : n = −i _ S 4 F • ndS = _ S 2 _ 2xzi+yj−z 2 k _ • (−i) dS = − _ S 4 2xzdS = 0 Hence, _ S F • ndS = _ S 1 F • ndS + _ S 2 F • ndS + _ S 3 F • ndS + _ S 4 F • ndS = 16 3 (44) (a) F = ¸x, y, z) ∇• F = ∂ ∂x (x) + ∂ ∂y (y) + ∂ ∂z (z) = 3 151 CHAPTER 6. VECTOR CALCULUS D = ¦(ρ, φ, θ) [0 ≤ ρ ≤ a, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π¦ _ _ D F • ndS = _ _ _ D (∇• F) dV = _ 2π 0 _ π 0 _ a 0 (3) ρ 2 sinφdρ dφdθ = 3 (Volume of a sphere with radius a) = 4πa 3 (b) F = ¸x, y, z) ∇• F = ∂ ∂x (x) + ∂ ∂y (y) + ∂ ∂z (z) = 3 D = ¦(x, y, z) [0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1¦ _ _ D F • ndS = _ _ _ D (∇• F) dV = _ 1 0 _ 1 0 _ 1 0 (3) dxdy dz = 3 (Volume of a unit cube) = 3 (c) F = ¸ x 2 , y 2 , z 2 _ ∇• F = ∂ ∂x _ x 2 _ + ∂ ∂y _ y 2 _ + ∂ ∂z _ z 2 _ = 2x + 2y + 2z D = ¦(ρ, φ, θ) [0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π¦ __ D F • ndS = _ _ _ D (∇• F) dV = _ 2π 0 _ π 0 _ 1 0 (2x + 2y + 2z) ρ 2 sinφdρ dφdθ = _ 2π 0 _ π 0 _ 1 0 2 (ρ sinφcos θ +ρ sinφsin θ +ρ cos φ) ρ 2 sinφdρ dφdθ = _ 1 0 ρ 3 dρ _ π 0 _ 2π 0 2 _ sin 2 φcos θ + sin 2 φsinθ + sinφcos φ _ dθ dφ = _ 1 4 ρ 4 _ 1 0 _ π 0 2 _ sin 2 φsin θ −sin 2 φcos θ +θ sinφcos φ ¸ 2π 0 dθ dφ = 1 4 _ π 0 4π sinφcos φdφ = π _ 1 2 sin 2 φ _ π 0 = 0 152 CHAPTER 6. VECTOR CALCULUS (45) The total force acting on D is − _ S ρgzndS • k = − _ S ρgzk • ndS = − _ D ∇• ρgzkdxdydz = −ρg _ D _ ∂ ∂x i+ ∂ ∂y j+ ∂ ∂z k _ • zkdxdydz = −ρg _ D ∂z ∂z dxdydz = −ρg _ D dxdydz = −ρg volume of D (46) (a) F = ¸2z, −y, x) ∇ F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z 2z −y x ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸0, 1, 0) The equation for the area within ∆ABC: unit normal vector = n = 1,1,1 √ 1 2 +1 2 +1 2 = _ 1 √ 3 , 1 √ 3 , 1 √ 3 _ (x −2) + (y −0) + (z −0) = 0 =⇒x +y +z = 2 find the integral by using surface integral S = ¦(x, y) [0 ≤ x ≤ 2, 0 ≤ y ≤ 2 −x¦ dS = 1 |n • k| dxdy = √ 3 dxdy _ ∂S F • dr = _ S (∇F) • ndS = _ 2 0 _ 2−x 0 _ 1 √ 3 _ √ 3 dy dx = _ 2 0 [y] 2−x 0 dx = _ 2 0 (2 −x) dx = _ 2x − 1 2 x 2 _ 2 0 = 2 Find the integral by using line integral Γ 1 (AB) : r (t) = ¸2 −2t, 2t, 0), 0 ≤ t ≤ 1 F(r (t)) = ¸0, −2t, 2 −2t) r ′ (t) = ¸−2, 2, 0) F(r (t)) • r ′ (t) = ¸0, −2t, 2 −2t) • ¸−2, 2, 0) = −4t _ Γ 1 F(r (t)) • r ′ (t) dt = _ 1 0 (−4t) dt = −2 Γ 2 (BC) : r (s) = ¸0, 2 −2s, 2s), 0 ≤ s ≤ 1 F(r (s)) = ¸4s, 2s −2, 0) r ′ (s) = ¸0, −2, 2) 153 CHAPTER 6. VECTOR CALCULUS F(r (s)) • r ′ (s) = ¸4s, 2s −2, 0) • ¸0, −2, 2) = 4 −4s _ Γ 2 F(r (s)) • r ′ (s) ds = _ 1 0 (4 −4s) ds = 2 Γ 3 (CA) : r (u) = ¸2u, 0, 2 −2u), 0 ≤ u ≤ 1 F(r (u)) = ¸4 −4u, 0, 2u) r ′ (u) = ¸2, 0, −2) F(r (u)) • r ′ (u) = ¸4 −4u, 0, 2u) • ¸2, 0, −2) = 8 −12u _ Γ 3 F(r (u)) • r ′ (u) du = _ 1 0 (8 −12u) du = 2 Therefore, _ ∂S F • dr = _ Γ 1 + _ Γ 2 + _ Γ 3 = −2 + 2 + 2 = 2. (b) F = ¸xz, −y, xy) ∇ F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z xz −y xy ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = ¸x, x −y, 0) The equation for the plane that including ∆ABC: unit normal vector = n = 1,1,1 √ 1 2 +1 2 +1 2 = _ 1 √ 3 , 1 √ 3 , 1 √ 3 _ (x −1) + (y −0) + (z −0) = 0 =⇒x +y +z = 1 Find the integral by using surface integral. S = ¦(x, y) [0 ≤ x ≤ 1, 0 ≤ y ≤ 1 −x¦ dS = 1 |n • k| dxdy = √ 3 dxdy _ ∂S F • dr = _ S (∇F) • ndS = _ 1 0 _ 1−x 0 _ 1 √ 3 (2x −y) _ √ 3 dy dx = _ 1 0 _ 2xy − 1 2 y 2 _ 1−x 0 dx = _ 1 0 _ 3x − 1 2 − 5 2 x 2 _ dx = 1 6 Find the integral by using line integral Γ 1 (AB) : r (t) = ¸1 −t, t, 0), 0 ≤ t ≤ 1 F(r (t)) = ¸ 0, −t, t −t 2 _ r ′ (t) = ¸−1, 1, 0) F(r (t)) • r ′ (t) = ¸ 0, −t, t −t 2 _ • ¸−1, 1, 0) = −t _ Γ 1 F(r (t)) • r ′ (t) dt = _ 1 0 (−t) dt = − 1 2 Γ 2 (BC) : r (s) = ¸0, 1 −s, s), 0 ≤ s ≤ 1 F(r (s)) = ¸0, s −1, 0) r ′ (s) = ¸0, −1, 1) F(r (s)) • r ′ (s) = ¸0, s −1, 0) • ¸0, −1, 1) = 1 −s _ Γ 2 F(r (s)) • r ′ (s) ds = _ 1 0 (1 −s) ds = 1 2 Γ 3 (CA) : r (u) = ¸u, 0, 1 −u), 0 ≤ u ≤ 1 F(r (u)) = ¸ u −u 2 , 0, 0 _ 154 CHAPTER 6. VECTOR CALCULUS r ′ (u) = ¸1, 0, −1) F(r (u)) • r ′ (u) = ¸ u −u 2 , 0, 0 _ • ¸1, 0, −1) = u −u 2 _ Γ 3 F(r (u)) • r ′ (u) du = _ 1 0 _ u −u 2 _ du = 1 6 Therefore, _ ∂S F • dr = _ Γ 1 + _ Γ 2 + _ Γ 3 = − 1 2 + 1 2 + 1 6 = 1 6 . (47) (a) ∇F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z x 2 2x z 2 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 2k By Stokes’ Theorem, _ C F • dr = __ S ∇F • ndS = __ 4x 2 +y 2 ≤4 2k • kdS = 2 __ 4x 2 +y 2 ≤4 dS = 2 2π = 4π (b) ∇F = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ i j k ∂ ∂x ∂ ∂y ∂ ∂z y zx x 2 ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = −xi −2xj+(z −1) k By Stokes’ Theorem, _ C F • dr = __ S ∇F • ndS = __ S (−xi −2xj+(z −1) k) • _ 1 √ 3 i + 1 √ 3 j+ 1 √ 3 k _ dS = 1 √ 3 __ S (−3x +z −1) dS = 1 √ 3 _ 1 0 _ 1−x 0 (−3x −x −y) _ 1 + (z x ) 2 + (z y ) 2 dydx = 1 √ 3 _ 1 0 _ 1−x 0 (−4x −y) _ 1 + (−1) 2 + (−1) 2 dydx = _ 1 0 _ 1−x 0 (−4x −y) dydx = _ −4xy − 1 2 y 2 _ 1−x 0 = _ 1 0 _ −3x + 7 2 x 2 − 1 2 _ dx = − 5 6 155 CHAPTER 6. VECTOR CALCULUS (48) By Stokes’ Theorem, __ S ∇(yi) • ndS = _ C (yi) • dr = _ C (yi) • (dxi +dyj +dzk) = _ C ydx = _ 2π 0 sintd (cos t) = − _ 2π 0 sin 2 tdt = −π 156 Chapter 7 Differential Equations (1) (a) dy dx = 4y x(y−3) =⇒ (y−3)dy y = 4 dx x =⇒ _ 1 − 3 y _ dy = 4 dx x _ _ 1 − 3 y _ dy = 4 _ dx x =⇒y −3 ln[y[ = 4 ln[x[ +C (b) dy dx = y 3 +4y x(y 2 +2y+4) =⇒ (y 2 +2y+4)dy y 3 +4y = dx x =⇒ _ 1 y + 2 y 2 +4 _ dy = dx x _ _ 1 y + 2 y 2 +4 _ dy = _ dx x =⇒ln[y[ + tan −1 y 2 = ln[x[ +C (c) dy dx − 2 x y = − 1 x 2 p (x) = − 2 x µ(x) = e _ p(x)dx = e − _ 2 x dx = e −2 ln|x| = 1 x 2 1 x 2 _ dy dx − 2 x y _ = − 1 x 4 =⇒ d dx _ 1 x 2 y _ = − 1 x 4 =⇒ 1 x 2 y = − _ 1 x 4 dx = 1 3x 3 +C y = 1 3x +Cx 2 (d) (x −1) dy dx = y + 2 (x −2) 3 =⇒ dy dx = y x−1 + 2(x−2) 3 x−1 =⇒ dy dx − y x−1 = 2(x−2) 3 x−1 p (x) = − 1 x−1 µ(x) = e − _ 1 x−1 dx = 1 x−1 1 x−1 _ dy dx − y x−1 _ = 2(x−2) 3 (x−1) 2 =⇒ d dx _ y x−1 _ = 2x −8 + 6 x−1 − 2 (x−1) 2 y x−1 = _ _ 2x −8 + 6 x−1 − 2 (x−1) 2 _ dx = x 2 −8x + 6 ln[x −1[ + 2 x−1 +C y = _ x 2 −8x + 6 ln[x −1[ + 2 x−1 +C _ (x −1) (e) x 3 dy dx + _ 2 −3x 2 _ y = x 3 =⇒ dy dx + (2−3x 2 ) x 3 y = 1 p (x) = (2−3x 2 ) x 3 µ(x) = e _ ( 2−3x 2 ) x 3 dx = e _ −3 ln x− 1 x 2 _ = 1 x 3 e − 1 x 2 1 x 3 e − 1 x 2 _ dy dx + (2−3x 2 ) x 3 y _ = 1 x 3 e − 1 x 2 =⇒ d dx _ 1 x 3 e − 1 x 2 y _ = 1 x 3 e − 1 x 2 1 x 3 e − 1 x 2 y = _ 1 x 3 e − 1 x 2 dx = 1 2 e − 1 x 2 +C =⇒y = 1 2 x 3 +Cx 3 e 1 x 2 (f) dy dx + (2 cos x) y = sin 2 xcos x 157 CHAPTER 7. DIFFERENTIAL EQUATIONS p (x) = 2 cos x µ(x) = e _ 2 cos xdx = e 2 sin x e 2 sin x _ dy dx + (2 cos x) y _ = e 2 sin x sin 2 xcos x =⇒ d dx _ e 2 sin x y _ = e 2 sin x sin 2 xcos x e 2 sin x y = _ e 2 sin x sin 2 xcos xdx = _ e 2 sin x sin 2 xd (sinx) Let u = sin 2 x and dv = e 2 sin x d (sinx), then du = 2 sinxcos xdx and v = 1 2 e 2 sin x . e 2 sin x y = 1 2 e 2 sin x sin 2 x − _ _ 1 2 e 2 sin x _ (2 sinxcos xdx) = 1 2 e 2 sin x sin 2 x − _ e 2 sin x sinxcos xdx = 1 2 e 2 sin x sin 2 x − _ e 2 sin x sinxd (sinx) = 1 2 e 2 sin x sin 2 x − 1 2 e 2 sin x sinx + 1 2 _ e 2 sin x cos xdx = 1 2 e 2 sin x sin 2 x − 1 2 e 2 sin x sinx + 1 4 e 2 sin x +C = e 2 sin x _ 1 2 sin 2 x − 1 2 sinx + 1 4 _ +C y = 1 2 sin 2 x − 1 2 sin x + 1 4 +Ce −2 sin x or y = Ce −2 sin x − 1 4 cos 2x − 1 2 sinx + 1 2 (2) (a) The substitution y = xv(x) reduces the differential equation to v +x dv dx = 2 +v 1 + 3v , or x dv dx = 2 −3v 2 1 + 3v . Therefore, one has _ 1 + 3v 2 −3v 2 dv = _ dx x . Integration yields √ 6 −6 12 ln _ y x + √ 6 3 _ − √ 6 + 6 12 ln _ y x − √ 6 3 _ = lnCx, where C is an arbitrary constant. (b) Re-writing the equation as y ′ + _ 3 x _ y = 4 + 2 x 2 and multiplying both sides of this equation by µ = e _ 3 x dx = x 3 , we obtain d dx _ x 3 y _ = 4x 3 + 2x. Integration yields x 3 y = x 4 +x 2 +C, or y = x + 1 x + C x 3 . 158 CHAPTER 7. DIFFERENTIAL EQUATIONS (c) Let v = y x , then dv dx = 1 x dy dx − y x 2 =⇒ dy dx = x _ dv dx + y x 2 _ . x 2 _ dv dx + y x 2 _ −y + 3x 3 y −x 4 = 0 =⇒x 2 dv dx + 3x 3 y −x 4 = 0 =⇒ dv dx + 3x 2 v = x 2 p (x) = 3x 2 µ(x) = e _ 3x 2 dx = e x 3 e x 3 _ dv dx + 3x 2 v _ = x 2 e x 3 =⇒ d dv _ e x 3 v _ = x 2 e x 3 =⇒e x 3 v = _ x 2 e x 3 dx = 1 3 e x 3 +C v = 1 3 +Ce −x 3 =⇒ y x = 1 3 +Ce −x 3 =⇒y = 1 3 x +Cxe −x 3 (d) Let v = y x , then dv dx = 1 x dy dx − y x 2 =⇒ dy dx = x _ dv dx + y x 2 _ . x _ dv dx + y x 2 _ = 2 y x + _ x y _ 3 = 2v + 1 v 3 =⇒x dv dx +v = 2v + 1 v 3 =⇒x dv dx = v + 1 v 3 =⇒x dv dx = v 4 +1 v 3 =⇒ v 3 dv v 4 +1 = dx x _ v 3 dv v 4 +1 = _ dx x =⇒ 1 4 ln ¸ ¸ v 4 + 1 ¸ ¸ = ln[x[ +C =⇒ 1 4 ln ¸ ¸ ¸ _ y x _ 4 + 1 ¸ ¸ ¸ = ln[x[ +C (e) Let v = 1 y , then dv = − 1 y 2 dy =⇒dy = −y 2 dv = − 1 v 2 dv. − dv v 2 dx + 1 v = 1 v 2 =⇒ dv v 2 dx = 1 v − 1 v 2 =⇒ dv dx = v −1 =⇒ dv v−1 = dx _ dv v−1 = _ dx =⇒ln[v −1[ = x +C =⇒ln ¸ ¸ ¸ 1 y −1 ¸ ¸ ¸ = x +C (f) Let v = 1 y 4 , then dv = − 4 y 5 dy =⇒dy = − 1 4 y 5 dv. − 1 4 y 5 dv dx − 1 2x y = −y 5 =⇒ dv dx + 2 xy 4 = 4 =⇒ dv dx + 2v x = 4 p (x) = 2 x µ(x) = e _ 2 x dx = e 2 ln x = x 2 x 2 _ dv dx + 2v x _ = 4x 2 =⇒ d dx _ x 2 v _ = 4x 2 =⇒x 2 v = _ 4x 2 dx = 4 3 x 3 +C v = 4 3 x +C 1 x 2 =⇒ 1 y 4 = 4 3 x +C 1 x 2 (g) Let v = y −4x, then dv dx = dy dx −4 =⇒ dy dx = dv dx + 4. dv dx + 4 = v 2 =⇒ dv dx = v 2 −4 =⇒ dv v 2 −4 = dx _ dv v 2 −4 = _ dx =⇒− _ dv 4−v 2 = _ dx =⇒− 1 4 ln ¸ ¸ ¸ v+2 v−2 ¸ ¸ ¸ = x +C Use the formula _ du a 2 −u 2 = 1 2a ln ¸ ¸ ¸ u+a u−a ¸ ¸ ¸ +C. (h) Putting v = 1 y , the equation is then transformed into − 1 v 2 dv dx + 1 xv = x v 2 , or dv dx − 1 x v = −x, which is a linear equation with unknown v. Multiplying the equation by µ = e _ − 1 x dx = 1 x , one obtains d dx _ 1 x v _ = −1, which implies 1 x v = −x + C, or 1 xy = −x +C. (i) Substituting y = w − 1 3 into the differential equation, we obtain x 3 3 w − 4 3 dw dx +x 2 w − 1 3 = w − 4 3 cos x, 159 CHAPTER 7. DIFFERENTIAL EQUATIONS or dw dx + _ 3 x _ w = 3 cos x x 3 , which is a linear equation with unknown w. Multiplying the equation by µ = x 3 , we have d dx (x 3 w) = 3 cos x. Integration then gives w = 3 sinx +C x 3 , or y = x 3 √ 3 sinx +C . (3) (a) The equation is separable and may be written as y dy = e x dx, which may be inte- grated to give y 2 = 2e x + C, where C is a constant of integration. Applying the initial condition y(0) = 1, we conclude that C = −1 and that the solution to the initial value problem satisfies y 2 = 2e x −1. (b) The change of variable y = xv(x) reduces the equation to dx x = dv cot v = tanv dv, or _ dx x = _ tanv dv, which implies lnx = −lncos v +lnC. Therefore, the solution of the differential equation satisfies xcos y x = C. By the initial condition y(π) = 0, we have C = π. Therefore, xcos y x = π. (c) The differential equation is separable, which may be integrated to give _ (1 −y 5 )dy = _ xe x 2 dx, or y − y 6 6 = 1 2 e x 2 + C. Since y(0) = 0, we conclude that C = − 1 2 . As such, the solution of the initial value problem satisfies y 6 −6y + 3e x 2 = 3. (d) dy dx = x 2 y−y y+1 = y(x 2 −1) y+1 =⇒ y+1 y dy = _ x 2 −1 _ dx =⇒ _ 1 + 1 y _ dy = _ x 2 −1 _ dx _ _ 1 + 1 y _ dy = _ _ x 2 −1 _ dx =⇒y + ln[y[ = x 3 3 −x +C x = 3, y = −1 =⇒−1 + ln[−1[ = 9 −3 +C =⇒C = −7 y + ln[y[ = x 3 3 −x −7 (e) dy dx + 2xy = 0 =⇒ dy dx = −2xy =⇒ dy y = −2xdx =⇒ln[y[ = −x 2 +C x = 0, y = 2 =⇒ln[2[ = C ln[y[ = −x 2 + ln2 =⇒y = e −x 2 +ln2 = 2e −x 2 (f) Let v = y x , then dv dx = 1 x dy dx − y x 2 =⇒ dy dx = x _ dv dx + y x 2 _ . x _ dv dx + y x 2 _ = 1+(y/x) 2 2(xy/x 2 ) = 1+v 2 2v =⇒x dv dx +v = 1+v 2 2v =⇒x dv dx = 1−v 2 2v =⇒ 2v 1−v 2 dv = dx x 160 CHAPTER 7. DIFFERENTIAL EQUATIONS _ 2v 1−v 2 dv = _ dx x =⇒−ln ¸ ¸ 1 −v 2 ¸ ¸ = ln[x[ +C =⇒−ln ¸ ¸ ¸1 − _ y x _ 2 ¸ ¸ ¸ = ln[x[ +C x = 1, y = −2 =⇒−ln ¸ ¸ ¸1 − _ −2 1 _ 2 ¸ ¸ ¸ = ln[1[ +C =⇒C = −ln3 −ln ¸ ¸ ¸1 − _ y x _ 2 ¸ ¸ ¸ = ln[x[ −ln3 (g) dy dx +xy = xy 2 =⇒ dy dx = xy (y −1) =⇒ dy y(y−1) = xdx =⇒ _ 1 y−1 − 1 y _ dy = xdx _ _ 1 y−1 − 1 y _ dy = _ xdx =⇒ln[y −1[ −ln[y[ = x 2 2 +C =⇒ln ¸ ¸ ¸ y−1 y ¸ ¸ ¸ = x 2 2 +C x = 0, y = 2 =⇒ln ¸ ¸ 2−1 2 ¸ ¸ = C =⇒C = −ln2 ln ¸ ¸ ¸ y−1 y ¸ ¸ ¸ = x 2 2 −ln2 (4) (a) With M = 2xy+cos x and N = x 2 +siny, a simple calculation gives ∂M ∂y = ∂N ∂x = 2x. So the equation is exact. Integrating the equation ∂f ∂x = M with respect to x results in f(x, y) = _ (2xy + cos x) dx = x 2 y + sinx +g(y), where g(y) is an arbitrary function of f. If we substitute this into the second equation ∂f ∂y = N, we obtain dg dy = siny, or g(y) = −cos y. We therefore conclude that the solution of the equation satisfies f(x, y) = C, or x 2 y + sinx −cos y = C. (b) Since ∂ ∂y _ y + 2xy 3 ¸ = ∂ ∂x _ 1 + 3x 2 y 2 +x ¸ = 1 + 6xy 2 , the equation is exact. We consider the equations _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ ∂f ∂x = y + 2xy 3 ∂f ∂y = 1 + 3x 2 y 2 +x Integrating the first equation with respect to x, we obtain f(x, y) = _ (y + 2xy 3 )dx = xy +x 2 y 3 +g(y). Substituting this into the second equation, we conclude that x + 3x 2 y 2 + dg dy = 1 + 3x 2 y 2 +x. Thus dg dy = 1 and g(y) = y. Therefore, f(x, y) = xy + x 2 y 3 + y, which implies that the solution of the equation satisfies xy +x 2 y 3 +y = C. 161 CHAPTER 7. DIFFERENTIAL EQUATIONS (5) Let y (t) be the number of bacteria at time t (hour) and dy dt be the growth rate of the bacteria. So dy dt = ay, where a is a constant. dy dt = ay =⇒ dy y = adt =⇒ ln[y[ = at + C =⇒ y (t) = e at e C = A 0 e at , where A 0 is a constant. Let y (0) = A 0 , then y (6) = A 0 e 6a = 2A 0 . =⇒e 6a = 2 =⇒a = 1 6 ln2 Thus, y (t) = A 0 e (t ln 2)/6 . y (24) = A 0 e (24 ln 2)/6 = 16A 0 So the number of bacteria is 16 times as much as the number at the beginning. (6) Let x(t) be the weight of the radioactive material at time t (year) and dx dt be the rate of change of the weight. So dx dt = −bx, where b is a constant. dx dt = −bx =⇒x(t) = B 0 e −bt , where B 0 is a constant. Let x(0) = 100 = B 0 , then x(2) = B 0 e −2b = 0.95 100. =⇒e −2b = 0.95 =⇒b = − 1 2 ln0.95 Thus, x(t) = 100e (t ln 0.95)/2 . x(t) = 100e (t ln 0.95)/2 = 0.9 100 =⇒e (t ln0.95)/2 = 0.9 =⇒t = 2 ln 0.9 ln 0.95 = 4.1082 So it takes 4.11 years for 10% of the original mass to decay. (7) Let z (t) be the amount of the deposit at time t (year) and dz dt be the rate of change of the deposit. So dz dt = 0.07z and z (t) = C 0 e 0.07t , where C 0 is a constant. Let z (0) = C 0 = 10000, then z (t) = 10000e 0.07t . z (2) = 10000e 0.07(2) = 11503 So there will be $11503 in the account after 2 years. (8) Let w(t) be the temperature of the bar at time t (hour) and dw dt be the rate of change of the temperature. So dw dt = d (w(t) −S), where d is a constant and S is the surrounding temperature. dw w(t)−S = (d) dt =⇒ln[w(t) −S[ = dt +c =⇒w(t) = S +D o e dt , where D 0 is a constant. Let w(0) = 100 and S = 20, then 20 +D 0 = 100 =⇒D 0 = 80. w _ 1 3 _ = 20 + 80e d/3 = 50 =⇒e d/3 = 3 8 =⇒d = 3 ln 3 8 Thus, w(t) = 20 + 80e (3 ln 3 8 )t . 162 CHAPTER 7. DIFFERENTIAL EQUATIONS (9) Let T(t) be the temperature of the copper ball at time t. By Newton’s Law of Cooling, we have dT dt = k(T −30), where k is a physical constant, with initial condition T(0) = 100. The equation is separable, and an integration gives T(t)−30 = C e kt for any t > 0. T(0) = 100 ⇒C = 70, and T(3) = 70 ⇒e 3k = 40 70 ⇒ k = − 1 3 ln _ 7 4 _ ∼ = −0.187. Thus T(t) = 30 + 70 e − 1 3 ln( 7 4 )t for any t > 0. If T(t) = 40, a simple calculation shows that t = 3 ln 7 ln(7/4) ∼ = 10.4 minutes. (10) Dividing the equation by L and then multiplying both sides by µ = e R L t , we obtain d dt _ e R L t i _ = E 0 L e R L t cos ωt. Integration yields e R L t i = E 0 L _ e R L t cos ωt dt, which implies i(t) = E 0 R 2 +ω 2 L 2 (Rcos ωt +ωLsin ωt) +C e − R L t , where C is an arbitrary constant. Using the initial condition i(0) = 0, we conclude that C = −E 0 R R 2 +ω 2 L 2 . Therefore, i(t) = E 0 R 2 +ω 2 L 2 _ Rcos ωt +ωLsin ωt −Re − R L t _ . (11) Let S(t) be the amount of pollutants at any instant t, S(0) = S 0 be the initial amount of pollutants, V be the volume of water in the tank and r be the rate of influx / outflux. Since the influx has no pollutants at all, we have dS dt = _ dS dt _ in − _ dS dt _ out = − rS V Integration gives S(t) = S 0 e −( r V )t for any t > 0. (a) If S(t 50% ) = 1 2 S 0 , then e −( r V )t 50% = 1 2 , or _ r V _ t 50% = ln2. Thus t 50% = V r ln 2. (b) If S(t 10% ) = 1 10 S 0 , then e −( r V )t 10% = 1 10 , or _ r V _ t 10% = ln10. Thus t 10% = V r ln10. As V = 5000 and r = 4000, t 50% = 5000 4000 ln2 ∼ = 0.866 hours (or 52 minutes) and 163 CHAPTER 7. DIFFERENTIAL EQUATIONS t 10% = 5000 4000 ln10 ∼ = 2.88 hours (or 2 hours 53 minutes). (12) The equation of motion is m¨ y = mg, or ¨ y = g, where g ∼ = 9.8 m/sec 2 is the acceleration due to gravity. The general solution is y(t) = 1 2 gt 2 +c 1 t+c 2 , where c 1 and c 2 are arbitrary constants. Using the initial conditions y(0) = 0 and ˙ y(0) = −5, we obtain c 2 = 0 and c 1 = −5 and therefore y(t) = 1 2 gt 2 − 5t. Since ˙ y(t) = 0 if and only if t = 5 g ∼ = 0.51 sec, the minimum value of y(t) is equal to − 25 2g ∼ = −1.276. As such, the maximum height reached by the stone is 1.276 meters above the point from which the stone was released. When y(t) = 650, we solve 1 2 gt 2 − 5t − 650 = 0 to obtain t = 5 + √ 25 + 1300g g ∼ = 12.04 sec, meaning that the stone will strike the ground approximately 12.04 seconds after release. Since ˙ y = gt − 5, the velocity when the stone strikes the ground is given by √ 25 + 1300g ∼ = 112.98 m/sec. (13) (a) Auxiliary equation: m 2 + 4m+ 4 = 0. ∴ m 1 = m 2 = −2. The general solution is: y = e −2x (c 1 +c 2 x) (b) Auxiliary equation: m 2 −7m+ 12 = 0. ∴ m 1 = 3, m 2 = 4. The general solution is: y = c 1 e 3x +c 2 e 4x . (c) Auxiliary equation: m 2 + 4m+ 9 = 0. ∴ m = −2 ± √ 5i. The general solution is: y = e −2x (c 1 cos √ 5x +c 2 sin √ 5x) (d) Auxiliary equation: m 2 −2m+ 10 = 0. ∴ m = 1 ±3i. The general solution is: y = e x (Acos 3x +Bsin 3x). (e) Auxiliary equation: m 2 −2m+ 1 = 0. ∴ m = 1, 1. The general solution is: y = c 1 e x +c 2 xe x . (f) Auxiliary equation: m 2 + 2m+ 2 = 0. ∴ m = −1 ±i. The general solution is: y = e −x (c 1 cos x +c 2 sinx). (14) (a) Let y p (x) = Ke 3x , then y ′ p (x) = 3Ke 3x . y ′ −5y = e 3x =⇒3Ke 3x −5 _ Ke 3x _ = e 3x =⇒−2K = 1 =⇒K = − 1 2 Therefore, y p (x) = − 1 2 e 3x (b) Let y p (x) = Ke −5x , then y ′ p (x) = −5Ke −5x . y ′ + 6y = 4e −5x =⇒−5Ke −5x + 6 _ Ke −5x _ = 4e −5x =⇒K = 4 Therefore, y p (x) = 4e −5x (c) Let y p (x) = (Ax +B) e 2x , then y ′ p (x) = Ae 2x + 2 (Ax +B) e 2x = [2Ax + (A+ 2B)] e 2x . 164 CHAPTER 7. DIFFERENTIAL EQUATIONS y ′ −5y = xe 2x [2Ax + (A+ 2B)] e 2x −5 _ (Ax +B) e 2x ¸ = xe 2x [−3Ax + (A−3B)] = x Comparing the coefficients of the polynormial, _ −3A = 1 A−3B = 0 =⇒ _ A = − 1 3 B = − 1 9 Therefore, y p (x) = _ − 1 3 x − 1 9 _ e 2x . (d) Let y p (x) = (Ax +B) e 2x , then y ′ p (x) = [2Ax + (A+ 2B)] e 2x . y ′ + 6y = (2x −1) e 2x =⇒[2Ax + (A+ 2B)] e 2x + 6 (Ax +B) e 2x = (2x −1) e 2x =⇒[8Ax + (A+ 8B)] = 2x −1 Comparing the coefficients of the polynormial, _ 8A = 2 A+ 8B = −1 =⇒ _ A = 1 4 B = − 5 32 Therefore, y p (x) = _ 1 4 x − 5 32 _ e 2x . (e) Let y p (x) = _ Ax 2 +Bx +C _ e 2x , then y ′ p (x) = (2Ax +B) e 2x + 2e 2x _ Ax 2 +Bx +C _ = _ 2Ax 2 + (2A+ 2B) x + (B + 2C) ¸ e 2x . y ′ −5y = _ −9x 2 + 6x _ e 2x =⇒ _ 2Ax 2 + (2A+ 2B) x + (B + 2C) ¸ e 2x −5 _ Ax 2 +Bx +C _ e 2x = _ −9x 2 + 6x _ e 2x =⇒y ′ p (x) = _ −3Ax 2 + (2A−3B) x + (B −3C) ¸ e 2x = _ −9x 2 + 6x _ e 2x Comparing the coefficients of the polynormial, _ ¸ ¸ _ ¸ ¸ _ −3A = −9 2A−3B = 6 B −3C = 0 =⇒ _ ¸ ¸ _ ¸ ¸ _ A = 3 B = 0 C = 0 Therefore, y p (x) = 3x 2 e 2x . (15) (a) Putting y p = Ax 2 +Bx+C, where A, B and C are coefficients to be determined, we obtain y ′ p = 2Ax + B and y ′′ p = 2A. Substituting into the equation and comparing coefficients, we have A = 1 2 , B = 3 2 , C = 7 4 Therefore, y p = 1 4 (2x 2 + 6x + 7). 165 CHAPTER 7. DIFFERENTIAL EQUATIONS (b) Putting y p = Acos 4x +Bsin4x into the equation, we have y ′ p = −Asin4x + 4Bcos 4x and y ′′ p = −16Acos 4x − 16Bsin 4x. Comparing coeffi- cients, we obtain A = 2 25 , B = − 1 25 . Therefore, y p = 1 25 (2 cos 4x −sin4x). (c) Solve for y h : m−5 = 0 =⇒m = 5 =⇒y h = c 1 e 5x . Let y p = Ax 2 +Bx +C, then y ′ p = 2Ax +B. (2Ax +B) −5 _ Ax 2 +Bx +C _ = −5Ax 2 + (2A−5B) x + (B −5C) = 8x 2 −2 Comparing the coefficients of the polynormial, _ ¸ ¸ _ ¸ ¸ _ −5A = 8 2A−5B = 0 B −5C = −2 =⇒ _ ¸ ¸ _ ¸ ¸ _ A = − 8 5 B = − 16 25 C = 34 125 Therefore, y p = − 8 5 x 2 − 16 25 x + 34 125 and y = c 1 e 5x − 8 5 x 2 +− 16 25 x + 34 125 . (d) y h = c 1 e 5x Let y p = Ke 3x , then y ′ p = 3Ke 3x . 3Ke 3x −5Ke 3x = e 3x =⇒K = − 1 2 Therefore, y p = − 1 2 e 3x and y = c 1 e 5x − 1 2 e 3x . (e) y h = c 1 e 5x Let y p = Asinx +Bcos x, then y ′ p = Acos x −Bsinx. (Acos x −Bsinx)−5 (Asinx +Bcos x) = (A−5B) cos x+(−B −5A) sinx = cos x Comparing the coefficients of sine and cosine, _ A−5B = 1 −B −5A = 0 =⇒ _ A = 1 26 B = − 5 26 Therefore, y p = 1 26 sinx − 5 26 cos x and y = c 1 e 5x + 1 26 sinx − 5 26 cos x. (f) Let y p = Ae 3x + Bx 2 + Cx + D, where A, B, C and D are constants to be deter- mined. Substituting y p into the equation y ′′ +3y ′ +2y = 10e 3x +4x 2 and comparing coefficients, we obtain A = 1 2 , B = 2, C = −6 and D = 7. Hence y p = 1 2 e 3x + 2x 2 −6x + 7. (g) Putting y p = Acos 2x +Bsin2x +Ce x , we obtain y ′ p = 2Bcos 2x −2Asin2x +Ce x and y ′′ p = −4Acos 2x −4Bsin2x +Ce x . By comparing coefficients, we have −3A+ 4B = 2, −4A−3B = 0, 4C = 3. Solving the equations, we conclude that A = − 6 25 , B = 8 25 and C = 3 4 . Therefore, y p = − 6 25 cos 2x + 8 25 sin2x + 3 4 e x . 166 CHAPTER 7. DIFFERENTIAL EQUATIONS (h) y h = c 1 e 5x Let y p = Asinx +Bcos x +Ce 3x , then y ′ p = Acos x −Bsinx + 3Ce x . (Acos x −Bsinx + 3Ce x ) −5 _ Asin x +Bcos x +Ce 3x _ = e 3x + cos x =⇒(A−5B) cos x −(5A+B) sinx −2Ce 3x = e 3x + cos x Comparing the coefficients of each function term, _ ¸ ¸ _ ¸ ¸ _ A−5B = 1 5A+B = 0 −2C = 1 =⇒ _ ¸ ¸ _ ¸ ¸ _ A = 1 26 B = − 5 26 C = − 1 2 Therefore, y p = 1 26 sinx− 5 26 cos x− 1 2 e 3x and y = c 1 e 5x + 1 26 sinx− 5 26 cos x− 1 2 e 3x . (i) Solve for y h : m 2 + 4m+ 8 = 0 =⇒m = −2 ±2i. With α = −2 and β = 2, y h = e −2x (c 1 sin2x +c 2 cos 2x). Let y p = Ax 2 +Bx +C, then y ′ p = 2Ax +B and y ′′ p = 2A. 2A + 4 (2Ax +B) + 8 _ Ax 2 +Bx +C _ = 8Ax 2 + (8A+ 8B) x + (2A+ 4B + 8C) = 8x 2 + 8x + 18 Comparing the coefficients of the polynormial, _ ¸ ¸ _ ¸ ¸ _ 8A = 8 8A+ 8B = 8 2A+ 4B + 8C = 18 =⇒ _ ¸ ¸ _ ¸ ¸ _ A = 1 B = 0 C = 2 Therefore, y p = x 2 + 2 and y = e −2x (c 1 sin2x +c 2 cos 2x) +x 2 + 2. (j) Solve for y h : m 2 + 4m+ 5 = 0 =⇒m = −2 ±i. With α = −2 and β = 1, y h = e −2x (c 1 sinx +c 2 cos x). Let y p = Asinx +Bcos x, then y ′ p = Acos x −Bsinx and y ′′ p = −Asinx −Bcos x. 4 (A−B) sinx + 4 (A+B) cos x = (−Asinx −Bcos x) + 4 (Acos x −Bsinx) + 5 (Asinx +Bcos x) Comparing the coefficients of the sine and cosine, _ 4 (A−B) = −2 4 (A+B) = 2 =⇒ _ A = 0 B = 1 2 Therefore, y p = 1 2 cos x and y = e −2x (c 1 sinx +c 2 cos x) + 1 2 cos x. (k) Solve for y h : m 2 −2m = 0 =⇒m = 0, 2. Then y h = c 1 +c 2 e 2x . 167 CHAPTER 7. DIFFERENTIAL EQUATIONS Let y p = Ae x sinx +Be x cos x, then y ′ p = Ae x cos x +Ae x sinx +Be x cos x −Be x sinx = [(A−B) sinx + (A+B) cos x] e x y ′′ p = [(A−B) sinx + (A+B) cos x] e x +e x [(A−B) cos x −(A+B) sin x] = e x (2Acos x −2Bsinx) . and e x (2Acos x −2Bsinx) −2 [(A−B) sinx + (A+B) cos x] e x = e x sin x e x (−2Asinx −2Bcos x) = e x sin x Comparing the coefficients of the sine and cosine, _ −2A = 1 −2B = 0 =⇒ _ A = − 1 2 B = 0 Therefore, y p = − 1 2 e x sin x and y = c 1 +c 2 e 2x − 1 2 e x sin x. (l) Solve for y h : m 2 −m+ 2 = 0 =⇒m = 1 2 ± √ 7 2 i. With α = 1 2 and β = √ 7 2 , y h = e x 2 _ c 1 sin √ 7 2 x +c 2 cos √ 7 2 x _ . Let y p = _ Ax 2 +Bx +C _ e x , then y ′ p = _ Ax 2 + (2A+B) x + (B +C) ¸ e x y ′′ p = e x _ Ax 2 + (4A+B) x + (2A+ 2B +C) ¸ . Substituting into the differential equation and simplifying yield e x _ 2Ax 2 + (2A+ 2B) x + (2A+B +C) ¸ = _ 6x 2 + 8x + 7 _ e x Comparing the coefficients of polynomial, _ ¸ ¸ _ ¸ ¸ _ 2A = 6 2A+ 2B = 8 2A+B + 2C = 7 =⇒ _ ¸ ¸ _ ¸ ¸ _ A = 3 B = 1 C = 0 Therefore, y p = _ 3x 2 +x _ e x and y = e x 2 _ c 1 sin √ 7 2 x +c 2 cos √ 7 2 x _ + _ 3x 2 +x _ e x . (m) Solve for y h : m 2 + 2m+ 3 = 0 =⇒m = −1 ± √ 2i. With α = −1 and β = √ 2, y h = e −x _ c 1 sin √ 2x +c 2 cos √ 2x _ . 168 CHAPTER 7. DIFFERENTIAL EQUATIONS Let y p = Ax 2 +Bx +C +Dsin x +E cos x, then y ′ p = 2Ax +B +Dcos x −E sinx y ′′ p = 2A−Dsinx −E cos x. Substituting into the differential equation and simplifying yield 3Ax 2 +(4A+ 3B) x+(2A+ 2B + 3C)+(2D −2E) sinx+(2D + 2E) cos x = x 2 +sinx Comparing the coefficients of each function term, _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 3A = 1 4A+ 3B = 0 2A+ 2B + 3C = 0 2D −2E = 1 2D + 2E = 0 =⇒ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ A = 1 3 B = − 4 9 C = 2 27 D = 1 4 E = − 1 4 Therefore, y p = 1 3 x 2 − 4 9 x+ 2 27 + 1 4 sin x− 1 4 cos x and y = e −x _ c 1 sin √ 2x +c 2 cos √ 2x _ + 1 3 x 2 − 4 9 x + 2 27 + 1 4 sinx − 1 4 cos x. (16) Substitution of y p = K(x) e kx , where K(x) is a function to be determined, into the differential equation yields K ′′ (x) + (2k +p)K ′ (x) + (k 2 +pk +q)K(x) = A. Since k 2 +pk +q = 0, one has K ′′ (x) + (2k +p)K ′ (x) = A. (i) If 2k +p ,= 0 (distinct roots), then K(x) = Ax 2k +p . (ii) If 2k +p = 0 (repeated root), we have K(x) = Ax 2 2 . (17) Putting y p = w(x) e kx ,where w(x) is a function to be determined, we obtain w ′′ + (2k +p)w ′ + (k 2 +pk +q)w = f(x). (a) Substitution of y = e −x w into the equation y ′′ + 2y ′ + y = 3x 2 e −x yields w ′′ = 3x 2 . Therefore, w = x 4 4 and thus y p = x 4 4 e −x is a particular solution. (b) Substitution of y = e 4x w into the equation y ′′ −8y ′ = 2e 4x sin 3x yields w ′′ −16w = 2 sin3x. Putting w = Acos 3x + Bsin3x and using the method of undetermined coefficients, we conclude that w = − 2 25 sin 3x. Therefore y p = − 2 25 sin3xe 4x is a particular solution of the equation. 169 CHAPTER 7. DIFFERENTIAL EQUATIONS (18) (a) W _ e t sint, e t cos t _ = det _ e t sint e t cos t e t (sint + cos t) e t (cos t −sint) _ = _ e t sint _ _ e t (cos t −sint) ¸ − _ e t cos t _ _ e t (sint + cos t) ¸ = −e 2t (b) W (y 1 , y 2 , y 3 ) = det _ ¸ ¸ _ 1 sin2t cos 2t (1) ′ (sin2t) ′ (cos 2t) ′ (1) ′′ (sin2t) ′′ (cos 2t) ′′ _ ¸ ¸ _ = det _ ¸ ¸ _ 1 sin2t cos 2t 0 2 cos 2t −2 sin 2t 0 −4 sin2t −4 cos 2t _ ¸ ¸ _ = det _ 2 cos 2t −2 sin2t −4 sin2t −4 cos 2t _ = −8 cos 2 2t −8 sin 2 2t = −8 (c) W (y 1 , y 2 , y 3 , y 4 ) = det _ ¸ ¸ ¸ ¸ _ 1 x x 2 x 3 0 1 2x 3x 2 0 0 2 6x 0 0 0 6 _ ¸ ¸ ¸ ¸ _ = 12 (19) (a) The homogeneous equation has two independent solutions y 1 = e −2x cos 2x and y 2 = e −2x sin2x. Therefore, W(x) = ¸ ¸ ¸ ¸ ¸ e −2x cos 2x e −2x sin2x −2e −2x (cos 2x + 2 sin2x) 2e −2x (−sin2x + cos 2x) ¸ ¸ ¸ ¸ ¸ = 2e −4x . Using variation of parameters, one has a particular solution of the form y p = v 1 e −2x cos 2x +v 2 e −2x sin 2x, where v 1 = _ −y 2 R(x) W(x) dx = − _ e −2x sin 2xe −2x 2e −4x dx = cos 2x 4 , v 2 = _ y 1 R(x) W(x) dx = _ e −2x cos 2xe −2x 2e −4x dx = sin2x 4 . Therefore, y p = cos 2x 4 e −2x cos 2x + sin2x 4 e −2x sin2x. 170 CHAPTER 7. DIFFERENTIAL EQUATIONS (b) The homogeneous equations has independent solutions y 1 = e −x and y 2 = e 2x . Therefore, W(x) = 3e x . The equation has a particular solution of the form y p = v 1 e −x +v 2 e 2x , where v 1 = _ −e 2x sin2x 3e x dx = − 1 3 _ e x sin2xdx = 1 15 (−e x sin2x + 2e x cos 2x) and v 2 = _ e −x sin2x 3e x dx = 1 3 _ e −2x sin2xdx = e −2x 12 (−cos 2x −sin2x) . Therefore, y p = 1 15 (−sin 2x + 2 cos 2x) + 1 12 (−sin2x −cos 2x) = − 3 20 sin2x + 1 20 cos 2x. (c) We have y 1 = e x , y 2 = e 3x and W(x) = 2e 4x . The equation has a particular solution of the form y p = v 1 e x +v 2 e 3x , where v 1 = _ −e 3x 2e 4x 1 1 +e −x dx = 1 2 ln(1 +e −x ) and v 2 = _ e x 2e 4x 1 1 +e −x dx = − 1 4 e −2x + 1 2 e −x − 1 2 ln(1 +e −x ). Then y p = 1 2 (e x −e 3x ) ln(1 +e −x ) − 1 4 e x + 1 2 e 2x . (20) (a) For y ′ + 1 x y = 0, dy dx = − y x =⇒ dy y = − dx x =⇒ln[y[ = −ln[x[ +C =⇒y h = e C e −ln|x| = c 1 1 x . Let y p = v (x) y h = v (x) 1 x , then y ′ p = 1 x v ′ (x) − 1 x 2 v (x) and v ′ (x) = ln x x 1 y h = lnx. Thus, v (x) = _ lnxdx = x(lnx −1) and y p = lnx−1. Therefore, y = c 1 1 x +lnx−1. (b) y h = c 1 e 5x v ′ (x) = e −5x _ 8x 2 −2 _ v (x) = _ e −5x _ 8x 2 −2 _ dx = e −5x _ − 8 5 x 2 − 16 25 x + 34 125 _ y p = − 8 5 x 2 − 16 25 x + 34 125 y = c 1 e 5x − 8 5 x 2 − 16 25 x + 34 125 (c) y h = c 1 e 5x v ′ (x) = e −5x _ e 3x _ = e −2x v (x) = _ e −2x dx = − 1 2 e −2x y p = − 1 2 e 3x y = c 1 e 5x − 1 2 e 3x 171 CHAPTER 7. DIFFERENTIAL EQUATIONS (d) y h = c 1 e 5x v ′ (x) = e −5x (cos x) v (x) = _ e −5x (cos x) dx = 1 26 e −5x (sinx −5 cos x) y p = 1 26 (sinx −5 cos x) y = c 1 e 5x + 1 26 sinx − 5 26 cos x (e) Solve for y h : m 2 −m−2 = 0 =⇒m = −1, 2. Thus, y h = c 1 e −x +c 2 e 2x . Let y 1 = e −x and y 2 = e 2x , then W (y 1 , y 2 ) = 3e x . v 1 = _ − e 2x (4x 2 ) 3e x dx = − 4 3 e x _ x 2 −2x + 2 _ v 2 = _ e −x (4x 2 ) 3e x dx = − 1 3 e −2x _ 2x 2 + 2x + 1 _ y p = − 4 3 e x _ x 2 −2x + 2 _ (e −x ) − 1 3 e −2x _ 2x 2 + 2x + 1 _ _ e 2x _ = −2x 2 + 2x −3 y = c 1 e −x +c 2 e 2x −2x 2 + 2x −3 (f) y h = c 1 e −x +c 2 e 2x Let y 1 = e −x and y 2 = e 2x , then W (y 1 , y 2 ) = 3e x . v 1 = _ − e 2x sin 2x 3e x dx = 2 15 e x cos 2x − 1 15 e x sin 2x v 2 = _ e −x sin 2x 3e x dx = − 1 12 e −2x (cos 2x + sin2x) y p = 1 15 (2 cos 2x −sin2x) − 1 12 (cos 2x + sin2x) = 1 20 cos 2x − 3 20 sin2x y = c 1 e −x +c 2 e 2x + 1 20 cos 2x − 3 20 sin2x (g) y h = c 1 e −x +c 2 e 2x Let y 1 = e −x and y 2 = e 2x , then W (y 1 , y 2 ) = 3e x . v 1 = _ − e 2x (e 2x ) 3e x dx = − 1 9 e 3x v 2 = _ e −x (e 2x ) 3e x dx = 1 3 x y p = − 1 9 e 2x + 1 3 xe 2x y = c 1 e −x +c 2 e 2x − 1 9 e 2x + 1 3 xe 2x (h) Solve for y h : m 2 −2m+ 1 = 0 =⇒m = 1, 1. Then y h = c 1 e x +c 2 xe x . Let y 1 = e x and y 2 = xe x . Then W (y 1 , y 2 ) = e 2x . v 1 = _ − xe x _ e x x _ e 2x dx = −x v 2 = _ e x _ e x x _ e 2x dx = lnx y p = −xe x +xe x lnx y = c 1 e x +c 2 xe x −xe x +xe x lnx (i) Solve for y h : m 2 + 4 = 0 =⇒ m = ±2i. Then with α = 0 and β = 2, y h = c 1 cos 2x +c 2 sin2x. 172 CHAPTER 7. DIFFERENTIAL EQUATIONS Let y 1 = cos 2x and y 2 = sin2x, then W (y 1 , y 2 ) = 2. v 1 = _ − sin2x(4 sec 2 2x) 2 dx = −sec 2x v 2 = _ cos 2x(4 sec 2 2x) 2 dx = ln[sec 2x + tan 2x[ y p = 2 sin 2xln[sec 2x + tan 2x[ −2 y = c 1 cos 2x +c 2 sin2x + sin 2xln[sec 2x + tan 2x[ −1 (21) By Hooke’s Law, the spring constant k satisfies the equation −98 = k(0.7), so that k = 140 N/m Then with m = 10 kg, y 0 = −0.05 m, and v 0 = −0.1 m/s, the motion of the mass is given by y = −0.05 cos √ 14x +−0.0267 sin √ 14x (22) Kirchoff’s loop law gives i ′′ (t)+20i ′ (t)+200i(t) = 0, and the solution is i = e −10t (c 1 cos 10t+ c 2 sin10t). The initial conditions are i(0) = 0 and i ′ (0) = E(0) L − R L i(0) − q(0) LC = 12 0.5 − 10 0.5 (0) − 0.5 10 −2 (0) = 24. This yields c 1 = 0 and c 2 = 12 5 ; thus i(t) = 12 5 e −10t sin 10t for any t > 0. (23) (a) Let A = _ 0 1 8 −2 _ , then λ 1 = 2, v 1 = _ 1 2 _ , λ 2 = −4 and v 2 = _ −1 4 _ . y = c 1 e 2x v 1 +c 2 e −4x v 2 Therefore, _ y 1 = c 1 e 2x −c 2 e −4x y 2 = 2c 1 e 2x + 4c 2 e −4x . (b) Let A = _ 0 1 8 −2 _ , then λ 1 = 2, v 1 = _ 1 2 _ , λ 2 = −4 and v 2 = _ −1 4 _ . P = _ 1 2 − 1 4 1 1 _ and P −1 = _ 4 3 1 3 − 4 3 2 3 _ . f (x) = _ 0 e x _ and g (x) = P −1 f (x) = _ 1 3 e x 2 3 e x _ . w 1 = e 2x __ e −2x 1 3 e x dx +c 1 _ = e 2x _ 1 3 _ e −x dx +c 1 _ = − 1 3 e x +c 1 e 2x w 2 = e −4x __ e 4x 2 3 e x dx +c 2 _ = e −4x _ 2 3 _ e 5x dx +c 2 _ = 2 15 e x +c 2 e −4x y = Pw = _ 1 2 − 1 4 1 1 __ − 1 3 e x +c 1 e 2x 2 15 e x +c 2 e −4x _ = _ 1 2 c 1 e 2x − 1 5 e x − 1 4 c 2 e −4x c 1 e 2x − 1 5 e x +c 2 e −4x _ . (c) With P = _ ¸ ¸ _ 1 1 2 1 3 4 2 4 5 _ ¸ ¸ _ and D = _ ¸ ¸ _ −2 0 0 0 4 0 0 0 1 _ ¸ ¸ _ , we have 173 CHAPTER 7. DIFFERENTIAL EQUATIONS A = PDP −1 , where A = _ ¸ ¸ _ −5 −15 −21 3 1 3 0 6 7 _ ¸ ¸ _ . Putting _ ¸ ¸ _ y 1 y 2 y 3 _ ¸ ¸ _ = P _ ¸ ¸ _ u 1 u 2 u 3 _ ¸ ¸ _ into the equation, one concludes that d dt _ ¸ ¸ _ u 1 u 2 u 3 _ ¸ ¸ _ = P −1 AP _ ¸ ¸ _ u 1 u 2 u 3 _ ¸ ¸ _ = D _ ¸ ¸ _ u 1 u 2 u 3 _ ¸ ¸ _ . This implies ˙ u 1 = −2u 1 , ˙ u 2 = 4u 2 and ˙ u 3 = u 3 . Integration yields u 1 = c 1 e −2t , u 2 = c 2 e 4t and u 3 = c 3 e t , where c 1 , c 2 and c 3 are arbitrary constants. We thus conclude that y = Pu = P _ ¸ ¸ _ c 1 e −2t c 2 e 4t c 3 e t _ ¸ ¸ _ = _ ¸ ¸ _ c 1 e −2t +c 2 e 4t +c 3 e t c 1 e −2t + 3c 2 e 4t + 4c 3 e t 2c 1 e −2t + 4c 2 e 4t + 5c 3 e t _ ¸ ¸ _ . (d) Let A = _ ¸ ¸ _ 1 −3 3 3 −5 3 6 −6 4 _ ¸ ¸ _ , then λ 1 = −2, v 1 = _ ¸ ¸ _ 1 1 0 _ ¸ ¸ _ , λ 2 = −2, v 2 = _ ¸ ¸ _ −1 0 1 _ ¸ ¸ _ , λ 3 = 4 and v 3 = _ ¸ ¸ _ 1 2 1 2 1 _ ¸ ¸ _ . P = _ ¸ ¸ _ 1 −1 1 2 1 0 1 2 0 1 1 _ ¸ ¸ _ and P −1 = _ ¸ ¸ _ − 1 2 3 2 − 1 2 −1 1 0 1 −1 1 _ ¸ ¸ _ . f (x) = _ ¸ ¸ _ 0 0 x 2 + 5 _ ¸ ¸ _ and g (x) = P −1 f (x) = _ ¸ ¸ _ − 1 2 3 2 − 1 2 −1 1 0 1 −1 1 _ ¸ ¸ _ _ ¸ ¸ _ 0 0 x 2 + 5 _ ¸ ¸ _ = _ ¸ ¸ _ − 1 2 x 2 − 5 2 0 x 2 + 5 _ ¸ ¸ _ . w 1 = e −2x __ e 2x _ − 1 2 x 2 − 5 2 _ dx +c 1 ¸ = − 1 8 _ 2x 2 −2x + 11 _ +c 1 e −2x w 2 = c 2 e −2x w 3 = e 4x __ e −4x _ x 2 + 5 _ dx _ = − 1 32 _ 8x 2 + 4x + 41 _ +c 3 e 4x 174 CHAPTER 7. DIFFERENTIAL EQUATIONS y = Pw = _ ¸ ¸ _ 3 16 x +c 1 e −2x −c 2 e −2x + 1 2 c 3 e 4x − 3 8 x 2 − 129 64 3 16 x +c 1 e −2x + 1 2 c 3 e 4x − 3 8 x 2 − 129 64 c 2 e −2x − 1 8 x +c 3 e 4x − 1 4 x 2 − 41 32 _ ¸ ¸ _ (e) Diagonalize A = _ 1 −2 −2 4 _ to obtain AP = PD, where P = _ 2 − 1 2 1 1 _ and D = _ 0 0 0 5 _ . Substituting y = Pz into the system y ′ = Ay + _ 1 t _ , one obtains z ′ = Dz +P −1 _ 1 t _ , or _ z ′ 1 z ′ 2 _ = _ 0 0 0 5 __ z 1 z 2 _ + _ 1 5 (t + 2) 1 5 (4t −2) _ We thus have z ′ 1 = 1 5 (t + 2) and z ′ 2 = 5z 2 + 1 5 (4t −2). Solving these equations, we conclude that z 1 = t 2 10 + 2t 5 +C 1 and e −5t z 2 = _ 1 5 e −5t (4t −2)dt +C 2 , or z 2 = C 2 e 5t − 4 25 t + 6 125 . Finally, using y = Pz, one has y 1 = − 1 2 C 2 e 5t + t 2 5 + 22 25 t + 2C 1 − 3 125 , y 2 = C 2 e 5t + t 2 10 + 6 25 t +C 1 + 6 125 . 175 Chapter 8 Laplace and Fourier Transformations (1) (a) L _ e 3t cos 2 2t ¸ = L _ e 3t _ 1 2 + 1 2 cos 4t __ = L _ e 3t 1 2 + 1 2 e 3t cos 4t _ = 1 2 L _ e 3t ¸ + 1 2 L _ e 3t cos 4t ¸ = 1 2 1 s −3 + 1 2 s −3 (s −3) 2 + 16 . (b) L _ t 2 sin 3t ¸ = d 2 ds 2 _ 3 s 2 + 9 _ = d ds _ − 6s (s 2 + 9) 2 _ = 18 (s 2 −3) (s 2 + 9) 3 . (2) (a) L −1 _ 1 s 2 −2s+9 _ = L −1 _ 1 (s−1) 2 +8 _ = 1 √ 8 L −1 _ √ 8 (s−1) 2 +8 _ = 1 √ 8 e t sin √ 8t (b) L −1 _ 3 s 2 +4s+6 _ = L −1 _ 3 (s+2) 2 +2 _ = 3 √ 2 L −1 _ √ 2 (s+2) 2 +2 _ = 3 √ 2 e −2t sin √ 2t (c) L −1 _ s + 4 s 2 + 4s + 8 _ = L −1 _ s + 4 (s + 2) 2 + 4 _ = L −1 _ s (s + 2) 2 + 4 + 4 (s + 2) 2 + 4 _ = L −1 _ s (s + 2) 2 + 4 _ + 2L −1 _ 2 (s + 2) 2 + 4 _ = e −2t (cos 2t −sin2t) + 2e −2t sin 2t = e −2t (sin2t + cos 2t) (d) L −1 _ 3s+7 s 2 −2s−3 _ = L −1 _ 3s+7 (s−3)(s+1) _ = 4L −1 _ 1 s−3 _ −L −1 _ 1 s+1 _ = 4e 3t −e −t 177 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS (e) L −1 _ 3s + 1 (9s −1) (s 2 + 1) _ = L −1 _ 54 41 (9s −1) _ −L −1 _ 6s −13 41 (s 2 + 1) _ = 6 41 L −1 _ 1 s −1/9 _ − 6 41 L −1 _ s s 2 + 1 _ + 13 41 L −1 _ 1 s 2 + 1 _ = 6 41 e t/9 − 6 41 cos t + 13 41 sint (f) L −1 _ s+1 s 2 +s _ = L −1 _ 1 s ¸ = 1 (3) (a) L −1 _ 1 s (s 2 −3s + 2) _ = L −1 _ 1 2s − 1 s −1 + 1 2 (s −2) _ = 1 2 −e t + 1 2 e 2t . (b) If a ,= b, then L −1 _ αs +β (s 2 +a 2 ) (s 2 +b 2 ) _ = αL −1 _ s (s 2 +a 2 ) (s 2 +b 2 ) _ +βL −1 _ 1 (s 2 +a 2 ) (s 2 +b 2 ) _ = α b 2 −a 2 (cos at −cos bt) + β b 2 −a 2 _ sin at a − sinbt b _ . If a = b, then L −1 _ αs +β (s 2 +a 2 ) (s 2 +b 2 ) _ = αL −1 _ s (s 2 +a 2 ) 2 _ +βL −1 _ 1 (s 2 +a 2 ) 2 _ = αt sin at 2a + β(sinat −at cos at) 2a 3 . (c) Since q − p 2 4 > 0, we may put ω 2 = q − p 2 4 and complete square to obtain L −1 _ αs +β (s 2 +ps +q) 2 _ = L −1 _ ¸ _ α(s +p/2) + (β −αp/2) _ (s +p/2) 2 +ω 2 _ 2 _ ¸ _ = e −pt/2 _ αcos ωt + β −αp/2 2ω 3/2 (sinωt −ωt cos ωt) _ . 178 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS (4) (a) L _ y ′ −5y ¸ = L _ 8t 2 −2 ¸ L _ y ′ ¸ −5L[y] = 8L _ t 2 ¸ −2L[1] [sY (s) −y (0)] −5Y (s) = 8 2! s 3 −2 1 s (s −5) Y (s) −1 = 16 −2s 2 s 3 Y (s) = s 3 −2s 2 + 16 s 3 (s −5) = 91 125 (s −5) + 34 125s − 16 25s 2 − 16 5s 3 y (t) = 91 125 L −1 _ 1 s −5 _ + 34 125 L −1 _ 1 s _ − 16 25 L −1 _ 1 s 2 _ − 8 5 L −1 _ 2 s 3 _ = 91 125 e 5t + 34 125 − 16 25 t − 8 5 t 2 (b) L _ y ′ −5y ¸ = L _ e 3t ¸ [sY (s) −y (0)] −5Y (s) = 1 s −3 Y (s) = s −2 (s −3) (s −5) = 3 2 (s −5) − 1 2 (s −3) y (t) = 3 2 L −1 _ 1 s −5 _ − 1 2 L −1 _ 1 s −3 _ = 3 2 e 5t − 1 2 e 3t (c) L _ y ′ −5y ¸ = L[cos t] [sY (s) −y (0)] −5Y (s) = s s 2 + 1 Y (s) = s 2 +s + 1 (s 2 + 1) (s −5) = 31 26 (s −5) − 5s −1 26 (s 2 + 1) y (t) = 31 26 L −1 _ 1 s −5 _ − 5 26 L −1 _ s s 2 + 1 _ + 1 26 L −1 _ 1 s 2 + 1 _ = 31 26 e 5t − 5 26 cos t + 1 26 sin t 179 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS (d) L _ y ′ −5y ¸ = L _ e 3t + cos t ¸ [sY (s) −y (0)] −5Y (s) = 1 s −3 + s s 2 + 1 Y (s) = 22 13 (s −5) − 1 2 (s −3) − 5s −1 26 (s 2 + 1) y (t) = 22 13 L −1 _ 1 s −5 _ − 1 2 L −1 _ 1 s −3 _ − 5 26 L −1 _ s s 2 + 1 _ + 1 26 L −1 _ 1 s 2 + 1 _ = 22 13 e 5t − 1 2 e 3t − 5 26 cos t + 1 26 sint (e) L _ y ′′ −y ′ −2y ¸ = L _ 4t 2 ¸ _ s 2 Y (s) −sy (0) −y ′ (0) ¸ −[sY (s) −y (0)] −2Y (s) = 8 s 3 _ s 2 −s −2 _ Y (s) = 8 s 3 + 1 Y (s) = 7 3 (s + 1) + 2 3 (s −2) − 3 s + 2 s 2 − 4 s 3 y (t) = 7 3 L −1 _ 1 s + 1 _ + 2 3 L −1 _ 1 s −2 _ −3L −1 _ 1 s _ + 2L −1 _ 1 s 2 _ −2L −1 _ 2 s 3 _ = 7 3 e −t + 2 3 e 2t −3 + 2t −2t 2 (f) L _ y ′′ −y ′ −2y ¸ = L[sin2t] _ s 2 Y (s) −sy (0) −y ′ (0) ¸ −[sY (s) −y (0)] −2Y (s) = 2 s 2 + 4 _ s 2 −s −2 _ Y (s) = 2 s 2 + 4 + 1 Y (s) = s −6 20 (s 2 + 4) − 7 15 (s + 1) + 5 12 (s −2) y (t) = 1 20 L −1 _ s s 2 + 4 _ − 3 20 L −1 _ 2 s 2 + 4 _ − 7 15 L −1 _ 1 s + 1 _ + 5 12 L −1 _ 1 s −2 _ = 1 20 cos 2t − 3 20 sin2t − 7 15 e −t + 5 12 e 2t 180 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS (g) L _ y ′′ −y ′ −2y ¸ = L _ e 2t ¸ _ s 2 Y (s) −sy (0) −y ′ (0) ¸ −[sY (s) −y (0)] −2Y (s) = 1 s −2 _ s 2 −s −2 _ Y (s) = 1 s −2 + 1 Y (s) = 2 9 (s −2) − 2 9 (s + 1) + 1 3 (s −2) 2 y (t) = 2 9 L −1 _ 1 s −2 _ − 2 9 L −1 _ 1 s + 1 _ + 1 3 L −1 _ 1 (s −2) 2 _ = 2 9 e 2t − 2 9 e −t + 1 3 te 2t (h) L _ y ′′ + 4y ′ + 8y ¸ = L _ 8t 2 + 8t + 18 ¸ _ s 2 Y (s) −sy (0) −y ′ (0) ¸ + 4 [sY (s) −y (0)] + 8Y (s) = 16 s 3 + 8 s 2 + 18 s _ s 2 + 4s + 8 _ Y (s) = 16 s 3 + 8 s 2 + 18 s + 1 Y (s) = − 2s + 7 s 2 + 4s + 8 + 2 s + 2 s 3 y (t) = −2L −1 _ s (s + 2) 2 + 4 _ − 7 2 L −1 _ 2 (s + 2) 2 + 4 _ + 2L −1 _ 1 s _ +L −1 _ 2 s 3 _ = −2e −2t (cos 2t −sin 2t) − 7 2 e −2t sin2t + 2 +t 2 = t 2 + 2 − 3 2 e −2t sin 2t −2e −2t cos 2t (i) L _ y ′′ + 4y ′ + 5y ¸ = L[2 cos t −2 sint] _ s 2 Y (s) −sy (0) −y ′ (0) ¸ + 4 [sY (s) −y (0)] + 5Y (s) = 2s s 2 + 1 − 2 s 2 + 1 _ s 2 + 4s + 5 _ Y (s) = 2s s 2 + 1 − 2 s 2 + 1 + 1 Y (s) = 1 2 s s 2 + 1 − 1 2 s s 2 + 4s + 5 − 1 s 2 + 4s + 5 181 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS y (t) = 1 2 L −1 _ s s 2 + 1 _ − 1 2 L −1 _ s (s + 2) 2 + 1 _ −L −1 _ 1 (s + 2) 2 + 1 _ = 1 2 cos t − 1 2 e −2t (cos t −2 sint) −e −2t sint = 1 2 cos t − 1 2 e −2t cos t (j) L _ y ′′ + 2y ′ + 3y ¸ = L _ t 2 + sint ¸ _ s 2 Y (s) −sy (0) −y ′ (0) ¸ + 2 [sY (s) −y (0)] + 3Y (s) = 2 s 3 + 1 s 2 + 1 _ s 2 + 2s + 3 _ Y (s) = 2 s 3 + 1 s 2 + 1 + 1 Y (s) = 19s + 167 108 (s 2 + 2s + 3) − s −1 4 (s 2 + 1) + 2 27s − 4 9s 2 + 2 3s 3 y (t) = 19 108 L −1 _ s (s + 1) 2 + 2 _ + 167 108 L −1 _ 1 (s + 1) 2 + 2 _ − 1 4 L −1 _ s s 2 + 1 _ + 1 4 L −1 _ 1 s 2 + 1 _ + 2 27 L −1 _ 1 s _ − 4 9 L −1 _ 1 s 2 _ + 1 3 L −1 _ 2 s 3 _ = 19 108 e −t _ cos √ 2t − √ 2 2 sin √ 2t _ + 167 108 √ 2 2 e −t sin √ 2t − 1 4 cos t + 1 4 sint + 2 27 − 4 9 t + 1 3 t 2 = 19 108 e −t cos √ 2t + 37 27 √ 2 2 e −t sin √ 2t − 1 4 cos t + 1 4 sint + 2 27 − 4 9 t + 1 3 t 2 (k) L _ y ′′ −2y ′ ¸ = L _ e t sint ¸ _ s 2 Y (s) −sy (0) −y ′ (0) ¸ −2 [sY (s) −y (0)] = 1 (s −1) 2 + 1 _ s 2 −2s _ Y (s) = 1 (s −1) 2 + 1 + 1 Y (s) = 3 4 (s −2) − 1 2 (s 2 −2s + 2) − 3 4s y (t) = 3 4 L −1 _ 1 s −2 _ − 1 2 L −1 _ 1 (s −1) 2 + 1 _ − 3 4 L −1 _ 1 s _ = 3 4 e 2t − 1 2 e t sint − 3 4 182 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS (l) Apply Laplace transform to the equation, i.e. L[y ′′ −y ′ + 2y] = L __ 6t 2 + 8t + 7 _ e t ¸ . Then _ s 2 Y (s) −sy (0) −y ′ (0) ¸ −[sY (s) −y (0)] + 2Y (s) = 6 2 (s −1) 3 + 8 1 (s −1) 2 + 7 1 s −1 Thus _ s 2 −s + 2 _ Y (s) = 6 2 (s −1) 3 + 8 1 (s −1) 2 + 7 1 s −1 + 1 Y (s) = 1 (s −1) 2 + 6 (s −1) 3 y (t) = L −1 _ 1 (s −1) 2 _ + 3L −1 _ 2 (s −1) 3 _ = te t + 3t 2 e t . (5) (a) Taking Laplace Transform on both sides of the differential equations and making use of the initial conditions, we have Y (s) = As (s 2 +ω 2 )(s 2 +k 2 ) + Bω (s 2 +ω 2 )(s 2 +k 2 ) . i. If ω = k, then y (t) = L −1 _ As (s 2 +ω 2 ) 2 + Bω (s 2 +ω 2 ) 2 _ = At sinωt 2ω +B sin ωt −ωt cos ωt 2ω 2 . ii. If ω ,= k, then y (t) = L −1 _ As (s 2 +ω 2 ) (s 2 +k 2 ) + Bω (s 2 +ω 2 ) (s 2 +k 2 ) _ = A k 2 −ω 2 (cos ωt −cos kt) + Bω k 2 −ω 2 _ sinkt k − sinωt ω _ . (b) Since L _ 5e −2t sin2t ¸ = 10 (s + 2) 2 + 4 , we obtain s 2 Y (s) −s +sY (s) −1 +Y (s) = 10 (s + 2) 2 + 4 . 183 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS Therefore, Y (s) = s + 1 s 2 +s + 1 + 1 s 2 +s + 1 10 (s + 2) 2 + 4 = s + 1 s 2 +s + 1 − 10 37 3s −4 s 2 +s + 1 + 10 37 3s + 5 (s + 2) 2 + 4 = 7 37 s (s + 1/2) 2 + 3/4 + 77 37 1 (s + 1/2) 2 + 3/4 + 30 37 s (s + 2) 2 + 4 + 50 37 1 (s + 2) 2 + 4 , from which it follows that y (t) = e −t/2 _ 7 37 cos √ 3 2 t − 70 37 √ 3 sin √ 3 2 t _ +e −2t _ 30 37 cos 2t − 5 37 sin 2t _ . (c) Take Laplace Transform to obtain sY (s) + 4Y (s) = 5s s 2 + 4 . Thus Y (s) = 5s (s + 4)(s 2 + 4) . Since 5s (s + 4)(s 2 + 4) = − 1 s + 4 + 1 +s s 2 + 4 , we conclude that y(t) = −e −4t + sin 2t 2 + cos 2t. (6) The circuit equation is given by 2 di dt + 20i + 1 0.08 _ t 0 i (τ) dτ = 200 e −t . Denoting L[i (t)] by I (s), we obtain 2sI (s) + 20I (s) + 25 2 I (s) s = 200 s + 1 , which implies I (s) = 400s (s + 1) (4s 2 + 40s + 25) = 100s (s + 1) _ s + 5 + 5 2 √ 3 _ _ s + 5 − 5 2 √ 3 _ = 400 11 (s + 1) + 100 33 _ √ 3 −6 s + 5 + 5 2 √ 3 _ − 100 33 _ √ 3 + 6 s + 5 − 5 2 √ 3 _ . 184 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS Taking inverse Laplace Transform, we obtain i (t) = 400 11 e −t + 100 _√ 3 −6 _ 33 e −(5+ 5 2 √ 3)t − 100 _√ 3 + 6 _ 33 e −(5− 5 2 √ 3)t . (7) (a) (f ∗ g) (t) = _ t 0 e aτ dτ = _ 1 a e aτ _ t 0 = e at −1 a . (b) (f ∗ g) (t) = _ t 0 (t −τ) sinωτdτ = t _ t 0 sinωτdτ − _ t 0 τ sin ωτdτ = t _ − cos ωt −1 ω _ t 0 − _ τ cos ωτ ω _ t 0 − _ t 0 _ − cos ωτ ω _ dτ = −sinωt +ωt ω 2 . (c) (f ∗ g) (t) = _ t 0 τ 2 e a(t−τ) dτ = _ − τ 2 a e a(t−τ) _ t 0 − _ t 0 _ − 2τ a e a(t−τ) _ dτ = _ − τ 2 a e a(t−τ) _ t 0 − _ 2τ a 2 e a(t−τ) _ t 0 + _ t 0 2 a 2 e a(t−τ) dτ = − 2at + 2 +a 2 t 2 −2e at a 3 . (8) (a) f (t) = _ 2t if 0 < t ≤ 1 t if t > 1 and f ′ (t) = _ 2 if 0 < t ≤ 1 1 if t > 1 . F (s) = L[f (t)] = _ ∞ 0 f (t) e −st dt = _ 1 0 2te −st dt + _ ∞ 1 te −st dt = _ ∞ 0 te −st dt + _ 1 0 te −st dt = _ − 1 s te −st _ ∞ 0 + 1 s lim x→∞ _ x 0 e −st dt + _ − 1 s te −st _ 1 0 + 1 s _ 1 0 e −st dt = 1 s lim x→∞ _ − 1 s e −st _ x 0 − 1 s e −s + 1 s _ − 1 s e −st _ 1 0 = 1 s 2 − 1 s e −s − 1 s 2 e −s + 1 s 2 = 2 s 2 − 1 s e −s − 1 s 2 e −s 185 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS L _ f ′ (t) ¸ = _ ∞ 0 f ′ (t) e −st dt = _ 1 0 2e −st dt + _ ∞ 1 e −st dt = _ ∞ 0 e −st dt + _ 1 0 e −st dt = lim x→∞ _ − 1 s e −st _ ∞ 0 + _ − 1 s e −st _ 1 0 = 2 s − 1 s e −s (b) f (t) = _ t 2 if 0 < t ≤ 1 0 if t > 1 , f ′ (t) = _ 2t if 0 < t < 1 0 if t > 1 , f ′′ (t) = _ 2 if 0 < t < 1 0 if t > 1 . L _ f ′′ (t) ¸ = _ ∞ 0 f ′′ (t) e −st dt = _ 1 0 2e −st dt = _ − 2 s e −st _ 1 0 = − 2 s e −s + 2 s = 2 s _ 1 −e −s _ (9) (a) L −1 _ s 2 (s 2 + 4) 2 _ = L −1 _ s (s 2 + 4) s (s 2 + 4) _ = _ t 0 cos 2xcos 2 (t −x) dx = 1 2 _ t 0 [cos (2t) + cos (4x −2t)] dx = 1 2 cos (2t) _ t 0 dx + 1 2 _ t 0 cos (4x −2t) dx = 1 4 sin 2t + 1 2 t cos 2t (b) L −1 _ 1 (s 2 + 1) 3 _ = L −1 _ 1 (s 2 + 1) 2 1 (s 2 + 1) _ = _ t 0 sinx −xcos x 2 sin(t −x) dx = 1 2 _ t 0 (sinx −xcos x) (sint cos x −cos t sinx) dx = 1 2 sint _ t 0 _ sin xcos x −xcos 2 x _ dx + 1 2 cos t _ t 0 _ xsinxcos x −sin 2 x _ dx = 3 8 sint − 1 8 t 2 sin t − 3 8 t cos t 186 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS (c) L −1 _ s (s 2 + 4) 3 _ = L −1 _ 1 (s 2 + 4) 2 s s 2 + 4 _ = _ t 0 sin2x −2xcos 2x 16 cos 2 (t −x) dx = 1 16 _ t 0 (sin2x −2xcos 2x) (cos 2t cos 2x + sin2t sin2x) dx = 1 16 cos 2t _ t 0 _ sin2xcos 2x −2xcos 2 2x _ dx + 1 16 sin2t _ t 0 _ sin 2 2x −2xsin2xcos 2x _ dx = 1 64 t sin2t − 1 32 t 2 cos 2t (10) (a) We take Laplace Transform of the equation of motion to obtain Y (s) = αsG(s) + (β + cα M )G(s) + 1 M F(s)G(s), where F(s) = L[f(t)] and G(s) = 1 s 2 + c M s + k M = 1 (s + c 2M ) 2 + 4kM−c 2 4M 2 . Suppose g(t) = L −1 [G(s)]. i. If c 2 < 4kM, define 4kM−c 2 4M 2 = ω 2 . Then g(t) = e − c 2M t sinωt ω and y(t) = αe − c 2M t (cos ωt − c 2Mω sinωt) + (β + cα M )g(t) + 1 M (f ∗ g)(t). ii. If c 2 = 4kM, then g(t) = t e − c 2M t and therefore y(t) = _ α − cα 2M t _ e − c 2M t + (β + cα M )g(t) + 1 M (f ∗ g)(t). iii. If c 2 > 4kM, put 4kM−c 2 4M 2 = −Ω 2 . Then g(t) = e − c 2M t e Ωt −e −Ωt 2Ω and therefore y(t) = αe − c 2M t _ e Ωt +e −Ωt 2 − ck 4M e Ωt −e −Ωt Ω _ + (β + cα M )g(t) + 1 M (f ∗ g)(t). 187 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS (11) Taking Laplace Transform, we obtain _ ¸ _ ¸ _ 2sI 1 + 2sI 2 = E 0 s 4sI 1 + _ 5s + 20 + 20 s _ I 2 = 0 . A simple calculation shows that I 2 = −2 E 0 s _ s + 20 + 20 1 s _ = −2 E 0 (s 2 + 20s + 20) = −2E 0 _ s + 10 + 4 √ 5 _ _ s + 10 −4 √ 5 _ = √ 5E 0 20 _ 1 s + 10 + 4 √ 5 − 1 s + 10 −4 √ 5 _ . Therefore, i 2 (t) = √ 5 E 0 20 _ e −(10+4 √ 5)t −e −(10−4 √ 5)t _ . Similarly, I 1 = E 0 2s 2 −I 2 implies i 1 (t) = L −1 _ E 0 2s 2 _ −i 2 (t) = E 0 t 2 − √ 5E 0 20 _ e −(10+4 √ 5)t −e −(10−4 √ 5)t _ . (12) (a) L[3 sin2t] = L _ y (t) + _ t 0 (t −τ) y (τ) dτ _ . Let G(s) = L[g (t −τ)] and g (t) = t, so G(s) = 1 s 2 3 _ 2 s 2 + 2 2 _ = Y (s) +Y (s) G(s) 6 s 2 + 4 = _ 1 + 1 s 2 _ Y (s) Y (s) = 6s 2 (s 2 + 4) (s 2 + 1) = 8 s 2 + 4 − 2 s 2 + 1 y (t) = L −1 [Y (s)] = L −1 _ 8 s 2 + 4 − 2 s 2 + 1 _ = 4 sin 2t −2 sint (b) L _ e −t ¸ = L _ y (t) + 2 _ t 0 cos (t −τ) y (τ) dτ _ . 188 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS Let G(s) = L[g (t −τ)] and g (t) = cos t, so G(s) = s s 2 +1 . 1 s + 1 = Y (s) + 2Y (s) G(s) 1 s + 1 = _ 1 + 2s s 2 + 1 _ Y (s) Y (s) = s 2 + 1 (s + 1) 3 = 1 s + 1 − 2 (s + 1) 2 + 2 (s + 1) 3 y (t) = L −1 [Y (s)] = L −1 _ 1 s + 1 − 2 (s + 1) 2 + 2 (s + 1) 3 _ = e −t −2te −t +t 2 e −t (13) Since m = 1, c = 2 and k = 15, the spring mass-system after the blow is given by y ′′ (t) + 2y ′ (t) + 5y(t) = δ(t) with initial conditions y(0) = y ′ (0) = 0. Apply Laplace transform to the system, we have s 2 Y (s) + 2sY (s) + 5Y (s) = 1 or Y (s) = 1 s 2 + 2s + 5 = 1 (s + 1) 2 + 2 2 Therefore, y(t) = 1 2 e −t sin(2t). (14) Apply Laplace transform to the equation y(t) = 1 4 e −3t sin(2t) to obtain Y (s) = 1 4 2 (s + 3) 2 + 2 2 = 1 2s 2 + 12s + 26 . Therefore, m = 2, c = 12, k = 26. (15) G(ω) = _ ∞ −∞ g(x)e −iωx dx = _ ∞ −∞ f(kx)e −iωx dx Let kx = t, then kdx = dt G(ω) = _ ∞ −∞ f(t)e −iω( t k ) dt k = 1 k _ ∞ −∞ f(t)e −i( ω k )t dt = 1 k F( w k ) 189 CHAPTER 8. LAPLACE AND FOURIER TRANSFORMATIONS (16) Determine the Fourier transform of the function f(x) = e −x 2 F(w) = _ ∞ −∞ f(x)e −iωx dx = _ ∞ −∞ e −x 2 e −iωx dx By table, if f = e −t 2 2σ 2 , then F(w) = σ √ 2πe −σ 2 ω 2 2 . Set σ 2 = 1 2 , then F(w) = _ ∞ −∞ e −x 2 e −iωx dx = _ 1 2 √ 2πe − 1 2 ω 2 2 = √ πe −1 4 ω 2 (17) Considering delta function, δ(t), with continuous g(t) and t 0 ∈ (a, b), then _ b a g(t)δ(t −t 0 )dt = g(t 0 ) where δ(t) = lim ǫ→0 f ǫ (t), f ǫ (t) = _ 1 ǫ , 0 ≤ t ≤ ǫ 0, otherwise . So, _ b a δ(t)dt = 1, δ(t) = lim ǫ→0 f ǫ (t) = 0 for t ,= 0 and δ(0) = lim ǫ→0 f ǫ (0) = ∞. So, _ b a g(t)δ(t −t 0 )dt = _ b a g(t) lim ǫ→0 f ǫ (t −t 0 )dt = g(t 0 ) ⇒ lim ǫ→0 _ b a g(t)f ǫ (t −t 0 )dt = g(t 0 ) 190 Chapter 9 Partial Differential Equations (1) Differentiation yields ∂ 2 u 1 ∂t 2 = cos x _ −c 2 cos ct ¸ = c 2 ∂ 2 u 1 ∂x 2 . Similar calculations for u 2 . Since u = cos(x ±ct) = cos ct cos x ∓sinct sinx, we conclude by the previous calculations and by the Principle of superposition that u satisfies the wave equation. (2) Write u(x, t) = X (x) T (t) and substitute into the partial differential equation: X (x) T ′ (t) = c 2 X ′′ (x) T (t) Hence, the two differential equations are: X ′′ (x) +λX (x) = 0, X (0) = 0, X (L) = 0 (SLP) T ′ (t) = −λc 2 T (t) (DET) Follow the same way as in Example 9.2.1, n 0 = 1, λ n = _ nπ L _ 2 , X n (x) = β n sin _ nπ L x _ Solving the DET, T (t) = ae −λc 2 t , and hence T n (t) = a n e −(nπc/L) 2 t 191 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS The solution u(x, t) now becomes: u(x, t) = ∞ n=1 B n e −(nπc/L) 2 t sin _ nπ L x _ By the initial condition, u(x, 0) = f (x) = ∞ n=1 B n e −(nπc/L) 2 (0) sin _ nπ L x _ = ∞ n=1 B n sin _ nπ L x _ _ L 0 sin 2 _ nπ L x _ dx = L 2 and B n = _ L 0 f (x) sin _ nπ L x _ dx _ L 0 sin 2 _ nπ L x _ dx = 2 L _ L 0 f (x) sin _ nπ L x _ dx For different function f (x) in part (a) to (f), find the corresponding B n : (a) f (x) = 1 B n = 2 L _ L 0 (1) sin _ nπ L x _ dx = 2 L _ L 0 sin _ nπ L x _ dx = 2 L _ −L nπ cos _ nπ L x _ _ L 0 = −2 nπ [cos (nπ) −1] = −2 nπ [(−1) n −1] = _ 4 nπ , n is odd, 0, n is even. Therefore, the solution is u(x, t) = ∞ n odd 4 nπ e −(nπc/L) 2 t sin _ nπ L x _ . (b) f (x) = x B n = 2 L _ L 0 xsin _ nπ L x _ dx = 2 L _ −L nπ xcos _ nπ L x _ _ L 0 − 2 L _ L 0 −L nπ cos _ nπ L x _ dx = −2 nπ Lcos (nπ) + 2 nπ _ L nπ sin _ nπ L x _ _ L 0 = 2L nπ (−1) n+1 + 2 nπ (0) Therefore, u(x, t) = ∞ n=1 2L(−1) n+1 nπ e −(nπc/L) 2 t sin _ nπ L x _ . 192 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS (c) f (x) = A+ B−A L x B n = 2 L _ L 0 _ A+ B −A L x _ sin _ nπ L x _ dx = A _ 2 L _ L 0 sin _ nπ L x _ dx _ + B −A L _ 2 L _ L 0 xsin _ nπ L x _ dx _ = A 2 nπ _ (−1) n+1 + 1 _ + B −A L 2L nπ (−1) n+1 = 2 nπ _ A+B(−1) n+1 _ (by (a) and (b)) Therefore, u(x, t) = ∞ n=1 2[A+B(−1) n+1 ] nπ e −(nπc/L) 2 t sin _ nπ L x _ . (d) f (x) = sin _ 3π L x _ B n = 2 L _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx = _ 0, n ,= 3, 2 L _ L 0 sin 2 _ 3π L x _ dx n = 3. B 3 = 2 L _ L 0 sin 2 _ 3π L x _ dx = 1 L _ L 0 _ 1 −cos _ 6π L x __ dx = 1 L _ x − L 6π sin _ 6π L x __ L 0 = 1 Therefore, u(x, t) = e −(3πc/L) 2 t sin _ 3π L x _ . (e) f (x) = cos _ 3π L x _ B n = 2 L _ L 0 cos _ 3π L x _ sin _ nπ L x _ dx = 1 L _ L 0 _ sin π L (3 +n) x −sin π L (3 −n) x _ dx = 1 L _ −L π (3 +n) cos π L (3 +n) x − −L π (3 −n) cos π L (3 −n) x _ L 0 = −1 π (3 +n) [cos π (3 +n) −1] − −1 π (3 −n) [cos π (3 −n) −1] (n ,= 3) = (−1) n+4 + 1 π (3 +n) − (−1) n−2 + 1 π (3 −n) = [(−1) n + 1] (3 −n) −[(−1) n + 1] (3 +n) π (9 −n 2 ) = 2n[(−1) n + 1] π (n 2 −9) = _ 0, n is odd, 4n π(n 2 −9) , n is even. 193 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS Therefore, u(x, t) = ∞ n even 4n π (n 2 −9) e −(nπc/L) 2 t sin _ nπ L x _ . (f) f (x) = _ x, 0 ≤ x ≤ L 2 , L −x, L 2 ≤ x ≤ L. B n = 2 L _ L 2 0 xsin _ nπ L x _ dx + 2 L _ L L 2 (L −x) sin _ nπ L x _ dx = 2 L _ L 0 xsin _ nπ L x _ dx + 2 L _ L L 2 (L −2x) sin _ nπ L x _ dx = 2L nπ (−1) n+1 − 2 nπ _ (L −2x) cos _ nπ L x __ L L 2 + 2 nπ _ L L 2 (−2) cos _ nπ L x _ dx = 2L nπ (−1) n+1 + 2L nπ (−1) n − 4 nπ _ L nπ sin _ nπ L x _ _ L L 2 = 4L n 2 π 2 sin _ nπ 2 _ = _ _ _ 4L(−1) k (2k+1) 2 π 2 , where n = 2k + 1, k = 0, 1, 2, ..., 0, otherwise. Therefore, u(x, t) = ∞ k=0 4L (2k + 1) 2 π 2 (−1) k e −[(2k+1)πc/L] 2 t sin _ (2k + 1) π L x _ . (3) Write u(x, t) = v (x, t) +l (x) and substitute into the partial differential equation: v t (x, t) = c 2 _ v xx (x, t) +l ′′ (x) ¸ The boundary conditions imply that u(0, t) = v (0, t) +l (0) = A, u(L, t) = v (L, t) +l (L) = B, 0 ≤ t ≤ ∞. Furthermore, the initial condition also implies that u(x, 0) = v (x, 0) +l (x) = f (x) , 0 ≤ x ≤ L. Note that l (x) = A+ B−A L x, l ′′ (x) = 0, l (0) = A and l (L) = B. 194 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS Then v (x, t) should satisfy the following initial boundary value problem _ ¸ ¸ _ ¸ ¸ _ ∂v ∂t = c 2 ∂ 2 v ∂x 2 , v (0, t) = 0, v (L, t) = 0, 0 ≤ t < ∞, v (x, 0) = f (x) −l (x) , 0 ≤ x ≤ L. Solve the initial boundary value problem in question 2: Write v (x, t) = X (x) T (t) and substitute into the partial differential equation: X (x) T ′ (t) = c 2 X ′′ (x) T (t) Hence, the two differential equations are: X ′′ (x) +λX (x) = 0, X (0) = 0, X (L) = 0 (SLP) T ′ (t) = −λc 2 T (t) (DET) Follow the same way as in Example 9.2.1, n 0 = 1, λ n = _ nπ L _ 2 , X n (x) = β n sin _ nπ L x _ Solving the DET,T (t) = Ae −λc 2 t , and hence T n (t) = A n e −(nπc/L) 2 t The solution v (x, t) now becomes: v (x, t) = ∞ n=1 B n e −(nπc/L) 2 t sin _ nπ L x _ By the initial condition, v (x, 0) = f (x) −l (x) = ∞ n=1 B n sin _ nπ L x _ Since _ L 0 sin 2 _ nπ L x _ dx = L 2 , B n = _ L 0 [f (x) −l (x)] sin _ nπ L x _ dx _ L 0 sin 2 _ nπ L x _ dx = 2 L _ L 0 [f (x) −l (x)] sin _ nπ L x _ dx 195 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS For different function f (x) in part (a) and (b), find the corresponding B n : (a) f (x) = sin _ 3π L x _ B n = 2 L _ L 0 _ sin _ 3π L x _ −A− B −A L x _ sin _ nπ L x _ dx = 2 L _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx − 2 L _ L 0 _ A+ B −A L x _ sin _ nπ L x _ dx By (2) (d), 2 L _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx = 2 L _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx = _ 0, n ,= 3, 1, n = 3. By (2) (c) 2 L _ L 0 _ A+ B −A L x _ sin _ nπ L x _ dx = 2 nπ _ A+ (−1) n+1 B _ Therefore, v (x, t) = e −( 3πc L ) 2 t sin _ 3π L x _ − ∞ n=1 2 nπ _ A+ (−1) n+1 B _ e −( nπc L ) 2 t sin _ nπ L x _ and u(x, t) = e −( 3πc L ) 2 t sin _ 3π L x _ − ∞ n=1 2 nπ _ A+ (−1) n+1 B _ e −( nπc L ) 2 t sin _ nπ L x _ + _ A+ B −A L x _ . (b) f (x) = Q(x) = _ x, 0 ≤ x ≤ L 2 , L −x, L 2 ≤ x ≤ L. B n = 2 L _ L 0 _ Q(x) −A− B −A L x _ sin _ nπ L x _ dx = 2 L _ L 0 Q(x) sin _ nπ L x _ dx − 2 L _ L 0 _ A+ B −A L x _ sin _ nπ L x _ dx 196 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS By (2) (f) 2 L _ L 0 Q(x) sin _ nπ L x _ dx = _ _ _ 4L(−1) k (2k+1) 2 π 2 , n = 2k + 1, k = 0, 1, 2, ..., 0, otherwise. By (2) (c) 2 L _ L 0 _ A+ B −A L x _ sin _ nπ L x _ dx = 2 nπ _ A+ (−1) n+1 B _ Therefore, v (x, t) = _ _ _ ∞ k=0 4L(−1) k (2k + 1) 2 π 2 − ∞ k=0 2 _ A+ (−1) 2k+3 B _ (2k + 1) π _ _ _ e −[(2k+1)πc/L] 2 t sin _ (2k + 1) π L x _ and u(x, t) = _ _ _ ∞ k=0 4L(−1) k (2k + 1) 2 π 2 − ∞ k=0 2 _ A+ (−1) 2k+3 B _ (2k + 1) π _ _ _ e −[(2k+1)πc/L] 2 t sin _ (2k + 1) π L x _ + _ A+ B −A L x _ . (4) (a) Separation of variables yields u(x, t) = ∞ n=1 b n e −4n 2 t sin nx, with b n = 2 π _ π 0 x(π −x) sinnxdx = 4[1 −(−1) n ] πn 3 . Thus u(x, t) = 8 π n odd e −4n 2 t n 3 sinnx. (b) The function w(x) = 30 + 70x satisfies w(0) = 30, w(1) = 100 and w ′′ (x) = 0. It is now clear that v(x, t) = u(x, t) − w(x) would satisfy the equation v t = c 2 v xx , with boundary conditions v(0, t) = v(1, t) = 0 and initial condition v(x, 0) = 20 − 70x. 197 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS We may now solve for v(x, t) to obtain u(x, t) = (30 + 70x) + ∞ n=1 _ 100 (−1) n + 40 nπ _ e −n 2 π 2 c 2 t sinnπx. (5) We substitute u(x, t) = φ(t) w(x, t) into u t = c 2 u xx −ku to obtain φ ′ (t) w +φ(t) w t = φ(t) _ c 2 w xx −kw ¸ If w(x, t) satisfies the standard heat equation w t = c 2 w xx , then φ ′ (t) = −k φ(t), or φ(t) = e −kt . For the given IBVP, we let u(x, t) = e −t w(x, t) to obtain _ ¸ ¸ _ ¸ ¸ _ w t = w xx for 0 < x < 1, t > 0; w(0, t) = 0 and w(1, t) = 0 for t ≥ 0; w(x, 0) = sinπx + 1 2 sin3πx for 0 ≤ x ≤ 1 It follows that w(x, t) = e −π 2 t sinπx + 1 2 e −9π 2 t sin 3πx and therefore u(x, t) = e −t w(x, t) = e −(π 2 +1)t sinπx + 1 2 e −(9π 2 +1)t sin 3πx. (6) 2 _ 1 0 sin p n xsinp m xdx = _ 1 0 [cos(p n −p m )x −cos(p n +p m )x] dx. For n ,= m, we have 2 _ 1 0 sinp n x sin p m xdx = sin(p n −p m ) p n −p m − sin(p n +p m ) p n +p m . Since p n is a positive root of the equation tanp = − p σ , it follows that sinp n = − _ p n σ _ cos p n . Therefore, 2 _ 1 0 sin p n x sinp m xdx = sinp n cos p m −cos p n sinp m p n −p m − sinp n cos p m + cos p n sin p m p n +p m = 0. Similarly, we have _ 1 0 sin 2 p n xdx = 1 2 _ 1 0 [1 −cos 2p n x] dx = 1 2 − sin 2p n 4p n = 1 2 − sinp n cos p n 2p n = 1 2 + cos 2 p n 2σ . 198 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS (7) (a) Solving the S-L Problem, one has λ n = (n− 1 2 ) 2 as eigenvalues and X n (x) = sin(n− 1 2 )x as the corresponding eigenfunctions . Therefore, u(x, t) = ∞ n=1 b n e −c 2 (n− 1 2 ) 2 t sin(n − 1 2 )x. Using the initial condition x(π −x) = u(x, 0), one has x(π −x) = ∞ n=1 b n sin(n − 1 2 )x, where b n = _ π 0 x(π −x) sin(n − 1 2 )xdx _ π 0 sin 2 (n − 1 2 )xdx = 8 π _ 4 + (−1) n (2n −1)π (2n −1) 3 _ Therefore, u(x, t) = 8 π ∞ n=1 _ 4 + (−1) n (2n −1)π (2n −1) 3 _ e −c 2 (n− 1 2 ) 2 t sin(n − 1 2 )x. (b) We have u(x, t) = a 0 + ∞ n=1 a n e −16n 2 t cos nx, where a 0 + ∞ n=1 a n cos nx = u(x, 0) = _ 2x π if 0 ≤ x ≤ π 2 2 π (π −x) if π 2 ≤ x ≤ π The arbitrary constants a 0 and a n ’s may now be calculated by Euler’s Formula. The result is u(x, t) = 1 2 + 8 π 2 n even _ cos nπ 2 −1 n 2 _ e −16n 2 t cos nx. (c) We substitute u(x, t) = ∞ n=1 T n (t) sin nπx, where T n (t)’s are to be determined, into the p.d.e. to obtain T ′′ n (t)+n 2 π 2 c 2 T n (t) = 0 for n ,= 3 and T ′′ 3 (t) = −9π 2 c 2 T 3 (t)+1. Initial conditions on u(x, t) imply T ′ n (0) = 0 for every n = 1, 2, 3, . . ., T 1 (0) = 3, T 4 (0) = −2, and T n (0) = 0 whenever n ,= 1, 4. Solving the ODE, we obtain T n (t) = 0 when n ,= 1, 3, 4; T 1 (t) = 3 cos πct, T 3 (t) = 1 9π 2 c 2 [1 −cos 3πct] and T 4 (t) = −2 cos 4πct. Therefore, u(x, t) = 3 cos πct sinπx + 1 _ 1 −cos 3πct 9π 2 c 2 _ sin3πx −2 cos 4πct sin4πx. 199 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS (d) We substitute u(x, t) = ∞ n=1 T n (t) sinp n x into the equation to obtain T ′ 1 (t) = −p 2 1 T 1 (t) + 1 and T ′ n (t) = −p 2 n T n (t) for n > 1, with T n (0) = 0 for n = 1, 2, 3, . . .. Here, p n ’s are positive roots of the equation tanp = −p. Thus T n (t) = 0 whenever n > 1 and T 1 (t) = 1 −e −p 2 1 t p 2 1 . Hence u(x, t) = _ 1 −e −p 2 1 t p 2 1 _ sinp 1 x. (e) We put u(x, t) = v(x, t) +x cos t into the equation to obtain ∂v ∂t = c 2 ∂ 2 v ∂x 2 +x sint, with boundary conditions v(0, t) = 0 and v(1, t) = 0 and initial condition v(x, 0) = 0 for 0 ≤ x ≤ 1. Substituting v(x, t) = ∞ n=1 T n (t) sinnπx into the equation and bearing in mind that x sint = ∞ n=1 2(−1) n+1 sint nπ sin nπx, we obtain T ′ n (t) +c 2 n 2 π 2 T n (t) = 2(−1) n+1 sint nπ . Solving for the equation, with initial condition T n (0) = 0, we have T n (t) = 2 (−1) n+1 nπ _ e −c 2 n 2 π 2 t c 4 n 4 π 4 + 1 − cos t c 4 n 4 π 4 + 1 + c 2 n 2 π 2 sint c 4 n 4 π 4 + 1 _ Hence u(x, t) = x cos t + 2 π ∞ n=1 (−1) n+1 _ e −c 2 n 2 π 2 t −cos t +c 2 n 2 π 2 sint n(c 4 n 4 π 4 + 1) _ sinnπx. (8) Write u(x, t) = X (x) T (t) and substitute into the partial differential equation: X (x) T ′′ (t) = c 2 X ′′ (x) T (t) 200 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS Hence, the two differential equations are: T ′′ (t) +λc 2 T (t) = 0 X ′′ (x) +λX (x) = 0, X (0) = 0, X (L) = 0 (SLP) Follow the same way as in Example 9.2.1, n 0 = 1, λ n = _ nπ L _ 2 , X n (x) = β n sin _ nπ L x _ Solving T ′′ (t) +λc 2 T (t) = 0 with λ n = _ nπ L _ 2 , T n (t) = a n cos _ nπc L t _ +b n sin _ nπc L t _ The solution u(x, t) now becomes: u(x, t) = ∞ n=1 _ A n cos _ nπc L t _ +B n sin _ nπc L t __ sin _ nπ L x _ By the initial condition, u(x, 0) = f (x) = ∞ n=1 A n sin _ nπ L x _ u t (x, t) = g (x) = nπc L ∞ n=1 _ −A n sin _ nπc L t _ +B n cos _ nπc L t __ sin _ nπ L x _ u t (x, 0) = g (x) = nπc L ∞ n=1 B n sin _ nπ L x _ Since _ L 0 sin 2 _ nπ L x _ dx = L 2 , A n = _ L 0 f (x) sin _ nπ L x _ dx _ L 0 sin 2 _ nπ L x _ dx = 2 L _ L 0 f (x) sin _ nπ L x _ dx and B n = L nπc _ L 0 g (x) sin _ nπ L x _ dx _ L 0 sin 2 _ nπ L x _ dx = 2 nπc _ L 0 g (x) sin _ nπ L x _ dx For different function f (x) in part (a) to (e), find the corresponding A n and B n : 201 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS (a) f (x) ≡ sin _ 3π L x _ and g (x) ≡ 0. By (2) (d) A n = 2 L _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx = _ 0, n ,= 3, 1, n = 3. B n = 2 nπc _ L 0 (0) sin _ nπ L x _ dx = 0 Therefore, the solution is u(x, t) = cos _ 3πc L t _ sin _ 3π L x _ . (b) f (x) ≡ 0 and g (x) ≡ sin _ 3π L x _ . By (2) (d) A n = 2 L _ L 0 (0) sin _ nπ L x _ dx = 0 B n = 2 nπc _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx = _ 0, n ,= 3, L 3πc , n = 3. Therefore, u(x, t) = L 3πc sin _ 3πc L t _ sin _ 3π L x _ . (c) f (x) ≡ sin _ 3π L x _ and g (x) ≡ sin _ 3π L x _ . By (2) (d) A n = 2 L _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx = _ 0, n ,= 3, 1, n = 3. B n = 2 nπc _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx = _ 0, n ,= 3, L 3πc , n = 3. Therefore, u(x, t) = _ cos _ 3πc L t _ + L 3πc sin _ 3πc L t _¸ sin _ 3π L x _ . (d) f (x) ≡ sin _ 3π L x _ and g (x) ≡ sin _ 4π L x _ . By (2) (d) A n = 2 L _ L 0 sin _ 3π L x _ sin _ nπ L x _ dx = _ 0, n ,= 3, 1, n = 3. B n = 2 nπc _ L 0 sin _ 4π L x _ sin _ nπ L x _ dx = _ 0, n ,= 4, 2 4πc _ L 0 sin 2 _ 4π L x _ dx, n = 4. B 4 = 1 4πc _ L 0 _ 1 −cos _ 8π L x __ dx = 1 4πc _ x − L 8π sin _ 8π L x __ L 0 = L 4πc Therefore, u(x, t) = cos _ 3πc L t _ sin _ 3π L x _ + L 4πc sin _ 4πc L t _ sin _ 4π L x _ . 202 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS (e) f (x) = ǫQ(x) = _ ǫx, 0 ≤ x ≤ L 2 , ǫ (L −x) , L 2 ≤ x ≤ L. and g (x) = 0. By (2) (f) A n = 2 L _ L 0 ǫQ(x) sin _ nπ L x _ dx = _ _ _ 4ǫL(−1) k (2k+1) 2 π 2 , where n = 2k + 1, k = 0, 1, 2, ..., 0, otherwise. B n = 2 nπc _ L 0 (0) sin _ nπ L x _ dx = 0 Therefore, u(x, t) = ∞ k=0 _ 4ǫL(−1) k (2k+1) 2 π 2 cos _ (2k+1)πc L t __ sin _ (2k+1)π L x _ . (9) Write u(x, t) = X (x) T (t) and substitute into the partial differential equation: X (x) T ′′ (t) = c 2 X ′′ (x) T (t) Hence, the two differential equations are: T ′′ (t) +λc 2 T (t) = 0 X ′′ (x) +λX (x) = 0, X ′ (0) = 0, X ′ (L) = 0 (SLP) Follow the same way as in Example 9.2.1, n 0 = 0, λ n = _ nπ L _ 2 , X n (x) = α n cos _ nπ L x _ Solving T ′′ (t) +λc 2 T (t) = 0 with λ n = _ nπ L _ 2 , T n (t) = a n cos _ nπc L t _ +b n sin _ nπc L t _ The solution u(x, t) now becomes: u(x, t) = ∞ n=0 _ A n cos _ nπc L t _ +B n sin _ nπc L t __ cos _ nπ L x _ 203 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS By the initial condition, u(x, 0) = A 0 + ∞ n=1 A n cos _ nπ L x _ u t (x, t) = B 0 + nπc L ∞ n=1 _ −A n sin _ nπc L t _ +B n cos _ nπc L t __ cos _ nπ L x _ u t (x, 0) = B 0 + nπc L ∞ n=1 B n cos _ nπ L x _ Since _ L 0 cos 2 _ nπ L x _ dx = _ L, n = 0, L 2 , n ≥ 1. We have A 0 = _ L 0 f (x) dx L = 1 L _ L 0 f (x) dx and A n = _ L 0 f (x) cos _ nπ L x _ dx _ L 0 cos 2 _ nπ L x _ dx = 2 L _ L 0 f (x) cos _ nπ L x _ dx and B 0 = _ L 0 g (x) dx L = 1 L _ L 0 g (x) dx and B n = L nπc _ L 0 g (x) cos _ nπ L x _ dx _ L 0 cos 2 _ nπ L x _ dx = 2 nπc _ L 0 g (x) cos _ nπ L x _ dx. Thus, u t (x, t) = B 0 + nπc L ∞ n=1 _ −A n sin _ nπc L t _ +B n cos _ nπc L t __ cos _ nπ L x _ and u(x, t) = A 0 +B 0 t + ∞ n=1 _ A n cos _ nπc L t _ +B n sin _ nπc L t __ cos _ nπ L x _ . (10) Separation of variables and the superposition principle give u(x, t) = ∞ n=1 [α n cos nπt +β n sinnπt] sinnπx Initial conditions are u(x, 0) = _ 0.02x if 0 ≤ x ≤ 1 2 0.02(1 −x) if 1 2 ≤ x ≤ 1 and u t (x, 0) = 0 204 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS Therefore, we have β n = 0 and α n = 2 _ _ 1/2 0 0.02x sinnπxdx + _ 1 1/2 0.02(1 −x) sinnπxdx _ . This implies (see Ex. 3 of Revision with L = 1 and K = 0.01) u(x, t) = 0.08 π 2 n odd (−1) n−1 2 n 2 cos nπt sin nπx. (11) We substitute u(x, t) = ∞ n=1 T n (t) sinnx into the wave equation u tt = u xx + sin 100x sin100t to obtain ordinary differential equations T ′′ n (t) +n 2 T n (t) = 0 when n ,= 100 and T ′′ 100 (t) + 10000 T 100 (t) = sin 100t, with initial conditions T 1 (0) = 2, T ′ 1 (0) = 0, T 4 (0) = −3, T ′ 4 (0) = 0 and T n (0) = T ′ n (0) = 0 for n ,= 1, 4 Solving these ODE, we obtain T 1 (t) = 2 cos t, T 4 (t) = −3 cos 4t, T 100 (t) = L −1 _ 100 (s 2 + 100 2 ) 2 _ = sin 100t −100t cos 100t 20000 . Therefore, u(x, t) = 2 cos t sinx −3 cos 4t sin4x + _ sin 100t −100t cos 100t 20000 _ sin100x (12) Differentiating E(t) with respect to t, one obtains E ′ (t) = _ L 0 [ρ u t (x, t) u tt (x, t) +T u x (x, t) u xx (x, t)] dx = _ L 0 _ ρ u t (x, t) c 2 u xx (x, t) +T u x (x, t) u xt (x, t) ¸ dx = T __ L 0 u t (x, t) u xx (x, t) dx + _ L 0 u x (x, t) u xt (x, t) dx _ 205 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS Integrating the second integral by parts, we have _ L 0 u x (x, t) u xt (x, t) dx = [u x (L, t) u t (L, t) −u x (0, t) u t (0, t)] − _ L 0 u xx (x, t) u t (x, t) dx = − _ L 0 u xx (x, t) u t (x, t) dx because u t (0, t) = u t (L, t) = 0 Therefore, E ′ (t) = 0, and the total energy is conserved. (13) Let u(x, y) = X(x)Y (y). Solving the following two ODEs: 1) X ′′ (x) +λX(x) = 0 where 0 < x < 2; 2) Y ′′ (y) −λY (y) = 0 where 0 < y < 1. So, with the given boundary conditions, λ n = n 2 π 2 4 , X n = sin nπx 2 , Y n = β[e nπ 2 y −e − nπ 2 y ], n = 1, 2, 3 . . . . Therefore, u n (x, y) = β n [e nπ 2 y −e − nπ 2 y ] sin nπx 2 , n = 1, 2, 3.... By Principle of Superposition, u(x, y) = ∞ n=1 β n [e nπ 2 y −e − nπ 2 y ] sin nπx 2 For boundary condition: u(x, 1) = x(2 −x), 0 < x < 2, β n = 2 2[e nπ 2 −e − nπ 2 ] _ 2 0 x(2 −x) sin nπx 2 dx, n = 1, 2, 3.... = 1 [e nπ 2 −e − nπ 2 ] −8(−2 + 2 cos nπ +nπ sinnπ) n 3 π 3 = 1 [e nπ 2 −e − nπ 2 ] 16(1 −(−1) n ) n 3 π 3 206 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS Finally, u(x, y) = ∞ n=1 _ 1 [e nπ 2 −e − nπ 2 ] 16(1 −(−1) n ) n 3 π 3 _ [e nπ 2 y −e − nπ 2 y ] sin nπx 2 = 32 π 3 n=odd 1 n 3 _ e nπ 2 y −e − nπ 2 y e nπ 2 −e − nπ 2 _ sin nπx 2 (14) Set Solution of u xx +u yy = 0 as u(x, y) = X(x)Y (y). Then Y ′′ X +X ′′ Y = 0 Y ′′ Y = − X ′′ X = −λ Solving the following two ODEs: (1) Y ′′ (x) +λY (x) = 0 where 0 < y < 1; (2) X ′′ (y) −λX(y) = 0 where 0 < x < 1. So, with the given boundary conditions, λ n = n 2 π 2 , Y n = sinnπy, X(x) = αe nπx +βe −nπx Since u(1, y) = 0, X(1) = αe nπ +βe −nπ = 0 ⇒β = −αe 2nπ Therefore, X n = α n [e nπx −e 2nπ e −nπx ] = α n [e nπx −e nπ(2−x) ], n = 1, 2, 3, . . . and u n (x, y) = α n [e nπx −e nπ(2−x) ] sinnπy, n = 1, 2, 3 . . . . By Principle of Superposition, u(x, y) = ∞ n=1 α n [e nπx −e nπ(2−x) ] sinnπy, n = 1, 2, 3.... From boundary condition, u(0, y) = 3 sinπy −4 sin 6πy for 0 ≤ y ≤ 1, we have α 1 = 3 (1 −e 2π ) , α 6 = − 4 (1 −e 12π ) and α n = 0, n ,= 1, 6 207 CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS Hence u(x, y) = 3 (1 −e 2π ) [e πx −e π(2−x) ] sinπy − 4 (1 −e 12π ) [e 6πx −e 6π(2−x) ] sin6πy. (15) Using polar coordinates, we substitute u(r, θ) = R(r)S(θ) into the equations and proceed as in the lecture notes. Since S(θ + 2π) = S(θ), we have S(θ) = a n cos nθ +b n sin nθ when n ≥ 1 and S(θ) = a 0 when n = 0, R(r) = c n r n +d n r −n when n ≥ 1 and R(r) = c 0 lnr +d 0 when n = 0 Consequently, the Principle of Superposition implies that u(r, θ) = (A 0 lnr +B 0 ) + ∞ n=1 r n [A n cos nθ +B n sinnθ] +r −n [C n cos nθ +D n sinnθ] , where A 0 , B 0 , A n , B n , C n and D n ’s are arbitrary constants. (a) To ensure that u(r, θ) is finite as r → 0, we must choose A 0 = C n = D n = 0. Furthermore, the boundary condition implies that B 0 + ∞ n=1 [A n cos nθ +B n sinnθ] = _ 1 for 0 ≤ θ ≤ π; 0 for π < θ < 2π Therefore, u(r, θ) = 1 2 + n odd _ 2 nπ _ r n sinnθ. (b) The requirement for lim r→∞ u(r, θ) to be finite implies that A 0 = A n = B n = 0 Therefore, u(r, θ) = B 0 + ∞ n=1 r −n [C n cos nθ +D n sin nθ]. Since u(1, θ) = 1 + cos 2θ − 3 sin 4θ, we may compare coefficients to obtain B 0 = 1, C 2 = 1, D 4 = −3 and all other coefficients = 0. Therefore, u(r, θ) = 1 + cos 2θ r 2 − 3 sin 4θ r 4 . 208