Calculus one and several variables 10E Salas solutions manual ch16

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBUJWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 788 SECTION 16.1 CHAPTER 16 SECTION 16.1 1. ∇f = (6x −y) i + (1 −x) j 2. ∇f = (2Ax +By)i + (Bx + 2Cy)j 3. ∇f = e xy [ (xy + 1) i +x 2 j] 4. ∇f = 1 (x 2 +y 2 ) 2 [(y 2 −x 2 + 2xy)i + (y 2 −x 2 −2xy)j] 5. ∇f = _ 2y 2 sin(x 2 + 1) + 4x 2 y 2 cos(x 2 + 1) ¸ i + 4xy sin(x 2 + 1) j 6. ∇f = 2x x 2 +y 2 i + 2y x 2 +y 2 j 7. ∇f = (e x−y +e y−x ) i + (−e x−y −e y−x ) j = (e x−y +e y−x )(i −j) 8. ∇f = AD −BC (Cx +Dy) 2 [ yi −xj ] 9. ∇f = (z 2 + 2xy) i + (x 2 + 2yz) j + (y 2 + 2zx) k 10. ∇f = x _ x 2 +y 2 +z 2 i + y _ x 2 +y 2 +z 2 j + z _ x 2 +y 2 +z 2 k 11. ∇f = e −z (2xy i +x 2 j −x 2 y k) 12. ∇f = _ xyz x +y +z +yz ln(x +y +z) _ i + _ xyz x +y +z +xz ln(x +y +z) _ j + _ xyz x +y +z +xy ln(x +y +z) _ k 13. ∇f = e x+2y cos _ z 2 + 1 _ i + 2e x+2y cos _ z 2 + 1 _ j −2ze x+2y sin _ z 2 + 1 _ k 14. ∇f = e yz 2 /x 3 _ − 3yz 2 x 4 i + z 2 x 3 j + 2yz x 3 k _ 15. ∇f = _ 2y cos(2xy) + 2 x _ i + 2xcos(2xy) j + 1 z k 16. ∇f = _ 2xy z −3z 4 _ i + x 2 z j − _ x 2 y z 2 + 12xz 3 _ k 17. ∇f = (4x −3y) i + (8y −3x) j; at (2, 3), ∇f = −i + 18j 18. ∇f = 1 (x −y) 2 (−2yi + 2xj), ∇f(3, 1) = − 1 2 i + 3 2 j 19. ∇f = 2x x 2 +y 2 i + 2y x 2 +y 2 j; at (2, 1), ∇f = 4 5 i + 2 5 j P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.1 789 20. ∇f = _ tan −1 (y/x) − xy x 2 +y 2 _ i + _ x 2 x 2 +y 2 _ j, ∇f(1, 1) = _ π 4 − 1 2 _ i + 1 2 j 21. ∇f = (sin xy +xy cos xy) i +x 2 cos xy j; at (1, π/2), ∇f = i 22. ∇f = e −(x 2 +y 2 ) [(y −2x 2 y)i + (x −2xy 2 )j], ∇f(1, −1) = e −2 (i −j) 23. ∇f = −e −x sin (z + 2y) i + 2e −x cos (z + 2y) j +e −x cos (z + 2y) k; at (0, π/4, π/4), ∇f = − 1 2 √ 2 (i + 2j +k) 24. ∇f = cos πzi −cos πzj −π(x −y) sin πzk, ∇f _ 1, 0, 1 2 _ = −πk 25. ∇f = i − y _ y 2 +z 2 j − z _ y 2 +z 2 k; at (2, −3, 4), ∇f = i + 3 5 j − 4 5 k 26. ∇f = −sin(xyz 2 )(yz 2 i +xz 2 j + 2xyzk), ∇f _ π, 1 4 , −1 _ = − √ 2 2 _ 1 4 i +π j − π 2 k _ 27. (a) ∇f(0, 2) = 4 i (b) ∇f _ 1 4 π, 1 6 π _ = _ −1 − −1 + √ 3 2 √ 2 _ i + _ − 1 2 − −1 + √ 3 √ 2 _ j (c) ∇f(1, e) = (1 −2e) i −2 j 28. (a) ∇f(1, 2, −3) = 1 8 √ 2 i + 1 2 √ 2 j − 27 8 √ 2 k (b) ∇f(1, −2, 3) = − 5 18 i + 1 9 j + 1 18 k (c) ∇f(1, e 2 , π/6) = √ 3 2 i + π 12e 2 j +k 29. For the function f(x, y) = 3x 2 −xy +y, we have f(x +h) −f(x) = f(x +h 1 , y +h 2 ) −f(x, y) = 3(x +h 1 ) 2 −(x +h 1 )(y +h 2 ) + (y +h 2 ) − _ 3x 2 −xy +y ¸ = [(6x −y) i + (1 −x) j] · (h 1 i +h 2 j) + 3h 2 1 −h 1 h 2 = [(6x −y) i + (1 −x) j] · h + 3h 2 1 −h 1 h 2 The remainder g(h) = 3h 2 1 −h 1 h 2 = (3h 1 i −h 1 j) · (h 1 i +h 2 j) , and |g(h)| h = 3h 1 i −h 1 j · h · cos θ h ≤ 3h 1 i −h 1 j Since 3h 1 i −h 1 j →0 as h →0 it follows that ∇f = (6x −y) i + (1 −x) j 30. f(x +h) −f(x) = [(x + 2y) i + (2x + 2y) j] · [h 1 i +h 2 j] + 1 2 h 2 1 + 2h 1 h 2 +h 2 2 ; g(h) = 1 2 h 2 1 + 2h 1 h 2 +h 2 2 is o(h). P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 790 SECTION 16.1 31. For the function f(x, y, z) = x 2 y +y 2 z +z 2 x, we have f(x +h) −f(x) = f(x +h 1 , y +h 2 , z +h 3 ) −f(x, y, z) = (x +h 1 ) 2 (y +h 2 ) + (y +h 2 ) 2 (z +h 3 ) + (z +h 3 ) 2 (x +h 1 ) − _ x 2 y +y 2 z +z 2 x _ = _ 2xy +z 2 _ h 1 + _ 2yz +x 2 _ h 2 + _ 2xz +y 2 _ h 3 + (2xh 2 +yh 1 +h 1 h 2 ) h 1 + (2yh 3 +zh 2 +h 2 h 3 ) h 2 + (2zh 1 +xh 3 +h 1 h 3 ) h 3 = __ 2xy +z 2 _ i + _ 2yz +x 2 _ j + _ 2xz +y 2 _ k ¸ · h +g(h) · h, where g(h) = (2xh 2 +yh 1 +h 1 h 2 ) i + (2yh 3 +zh 2 +h 2 h 3 ) j + (2zh 1 +xh 3 +h 1 h 3 ) k Since |g(h)| h →0 as h →0 it follows that ∇f = _ 2xy +z 2 _ i + _ 2yz +x 2 _ j + _ 2xz +y 2 _ k 32. f(x +h) −f(x) = _ (2xy + 2h 2 x +h 1 y) i + 2x 2 j + 1 z(z +h 3 ) k _ · (h 1 i +h 2 j +h 3 k) +h 2 1 ; g(h) = h 2 1 h 2 is o(h) and ∇f = 4xy i = 2x 2 j + 1 z 2 k. 33. ∇f = F(x, y) = 2xy i + _ 1 +x 2 _ j ⇒ ∂f ∂x = 2xy ⇒ f(x, y) = x 2 y +g(y) for some function g. Now, ∂f ∂y = x 2 +g (y) = 1 +x 2 ⇒ g (y) = 1 ⇒ g(y) = y +C, C a constant. Thus, f(x, y) = x 2 y +y +C 34. ∇f = (2xy +x)i + (x 2 +y)j =⇒ f x = 2xy +x =⇒ f(x, y) = x 2 y + 1 2 x 2 +g(y) Now, f y = x 2 +g (y) = x 2 +y =⇒ g (y) = y =⇒ g(y) = 1 2 y 2 +C Thus, f(x, y) = x 2 y + 1 2 x 2 + 1 2 y 2 +C 35. ∇f = F(x, y) = (x + sin y) i + (xcos y −2y) j ⇒ ∂f ∂x = x + sin y ⇒ f(x, y) = 1 2 x 2 +x sin y +g(y) for some function g. Now, ∂f ∂y = x cos y +g (y) = x cos y −2y ⇒ g (y) = −2y ⇒ g(y) = −y 2 +C, C a constant. Thus, f(x, y) = 1 2 x 2 +x sin y −y 2 +C. 36. ∇f = yzi + (xz + 2yz)j + (xy +y 2 )k =⇒ f x = yz =⇒ f(x, y, z) = xyz +g(y, z). f y = xz +g y = xz + 2yz =⇒ g y = 2yz =⇒ g(y, z) = y 2 z +h(z) =⇒ f(x, y, z) = xyz +y 2 z +h(z). f x = xy +y 2 +h (z) = xy +y 2 =⇒ h (z) = 0 =⇒ h(z) = C. Thus, f(x, y, z) = xyz +y 2 z +C. 37. With r = (x 2 +y 2 +z 2 ) 1/2 we have ∂r ∂x = x r , ∂r ∂y = y r , ∂r ∂z = z r . (a) ∇(ln r) = ∂ ∂x (ln r) i + ∂ ∂y (ln r) j + ∂ ∂z (ln r)k = 1 r ∂r ∂x i + 1 r ∂r ∂y j + 1 r ∂r ∂z k = x r 2 i + y r 2 j + z r 2 k = r r 2 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.1 791 (b) ∇(sin r) = ∂ ∂x (sin r) i + ∂ ∂y (sin r) j + ∂ ∂z (sin r)k = cos r ∂r ∂x i + cos r ∂r ∂y j + cos r ∂r ∂z k = (cos r) x r i + (cos r) y r j + (cos r) z r k = _ cos r r _ r (c) ∇e r = _ e r r _ r [ same method as in (a) and (b) ] 38. With r n = (x 2 +y 2 +z 2 ) n/2 we have ∂r n ∂x = n 2 (x 2 +y 2 +z 2 ) (n/2)−1 (2x) = n(x 2 +y 2 +z 2 ) (n−2)/2 x = nr n−2 x. Similarly ∂r n ∂y = nr n−2 y and ∂r n ∂z = nr n−2 z. Therefore ∇r n = nr n−2 xi +nr n−2 yj +nr n−2 zk = nr n−2 (xi +yj +zk) = nr n−2 r 39. (a) ∇f = 2xi + 2y j = 0 =⇒ x = y = 0; ∇f = 0 at (0, 0). (b) (c) f has an absolute minimum at (0, 0) 40. (a) ∇f = −1 _ 4 −x 2 −y 2 (xi +yj) = 0 at (0, 0) (b) (c) f has a maximum at (0, 0) P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 792 SECTION 16.2 41. (a) Let c = c 1 i +c 2 j +c 3 k. First, we take h = hi. Since c · h is o(h), 0 = lim h→0 c · h h = lim h→0 c 1 h h = c 1 . Similarly, c 2 = 0 and c 3 = 0. (b) (y −z) · h = [f(x +h) −f(x) −z · h] + [y · h −f(x +h) +f(x) ] = o(h) +o(h) = o(h), so that, by part (a), y −z = 0. 42. lim h→0 g(h) = lim h→0 _ h g(h) h _ = _ lim h→0 h __ lim h→0 g(h) h _ = (0)(0) = (0). 43. (a) In Section 15.6 we showed that f was not continuous at (0, 0). It is therefore not differentiable at (0, 0). (b) For (x, y) = (0, 0), ∂f ∂x = 2y(y 2 −x 2 ) (x 2 +y 2 ) 2 . As (x, y) tends to (0, 0) along the positive y-axis, ∂f ∂x = 2y 3 y 4 = 2 y tends to ∞. SECTION 16.2 1. ∇f = 2xi + 6yj, ∇f(1, 1) = 2i + 6j, u = 1 2 √ 2 (i −j), f u (1, 1) = ∇f(1, 1) · u = −2 √ 2 2. ∇f = [1 + cos(x +y)]i + cos(x +y)j, ∇f(0, 0) = 2i +j, u = 1 √ 5 (2i +j), f u (0, 0) = ∇f(0, 0) · u = √ 5 3. ∇f = (e y −ye x ) i + (xe y −e x ) j, ∇f(1, 0) = i + (1 −e)j, u = 1 5 (3i + 4j), f u (1, 0) = ∇f(1, 0) · u = 1 5 (7 −4e) 4. ∇f = 1 (x −y) 2 (−2yi + 2xj), ∇f(1, 0) = 2j, u = 1 2 (i − √ 3j), f u (1, 0) = ∇f(1, 0) · u = − √ 3 5. ∇f = (a −b)y (x +y) 2 i + (b −a)x (x +y) 2 j, ∇f(1, 1) = a −b 4 (i −j), u = 1 2 √ 2 (i −j), f u (1, 1) = ∇f(1, 1) · u = 1 4 √ 2 (a −b) 6. ∇f = 1 (cx +dy) 2 [(d −c)yi + (c −d)xj] , ∇f(1, 1) = d −c (c +d) 2 (i −j), u = 1 √ c 2 +d 2 (ci −dj), f u (1, 1) = ∇f(1, 1) · u = d −c (c +d) √ c 2 +d 2 7. ∇f = 2x x 2 +y 2 i + 2y x 2 +y 2 j, ∇f(0, 1) = 2 j, u = 1 √ 65 (8 i +j), f u (0, 1) = ∇f(0, 1) · u = 2 √ 65 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.2 793 8. ∇f = 2xyi + (x 2 + sec 2 y)j, ∇f(−1, π 4 ) = − π 2 i + 3j, u = 1 √ 5 (i −2j) f u _ −1, π 4 _ = ∇f _ −1, π 4 _ · u = − 1 √ 5 _ π 2 + 6 _ 9. ∇f = (y +z)i + (x +z)j + (y +x)k, ∇f(1, −1, 1) = 2j, u = 1 6 √ 6 (i + 2j +k), f u (1, −1, 1) = ∇f(1, −1, 1) · u = 2 3 √ 6 10. ∇f = (z 2 + 2xy)i + (x 2 + 2yz)j + (y 2 + 2zx)k, ∇f(1, 0, 1) = i +j + 2k, u = 1 √ 10 (3j −k) f u (1, 0, 1) = ∇f(1, 0, 1) · u = √ 10 10 11. ∇f = 2 _ x +y 2 +z 3 _ _ i + 2yj + 3z 2 k _ , ∇f(1, −1, 1) = 6(i −2j + 3k), u = 1 2 √ 2 (i +j), f u (1, −1, 1) = ∇f(1, −1, 1) · u = −3 √ 2 12. ∇f = (2Ax +Byz)i + (Bxz + 2Cy)j +Bxyk, ∇f(1, 2, 1) = 2(A+B)i + (B + 4C)j + 2Bk u = 1 √ A 2 +B 2 +C 2 (Ai +Bj +Ck); f u (1, 2, 1) = ∇f(1, 2, 1) · u = 2A 2 +B 2 + 2AB + 6BC √ A 2 +B 2 +C 2 13. ∇f = tan −1 (y +z) i + x 1 + (y +z) 2 j + x 1 + (y +z) 2 k, ∇f(1, 0, 1) = π 4 i + 1 2 j + 1 2 k, u = 1 √ 3 (i +j −k), f u (1, 0, 1) = ∇f(1, 0, 1) · u = π 4 √ 3 = √ 3 12 π 14. ∇f = (y 2 cos z −2πyz 2 cos πx + 6zx)i + (2xy cos z −2z 2 sin πx)j + (−xy 2 sin z −4yz sin πx + 3x 2 )k ∇f(0, −1, π) = (2π 3 −1)i; u = 1 3 (2i −j + 2k), f u (0, −1, π) = ∇f(0, −1, π) · u = 2 3 (2π 3 −1). 15. ∇f = x x 2 +y 2 i + y x 2 +y 2 j, u = 1 _ x 2 +y 2 (−xi −yj) , f u (x, y) = ∇f · u = − 1 _ x 2 +y 2 16. ∇f = e xy _ (y 2 +xy 3 −y 3 )i + (x −1)(2y +xy 2 )j ¸ , ∇f(0, 1) = −2j u = 1 √ 5 (−i + 2j), f u (0, 1) = ∇f(0, 1) · u = − 4 5 √ 5 17. ∇f = (2Ax + 2By) i + (2Bx + 2Cy) j, ∇f(a, b) = (2aA+ 2bB)i + (2aB + 2bC) j (a) u = 1 2 √ 2 (−i +j), f u (a, b) = ∇f(a, b) · u = √ 2 [a(B −A) +b(C −B)] (b) u = 1 2 √ 2 (i −j), f u (a, b) = ∇f(a, b) · u = √ 2 [a(A−B) +b(B −C)] 18. ∇f = z x i − z y j + ln _ x y _ k, ∇f(1, 1, 2) = 2i −2j u = 1 √ 3 (i +j −k); f u (1, 1, 2) = ∇f(1, 1, 2) · u = 0 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 794 SECTION 16.2 19. ∇f = e y 2 −z 2 (i + 2xyj −2xzk), ∇f(1, 2, −2) = i + 4j + 4k, r (t) = i −2 sin (t −1) j −2e t−1 k, at (1, 2, −2) t = 1, r (1) = i −2k, u = 1 5 √ 5 (i −2k), f u (1, 2, −2) = ∇f(1, 2, −2) · u = − 7 5 √ 5 20. ∇f = 2xi +zj +yk, ∇f(1, −3, 2) = 2i + 2j −3k Direction: r (−1) = −2i + 3j −3k, u = 1 √ 22 (−2i + 3j −3k), f u (1, −3, 2) = ∇f(1, −3, 2) · u = 1 2 √ 22 21. ∇f = (2x + 2yz) i + _ 2xz −z 2 _ j + (2xy −2yz) k, ∇f(1, 1, 2) = 6 i −2 k The vectors v = ±(2 i +j −3 k) are direction vectors for the given line; u = ± _ 1 √ 14 [2 i +j −3 k] _ are corresponding unit vectors; f u (1, 1, 2) = ∇f(1, 1, 2) · (±u) = ± 18 √ 14 22. ∇f = e x (cos πyzi −πz sin πyzj −πy sin πyzk), ∇f(0, 1, 1 2 ) = − π 2 j −π k The vectors v = ±(2 i + 3 j + 5 k) are direction vectors for the line; u = ± _ 1 √ 38 [2 i + 3 j + 5 k] _ are corresponding unit vectors; f u (0, 1, 1 2 ) = ∇f(0, 1, 1 2 ) · (±u) = ∓ 13π 2 √ 38 23. ∇f = 2y 2 e 2x i + 2ye 2x j, ∇f(0, 1) = 2 i + 2 j, ∇f = 2 √ 2, ∇f ∇f = 1 √ 2 (i +j) f increases most rapidly in the direction u = 1 √ 2 (i +j); the rate of change is 2 √ 2. f decreases most rapidly in the direction v = − 1 √ 2 (i +j); the rate of change is −2 √ 2. 24. ∇f = [1 + cos(x + 2y)]i + 2 cos(x + 2y)j, ∇f(0, 0) = 2i + 2j Fastest increase in direction u = 1 √ 2 (i +j), rate of change ∇f(0, 0) = 2 √ 2 Fastest decrease in direction v = − 1 √ 2 (i +j), rate of change −2 √ 2 25. ∇f = x _ x 2 +y 2 +z 2 i + y _ x 2 +y 2 +z 2 j + z _ x 2 +y 2 +z 2 k, ∇f(1, −2, 1) = 1 √ 6 (i −2 j +k), ∇f = 1 f increases most rapidly in the direction u = 1 √ 6 (i −2 j +k); the rate of change is 1. f decreases most rapidly in the direction v = − 1 √ 6 (i −2 j +k); the rate of change is −1. 26. ∇f = (2xze y +z 2 )i +x 2 ze y j + (x 2 e y + 2xz)k, ∇f(1, ln 2, 2) = 12i + 4j + 6k Fastest increase in direction u = 1 7 (6i + 2j + 3k), rate of change ∇f(1, ln 2, 2) = 14 Fastest decrease in direction v = − 1 7 (6i + 2j + 3k), rate of change −14 27. ∇f = f (x 0 ) i. If f (x 0 ) = 0, the gradient points in the direction in which f increases: to the right if f (x 0 ) > 0, to the left if f (x 0 ) < 0. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.2 795 28. 0; the vector c = ∂f ∂y (x 0 , y 0 )i − ∂f ∂x (x 0 , y 0 )j is perpendicular to the gradient ∇f(x 0 , y 0 ) and points along the level curve of f at (x 0 , y 0 ). 29. (a) lim h→0 f(h, 0) −f(0, 0) h = lim h→0 √ h 2 h = lim h→0 |h| h does not exist (b) no; by Theorem 16.2.5 f cannot be differentiable at (0, 0) 30. (a) g(x +h)o(h) h = g(x +h) o(h) h →g(x)(0) = 0 (b) |[g(x +h) −g(x)]∇f(x) · h| h ≤ [g(x +h) −g(x)]∇f(x)h h by Schwarz’s inequality = |g(x +h) −g(x)| · ∇f(x) →0 31. ∇λ(x, y) = − 8 3 xi −6yj (a) ∇λ(1, −1) = − 8 3 i = 6j, u = −∇λ(1, −1) ∇λ(1, −1) = 8 3 i −6j 2 3 √ 97 , λ u (1, −1) = ∇λ(1, −1) · u = − 2 3 √ 97 (b) u = i, λ u (1, 2) = ∇λ(1, 2) · u = _ − 8 3 i −12j _ · i = − 8 3 (c) u = 1 2 √ 2 (i +j), λ u (2, 2) = ∇λ(2, 2) · u = _ − 16 3 i −12 j _ · _ 1 2 √ 2 (i +j) ¸ = − 26 3 √ 2 32. ∇I = −4xi −2yj. We want the curve r(t) = x(t)i +y(t)j which begins at (−2, 1) and has tangent vector r (t) in the direction ∇I. We can satisfy these conditions by setting x (t) = −4x(t), x(0) = −2; y (t) = −2y(t), y(0) = 1. These equations imply that x(t) = −2e −4t , y(t) = e −2t . Eliminating the parameter, we get x = −2y 2 ; the particle will follow the parabolic path x = −2y 2 toward the origin. 33. (a) The projection of the path onto the xy-plane is the curve C: r(t) = x(t)i +y(t)j which begins at (1, 1) and at each point has its tangent vector in the direction of −∇f. Since ∇f = 2xi + 6yj, we have the initial-value problems x (t) = −2x(t), x(0) = 1 and y (t) = −6y(t), y(0) = 1. From Theorem 7.6.1 we find that x(t) = e −2t and y(t) = e −6t . Eliminating the parameter t, we find that C is the curve y = x 3 from (1, 1) to (0, 0). P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 796 SECTION 16.2 (b) Here x (t) = −2x(t), x(0) = 1 and y (t) = −6y(t), y(0) = −2 so that x(t) = e −2t and y(t) = −2e −6t . Eliminating the parameter t, we find that the projection of the path onto the xy-plane is the curve y = −2x 3 from (1, −2) to (0, 0). 34. z = f(x, y) = 1 2 x 2 −y 2 ; ∇f = xi −2yj, so we choose the projection r(t) = x(t)i +y(t)j of the path onto the xy-plane such that x (t) = x(t), y (t) = −2y(t) (a) With initial point (−1, 1, − 1 2 ), we get x(t) = −e t , y(t) = e −2t , or y = 1 x 2 from (−1, 1), in the direction of decreasing x. (b) With initial point (1, 0, 1 2 ), we get x(t) = e t , y(t) = 0, or the x-axis from (1, 0), in the direction of increasing x. 35. The projection of the path onto the xy-plane is the curve C: r(t) = x(t)i +y(t)j which begins at (a, b) and at each point has its tangent vector in the direction of −∇f = − _ 2a 2 xi + 2b 2 yj _ . We can satisfy these conditions by setting x (t) = −2a 2 x(t), x(0) = a 2 and y (t) = −2b 2 y(t), y(0) = b so that x(t) = ae −2a 2 t and y(t) = be −2b 2 t . Since _ x a _ b 2 = _ e −2a 2 t _ b 2 = _ y b _ a 2 , C is the curve (b) a 2 x b 2 = (a) b 2 y a 2 from (a, b) to (0, 0). 36. The particle must go in he direction −∇T = −e y cos xi −e y sin xj, so we set x (t) = −e y(t) cos x(t), y (t) = −e y(t) sin x(t). Dividing, we have y (t) x (t) = sin x(t) cos x(t) , or dy dx = tan x. With initial point (0, 0), we get y = ln | sec x|, in the direction of decreasing x (since x (0) < 0). 37. We want the curve C: r(t) = x(t)i +y(t)j which begins at (π/4, 0) and at each point has its tangent vector in the direction of ∇T = − √ 2 e −y sin xi − √ 2 e −y cos xj. From x (t) = − √ 2 e −y sin x and y (t) = − √ 2 e −y cos x P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.2 797 we obtain dy dx = y (t) x (t) = cot x so that y = ln | sin x| +C. Since y = 0 when x = π/4, we get C = ln √ 2 and y = ln | √ 2 sin x|. As ∇T(π/4, 0) = −i −j, the curve y = ln | √ 2 sin x| is followed in the direction of decreasing x. 38. ∇z = (1 −2x)i + (2 −6y)j, so the projection of the path onto the xy-plane satisfies x (t) = 1 −2x(t), y (t) = 2 −6y(t), or dy dx = 2 −6y 1 −2x . With initial point (0, 0), this gives the curve 3y = (2x −1) 3 + 1, in the direction of increasing x. 39. (a) lim h→0 f _ 2 +h, (2 +h) 2 _ −f(2, 4) h = lim h→0 3(2 +h) 2 + (2 +h) 2 −16 h = lim h→0 4 _ 4h +h 2 h _ = lim h→0 4(4 +h) = 16 (b) lim h→0 f _ h + 8 4 , 4 +h _ −f(2, 4) h = lim h→0 3 _ h + 8 4 _ 2 + (4 +h) −16 h = lim h→0 3 16 h 2 + 3h + 12 + 4 +h −16 h = lim h→0 _ 3 16 h + 4 _ = 4 (c) u = 1 17 √ 17 (i + 4j), ∇f(2, 4) = 12i +j; f u (2, 4) = ∇f(2, 4) · u = 16 17 √ 17 (d) The limits computed in (a) and (b) are not directional derivatives. In (a) and (b) we have, in essence, computed ∇f(2, 4) · r 0 taking r 0 = i + 4j in (a) and r 0 = 1 4 i +j in (b). In neither case is r 0 a unit vector. 40. ∇f = −GMm (x 2 +y 2 +z 2 ) 3/2 (xi +yj +zk) = −GMm r 3 r 41. (a) u = cos θ i + sin θ j, ∇f(x, y) = ∂f ∂x i + ∂f ∂y j; f u (x, y) = ∇f · u = _ ∂f ∂x i + ∂f ∂y j _ · (cos θ i + sin θ j) = ∂f ∂x cos θ + ∂f ∂y sin θ (b) ∇f = _ 3x 2 + 2y −y 2 _ i + (2x −2xy) j, ∇f(−1, 2) = 3 i + 2 j f u (−1, 2) = 3 cos(2π/3) + 2 sin(2π/3) = 2 √ 3 −3 2 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 798 SECTION 16.3 42. f u (x, y) = ∂f ∂x cos 5π 4 + ∂f ∂y sin 5π 4 = 2xe 2y _ − √ 2 2 _ + 2x 2 e 2y _ − √ 2 2 _ = − √ 2xe 2y (1 +x) f u (2, ln 2) = − √ 2 · 2 · e 2ln2 (1 + 2) = −24 √ 2 43. ∇(fg) = ∂(fg) ∂x i + ∂(fg) ∂y j + ∂(fg) ∂z k = _ f ∂g ∂x +g ∂f ∂x _ i + _ f ∂g ∂y +g ∂f ∂y _ j + _ f ∂g ∂z +g ∂f ∂z _ k = f _ ∂g ∂x i + ∂g ∂y j + ∂g ∂z k _ +g _ ∂f ∂x i + ∂f ∂y j + ∂f ∂z k _ = f ∇g +g ∇f 44. ∇ _ f g _ = ∂ ∂x _ f g _ i + ∂ ∂y _ f g _ j + ∂ ∂z _ f g _ k = ∂f ∂x g −f ∂g ∂x g 2 i + ∂f ∂y g −f ∂g ∂y g 2 j + ∂f ∂z g −f ∂g ∂z g 2 k = g(x)∇f(x) −f(x)∇g(x) g 2 (x) 45. ∇f n = ∂f n ∂x i + ∂f n ∂y j + ∂f n ∂z k = nf n−1 ∂f ∂x i +nf n−1 ∂f ∂y j +nf n−1 ∂f ∂z k = nf n−1 ∇f SECTION 16.3 1. f(b) = f(1, 3) = −2; f(a) = f(0, 1) = 0; f(b) −f(a) = −2 ∇f = _ 3x 2 −y _ i −xj; b −a = i + 2 j and ∇f · (b −a) = 3x 2 −y −2x The line segment joining a and b is parametrized by x = t, y = 1 + 2t, 0 ≤ t ≤ 1 Thus, we need to solve the equation 3t 2 −(1 + 2t) −2t = −2, which is the same as 3t 2 −4t + 1 = 0, 0 ≤ t ≤ 1 The solutions are: t = 1 3 , t = 1. Thus, c = ( 1 3 , 5 3 ) satisfies the equation. Note that the endpoint b also satisfies the equation. 2. ∇f = 4zi −2yj + (4x + 2z)k, f(a) = f(0, 1, 1) = 0, f(b) = f(1, 3, 2) = 3 b −a = i + 2j +k, so we want (x, y, z) such that ∇f · (b −a) = 4z −4y + 4x + 2z = 6z −4y + 4x = f(b) −f(a) = 3 Parametrizing the line segment from a to b by x(t) = t, y(t) = 1 + 2t, z(t) = 1 +t, we get t = 1 2 , or c = ( 1 2 , 2, 3 2 ) 3. (a) f(x, y, z) = a 1 x +a 2 y +a 3 z +C (b) f(x, y, z) = g(x, y, z) +a 1 x +a 2 y +a 3 z +C 4. Using the mean-value theorem 16.3.1, there exists c such that ∇f(c) · (b −a) = 0 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.3 799 5. (a) U is not connected (b) (i) g(x) = f(x) −1 (ii) g(x) = −f(x) 6. By the mean-value theorem f(x 1 ) −f(x 2 ) = ∇f(c) · (x 1 −x 2 ) for some point c on the line segment x 1 x 2 . Since Ω is convex, c is in Ω. Thus |f(x 1 ) −f(x 2 )| = |∇f(c) · (x 1 −x 2 )| ≤ ∇f(c)x 1 −x 2 ≤ Mx 1 −x 2 . by Schwarz’s inequality ∧ 7. ∇f = 2xyi +x 2 j; ∇f(r(t)) · r (t) = _ 2i +e 2t j _ · (e t i −e −t j) = e t 8. ∇f = i −j; ∇f(r(t)) · r (t) = (i −j) · (ai −ab sin atj) = a(1 +b sin at) 9. ∇f = −2x 1 + (y 2 −x 2 ) 2 i + 2y 1 + (y 2 −x 2 ) 2 j, ∇f(r(t)) = −2 sin t 1 + cos 2 2t i + 2 cos t 1 + cos 2 2t j ∇f(r(t)) · r (t) = _ −2 sin t 1 + cos 2 2t i + 2 cos t 1 + cos 2 2t j _ · (cos t i −sin t j) = −4 sin t cos t 1 + cos 2 2t = −2 sin 2t 1 + cos 2 2t 10. ∇f = 1 2x 2 +y 3 (4xi + 3y 2 j) ∇f(r(t)) · r (t) = 1 2e 4t +t (4e 2t i + 3t 2/3 j) · (2e 2t i + 1 3 t −2/3 j) = 8e 4t + 1 2e 4t +t 11. ∇f = (e y −ye −x ) i + (xe y +e −x ) j; ∇f(r(t)) = (t t −ln t) i + _ t t ln t + 1 t _ j ∇f(r(t)) · r (t) = _ (t t −ln t) i + _ t t ln t + 1 t _ j _ · _ 1 t i + [1 + ln t] j _ = t t _ 1 t + ln t + [ln t] 2 _ + 1 t 12. ∇f = 2 x 2 +y 2 +z 2 (xi +yj +zk) ∇f(r(t)) · r (t) = 2 1 +e 4t (sin ti + cos tj +e 2t k) · (cos ti −sin tj + 2e 2t k) = 4e 4t 1 +e 4t 13. ∇f = yi + (x −z)j −yk; ∇f(r(t)) · r (t) = _ t 2 i + _ t −t 3 _ j −t 2 k _ · _ i + 2tj + 3t 2 k _ = 3t 2 −5t 4 14. ∇f = 2xi + 2y j ∇f(r(t)) · r (t) = (2a cos ωti + 2b sin ωtj) · (−ωa sin ωti +ωb cos ωtj +bωk) = 2ω(b 2 −a 2 ) sin ωt cos ωt 15. ∇f = 2xi + 2yj +k; ∇f(r(t)) · r (t) = (2a cos ωt i + 2b sin ωt j +k) · (−aω sin ωt i +bω cos ωt j +bωk) = 2ω _ b 2 −a 2 _ sin ωt cos ωt +bω P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 800 SECTION 16.3 16. ∇f = y 2 cos(x +z)i + 2y sin(x +z)j +y 2 cos(x +z)k ∇f(r(t)) · r (t) = [cos 2 t cos(2t +t 3 )i + 2 cos t sin(2t +t 3 )j + cos 2 t cos(2t +t 3 )k] · (2i −sin tj + 3t 2 k) = cos t[(2 + 3t 2 ) cos t cos(2t +t 3 ) −2 sin t sin(2t +t 3 )] 17. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt = (2x −3y)(−sin t) + (4y −3x)(cos t) = 2 cos t sin t + 3 sin 2 t −3 cos 2 t = sin 2t −3 cos 2t 18. du dt = ∂u ∂x · dx dt + ∂u ∂y · dy dt = _ 1 + 2 _ y x _ 3t 2 + _ 2 _ x y −3 __ − 1 t 2 _ = _ 1 + 2 t 2 _ 3t 2 + (2t 2 −3) _ − 1 t 2 _ = 3t 2 + 4 + 3 t 2 19. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt = (e x sin y +e y cos x) _ 1 2 _ + (e x cos y +e y sin x) (2) = e t/2 _ 1 2 sin 2t + 2 cos 2t _ +e 2t _ 1 2 cos 1 2 t + 2 sin 1 2 t _ 20. du dt = ∂u ∂x · dx dt + ∂u ∂y · dy dt = (4x −y)(−2 sin 2t) + (2y −x) cos t = 2 sin 2t(sin t −4 cos 2t) + cos t(2 sin t −cos 2t) 21. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt = (e x sin y) (2t) + (e x cos y) (π) = e t 2 [2t sin(πt) +π cos(πt)] 22. du dt = ∂u ∂x · dx dt + ∂u ∂y · dy dt + ∂u ∂z · dz dt = − z x 2t + z y 1 2 √ t + ln _ y x _ e t (1 +t) = − 2t 2 e t t 2 + 1 + e t 2 + ln _ √ t t 2 + 1 _ e t (1 +t) 23. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt + ∂u ∂z dz dt = (y +z)(2t) + (x +z)(1 −2t) + (y +x)(2t −2) = (1 −t)(2t) + (2t 2 −2t + 1)(1 −2t) +t(2t −2) = 1 −4t + 6t 2 −4t 3 24. du dt = ∂u ∂x · dx dt + ∂u ∂y · dy dt + ∂u ∂z · dz dt = (sin πy +πz sin πx)2t +πxcos πy(−1) −cos πx(−2t) = 2t _ sin[π(1 −t)] +π(1 −t 2 ) sin(πt 2 ) ¸ −πt 2 cos[π(1 −t)] + 2t cos(πt 2 ) 25. V = 1 3 πr 2 h, dV dt = ∂V ∂r dr dt + ∂V ∂h dh dt = _ 2 3 πrh _ dr dt + _ 1 3 πr 2 _ dh dt . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.3 801 At the given instant, dV dt = 2 3 π(280)(3) + 1 3 π(196)(−2) = 1288 3 π. The volume is increasing at the rate of 1288 3 π in. 3 / sec . 26. v = πr 2 h, dv dt = ∂v ∂r · dr dt + ∂v ∂h · dh dt = 2πrh dr dt +πr 2 dh dt dr dt = −2, dh dt = 3, r = 13, h = 18 =⇒ dv dt = −429π : decreasing at the rate of 429π cm 3 /sec. 27. A = 1 2 xy sin θ; dA dt = ∂A ∂x dx dt + ∂A ∂y dy dt + ∂A ∂θ dθ dt = 1 2 _ (y sin θ) dx dt + (xsin θ) dy dt + (xy cos θ) dθ dt _ . At the given instant dA dt = 1 2 [(2 sin 1) (0.25) + (1.5 sin 1) (0.25) −(2(1.5) cos 1) (0.1)] ∼ = 0.2871 ft 2 /s ∼ = 41.34 in 2 /s 28. dz dt = 2x dx dt + y 2 dy dt . But x 2 +y 2 = 13 =⇒ 2x dx dt + 2y dy dt = 0 =⇒ dy dt = − x y dx dt =⇒ dz dt = 2x dx dt + y 2 _ − x y dx dt _ = 3x 2 dx dt = 15. z is increasing 15 centimeters per second 29. ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s = (2x −y)(cos t) + (−x)(t cos s) = 2s cos 2 t −t sin s cos t −st cos s cos t ∂u ∂t = ∂u ∂x ∂x ∂t + ∂u ∂y ∂y ∂t = (2x −y)(−s sin t) + (−x)(sin s) = −2s 2 cos t sin t +st sin s sin t −s cos t sin s 30. ∂u ∂s = ∂u ∂x · ∂x ∂s + ∂u ∂y · ∂y ∂s = [cos(x −y) −sin(x +y)]t + [−cos(x −y) −sin(x +y)]2s = (t −2s) cos(st −s 2 +t 2 ) −(t + 2s) sin(st +s 2 −t 2 ) ∂u ∂t = ∂u ∂x · ∂x ∂t + ∂u ∂y · ∂y ∂t = [cos(x −y) −sin(x +y)]s + [−cos(x −y) −sin(x +y)](−2t) = (s + 2t) cos(st −s 2 +t 2 ) −(s −2t) sin(st +s 2 −t 2 ) 31. ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s = (2xtan y)(2st) + _ x 2 sec 2 y _ (1) = 4s 3 t 2 tan _ s +t 2 _ +s 4 t 2 sec 2 _ s +t 2 _ ∂u ∂t = ∂u ∂x ∂x ∂t + ∂u ∂y ∂y ∂t = (2xtan y) _ s 2 _ + _ x 2 sec 2 y _ (2t) = 2s 4 t tan _ s +t 2 _ + 2s 4 t 3 sec 2 _ s +t 2 _ P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 802 SECTION 16.3 32. ∂u ∂s = ∂u ∂x · ∂x ∂s + ∂u ∂y · ∂y ∂s + ∂u ∂z · ∂z ∂s = z 2 y sec xy tan xy(2t) +z 2 xsec xy tan xy + 2z sec xy(2st) = sec[2st(s −t 2 )] _ 2s 4 t 3 (s −t 2 ) tan[2st(s −t 2 )] + 2s 3 t 2 tan[2st(s −t 2 )] + 4s 3 t 2 _ ∂u ∂t = z 2 y sec xy tan xy(2s) +z 2 xsec xy tan xy(−2t) + 2z sec xy(s 2 ) = sec[2st(s −t 2 )] _ 2s 5 t 2 (s −t 2 ) tan[2st(s −t 2 )] −4s 5 t 4 tan[2st(s −t 2 )] + 2s 4 t _ 33. ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s + ∂u ∂z ∂z ∂s = (2x −y)(cos t) + (−x)(−cos (t −s)) + 2z(t cos s) = 2s cos 2 t −sin (t −s) cos t +s cos t cos (t −s) + 2t 2 sin s cos s ∂u ∂t = ∂u ∂x ∂x ∂t + ∂u ∂y ∂y ∂t + ∂u ∂z ∂z ∂t = (2x −y)(−s sin t) + (−x)(cos (t −s)) + 2z(sin s) = −2s 2 cos t sin t +s sin (t −s) sin t −s cos t cos (t −s) + 2t sin 2 s 34. ∂u ∂s = ∂u ∂x · ∂x ∂s + ∂u ∂y · ∂y ∂s + ∂u ∂z · ∂z ∂s = e yz 2 1 s +xz 2 e yz 2 · 0 + 2xyze yz 2 2s = 1 s e t 3 (s 2 +t 2 ) 2 + 4st 3 (s 2 +t 2 ) ln(st)e t 3 (s 2 +t 2 ) 2 ∂u ∂t = ∂u ∂x · ∂x ∂t + ∂u ∂y · ∂y ∂t + ∂u ∂z · ∂z ∂t = e yz 2 1 t +xz 2 e yz 2 3t 2 + 2xyze yz 2 2t = 1 t e t 3 (s 2 +t 2 ) 2 +t 2 (s 2 +t 2 )(3s 2 + 7t 2 ) ln(st)e t 3 (s 2 +t 2 ) 2 35. d dt [f(r(t) ) ] = _ ∇f(r(t) ) · r (t) r (t) _ r (t) = f u(t) (r(t)) r (t) where u(t) = r (t) r (t) 36. ∂ ∂x [f(r)] = d dr [f(r)] ∂r ∂x = f (r) ∂r ∂x = f (r) x r ; similarly ∂ ∂y [f(r)] = f (r) y r and ∂ ∂z [f(r)] = f (r) z r . Therefore ∇f(r) = f (r) x r i +f (r) y r j +f (r) z r k = f (r) r r . 37. (a) (cos r) r r (b) (r cos r + sin r) r r 38. (a) ∇(r ln r) = (1 + ln r) r r (b) ∇(e 1−r 2 ) = −2re 1−r 2 r r = −2e 1−r 2 r P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.3 803 39. (a) (r cos r − sin r) r r 3 (b) _ sin r −r cos r sin 2 r _ r r 40. (a) (b) du dt = ∂u ∂x dx ds ds dt + ∂u ∂y dy ds ds dt 41. (a) (b) ∂u ∂r = ∂u ∂x _ ∂x ∂w ∂w ∂r + ∂x ∂t ∂t ∂r _ + ∂u ∂y _ ∂y ∂w ∂w ∂r + ∂y ∂t ∂t ∂r _ + ∂u ∂z _ ∂z ∂w ∂w ∂r + ∂z ∂t ∂t ∂r _ . To obtain ∂u/∂s, replace each r by s. 42. (a) (b) ∂u ∂r = ∂u ∂x ∂x ∂r + ∂u ∂z ∂z ∂r + ∂u ∂w ∂w ∂r , ∂u ∂v = ∂u ∂y ∂y ∂v + ∂u ∂w ∂w ∂v P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 804 SECTION 16.3 43. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt d 2 u dt 2 = ∂u ∂x d 2 x dt 2 + dx dt _ ∂ 2 u ∂x 2 dx dt + ∂ 2 u ∂y∂x dy dt _ + ∂u ∂y d 2 y dt 2 + dy dt _ ∂ 2 u ∂x∂y dx dt + ∂ 2 u ∂y 2 dy dt _ and the result follows. 44. ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s ∂ 2 u ∂s 2 = ∂u ∂x ∂ 2 x ∂s 2 + ∂x ∂s _ ∂ 2 u ∂x 2 ∂x ∂s + ∂ 2 u ∂y∂x ∂y ∂s _ + ∂u ∂y ∂ 2 y ∂s 2 + ∂y ∂s _ ∂ 2 u ∂x∂y ∂x ∂s + ∂ 2 u ∂y 2 ∂y ∂s _ = ∂ 2 u ∂x 2 _ ∂x ∂s _ 2 + 2 ∂ 2 u ∂x∂y ∂x ∂s ∂y ∂s + ∂ 2 u ∂y 2 _ ∂y ∂s _ 2 + ∂u ∂x ∂ 2 x ∂s 2 + ∂u ∂y ∂ 2 y ∂s 2 45. (a) ∂u ∂r = ∂u ∂x ∂x ∂r + ∂u ∂y ∂y ∂r = ∂u ∂x cos θ + ∂u ∂y sin θ ∂u ∂θ = ∂u ∂x ∂x ∂θ + ∂u ∂y ∂y ∂θ = ∂u ∂x (−r sin θ) + ∂u ∂y (r cos θ) (b) _ ∂u ∂r _ 2 = _ ∂u ∂x _ 2 cos 2 θ + 2 ∂u ∂x ∂u ∂y cos θ sin θ + _ ∂u ∂y _ 2 sin 2 θ, 1 r 2 _ ∂u ∂θ _ 2 = _ ∂u ∂x _ 2 sin 2 θ −2 ∂u ∂x ∂u ∂y cos θ sin θ + _ ∂u ∂y _ 2 cos 2 θ, _ ∂u ∂r _ 2 + 1 r 2 _ ∂u ∂θ _ 2 = _ ∂u ∂x _ 2 _ cos 2 θ + sin 2 θ _ + _ ∂u ∂y _ 2 _ sin 2 θ + cos 2 θ _ = _ ∂u ∂x _ 2 + _ ∂u ∂y _ 2 46. (a) By Exercise 45 (a) ∂w ∂r = ∂w ∂x cos θ + ∂w ∂y sin θ, ∂w ∂θ = − ∂w ∂x r sin θ + ∂w ∂y r cos θ. Solve these equations simultaneously for ∂w ∂x and ∂w ∂y . (b) To obtain the first pair of equations set w = r; to obtain the second pair of equations set w = θ. (c) θ is not independent of x; r = _ x 2 +y 2 gives ∂r ∂x = x _ x 2 +y 2 = r cos θ r = cos θ 47. Solve the equations in Exercise 45 (a) for ∂u ∂x and ∂u ∂y : ∂u ∂x = ∂u ∂r cos θ − 1 r ∂u ∂θ sin θ, ∂u ∂y = ∂u ∂r sin θ + 1 r ∂u ∂θ cos θ Then ∇u = ∂u ∂x i + ∂u ∂y j = ∂u ∂r (cos θ i + sin θ j) + 1 r ∂u ∂θ (−sin θ i + cos θ j) 48. u(r, θ) = r 2 =⇒ ∇u = 2re r P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.3 805 49. u(x, y) = x 2 −xy +y 2 = r 2 −r 2 cos θ sin θ = r 2 _ 1 − 1 2 sin 2θ _ ∂u ∂r = r(2 −sin 2θ), ∂u ∂θ = −r 2 cos 2θ ∇u = ∂u ∂r e r + 1 r ∂u ∂θ e θ = r(2 −sin 2θ)e r −r cos 2θ e θ 50. ∂u ∂θ = ∂u ∂x ∂x ∂θ + ∂u ∂y ∂y ∂θ = −r sin θ ∂u ∂x +r cos θ ∂u ∂y ∂ 2 u ∂r∂θ = −sin θ ∂u ∂x −r sin θ _ ∂ 2 u ∂x 2 ∂x ∂r + ∂ 2 u ∂y∂x ∂y ∂r _ + cos θ ∂u ∂y +r cos θ _ ∂ 2 u ∂x∂y ∂x ∂r + ∂ 2 u ∂y 2 ∂y ∂r _ = −sin θ ∂u ∂x + cos θ ∂u ∂y +r sin θ cos θ _ ∂ 2 u ∂y 2 − ∂ 2 u ∂x 2 _ +r(cos 2 θ −sin 2 θ) ∂ 2 u ∂x∂y 51. From Exercise 45 (a), ∂ 2 u ∂r 2 = ∂ 2 u ∂x 2 cos 2 θ + 2 ∂ 2 u ∂y ∂x sin θ cos θ + ∂ 2 u ∂y 2 sin 2 θ ∂ 2 u ∂θ 2 = ∂ 2 u ∂x 2 r 2 sin 2 θ −2 ∂ 2 u ∂y ∂x r 2 sin θ cos θ + ∂ 2 u ∂y 2 r 2 cos 2 θ −r _ ∂u ∂x cos θ + ∂u ∂y sin θ _ . The term in parentheses is ∂u ∂r . Now divide the second equation by r 2 and add the two equations. The result follows. 52. u(x, y) = x 2 −2xy +y 4 −4, ∂u ∂x = 2x −2y, ∂u ∂y = −2x + 4y 3 dy dx = − ∂u ∂x ∂u ∂y = 2y −2x 4y 3 −2x = y −x 2y 3 −x 53. Set u = xe y +ye x −2x 2 y. Then ∂u ∂x = e y +ye x −4xy, ∂u ∂y = xe y +e x −2x 2 dy dx = − ∂u/∂x ∂u/∂y = − e y +ye x −4xy xe y +e x −2x 2 . 54. u(x, y) = x 2/3 +y 2/3 , ∂u ∂x = 2 3 x −1/3 , ∂u ∂y = 2 3 y −1/3 =⇒ dy dx = − ∂u ∂x ∂u ∂y = − _ y x _ 1/3 55. Set u = x cos xy +y cos x −2. Then ∂u ∂x = cos xy −xy sin xy −y sin x, ∂u ∂y = −x 2 sin xy + cos x dy dx = − ∂u/∂x ∂u/∂y = cos xy −xy sin xy −y sin x x 2 sin xy −cos x . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 806 SECTION 16.4 56. Set u(x, y, z) = z 4 +x 2 z 3 +y 2 +xy −2. Then ∂u ∂x = 2xz 3 +y, ∂u ∂y = 2y +x, ∂u ∂z = 4z 3 + 3x 2 z 2 ∂z ∂x = − ∂u ∂x ∂u ∂z = − 2xz 3 +y 4z 3 + 3x 2 z 2 , ∂z ∂y = − ∂u ∂y ∂u ∂z = − 2y +x 4z 3 + 3x 2 z 2 57. Set u = cos xyz + ln _ x 2 +y 2 +z 2 _ . Then ∂u ∂x = −yz sin xyz + 2x x 2 +y 2 +z 2 , ∂u ∂y = −xz sin xyz + 2y x 2 +y 2 +z 2 , and ∂u ∂z = −xy sin xyz + 2z x 2 +y 2 +z 2 . ∂z ∂x = − ∂u/∂x ∂u/∂z = − 2x −yz _ x 2 +y 2 +z 2 _ sin xyz 2z −xy (x 2 +y 2 +z 2 ) sin xyz , ∂z ∂y = − ∂u/∂y ∂u/∂z = − 2y −xz _ x 2 +y 2 +z 2 _ sin xyz 2z −xy (x 2 +y 2 +z 2 ) sin xyz . 58. (a) Use du dt = du 1 dt i + du 2 dt j and apply the chain rule to u 1 , u 2 . (b) (i) du dt = t(e x cos yi +e x sin yj) +π(−e x sin yi +e x cos yj) = te t 2 /2 (cos πti + sin πtj) +πe t 2 /2 (−sin πti + cos πtj) (ii) u(t) = e t 2 /2 cos πti +e t 2 /2 sin πtj du dt = (−πe t 2 /2 sin πt +te t 2 /2 cos πt)i + (πe t 2 /2 cos πt +te t 2 /2 sin πt)j 59. ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s , ∂u ∂t = ∂u ∂x ∂x ∂t + ∂u ∂y ∂y ∂t 60. du dt = ∂u ∂x dx dt + ∂u ∂y dy dt + ∂u ∂z dz dt where ∂u ∂x = ∂u 1 ∂x i + ∂u 2 ∂x j + ∂u 3 ∂x k, ∂u ∂y = ∂u 1 ∂y i + ∂u 2 ∂y j + ∂u 3 ∂y k, ∂u ∂z = ∂u 1 ∂z i + ∂u 2 ∂z j + ∂u 3 ∂z k. SECTION 16.4 1. Set f(x, y) = x 2 +xy +y 2 . Then, ∇f = (2x +y)i + (x + 2y)j, ∇f(−1, −1) = −3i −3j. normal vector i +j; tangent vector i −j tangent line x +y + 2 = 0; normal line x −y = 0 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.4 807 2. Set f(x, y) = (y −x) 2 −2x, ∇f = −2(y −x + 1)i + 2(y −x)j, ∇f(2, 4) = −6i + 4j normal vector −3i + 2j; tangent vector 2i + 3j tangent line 3x −2y + 2 = 0; normal line 2x + 3y −16 = 0 3. Set f(x, y) = _ x 2 +y 2 _ 2 −9 _ x 2 −y 2 _ . Then, ∇f = [4x(x 2 +y 2 ) −18x]i + _ 4y _ x 2 +y 2 _ + 18y ¸ j, ∇f _ √ 2, 1 _ = −6 √ 2 i + 30j. normal vector √ 2 i −5 j; tangent vector 5i + √ 2 j tangent line √ 2x −5y + 3 = 0; normal line 5x + √ 2 y −6 √ 2 = 0 4. Set f(x, y) = x 3 +y 3 , ∇f = 3x 2 i + 3y 2 j, ∇f(1, 2) = 3i + 12j normal vector i + 4j; tangent vector 4i −j tangent line x + 4y −9 = 0; normal line 4x −y −2 = 0 5. Set f(x, y) = xy 2 −2x 2 +y + 5x. Then, ∇f = (y 2 −4x + 5) i + (2xy + 1) j, ∇f(4, 2) = −7i + 17j. normal vector 7i −17j; tangent vector 17i + 7j tangent line 7x −17y + 6 = 0; normal line 17x + 7y −82 = 0 6. Set f(x, y) = x 5 +y 5 −2x 3 . ∇f = (5x 4 −6x 2 )i + 5y 4 j, ∇f(1, 1) = −i + 5j normal vector i −5j; tangent vector 5i +j tangent line x −5y + 4 = 0; normal line 5x +y −6 = 0 7. Set f(x, y) = 2x 3 −x 2 y 2 −3x +y. Then, ∇f = (6x 2 −2xy 2 −3) i + (−2x 2 y + 1) j, ∇f(1, −2) = −5i + 5j. normal vector i −j; tangent vector i +j tangent line x −y −3 = 0; normal line x +y + 1 = 0 8. Set f(x, y) = x 3 +y 2 + 2x. ∇f = (3x 2 + 2)i + 2yj, ∇f(−1, 3) = 5i + 6j normal vector 5i + 6j; tangent vector 6i −5j tangent line 5x + 6y −13 = 0; normal line 6x −5y + 21 = 0 9. Set f(x, y) = x 2 y +a 2 y. By (15.4.4) m = − ∂f/∂x ∂f/∂y = − 2xy x 2 +a 2 . At (0, a) the slope is 0. 10. Set f(x, y, z) = (x 2 +y 2 ) 2 −z. ∇f = 4x(x 2 +y 2 )i + 4y(x 2 +y 2 )j −k, ∇f(1, 1, 4) = 8i + 8j −k Tangent plane: 8x + 8y −z −12 = 0 Normal: x = 1 + 8t, y = 1 + 8t, z = 4 −t P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 808 SECTION 16.4 11. Set f(x, y, z) = x 3 +y 3 −3xyz. Then, ∇f = (3x 2 −3yz) i + (3y 2 −3xz) j −3xyk, ∇f _ 1, 2, 3 2 _ = −6i + 15 2 j −6k; tangent plane at _ 1, 2, 3 2 _ : −6(x −1) + 15 2 (y −2) −6 _ z − 3 2 _ = 0, which reduces to 4x −5y + 4z = 0. Normal: x = 1 + 4t, y = 2 −5t, z = 3 2 + 4t 12. Set f(x, y, z) = xy 2 + 2z 2 . ∇f = y 2 i + 2xyj + 4zk, ∇f(1, 2, 2) = 4i + 4j + 8k Tangent plane: x +y + 2z −7 = 0 Normal: x = 1 +t, y = 2 +t, z = 2 + 2t 13. Set z = g(x, y) = axy. Then, ∇g = ayi +axj, ∇g _ 1, 1 a _ = i +aj. tangent plane at _ 1, 1 a , 1 _ : z −1 = 1(x −1) +a _ y − 1 a _ , which reduces to x +ay −z −1 = 0 Normal: x = 1 +t, y = 1 a +at, z = 1 −t 14. Set f(x, y, z) = √ x + √ y + √ z. ∇f = 1 2 √ x i + 1 2 √ y j + 1 2 √ z k, ∇f(1, 4, 1) = 1 2 i + 1 4 j + 1 2 k Tangent plane: 2x +y + 2z −8 = 0 Normal: x = 1 + 2t, y = 4 +t, z = 1 + 2t 15. Set z = g(x, y) = sin x + sin y + sin (x +y). Then, ∇g = [cos x + cos (x +y)] i + [cos y + cos (x +y)] j, ∇g(0, 0) = 2i + 2j; tangent plane at (0, 0, 0): z −0 = 2(x −0) + 2(y −0), 2x + 2y −z = 0. Normal: x = 2t, y = 2t, z = −t 16. Set f(x, y, z) = x 2 +xy +y 2 −6x + 2 −z. ∇f = (2x +y −6)i + (x + 2y)j −k, ∇f(4, −2, −10) = −k Tangent plane: z = −10 Normal: x = 4, y = −2, z = −10 +t 17. Set f(x, y, z) = b 2 c 2 x 2 −a 2 c 2 y 2 −a 2 b 2 z 2 . Then, ∇f (x 0 , y 0 , z 0 ) = 2b 2 c 2 x 0 i −2a 2 c 2 y 0 j −2a 2 b 2 z 0 k; tangent plane at (x 0 , y 0 , z 0 ): 2b 2 c 2 x 0 (x −x 0 ) −2a 2 c 2 y 0 (y −y 0 ) −2a 2 b 2 z 0 (z −z 0 ) = 0, which can be rewritten as follows: b 2 c 2 x 0 x −a 2 c 2 y 0 y −a 2 b 2 z 0 z = b 2 c 2 x 0 2 −a 2 c 2 y 0 2 −a 2 b 2 z 0 2 = f (x 0 , y 0 , z 0 ) = a 2 b 2 c 2 . Normal: x = x 0 + 2b 2 c 2 x 0 t, y = y 0 −2a 2 c 2 y 0 t, z = z 0 −2a 2 b 2 z 0 t P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.4 809 18. Set f(x, y, z) = sin(xcos y) −z. ∇f = cos y cos(xcos y)i −xsin y cos(xcos y)j −k, ∇f(1, π 2 , 0) = j −k Tangent plane: y +z = π 2 Normal: x = 1, y = π 2 +t, z = t 19. Set z = g(x, y) = xy +a 3 x −1 +b 3 y −1 . ∇g = _ y −a 3 x −2 _ i + _ x −b 3 y −2 _ j, ∇g = 0 =⇒ y = a 3 x −2 and x = b 3 y −2 . Thus, y = a 3 b −6 y 4 , y 3 = b 6 a −3 , y = b 2 /a, x = b 3 y −2 = a 2 /b and g _ a 2 /b, b 2 /a _ = 3ab. The tangent plane is horizontal at _ a 2 /b, b 2 /a, 3ab _ . 20. z = g(x, y) = 4x + 2y −x 2 +xy −y 2 . ∇g = (4 −2x +y)i + (2 +x −2y)j ∇g = 0 =⇒ 4 −2x +y = 0, 2 +x −2y = 0 =⇒ x = 10 3 , y = 8 3 The tangent plane is horizontal at ( 10 3 , 8 3 , 28 3 ). 21. Set z = g(x, y) = xy. Then, ∇g = yi +xj. ∇g = 0 =⇒ x = y = 0. The tangent plane is horizontal at (0, 0, 0). 22. z = g(x, y) = x 2 +y 2 −x −y −xy. ∇g = (2x −1 −y)i + (2y −1 −x)j ∇g = 0 =⇒ 2x −1 −y = 0 = 2y −1 −x = 0 =⇒ x = 1, y = 1 The tangent plane is horizontal at (1, 1, −1). 23. Set z = g(x, y) = 2x 2 + 2xy −y 2 −5x + 3y −2. Then, ∇g = (4x + 2y −5) i + (2x −2y + 3) j. ∇g = 0 =⇒ 4x + 2y −5 = 0 = 2x −2y + 3 =⇒ x = 1 3 , y = 11 6 . The tangent plane is horizontal at _ 1 3 , 11 6 , − 1 12 _ . 24. (a) Set f(x, y, z) = xy −z. ∇f = yi +xj −k, ∇f(1, 1, 1) = i +j −k upper unit normal = √ 3 3 (−i −j +k) (b) Set f(x, y, z) = 1 x − 1 y −z. ∇f = − 1 x 2 i + 1 y 2 j −k, ∇f(1, 1, 0) = −i +j −k lower unit normal: = √ 3 3 (−i +j −k) 25. x −x 0 (∂f/∂x)(x 0 , y 0 , z 0 ) = y −y 0 (∂f/∂y)(x 0 , y 0 , z 0 ) = z −z 0 (∂f/∂z)(x 0 , y 0 , z 0 ) P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 810 SECTION 16.4 26. All the tangent planes pass through the origin. To see this, write the equation of the surface as xf(x/y) −z = 0. The tangent plane at (x 0 , y 0 , z 0 ) has equation (x −x 0 ) _ x 0 y 0 f _ x 0 y 0 _ +f _ x 0 y 0 __ −(y −y 0 ) _ x 0 2 y 0 2 f _ x 0 y 0 __ −(z −z 0 ) = 0. The plane passes through the origin: − x 0 2 y 0 f _ x 0 y 0 _ −x 0 f _ x 0 y 0 _ + x 0 2 y 0 f _ x 0 y 0 _ +z 0 = z 0 −x 0 f _ x 0 y 0 _ = 0. 27. Since the tangent planes meet at right angles, the normals ∇F and ∇G meet at right angles: ∂F ∂x ∂G ∂x + ∂F ∂y ∂G ∂y + ∂F ∂z ∂G ∂z = 0. 28. The sum of the intercepts is a. To see this, note that the equation of the tangent plane at (x 0 , y 0 , z 0 ) can be written x −x 0 √ x 0 + y −y 0 √ y 0 + z −z 0 √ z 0 = 0. Setting y = z = 0 we have x −x 0 √ x 0 = √ y 0 + √ z 0 . Therefore the x-intercept is given by x = x 0 + √ x 0 ( √ y 0 + √ z 0 ) = x 0 + √ x 0 ( √ a − √ x 0 ) = √ x 0 √ a. Similarly the y-intercept is √ y 0 √ a and the z-intercept is √ z 0 √ a. The sum of the intercepts is ( √ x 0 + √ y 0 + √ z 0 ) √ a = √ a √ a = a. 29. The tangent plane at an arbitrary point (x 0 , y 0 , z 0 ) has equation y 0 z 0 (x −x 0 ) +x 0 z 0 (y −y 0 ) +x 0 y 0 (z −z 0 ) = 0, which simplifies to y 0 z 0 x +x 0 z 0 y +x 0 y 0 z = 3x 0 y 0 z 0 and thus to x 3x 0 + y 3y 0 + z 3z 0 = 1. The volume of the pyramid is V = 1 3 Bh = 1 3 _ (3x 0 ) (3y 0 ) 2 _ (3z 0 ) = 9 2 x 0 y 0 z 0 = 9 2 a 3 . 30. The equation of the tangent plane at (x 0 , y 0 , z 0 ) can be written x 0 −1/3 (x −x 0 ) +y 0 −1/3 (y −y 0 ) +z 0 −1/3 (z −z 0 ) = 0 Setting y = z = 0, we get the x-intercept x = x 0 +x 0 1/3 (y 0 2/3 +z 0 2/3 ) = x 0 +x 0 1/3 (a 2/3 −x 0 2/3 ) =⇒ x = x 0 1/3 a 2/3 Similarly, the y-intercept is y 0 1/3 a 2/3 and the z-intercept is z 0 1/3 a 2/3 . The sum of the squares of the intercepts is (x 0 2/3 +y 0 2/3 +z 0 2/3 )a 4/3 = a 2/3 a 4/3 = a 2 . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.4 811 31. The point (2, 3, −2) is the tip of r(1). Since r (t) = 2i − 3 t 2 j −4tk, we have r (1) = 2i −3j −4k. Now set f(x, y, z) = x 2 +y 2 + 3z 2 −25. The function has gradient 2xi + 2yj + 6zk. At the point (2, 3, −2), ∇f = 2(2i + 3j −6k). The angle θ between r (1) and the gradient gives cos θ = (2i −3j −4k) √ 29 · (2i + 3j −6k) 7 = 19 7 √ 29 ∼ = 0.504. Therefore θ ∼ = 1.043 radians. The angle between the curve and the plane is π 2 −θ ∼ = 1.571 −1.043 ∼ = 0.528 radians. 32. The curve passes through the point (3, 2, 1) at t = 1, and its tangent vector is r (1) = 3i + 4j + 3k. For the ellipsoid, set f(x, y, z) = x 2 + 2y 2 + 3z 2 . ∇f = 2xi + 4yj + 6zk, ∇f(3, 2, 1) = 6i + 8j + 6k, which is parallel to r (1). 33. Set f(x, y, z) = x 2 y 2 + 2x +z 3 . Then, ∇f = (2xy 2 + 2) i + 2x 2 yj + 3z 2 k, ∇f(2, 1, 2) = 6i + 8j + 12k. The plane tangent to f(x, y, z) = 16 at (2, 1, 2) has equation 6(x −2) + 8(y −1) + 12(z −2) = 0, or 3x + 4y + 6z = 22. Next, set g(x, y, z) = 3x 2 +y 2 −2z. Then, ∇g = 6xi + 2yj −2k, ∇g(2, 1, 2) = 12i + 2j −2k. The plane tangent to g(x, y, z) = 9 at (2, 1, 2) is 12(x −2) + 2(y −1) −2(z −2) = 0, or 6x +y −z = 11. 34. Sphere: f(x, y, z) = x 2 +y 2 +z 2 −8x −8y −6z + 24, ∇f = (2x −8)i + (2y −8)j + (2z −6)k ∇f(2, 1, 1) = −4i −6j −4k Ellipsoid: g(x, y, z) = x 2 + 3y 2 + 2z 2 , ∇g = 2xi + 6yj + 4zk ∇g(2, 1, 1) = 4i + 6j + 4k Since their normal vectors are parallel, the surfaces are tangent. 35. A normal vector to the sphere at (1, 1, 2) is 2xi + (2y −4) j + (2z −2)k = 2i −2j + 2k. A normal vector to the paraboloid at (1, 1, 2) is 6xi + 4yj −2k = 6i + 4j −2k. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 812 SECTION 16.4 Since (2i −2j + 2k) · (6i + 4j −2k) = 0, the surfaces intersect at right angles. 36. Surface A: Set f(x, y, z) = xy −az 2 , ∇f = yi +xj −2azk Surface B: Set g(x, y, z) = x 2 +y 2 +z 2 , ∇g = 2xi + 2yj + 2zk Surface C: Set h(x, y, z) = z 2 + 2x 2 −c(z 2 + 2y 2 ). ∇h = 4xi −4cyj + 2(1 −c)zk Where surface A and surface B intersect, ∇f · ∇g = 4(xy −az 2 ) = 0 Where surface A and surface C intersect, ∇f · ∇h = 4(1 −c)(xy −az 2 ) = 0 Where surface B and surface C intersect, ∇g · ∇h = 4[2x 2 −2cy 2 + (1 −c)z 2 ] = 0 37. (a) 3x + 4y + 6 = 0 since plane p is vertical. (b) y = − 1 4 (3x + 6) = − 1 4 [3(4t −2) + 6] = −3t z = x 2 + 3y 2 + 2 = (4t −2) 2 + 3(−3t) 2 + 2 = 43t 2 −16t + 6 r(t) = (4t −2)i −3tj + (43t 2 −16t + 6)k (c) From part (b) the tip of r(1) is (2, −3, 33). We take r (1) = 4i −3j + 70j as d to write R(s) = (2i −3j + 33k) +s(4i −3j + 70k). (d) Set g(x, y) = x 2 + 3y 2 + 2. Then, ∇g = 2xi + 6yj and ∇g(2, −3) = 4i −18j. An equation for the plane tangent to z = g(x, y) at (2, −3, 33) is z −33 = 4(x −2) −18(y + 3) which reduces to 4x −18y −z = 29. (e) Substituting t for x in the equations for p and p 1 , we obtain 3t + 4y + 6 = 0 and 4t −18y −z = 29. From the first equation y = − 3 4 (t + 2) and then from the second equation z = 4t −18 _ − 3 4 (t + 2) ¸ −29 = 35 2 t −2. Thus, (∗) r(t) = ti −( 3 4 t + 3 2 )j + _ 35 2 t −2 _ k. Lines l and l are the same. To see this, consider how l and l are formed; to assure yourself, replace t in (∗) by 4s + 2 to obtain R(s) found in part (c). P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.4 813 38. (a) normal vector: 12 25 i − 14 25 j; normal line: x = 2 + 12 25 t, y = 1 − 14 25 t (b) tangent line: x = 2 + 14 25 t, y = 1 + 12 25 t (c) 1 2 3 x 1 2 y 39. (a) normal vector: 2 i + 2 j + 4 k; normal line: x = 1 + 2t, y = 2 + 2t, z = 2 + 4t (b) tangent plane: 2(x −1) + 2(y −2) + 4(z −2) = 0 or x +y + 2z −7 = 0 (c) 40. (a) -1 0 1 -1 0 1 0 2 -1 0 1 (b) -1.5 -1 -0.5 0 0.5 1 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 (c) ∇f = 0 at (±1, 0), _ 0, ± _ 3/2 _ P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 814 SECTION 16.5 41. (a) -1 0 1 -1 0 1 0 1 2 3 -1 0 1 (b) -1.5 -1 -0.5 0 0.5 1 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 (c) ∇f = _ 4x 3 −4x _ i − _ 4y 3 −4y _ j; ∇f = 0 : 4x 3 −4x = 0 =⇒ x = 0, ±1; 4y 3 −4y = 0 =⇒ y = 0, ±1 ∇f = 0 at (0, 0), (±1, 0), (0, ±1), (±1, ±1) 42. (a) -2 -1 0 1 2 -2 -1 0 1 2 -1 0 1 -2 -1 0 1 (b) -2 -1 0 1 2 -2 -1 0 1 2 (c) ∇f = 0 at (0, 0), _ ± √ 2/2, ± √ 2/2 _ SECTION 16.5 1. ∇f = (2 −2x) i −2y j = 0 only at (1, 0). The difference f(1 +h, k) −f(1, 0) = _ 2(1 +h) −(1 +h) 2 −k 2 ¸ −1 = −h 2 −k 2 ≤ 0 for all small h and k; there is a local maximum of 1 at (1, 0). 2. ∇f = (2 −2x) i + (2 + 2y) j = 0 only at (1, −1). The difference f(1 +h, −1 +k) −f(1, −1) = [2(1 +h) + 2(−1 +k) −(1 +h) 2 + (−1 +k) 2 + 5] −5 = −h 2 +k 2 does not keep a constant sign for small h and k; (1, −1) is a saddle point. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.5 815 3. ∇f = (2x +y + 3) i + (x + 2y) j = 0 only at (−2, 1). The difference f(−2 +h, 1 +k) −f(−2, 1) = [(−2 +h) 2 + (−2 +h)(1 +k) + (1 +k) 2 + 3(−2 +h) + 1] −(−2) = h 2 +hk +k 2 is nonnegative for all small h and k. To see this, note that h 2 +hk +k 2 ≥ h 2 −2|h||k| +k 2 = (|h| −|k|) 2 ≥ 0; there is a local minimum of −2 at (−2, 1). 4. ∇f = (3x 2 −3) i +j is never 0; there are no stationary points and no local extreme values. 5. ∇f = (2x +y −6) i + (x + 2y) j = 0 only at (4, −2). f xx = 2, f xy = 1, f yy = 2. At (4, −2), D = 3 > 0 and A = 2 > 0 so we have a local min; the value is −10. 6. ∇f = (2x + 2y + 2) i + (2x + 6y + 10) j = 0 only at (1, −2). ∂ 2 f ∂x 2 = 2, ∂ 2 f ∂y∂x = 2, ∂ 2 f ∂y 2 = 6; D = 2 · 6 −2 2 > 0, A = 2 =⇒ local min; the value is −8. 7. ∇f = (3x 2 −6y) i + _ 3y 2 −6x _ j = 0 at (2, 2) and (0, 0). f xx = 6x, f xy = −6, f yy = 6y, D = 36xy −36. At (2, 2), D = 108 > 0 and A = 12 > 0 so we have a local min; the value is −8. At (0, 0), D = −36 < 0 so we have a saddle point. 8. ∇f = (6x +y + 5) i + (x −2y −5) j = 0 at _ − 5 13 , − 35 13 _ . ∂ 2 f ∂x 2 = 6, ∂ 2 f ∂y∂x = 1, ∂ 2 f ∂y 2 = −2; D = 6 · (−2) −1 2 < 0; (−5/13, −35/13) is a saddle point. 9. ∇f = (3x 2 −6y + 6) i + (2y −6x + 3) j = 0 at (5, 27 2 ) and (1, 3 2 ). f xx = 6x, f xy = −6, f yy = 2, D = 12x −36. At (5, 27 2 ), D = 24 > 0 and A = 30 > 0 so we have a local min; the value is − 117 4 . At (1, 3 2 ), D = −24 < 0 so we have a saddle point. 10. ∇f = (2x −2y −3) i + (−2x + 4y + 5) j = 0 at ( 1 2 , −1). ∂ 2 f ∂x 2 = 2, ∂ 2 f ∂y∂x = −2, ∂ 2 f ∂y 2 = 4; D = 2 · 4 −(−2) 2 > 0, A = 2 =⇒ local minimum; the value is − 13 4 . 11. ∇f = sin y i +xcos y j = 0 at (0, nπ) for all integral n. f xx = 0, f xy = cos y, f yy = −xsin y. Since D = −cos 2 nπ = −1 < 0, each stationary point is a saddle point. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 816 SECTION 16.5 12. ∇f = sin y i + (1 +xcos y) j = 0 at (−1, 2nπ) and (1, (2n + 1)π) for all integral n. ∂ 2 f ∂x 2 = 0, ∂ 2 f ∂y∂x = cos y, ∂ 2 f ∂y 2 = −xsin y; D = 0 · (−xsin y) −cos 2 y < 0 at the above points so they are all saddle points 13. ∇f = (2xy + 1 +y 2 ) i + _ x 2 + 2xy + 1 _ j = 0 at (1, −1) and (−1, 1). f xx = 2y, f xy = 2x + 2y, f yy = 2x, D = 4xy −4(x +y) 2 . At both (1, −1) and (−1, 1) we have saddle points since D = −4 < 0. 14. ∇f = _ 1 y + y x 2 _ i + _ − x y 2 − 1 x _ j = x 2 +y 2 x 2 y i − x 2 +y 2 xy 2 j is never 0; no stationary points, no local extreme values. 15. ∇f = (y −x −2 ) i + _ x −8y −2 _ j = 0 only at _ 1 2 , 4 _ . f xx = 2x −3 , f xy = 1, f yy = 16y −3 , D = 32x −3 y −3 −1. At _ 1 2 , 4 _ , D = 3 > 0 and A = 16 > 0 so we have a local min; the value is 6. 16. ∇f = (2x −2y) i + (−2x −2y) j = 0 only at (0, 0) ∂ 2 f ∂x 2 = 2, ∂ 2 f ∂y∂x = −2, ∂ 2 f ∂y 2 = −2; D = 2 · (−2) −(−2) 2 < 0; (0, 0) is a saddle point. 17. ∇f = (y −x −2 ) i + _ x −y −2 _ j = 0 only at (1, 1). f xx = 2x −3 , f xy = 1, f yy = 2y −3 , D = 4x −3 y −3 −1. At (1, 1), D = 3 > 0 and A = 2 > 0 so we have a local min; the value is 3. 18. ∇f = (2xy −y 2 −1) i + (x 2 −2xy + 1) j = 0 at (1, 1), (−1, −1) ∂ 2 f ∂x 2 = 2y, ∂ 2 f ∂y∂x = 2(x −y), ∂ 2 f ∂y 2 = −2x; D = −4xy −4(x −y) 2 < 0 at the above points; (1, 1) and (−1, −1) are saddle points. 19. ∇f = 2 _ x 2 −y 2 −1 _ (x 2 +y 2 + 1) 2 i + 4xy (x 2 +y 2 + 1) 2 j = 0 at (1, 0) and (−1, 0). f xx = −4x 3 + 12xy 2 + 12x (x 2 +y 2 + 1) 3 , f xy = 4y 3 + 4y −12x 2 y (x 2 +y 2 + 1) 3 , f yy = 4x 3 + 4x −12xy 2 (x 2 +y 2 + 1) 3 . point A B C D result (1, 0) 1 0 1 1 loc. min. (−1, 0) −1 0 −1 1 loc. max. f(1, 0) = −1; f(−1, 0) = 1 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.5 817 20. ∇f = _ ln xy + 1 − 3 x _ i + x −3 y j = 0 at (3, 1/3) ∂ 2 f ∂x 2 = 1 x + 3 x 2 , ∂ 2 f ∂y∂x = 1 y , ∂ 2 f ∂y 2 = 3 −x y 2 At (3, 1/3), ∂ 2 f ∂x 2 = 2 3 , ∂ 2 f ∂y∂x = 3, ∂ 2 f ∂y 2 = 0 and D = −9 < 0 =⇒ saddle point. 21. ∇f = _ 4x 3 −4x _ i + 2y j = 0 at (0, 0), (1, 0), and (−1, 0). f xx = 12x 2 −4, f xy = 0, f yy = 2. point A B C D result (0, 0) −4 0 2 −8 saddle (1, 0) 8 0 2 16 loc. min. (−1, 0) 8 0 2 16 loc. min. f(±1, 0) = −3. 22. ∇f = 2xe x 2 −y 2 (1 +x 2 +y 2 ) i + 2ye x 2 −y 2 (1 −x 2 −y 2 ) j = 0 at (0, 0), (0, 1), (0, −1) A = ∂ 2 f ∂x 2 = 2xe x 2 −y 2 (2x + 2x 3 + 2xy 2 ) +e x 2 −y 2 (2 + 6x 2 + 2y 2 ) B = ∂ 2 f ∂y∂x = −2ye x 2 −y 2 (2x + 2x 3 + 2xy 2 ) +e x 2 −y 2 (4xy) C = ∂ 2 f ∂y 2 = −2ye x 2 −y 2 (2y −2yx 2 −2y 3 ) +e x 2 −y 2 (2 −2x 2 −6y 2 ) At (0, 0), AC −B 2 = (2)(2) = 4 > 0, A > 0 local minimum; the value is 0. At (0, ±1), AC −B 2 = (4e −1 )(−4e −1 ) = −8e −2 > 0, saddle points 23. ∇f = cos x sin y i + sin x cos y j = 0 at _ 1 2 π, 1 2 π _ , _ 1 2 π, 3 2 π _ , (π, π), _ 3 2 π, 1 2 π _ , _ 3 2 π, 3 2 π _ . f xx = −sin x sin y, f xy = cos x cos y, f yy = −sin x sin y point A B C D result _ 1 2 π, 1 2 π _ −1 0 −1 1 loc. max. _ 1 2 π, 3 2 π _ 1 0 1 1 loc. min. (π, π) 0 1 0 −1 saddle _ 3 2 π, 1 2 π _ 1 0 1 1 loc. min. _ 3 2 π, 3 2 π _ −1 0 −1 1 loc. max. f _ 1 2 π, 1 2 π _ = f _ 3 2 π, 3 2 π _ = 1; f _ 1 2 π, 3 2 π _ = f _ 3 2 π, 1 2 π _ = −1 24. ∇f = −sin x cosh y i + cos x sinh y j = 0 at (−π, 0) , (0, 0) , (π, 0). f xx = −cos x cosh y, f xy = −sin x sinh y, f yy = cos x cosh y D = −cos 2 xcosh 2 y −sin 2 xsinh 2 y < 0; (−π, 0) , (0, 0) , (π, 0) are saddle points. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 818 SECTION 16.5 25. (a) ∇f = (2x +ky) i + (2y +kx) j and ∇f(0, 0) = 0 independent of the value of k. (b) f xx = 2, f xy = k, f yy = 2, D = 4 −k 2 . Thus, D < 0 for |k| > 2 and (0, 0) is a saddle point (c) D = 4 −k 2 > 0 for |k| < 2. Since A = f xx = 2 > 0, (0, 0) is a local minimum. (d) The test is inconclusive when D = 4 −k 2 = 0 i.e., for k = ±2. (If k = ±2, f(x, y) = (x ±y) 2 and (0, 0) is a minimum.) 26. (a) ∇f = (2x +ky) i + (kx + 8y) j = 0 at (0, 0). (b) ∂ 2 f ∂x 2 = 2, ∂ 2 f ∂y∂x = k, ∂ 2 f ∂y 2 = 8; we want 16 −k 2 < 0, or |k| > 4 (c) We want 16 −k 2 > 0, or |k| < 4 (d) k = ±4. (If k = ±4, f(x, y) = (x ±2y) 2 and (0, 0) is a minimum.) 27. Let P (x, y, z) be a point in the plane. We want to find the minimum of f(x, y, z) = _ x 2 +y 2 +z 2 . However, it is sufficient to minimize the square of the distance: F(x, y, z) = x 2 +y 2 +z 2 . It is clear that F has a minimum value, but no maximum value. Since P lies in the plane, 2x −y + 2z = 16 which implies y = 2x + 2z −16 = 2(x +z −8). Thus, we want to find the minimum value of F(x, z) = x 2 + 4(x +z −8) 2 +z 2 Now, ∇F = [2x + 8(x +z −8)] i + [8(x +z −8) + 2z] k The gradient is 0 when 2x + 8(x +z −8) = 0 and 8(x +z −8) + 2z = 0 The only solution to this pair of equations is: x = z = 32 9 , from which it follows that y = − 16 9 . The point in the plane that is closest to the origin is P _ 32 9 , − 16 9 , 32 9 _ . The distance from the origin to the plane is: F(P) = 16 3 . Check using (13.6.5): d(P, 0) = |2 · 0 −0 + 2 · 0 −16| _ 2 2 + (−1) 2 + 2 2 = 16 3 . 28. We want to minimize (x + 1) 2 + (y −2) 2 + (z −4) 2 on the plane. Since z = −16 − 3 2 x + 2y, we need to minimize f(x, y) = (x + 1) 2 + (y −2) 2 + (−20 − 3 2 x + 2y) 2 ; ∇f = _ 13 2 x −6y + 62 _ i + (−84 −6x + 10y) j = 0 at (−4, 6) Closest point (−4, 6, 2), distance= _ (−1 −(−4)) 2 + (2 −6) 2 + (4 −2) 2 = √ 29 29. f(x, y) = (x −1) 2 + (y −2) 2 +z 2 = (x −1) 2 + (y −2) 2 +x 2 + 2y 2 _ since z = _ x 2 + 2y 2 _ ∇f = [2(x −1) + 2x] i + [2(y −2) + 4y] j = 0 =⇒ x = 1 2 , y = 2 3 . f xx = 4 > 0, f xy = 0, f yy = 6, D = 24 > 0. Thus, f has a local minimum at (1/2, 2/3). The shortest distance from (1, 2, 0) to the cone is _ f _ 1 2 , 2 3 _ = 1 6 √ 114 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.5 819 30. V = 8xyz, x 2 +y 2 +z 2 = a 2 =⇒ V (x, y) = 8xy _ a 2 −x 2 −y 2 , x > 0, y > 0, x 2 +y 2 < a 2 ∇V = 8y(a 2 −x 2 −y 2 ) −8x 2 y _ a 2 −x 2 −y 2 i + 8x(a 2 −x 2 −y 2 ) −8xy 2 _ a 2 −x 2 −y 2 j = 0 at _ a/ √ 3, a/ √ 3 _ dimensions: 2a √ 3 × 2a √ 3 × 2a √ 3 , maximum volume: 8 9 a 3 √ 3 31. (a) -2 -1 0 1 2 -2 -1 0 1 2 0 1 2 3 4 0 1 2 3 (b) -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 (c) ∇f = (4y −4x 3 ) i + (4x −4y 3 ) j = 0 at (0, 0), (1, 1), (−1, −1). f xx = −12x 2 , f xy = 4, f yy = −12y 2 , D = 144x 2 y 2 −16 point A B C D result (0, 0) 0 4 0 −16 saddle (1, 1) −12 4 −12 128 loc. max. (−1, −1) −12 4 −12 128 loc. max. f(1, 1) = f(−1, −1) = 3 32. (a) -1 0 1 -1 0 1 1 2 3 -1 0 1 (b) -1.5 -1 -0.5 0 0.5 1 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 (c) ∇f = (4x 3 −4x) i + 2y j = 0 at (0, 0), (1, 0), (−1, 0). f xx = 12x 2 −4, f x,y = 0, f yy = 2, D = 24x 2 −8 point A B C D result (0, 0) −8 0 2 −8 saddle (1, 0) 8 0 2 16 loc. min. (−1, 0) 8 0 2 16 loc. min. f(1, 0) = f(−1, 0) = 0 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 820 SECTION 16.5 33. (a) -2 0 2 0 2 4 (b) -1.5 -1 -0.5 0 0.5 1 1.5 -1.5 -1 -0.5 0 0.5 1 1.5 f(1, 1) = 3 is a local max.; f has a saddle at (0, 0). 34. (a) -2 -1 0 1 2 -2 -1 0 1 2 0 0.2 0.4 0.6 -2 -1 0 1 2 (b) -2 -1 0 1 2 -2 -1 0 1 2 f(0, 0) = 0 is a local min.; f(0, 1) = f(0, −1) = 2e −1 are local maxima; f has a saddle at (±1, 0). 35. (a) -2 0 2 -2 0 2 -1 0 1 (b) -2 -1 0 1 2 -2 -1 0 1 2 f(1, 0) = −1 is a local min.; f(−1, 0) = 1 is a loc. max. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.6 821 36. (a) 0 2 4 6 8 10 0 2 4 6 8 10 -1 0 1 2 3 0 2 4 6 8 (b) 0 2 4 6 8 10 0 2 4 6 8 10 f(π/2, π/2) = f(π/2, 5π/2) = f(5π/2, π/2) = f(5π/2, 5π/2) = 3 are local maxima; (π/2, 3π/2), (3π/2, π/2), (3π/2, 3π/2), (3π/2, 5π/2), (5π/2, 3π/2) are saddle points of f; f(7π/6, 7π/6) = f(11π/6, 11π/6) = − 3 2 are local minima. SECTION 16.6 1. ∇f = (4x −4) i + (2y −2) j = 0 at (1, 1) in D; f(1, 1) = −1 Next we consider the boundary of D. We parametrize each side of the triangle: 1 2 x 4 y C 1 : r 1 (t) = t i, t ∈ [ 0, 2 ], C 2 : r 2 (t) = 2 i +t j, t ∈ [ 0, 4 ], C 3 : r 3 (t) = t i + 2t j, t ∈ [ 0, 2 ], Now, f 1 (t) = f(r 1 (t)) = 2(t −1) 2 , t ∈ [ 0, 2 ]; critical number: t = 1, f 2 (t) = f(r 2 (t)) = (t −1) 2 + 1, t ∈ [ 0, 4 ]; critical number: t = 1, f 3 (t) = f(r 3 (t)) = 6t 2 −8t + 2, t ∈ [ 0, 2 ]; critical number: t = 2 3 . Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 1 (0) = f 3 (0) = f(0, 0) = 2; f 1 (1) = f(1, 0) = 0; f 1 (2) = f 2 (0) = f(2, 0) = 2; f 2 (1) = f(2, 1) = 1; f 2 (4) = f 3 (2) = f(2, 4) = 10; f 3 (2/3) = f(2/3, 4/3) = − 2 3 . f takes on its absolute maximum of 10 at (2, 4) and its absolute minimum of −1 at (1, 1). 2. ∇f = −3 i + 2 j = 0; no stationary points in D; Next we consider the boundary of D. We parametrize each side of the triangle: C 1 : r 1 (t) = t i, t ∈ [ 0, 4 ], C 2 : r 2 (t) = t i + _ − 3 2 t + 6 _ j, t ∈ [ 0, 4 ], C 3 : r 3 (t) = t j, t ∈ [ 0, 6 ], P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 822 SECTION 16.6 and evaluate f: f 1 (t) = f(r 1 (t)) = 2 −3t, t ∈ [ 0, 4 ]; no critical numbers, f 2 (t) = f(r 2 (t)) = −6t + 14, t ∈ [ 0, 4 ]; no critical numbers, f 3 (t) = f(r 3 (t)) = 2 + 2t, t ∈ [ 0, 6 ]; no critical numbers. Evaluating these functions at the endpoints of their domains, we find that: f 1 (0) = f 3 (0) = f(0, 0) = 2; f 1 (4) = f 2 (4) = f(4, 0) = −10; f 2 (0) = f 3 (6) = f(0, 6) = 14; f takes on its absolute maximum of 14 at (0, 6) and its absolute minimum of −10 at (4, 0). 3. ∇f = (2x +y −6) i + (x + 2y) j = 0 at (4, −2) in D; f(4, −2) = −13 Next we consider the boundary of D. We parametrize each side of the rectangle: 1 2 3 4 5 x -3 -2 -1 y C 1 : r 1 (t) = −t j, t ∈ [ 0, 3 ] C 2 : r 2 (t) = t i −3 j, t ∈ [ 0, 5 ] C 3 : r 3 (t) = 5 i −t j, t ∈ [ 0, 3 ] C 4 : r 4 (t) = t i, t ∈ [ 0, 5 ] Now, f 1 (t) = f(r 1 (t)) = t 2 −1, t ∈ [ 0, 3 ]; no critical numbers f 2 (t) = f(r 2 (t)) = t 2 −9t + 8, t ∈ [ 0, 5 ]; critical number: t = 9 2 f 3 (t) = f(r 3 (t)) = t 2 −5t −6, t ∈ [ 0, 3 ]; critical number: t = 5 2 f 4 (t) = f(r 4 (t)) = t 2 −6t −1, t ∈ [ 0, 5 ]; critical number: t = 3 Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 1 (0) = f 4 (0) = f(0, 0) = −1; f 1 (−3) = f 2 (0) = f(0, −3) = 8; f 2 (9/2) = f(9/2, −3) = − 49 4 ; f 2 (5) = f 3 (3) = f(5, −3) = −12; f 3 (5/2) = f(5, −5/2) = − 49 4 ; f 3 (0) = f 4 (5) = f(5, 0) = −6. f 4 (3) = f(3, 0) = −10 f takes on its absolute maximum of 8 at (0, −3) and its absolute minimum of −13 at (4, −2). 4. ∇f = (2x + 2y) i + (2x + 6y) j = 0 at (0, 0) in D; f(0, 0) = 0 Next we consider the boundary of D. We parametrize each side of the square: C 1 : r 1 (t) = t i −2 j, t ∈ [ −2, 2 ], C 2 : r 2 (t) = 2 i +t j, t ∈ [ −2, 2 ], C 3 : r 3 (t) = t i + 2 j, t ∈ [ −2, 2 ], C 4 : r 4 (t) = −2 i +t j, t ∈ [ −2, 2 ], P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.6 823 and evaluate f: f 1 (t) = f(r 1 (t)) = t 2 −4t + 12, t ∈ [ −2, 2 ]; no critical numbers, f 2 (t) = f(r 2 (t)) = 4 + 4t + 3t 2 , t ∈ [ −2, 2 ]; critical number: t = − 2 3 , f 3 (t) = f(r 3 (t)) = t 2 + 4t + 12, t ∈ [ −2, 2 ]; no critical numbers, f 4 (t) = f(r 4 (t)) = 4 −4t + 3t 2 , t ∈ [ −2, 2 ]; critical number: t = 2 3 . Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 1 (−2) = f 4 (−2) = f(−2, −2) = 24; f 1 (2) = f 2 (−2) = f(2, −2) = 8; f 2 (−2/3) = f(2, −2/3) = 8 3 ; f 2 (2) = f 3 (2) = f(2, 2) = 24; f 3 (−2) = f 4 (2) = f(−2, 2) = 8; f 4 (2/3) = f(−2, 2/3) = 8 3 . f takes on its absolute maximum of 24 at (−2, −2) and (2, 2) and its absolute minimum of 0 at (0, 0). Note that x 2 + 2xy + 3y 2 = (x +y) 2 + 2y 2 . The results follow immediately from this observation. 5. ∇f = (2x + 3y) i + (2y + 3x) j = 0 at (0, 0) in D; f(0, 0) = 2 Next we consider the boundary of D. We parametrize the circle by: C : r(t) = 2 cos t i + 2 sin t j, t ∈ [ 0, 2π ] The values of f on the boundary are given by the function F(t) = f(r(t)) = 6 + 12 sin t cos t, t ∈ [ 0, 2π ] F (t) = 12 cos 2 t −12 sin 2 t : F (t) = 0 =⇒ cos t = ± sin t =⇒ t = 1 4 π, 3 4 π, 5 4 π, 7 4 π Evaluating F at the endpoints and critical numbers, we have: F(0) = F(2π) = f(2, 0) = 6; F _ 1 4 π _ = F _ 5 4 π _ = f _ √ 2, √ 2 _ = f( _ − √ 2, − √ 2 _ = 12; F _ 3 4 π _ = f _ − √ 2, √ 2 _ = F _ 7 4 π _ = f _ √ 2, − √ 2 _ = 0 f takes on its absolute maximum of 12 at _√ 2, √ 2 _ and at _ − √ 2, − √ 2 _ ; f takes on its absolute minimum of 0 at _ − √ 2, √ 2 _ and at _√ 2, − √ 2 _ . 6. ∇f = yi + (x −3)j = 0 at (3, 0), which is not in the interior of D. The boundary is r(t) = 3 cos t i + 3 sin t j. f(r(t)) = 3 sin t(3 cos t −3) = 9 sin t(cos t −1), t ∈ [0, 2π]. df dt = 9(2 cos 2 t −cos t −1); df dt = 0 =⇒ cos t = 1, − 1 2 which yields the points A(3, 0), B(− 3 2 , 3 √ 3 2 ), C (− 3 2 , − 3 √ 3 2 ) : f(A) = 0, f(B) = − 27 √ 3 4 min, f(C) = 27 √ 3 4 max 7. ∇f = 2(x −1)i + 2(y −1) j = 0 only at (1, 1) in D. As the sum of two squares, f(x, y) ≥ 0. Thus, f(1, 1) = 0 is a minimum. To examine the behavior of f on the boundary of D, we note that f represents the square of the distance between (x, y) and (1, 1). Thus, f is maximal at the point of the boundary furthest from (1, 1). This is the point _ − √ 2, − √ 2 _ ; the maximum value of f is f _ − √ 2, − √ 2 _ = 6 + 4 √ 2. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 824 SECTION 16.6 8. ∇f = (y + 1) i + (x −1) j = 0 at (1, −1) which is not in the interior of D. Next we consider the boundary of D. We parametrize the boundary by: C 1 : r 1 (t) = t j +t 2 j, t ∈ [ −2, 2 ], C 2 : r 2 (t) = t i + 4 j, t ∈ [ −2, 2 ], and evaluate f: f 1 (t) = f(r 1 (t)) = t 3 −t 2 +t + 3, t ∈ [ −2, 2 ]; no critical numbers, f 2 (t) = f(r 2 (t)) = 5t −1, t ∈ [ −2, 2 ]; no critical numbers. Evaluating these functions at the endpoints of their domains, we find that: f 1 (−2) = f 2 (−2) = f(−2, 4) = −11; f 1 (2) = f 2 (2) = f(2, 4) = 9. f takes on its absolute maximum of 9 at (2, 4) and its absolute minimum of −11 at (−2, 4). 9. ∇f = 2x 2 −2y 2 −2 (x 2 +y 2 + 1) 2 i + 4xy (x 2 +y 2 + 1) 2 j = 0 at (1, 0) and (−1, 0) in D; f(1, 0) = −1, f(−1, 0) = 1. Next we consider the boundary of D. We parametrize each side of the squre: C 1 : r 1 (t) = −2 i +t j, t ∈ [ −2, 2 ] C 2 : r 2 (t) = t i + 2 j, t ∈ [ −2, 2 ] C 3 : r 3 (t) = 2 i +t j, t ∈ [ −2, 2 ] C 4 : r 4 (t) = t i, t ∈ [ −2, 2 ] Now, f 1 (t) = f(r 1 (t)) = 4 t 2 + 5 , t ∈ [ −2, 2 ]; critical number: t = 0 f 2 (t) = f(r 2 (t)) = −2t t 2 + 5 , t ∈ [ −2, 2 ]; no critical numbers f 3 (t) = f(r 3 (t)) = −4 t 2 + 5 , t ∈ [ −2, 2 ]; critical number: t = 0 f 4 (t) = f(r 4 (t)) = −2t t 2 + 5 , t ∈ [ −2, 2 ]; no critical numbers Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 1 (−2) = f 4 (−2) = f(−2, −2) = 4 9 ; f 1 (0) = f(−2, 0) = 4 5 ; f 1 (2) = f 2 (−2) = f(−2, 2) = 4 9 ; f 4 (2) = f 3 (−2) = f(2, −2) = − 4 9 ; f 3 (0) = f(2, 0) = − 4 5 ; f 2 (2) = f 3 (2) = f(2, 2) = − 4 9 . f takes on its absolute maximum of 1 at (−1, 0) and its absolute minimum of −1 at (1, 0). 10. ∇f = 2x 2 −2y 2 −2 (x 2 +y 2 + 1) 2 i + 4xy (x 2 +y 2 + 1) 2 j = 0 at (1, 0) in D; f(1, 0) = −1. Next we consider the boundary of D. We parametrize each side of the triangle: C 1 : r 1 (t) = t i −t j, t ∈ [ 0, 2 ] C 2 : r 2 (t) = 2 i +t j, t ∈ [ −2, 2 ] C 3 : r 3 (t) = t i +t j, t ∈ [ 0, 2 ], P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.6 825 and evaluate f: f 1 (t) = f(r 1 (t)) = −2t 2t 2 + 1 , t ∈ [ 0, 2 ]; critical number: t = 1/ √ 2, f 2 (t) = f(r 2 (t)) = −4 t 2 + 5 , t ∈ [ −2, 2 ]; critical number: t = 0 f 3 (t) = f(r 3 (t)) = −2t 2t 2 + 1 , t ∈ [ 0, 2 ]; critical number: t = 1/ √ 2. Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 1 (0) = f 3 (0) = f(0, 0) = 0; f 1 (1/ √ 2) = f(1/ √ 2, −1/ √ 2) = −1/ √ 2; f 1 (2) = f 2 (−2) = f(2, −2) = − 4 9 ; f 2 (0) = f(2, 0) = − 4 5 ; f 2 (2) = f 3 (2) = f(2, 2) = − 4 9 ; f 3 (1/ √ 2) = f(1/ √ 2, 1/ √ 2) = −1/ √ 2. f takes on its absolute maximum of 0 at (0, 0) and its absolute minimum of −1 at (1, 0). 11. ∇f = (4 −4x) cos y i −(4x −2x 2 ) sin y j = 0 at (1, 0) in D: f(1, 0) = 2 Next we consider the boundary of D. We parametrize each side of the rectangle: C 1 : r 1 (t) = t j, t ∈ _ − 1 4 π, 1 4 π ¸ C 2 : r 2 (t) = t i − 1 4 π j, t ∈ [ 0, 2 ] C 3 : r 3 (t) = 2 i +t j, t ∈ _ − 1 4 π, 1 4 π ¸ C 4 : r 4 (t) = t i + 1 4 π j, t ∈ [ 0, 2 ] Now, f 1 (t) = f(r 1 (t)) = 0; f 2 (t) = f(r 2 (t)) = √ 2 2 (4t −2t 2 ), t ∈ [ 0, 2 ]; critical number: t = 1; f 3 (t) = f(r 3 (t)) = 0; f 4 (t) = f(r 4 (t)) = √ 2 2 (4t −2t 2 ), t ∈ [ 0, 2 ]; critical number: t = 1; f at the vertices of the rectangle has the value 0; f 2 (1) = f 4 (1) = f _ 1, − 1 4 π _ = f _ 1, 1 4 π _ = √ 2. f takes on its absolute maximum of 2 at (1, 0) and its absolute minimum of 0 along the lines x = 0 and x = 2. 12. ∇f = 2(x −3)i + 2yj = 0 at (3, 0) which is not in the interior of D. Boundary: On y = x 2 , f = (x −3) 2 +x 4 , df dx = 2(x −3) + 4x 3 = 0 at x = 1 =⇒ (1, 1) On y = 4x, f = (x −3) 2 + 16x 2 , df dx = 2(x −3) + 32x = 0 at x = 3 17 =⇒ ( 3 17 , 12 17 ). So the maximum and minimum occur at one or more of the following points: (0, 0), (4, 16), (1, 1), ( 3 17 , 12 17 ). Evaluating f at these points, we find that f(1, 1) = 5 is the absolute minimum of f; f(4, 16) = 257 is the absolute maximum of f. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 826 SECTION 16.6 13. ∇f = (3x 2 −3y) i + (−3x −3y 2 ) j = 0 at (−1, 1) in D; f(−1, 1) = 1 Next we consider the boundary of D. We parametrize each side of the triangle: -2 -1 1 2 x -2 2 y C 1 : r 1 (t) = −2 i +t j, t ∈ [ −2, 2 ], C 2 : r 2 (t) = t i +t j, t ∈ [ −2, 2 ], C 3 : r 3 (t) = t i + 2 j, t ∈ [ −2, 2 ], and evaluate f: f 1 (t) = f(r 1 (t)) = −8 + 6t −t 3 , t ∈ [ −2, 2 ]; critical numbers: t = ± √ 2, f 2 (t) = f(r 2 (t)) = −3t 2 , t ∈ [ −2, 2 ]; critical number: t = 0, f 3 (t) = f(r 3 (t)) = t 3 −6t −8, t ∈ [ −2, 2 ]; critical numbers: t = ± √ 2. Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 1 (−2) = f 2 (−2) = f(−2, −2) = −12; f 1 (− √ 2) = f(−2, − √ 2) = −8 −4 √ 2 ∼ = −13.66; f 1 ( √ 2) = f(−2, √ 2) = −8 + 4 √ 2 ∼ = −2.34; f 1 (2) = f 3 (−2) = f(−2, 2) = −4; f 2 (0) = f(0, 0) = 0; f 2 (2) = f 3 (2) = f(2, 2) = −12; f 3 (− √ 2) = f(− √ 2, 2) = −8 + 4 √ 2; f 3 ( √ 2) = f( √ 2, 2) = −8 −4 √ 2 f takes on its absolute maximum of 1 at (−1, 1) and its absolute minimum of −8 −4 √ 2 at ( √ 2, 2) and (−2, − √ 2). 14. ∇f = 2(x −4)i + 2yj = 0 at (4, 0) which is not in the interior of D. Next we examine f on the boundary of D: C 1 : r 1 (t) = ti + 4tj, t ∈ [ 0, 2, ], C 2 : r 2 (t) = ti +t 3 j, t ∈ [ 0, 2 ]. Note that f 1 (t) = f (r 1 (t)) = 17t 2 −8t + 16, f 2 (t) = f (r 2 (t)) = (t −4) 2 +t 6 . Next f 1 (t) = 34t −8 = 0 =⇒ t = 4/17 and gives x = 4/17, y = 16/17 and f 2 (t) = 6t 5 + 2t −8 = 0 =⇒ t = 1 and gives x = 1, y = 1. The extreme values of f can be culled from the following list: f(0, 0) = 16, f(2, 8) = 68, f _ 4 17 , 16 17 _ = 256 17 , f(1, 1) = 10. We see that f(1, 1) = 10 is the absolute minimum and f(2, 8) = 68 is the absolute maximum. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.6 827 15. ∇f = 4xy (x 2 +y 2 + 1) 2 i + 2y 2 −2x 2 −2 (x 2 +y 2 + 1) 2 j = 0 at (0, 1) and (0, −1) in D; f(0, 1) = −1, f(0, −1) = 1 Next we consider the boundary of D. We parametrize the circle by: C : r(t) = 2 cos t i + 2 sin t j, t ∈ [ 0, 2π ] The values of f on the boundary are given by the function F(t) = f(r(t)) = − 4 5 sin t, t ∈ [ 0, 2π ] F (t) = − 4 5 cos t : F (t) = 0 =⇒ cos t = 0 =⇒ t = 1 2 π, 3 2 π. Evaluating F at the endpoints and critical numbers, we have: F(0) = F(2π) = f(2, 0) = 0; F _ 1 2 π _ = f(0, 2) = − 4 5 ; F _ 3 2 π _ = f(0, −2) = 4 5 f takes on its absolute maximum of 1 at (0, −1) and its absolute minimum of −1 at (0, 1). 16. ∇f = (2x + 1) i + (8y −2) j = 0 at _ − 1 2 , 1 4 _ in D; f _ − 1 2 , 1 4 _ = − 1 2 Next we consider the boundary of D. We parametrize the ellipse by: C : r(t) = 2 cos t i + sin t j, t ∈ [ 0, 2π ] The values of f on the boundary are given by the function F(t) = f(r(t)) = 4 cos 2 t + 4 sin 2 t + 2 cos t −2 sin t = 4 + 2 cos t −2 sin t, t ∈ [ 0, 2π ] F (t) = −2 sin t −2 cos t : F (t) = 0 =⇒ cos t = −sin t =⇒ t = 3 4 π, or 7 4 π Evaluating F at the endpoints and critical numbers, we have: F(0) = F(2π) = f(2, 0) = 6; F _ 3 4 π _ = f _ − √ 2, √ 2/2 _ = 4 −2 √ 2; F _ 7 4 π _ = f _ √ 2, − √ 2/ _ = 4 + 2 √ 2 f takes on its absolute maximum of 4 + 2 √ 2 at _√ 2, − √ 2/2 _ ; f takes on its absolute minimum of − 1 2 at _ − 1 2 , 1 4 _ . 17. ∇f = 2(x −y)i −2(x −y) j = 0 at each point of the line segment y = x from (0, 0) to (4, 4). Since f(x, x) = 0 and f(x, y) ≥ 0, f takes on its minimum of 0 at each of these points. Next we consider the boundary of D. We parametrize each side of the triangle: C 1 : r 1 (t) = tj, t ∈ [ 0, 12 ] C 2 : r 2 (t) = ti, t ∈ [ 0, 6 ] C 3 : r 3 (t) = ti + (12 −2t) j, t ∈ [ 0, 6 ] P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 828 SECTION 16.6 and observe from f(r 1 (t)) = t 2 , t ∈ [ 0, 12 ] f(r 2 (t)) = t 2 , t ∈ [ 0, 6 ] f(r 3 (t)) = (3t −12) 2 , t ∈ [ 0, 6 ] that f takes on its maximum of 144 at the point (0, 12). 18. ∇f = 1 (x 2 +y 2 ) 3/2 (−xi −yj) = 0 in D. Note that f(x, y) is the reciprocal of the distance of (x, y) from the origin. The point of D closest to the origin (draw a figure) is (1, 1). Therefore f(1, 1) = 1/ √ 2 is the maximum value of f. The point of D furthest from the origin is (3, 4). Therefore f(3, 4) = 1/5 is the least value taken on by f. 19. Using the hint, we want to find the maximum value of f(x, y) = 18xy −x 2 y −xy 2 in the triangular region. The gradient of f is: ∇D = _ 18y −2xy −y 2 _ i + _ 18x −x 2 −2xy _ j The gradient is 0 when 18y −2xy −y 2 = 0 and 18x −x 2 −2xy = 0 The solution set of this pair of equations is: (0, 0), (18, 0), (0, 18), (6, 6). It is easy to verify that f is a maximum when x = y = 6. The three numbers that satisfy x +y +z = 18 and maximize the product xyz are: x = 6, y = 6, z = 6. 20. f(y, z) = 30yz 2 −y 2 z 2 −yz 3 , ∇f = (30z 2 −2yz 2 −z 3 )j + (60yz −2y 2 z −3yz 2 )k = 0 at _ 15 2 , 15 _ (other points are not in the interior); f _ 15 2 , 15 _ = 15 4 4 . On the line y +z = 30, f(y, z) = 0 so the maximum of xyz 2 occurs at x = y = 15 2 , z = 15 21. f(x, y) = xy(1 −x −y), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 −x. [ dom(f) is the triangle with vertices (0, 0), (1, 0), (0, 1).] ∇f = (y −2xy −y 2 )i + _ x −2xy −x 2 _ j = 0 =⇒ x = y = 0, x = 1, y = 0, x = 0, y = 1, x=y = 1 3 . (Note that [ 0, 0 ] is not an interior point of the domain of f.) f xx = −2y, f xy = 1 −2x −2y, f yy = −2x. At _ 1 3 , 1 3 _ , D = 1 3 > 0 and A < 0 so we have a local max; the value is 1/27. Since f(x, y) = 0 at each point on the boundary of the domain, the local max of 1/27 is also the absolute max. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.6 829 22. V = xyz, x a + y b + z c = 1 =⇒ V (x, y) = xyc _ 1 − x a − y b _ , x > 0, y > 0, x a + y b < 1 ∇V = yc _ 1 − 2x a − y b _ i +xc _ 1 − x a − 2y b _ j = 0 at _ a 3 , b 3 _ Maximum volume V = a 3 · b 3 · c 3 = abc 27 23. (a) ∇f = 1 2 xi − 2 9 y j = 0 only at (0, 0). (b) The difference f(h, k) −f(0, 0) = 1 4 h 2 − 1 9 k 2 does not keep a constant sign for all small h and k; (0, 0) is a saddle point. The function has no local extreme values. (c) Being the difference of two squares, f can be maximized by maximizing 1 4 x 2 and minimizing 1 9 y 2 ; (1, 0) and (−1, 0) give absolute maximum value 1 4 . Similarly, (0, 1) and (0, −1) give ab- solute minimum value − 1 9 . 24. (a) ∇f = anx n−1 i +cny n−1 j = 0 at (0, 0) (b) ∂ 2 f ∂x 2 = an(n −1)x n−2 , ∂ 2 f ∂y∂x = 0, ∂ 2 f ∂y 2 = cn(n −1)y n−2 ; at (0, 0), D = 0. (c) (i) (0, 0) gives absolute min of 0 if n is even; no extreme value if n is odd. (ii) (0, 0) gives absolute max of 0 if n is even; no extreme value if n is odd. (iii) no extreme values. 25. Let x, y and z be the length, width and height of the box. The surface area is given by S = 2xy + 2xz + 2yz, so z = S −2xy 2(x +y) , where S is a constant, and x, y, z > 0. Now, the volume V = xyz is given by: V (x, y) = xy _ S −2xy 2(x +y) _ and ∇V = _ y _ S −2xy 2(x +y) _ +xy 2(x +y)(−2y) −(S −2xy)(2) 4(x +y) 2 _ i + _ x _ S −2xy 2(x +y) _ +xy 2(x +y)(−2x) −(S −2xy)(2) 4(x +y) 2 _ j Setting ∂V ∂x = ∂V ∂y = 0 and simplifying, we get the pair of equations 2S −4x 2 −8xy = 0 2S −4y 2 −8xy = 0 from which it follows that x = y = _ S/6. From practical considerations, we conclude that V has a maximum value at ( _ S/6, _ S/6). Substituting these values into the equation for z, we get z = _ S/6 and so the box of maximum volume is a cube. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 830 SECTION 16.6 26. V = xyz, S = xy + 2xz + 2yz =⇒ V (x, y) = xy (S −xy) 2(x +y) , x > 0, y > 0, xy < S. ∇V = y 2 (S −x 2 −2xy) 2(x +y) 2 i + x 2 (S −y 2 −2xy) 2(x +y) 2 j ∇V = 0 =⇒ x = _ S 3 , y = _ S 3 ; dimensions for maximum volume: _ S 3 × _ S 3 × 1 2 _ S 3 27. f(x, y) = 3 i=1 _ (x −x i ) 2 + (y −y i ) 2 _ ∇f(x, y) = 2 [(3x −x 1 −x 2 −x 3 ) i + (3y −y 1 −y 2 −y 3 ) j] ∇f = 0 only at _ x 1 +x 2 +x 3 3 , y 1 +y 2 +y 3 3 _ = (x 0 , y 0 ) . The difference f(x 0 +h, y 0 +k) −f (x 0 , y 0 ) = 3 i=1 _ (x 0 +h −x i ) 2 + (y 0 +k −y i ) 2 −(x 0 −x i ) 2 −(y 0 −y i ) 2 _ = 3 i=1 _ 2h(x 0 −x i ) +h 2 + 2k (y 0 −y i ) +k 2 ¸ = 2h(3x 0 −x 1 −x 2 −x 3 ) + 2k (3y 0 −y 1 −y 2 −y 3 ) + 3h 2 + 3k 2 = 3h 2 + 3k 2 is nonnegative for all h and k. Thus, f has its absolute minimum at (x 0 , y 0 ) . 28. Profit P(x, y) = N 1 (x −50) +N 2 (y −60) = 250(y −x)(x −50) + [32, 000 + 250(x −2y)](y −60) ∇P = 250(2y −2x −10)i + [32, 000 + 250(2x + 70 −4y)]j = 0 =⇒ x = 89, y = 94 29. A = xy + 1 2 x _ x 2 tan θ _ , P = x + 2y + 2 _ x 2 sec θ _ , 0 < θ < 1 2 π, 0 < x < P 1 + sec θ . A(x, θ) = 1 2 x(P −x −xsec θ) + 1 4 x 2 tan θ, ∇A = _ P 2 −x −xsec θ + x 2 tan θ _ i + _ x 2 4 sec 2 θ − x 2 2 sec θ tan θ _ j, (Here j is the unit vector in the direction of increasing θ.) ∇A = 1 2 [P +x(tan θ −2 sec θ −2)] i + x 2 4 sec θ (sec θ −2 tan θ) j. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.6 831 From ∂A ∂θ = 0 we get θ = 1 6 π and then from ∂A ∂x = 0 we get P +x _ 1 3 √ 3 − 4 3 √ 3 −2 _ = 0 so that x = (2 − √ 3)P. Next, A xx = 1 2 (tan θ −2 sec θ −2), A xθ = x 2 sec θ (sec θ −2 tan θ), A θθ = x 2 2 sec θ _ sec θ tan θ −sec 2 θ −tan 2 θ _ . Apply the second-partials test: A = − 1 2 (2 + √ 3 ), B = 0, C = − 1 3 P 2 √ 3 (2 − √ 3 ) 2 , D < 0. Since, D > 0 and A < 0, the area is a maximum when θ = 1 6 π, x = (2 − √ 3 ) P and y = 1 6 (3 − √ 3 )P. 30. (a) ∇f = (2ax +by)i + (bx + 2cy)j ∂ 2 f ∂x 2 = 2a, ∂ 2 f ∂y∂x = b, ∂ 2 f ∂y 2 = 2c; D = 4ac −b 2 . (b) The point (0, 0) is the only stationary point. If D < 0, (0, 0) is a saddle point; if D > 0, (0, 0) is a local minimum if a > 0 and a local maximum if a < 0. (c) (i) if b > 0, f(x, y) = ( √ ax + √ cy) 2 ; every point on the line √ ax + √ cy = 0 is a stationary point and at each such point f takes on a local and absolute min of 0 if b < 0, f(x, y) = ( √ ax − √ cy) 2 ; every point on the line √ ax − √ cy = 0 is a stationary point and at each such point f takes on a local and absolute min of 0 (ii) if b > 0, f(x, y) = −( _ |a|x − _ |c|y) 2 ; every point on the line _ |a|x − _ |c|y = 0 is a stationary point and at each such point f takes on a local and absolute max of 0 if b < 0, f(x, y) = −( _ |a|x + _ |c|y) 2 ; every point on the line _ |a|x + _ |c|y = 0 is a stationary point and at each such point f takes on a local and absolute max of 0 31. From x = 1 2 y = 1 3 z = t and x = y −2 = z = s we take (t, 2t, 3t) and (s, 2 +s, s) as arbitrary points on the lines. It suffices to minimize the square of the distance between these points: f(t, s) = (t −s) 2 + (2t −2 −s) 2 + (3t −s) 2 = 14t 2 −12ts + 3s 2 −8t + 4s + 4, t, s real. Let i and k be the unit vectors in the direction of increasing t and s, respectively. ∇f = (28t −12s −8)i + (−12t + 6s + 4) j; ∇f = 0 =⇒ t = 0, s = −2/3. f tt = 28, f ts = −12, f ss = 6, D = 6(28) −(−12) 2 = −24 < 0. By the second-partials test, the distance is a minimum when t = 0, s = −2/3; the nature of the problem tells us the minimum is absolute. The distance is _ f(0, −2/3) = 2 3 √ 6. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 832 SECTION 16.6 32. We want to minimize S = 4πr 2 + 2πrh given that V = 4 3 πr 3 +πr 2 h = 10, 000. S(r) = 4πr 2 + 2πr _ V πr 2 − 4 3 r _ = 4 3 πr 2 + 2V r S (r) = 8πr 3 −6V 3r 2 = 0 =⇒ r = 3 _ 6V 8π , h = V πr 2 − 4 3 r = 0 The optimal container is a sphere of radius r = 3 _ 7500/π meters. 33. (a) Let x and y be the cross-sectional measurements of the box, and let l be its length. Then V = xyl, where 2x + 2y +l ≤ 108, x, y > 0 To maximize V we will obviously take 2x+2y+l =108. Therefore, V (x, y)=xy(108−2x−2y) and ∇V = [y(108 −2x −2y) −2xy] i + [x(108 −2x −2y) −2xy] j Setting ∂V ∂x = ∂V ∂y = 0, we get the pair of equations ∂V ∂x = 108y −4xy −2y 2 = 0 ∂V ∂y = 108x −4xy −2x 2 = 0 from which it follows that x = y = 18 =⇒ l = 36. Now, at (18, 18), we have A = V xx = −4y = −72 < 0, B = V xy = 108 −4x −4y = −36, C = V yy = −4x = −72, and D = (36) 2 −(72) 2 < 0. Thus, V is a maximum when x = y = 18 inches and l = 36 inches. (b) Let r be the radius of the tube and let l be its length. Then V = π r 2 l, where 2π r +l ≤ 108, r > 0 To maximize V we take 2π r +l = 108. Then V (r) = π r 2 (108 −2π r) = 108π r 2 −2π 2 r 3 . Now dV dr = 216π r −6π 2 r 2 Setting dV dr = 0, we get 216π r −6π 2 r 2 = 0 =⇒ r = 36 π =⇒ l = 36 Now, at r = 36/π, we have d 2 V dr 2 = 216π −12π 2 36 π = −216π < 0 Thus, V is a maximum when r = 36/π inches and l = 36 inches. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.6 833 34. Let (x, y, z) be on the ellipsoid, x > 0, y > 0, z > 0. Then V = 2x · 2y · 2z = 8xyz. Note that V achieves its maximum ⇐⇒ x 2 y 2 z 2 achieves its maximum. Let s = x 2 y 2 z 2 , then s = x 2 y 2 _ 1 − x 2 9 − y 2 4 _ , ∂s ∂x = 2xy 2 _ 1 − 2x 2 9 − y 2 4 _ = 0 ∂s ∂y = 2x 2 y _ 1 − x 2 9 − 2y 2 4 _ = 0 =⇒ 2x 2 9 + y 2 4 = 1, x 2 9 + 2y 2 4 = 1 =⇒ x = 3 √ 3 , y = 2 √ 3 , z = 1 √ 3 Thus, V max = 8xyz = 8 · 3 √ 3 · 2 √ 3 · 1 √ 3 = 16 √ 3 3 . 35. Let S denote the cross-sectional area. Then S = 1 2 (12 −2x + 12 −2x + 2x cos θ) x sin θ = 12x sin θ −2x 2 sin θ + 1 2 x 2 sin 2θ, where 0 < x < 6, 0 < θ < π/2 Now, with j in the direction of increasing θ, ∇S = (12 sin θ −4x sin θ +x sin 2θ) i + (12x cos θ −2x 2 cos θ +x 2 cos 2θ) j Setting ∂S ∂x = ∂S ∂θ = 0, we get the pair of equations 12 sin θ −4x sin θ +x sin 2θ = 0 12x cos θ −2x 2 cos θ +x 2 cos 2θ = 0 from which it follows that x = 4, θ = π/3. Now, at (4, π/3), we have A = S xx = −4 sin θ + sin 2θ = − 3 2 √ 3, B = S xθ = 12 cos θ −4x cos θ + 2x cos 2θ = −6, C = S θθ = −12x sin θ + 2x 2 sin θ −2x 2 sin 2θ = −24 √ 3 and D = 108 −36 > 0. Thus, S is a maximum when x = 4 inches and θ = π/3. 36. 96 = xyz, C = 30xy + 10(2xz + 2yz) = 30xy + 20(x +y) 96 xy . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 834 SECTION 16.7 C(x, y) = 30 _ xy + 64 x + 64 y _ , ∇C = 30(y −64x −2 )i + 30(x −64y −2 ) j = 0 =⇒ x = y = 4. C xx = 128x −3 , C xy = 1, C yy = 128y −3 . When x = y = 4, we have D = 3 > 0 and A = 2 > 0 so the cost is minimized by making the dimensions of the crate 4 ×4 ×6 meters. 37. (a) f(m, b) = [2 −b] 2 + [−5 −(m+b)] 2 + [4 −(2m+b)] 2 . f m = 10m+ 6b −6, f b = 6m+ 6b −2; f m = f b = 0 =⇒ m = 1, b = − 2 3 . f mm = 10, f mb = 6, f bb = 6, D = 24 > 0 =⇒ a minimum. Answer: the line y = x − 2 3 . (b) f(α, β) = [2 −β] 2 + [−5 −(α +β)] 2 + [4 −(4α +β)] 2 . f α = 34α + 10β −22, f β = 10α + 6β −2; f α = f β = 0 =⇒ ⎧ ⎨ ⎩ α = 14 13 β = − 19 13 ⎤ ⎦ . f αα = 34, f αβ = 10, f ββ = 6, D = 104 > 0 =⇒ a minimun. Answer: the parabola y = 1 13 _ 14x 2 −19 _ . 38. (a) f(m, b) = [2 −(−m+b)] 2 + [−1 −b] 2 + [1 −(m+b)] 2 f m = 4m+ 2, f b = 6b −4, f m = f b = 0 =⇒ m = − 1 2 , b = 2 3 f mm = 4, f mb = 0, f bb = 6, D = 24 > 0 =⇒ minimum Answer: the line y = − 1 2 x + 2 3 (b) f(α, β) = [2 −(α +β)] 2 + [−1 −β] 2 + [1 −(α +β)] 2 f α = 4α + 4β −6, f β = 4α + 6β −4; f α = f β = 0 =⇒ α = 5 2 , β = −1 f αα = 4, f αβ = 4, f ββ = 6, D = 8 > 0 =⇒ minimum Answer: the parabola y = 5 2 x 2 −1 SECTION 16.7 1. f(x, y) = x 2 +y 2 , g(x, y) = xy −1 ∇f = 2xi + 2yj, ∇g = yi +xj. ∇f = λ∇g =⇒ 2x = λy and 2y = λx. Multiplying the first equation by x and the second equation by y, we get 2x 2 = λxy = 2y 2 . Thus, x = ±y. From g(x, y) = 0 we conclude that x = y = ±1. The points (1, 1) and (−1, −1) clearly give a minimum, since f represents the square of the distance of a point on the hyperbola from the origin. The minimum is 2. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.7 835 2. f(x, y) = xy, g(x, y) = b 2 x 2 +a 2 y 2 −a 2 b 2 ∇f = yi +xj, ∇g = 2b 2 xi + 2a 2 yj ∇f = λ∇g =⇒ y = 2λb 2 x, x = 2λa 2 y =⇒ a 2 y 2 = b 2 x 2 From g(x, y) = 0 we get 2b 2 x 2 = a 2 b 2 =⇒ x = ± a √ 2 , y = ± b √ 2 The maximum value of xy is 1 2 ab, achieved at (a/ √ 2, b √ 2) and at (−a/ √ 2, −b √ 2). 3. f(x, y) = xy, g(x, y) = b 2 x 2 +a 2 y 2 −a 2 b 2 ∇f = yi +xj, ∇g = 2b 2 xi + 2a 2 yj. ∇f = λ∇g =⇒ y = 2λb 2 x and x = 2λa 2 y. Multiplying the first equation by a 2 y and the second equation by b 2 x, we get a 2 y 2 = 2λa 2 b 2 xy = b 2 x 2 . Thus, ay = ±bx. From g(x, y) = 0 we conclude that x = ± 1 2 a √ 2 and y = ± 1 2 b √ 2. Since f is continuous and the ellipse is closed and bounded, the minimum exists. It occurs at _ 1 2 a √ 2, − 1 2 b √ 2 _ and _ − 1 2 a √ 2, 1 2 b √ 2 _ ; the minimum is − 1 2 ab. 4. f(x, y) = xy 2 , g(x, y) = x 2 +y 2 −1 ∇f = y 2 i + 2xyj, ∇g = 2xi + 2yj ∇f = λ∇g =⇒ y 2 = 2λx, 2xy = 2λy =⇒ y = 0 or y 2 = 2x 2 y = 0: From g(x, y) = 0 we get x = ±1; f(±1, 0) = 0 y 2 = 2x 2 : From g(x, y) = 0 we get 3x 2 = 1, =⇒ x = ± 1 √ 3 , y = ± √ 2 √ 3 The Minimum of xy 2 is: − 2 9 √ 3 at (−1/ √ 3, ± √ 2 √ 3) 5. Since f is continuous and the ellipse is closed and bounded, the maximum exists. f(x, y) = xy 2 , g(x, y) = b 2 x 2 +a 2 y 2 −a 2 b 2 ∇f = y 2 i + 2xyj, ∇g = 2b 2 xi + 2a 2 yj. ∇f = λ∇g =⇒ y 2 = 2λb 2 x and 2xy = 2λa 2 y. Multiplying the first equation by a 2 y and the second equation by b 2 x, we get a 2 y 3 = 2λa 2 b 2 xy = 2b 2 x 2 y. We can exclude y = 0; it clearly cannot produce the maximum. Thus, a 2 y 2 = 2b 2 x 2 and, fromg(x, y) = 0, 3b 2 x 2 = a 2 b 2 . This gives us x = ± 1 3 √ 3 a and y = ± 1 3 √ 6 b. The maximum occurs at x = 1 3 √ 3 a, y = ± 1 3 √ 6 b; the value there is 2 9 √ 3 ab 2 . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 836 SECTION 16.7 6. f(x, y) = x +y, g(x, y) = x 4 +y 4 −1 ∇f = i +j, ∇g = 4x 3 i + 4y 3 j ∇f = λ∇g =⇒ 1 = 4λx 3 , 1 = 4λy 3 =⇒ x = y From g(x, y) = 0 we get 2x 4 = 1 =⇒ x = y = ±2 −1/4 The maximum of x +y is: 2 · 2 −1/4 = 2 3/4 , achieved at (2 −1/4 , 2 −1/4 ). 7. The given curve is closed and bounded. Since x 2 +y 2 represents the square of the distance from points on this curve to the origin, the maximum exists. f(x, y) = x 2 +y 2 , g(x, y) = x 4 + 7x 2 y 2 +y 4 −1 ∇f = 2xi + 2yj, ∇g = _ 4x 3 + 14xy 2 _ i + _ 4y 3 + 14x 2 y _ j. We use the cross-product equation (16.7.4): 2x(4y 3 + 14x 2 y) −2y(4x 3 + 14xy 2 ) = 0, 20x 3 y −20xy 3 = 0, xy(x 2 −y 2 ) = 0. Thus, x = 0, y = 0, or x = ±y. From g(x, y) = 0 we conclude that the points to examine are (0, ±1), (±1, 0), _ ± 1 3 √ 3, ± 1 3 √ 3 _ . The value of f at each of the first four points is 1; the value at the last four points is 2/3. The maximum is 1. 8. f(x, y, z) = xyz, g(x, y, z) = x 2 +y 2 +z 2 −1 ∇f = yzi +xzj +xyk, ∇g = 2xi + 2yj + 2zk ∇f = λ∇g =⇒ yz = 2λx, xz = 2λy, xy = 2λz =⇒ x 2 = y 2 = z 2 or λ = 0. λ = 0: In this case, at least two of x, y, z are 0 and f = 0. x 2 = y 2 = z 2 From g(x, y, z) = 0 we get 3x 2 = 1 =⇒ x = ± 1 √ 3 , y = ± 1 √ 3 , z = ± 1 √ 3 The minimum of xyz is: − 1 9 √ 3 at (−1/ √ 3, −1/ √ 3, −1/ √ 3), (−1/ √ 3, 1/ √ 3, 1/ √ 3), (1/ √ 3, −1/ √ 3, 1/ √ 3), (1/ √ 3, 1/ √ 3, −1/ √ 3). 9. The maximum exists since xyz is continuous and the ellipsoid is closed and bounded. f(x, y, z) = xyz, g(x, y, z) = x 2 a 2 + y 2 b 2 + z 2 c 2 −1 ∇f = yzi +xzj +xyk, ∇g = 2x a 2 i + 2y b 2 j + 2z c 2 k. ∇f = λ∇g =⇒ yz = 2x a 2 λ, xz = 2y b 2 λ, xy = 2z c 2 λ. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.7 837 We can assume x, y, z are non-zero, for otherwise f(x, y, z) = 0, which is clearly not a maximum. Then from the first two equations yza 2 x = 2λ = xzb 2 y so that a 2 y 2 = b 2 x 2 or x 2 a 2 = y 2 b 2 . Similarly from the second and third equations we get b 2 z 2 = c 2 y 2 or y 2 b 2 = z 2 c 2 . From g(x, y, z) = 0, we get 3x 2 a 2 = 1 =⇒ x ± a √ 3 , from which it follows that y = ± b √ 3 , z = ± c √ 3 . The maximum value is 1 9 √ 3 abc. 10. f(x, y, z) = x + 2y + 4z, g(x, y, z) = x 2 +y 2 +z 2 −7 ∇f = i + 2j + 4k, ∇g = 2xi + 2yj + 2zk ∇f = λ∇g =⇒ 1 = 2λx, 2 = 2λy, 4 = 2λz =⇒ y = 2x, z = 4x From g(x, y, z) = 0 we get 21x 2 = 7 =⇒ x = ± 1 √ 3 Minimum of x + 2y + 4z is: −7 √ 3, achieved at (−1/ √ 3, −2/ √ 3, −4/ √ 3). 11. Since the sphere is closed and bounded and 2x + 3y + 5z is continuous, the maximum exists. f(x, y, z) = 2x + 3y + 5z, g(x, y, z) = x 2 +y 2 +z 2 −19 ∇f = 2i + 3j + 5k, ∇g = 2xi + 2yj + 2zk. ∇f = λ∇g =⇒ 2 = 2λx, 3 = 2λy, 5 = 2λz. Since λ = 0 here, we solve the equations for x, y and z: x = 1 λ , y = 3 2λ , z = 5 2λ , and substitute these results in g(x, y, z) = 0 to obtain 1 λ 2 + 9 4λ 2 + 25 4λ 2 −19 = 0, 38 4λ 2 −19 = 0, λ = ± 1 2 √ 2. The positive value of λ will produce positive values for x, y, z and thus the maximum for f. We get x = √ 2, y = 3 2 √ 2, z = 5 2 √ 2, and 2x + 3y + 5z = 19 √ 2. 12. f(x, y, z) = x 4 +y 4 +z 4 , g(x, y, z) = x +y +z −1 ∇f = 4x 3 i + 4y 3 j + 4z 3 k, ∇g = i +j +k ∇f = λ∇g =⇒ 4x 3 = λ, 4y 3 = λ, 4z 3 = λ =⇒ x = y = z From g(x, y, z) = 0 we get 3x = 1, =⇒ x = 1 3 = y = z Minimum is: 1 27 13. f(x, y, z) = xyz, g(x, y, z) = x a + y b + z c −1 ∇f = yzi +xzj +xyk, ∇g = 1 a i + 1 b j + 1 c k. ∇f = λ∇g =⇒ yz = λ a , xz = λ b , xy = λ c . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 838 SECTION 16.7 Multiplying these equations by x, y, z respectively, we obtain xyz = λx a , xyz = λy b , xyz = λz c . Adding these equations and using the fact that g(x, y, z) = 0, we have 3xyz = λ _ x a + y b + z c _ = λ. Since x, y, z are non-zero, yz = λ a = 3xyz a , 1 = 3x a , x = a 3 . Similarly, y = b 3 and z = c 3 . The maximum is 1 27 abc. 14. Maximize area A = xy given that the perimeter P = 2x + 2y f(x, y) = xy, g(x, y) = 2x + 2y −P ∇f = yi +xj, ∇g = 2i + 2j; ∇f = λ∇g =⇒ y = 2λ, x = 2λ =⇒ x = y. The rectangle of maximum area is a square. 15. It suffices to minimize the square of the distance from (0, 1) to a point on the parabola. Clearly, the minimum exists. f(x, y) = x 2 + (y −1) 2 , g(x, y) = x 2 −4y ∇f = 2xi + 2(y −1)j, ∇g = 2xi −4j. We use the cross-product equation (16.7.4): 2x(−4) −2x(2y −2) = 0, 4x + 4xy = 0, x(y + 1) = 0. Since y ≥ 0, we have x = 0 and thus y = 0. The minimum is 1. 16. Minimize f(x, y) = (x −p) 2 + (y −4p) 2 subject to g(x, y) = 2px −y 2 = 0 ∇f = 2(x −p)i + 2(y −4p)j, ∇g = 2pi −2yj ∇f = λ∇g =⇒ 2(x −p) = 2λp, 2(y −4p) = −2λy =⇒ x = 4p 2 y From g(x, y) = 0 we get 8p 3 y = y 2 =⇒ y = 2p, x = 2p Distance to parabola is: _ f(x, y) = √ 5p 17. It suffices to maximize and minimize the square of the distance from (2, 1, 2) to a point on the sphere. Clearly, these extreme values exist. f(x, y, z) = (x −2) 2 + (y −1) 2 + (z −2) 2 , g(x, y, z) = x 2 +y 2 +z 2 −1 ∇f = 2(x −2) i + 2(y −1) j + 2(z −2) k, ∇g = 2xi + 2y j + 2z k. ∇f = λ∇g =⇒ 2(x −2) = 2xλ, 2(y −1) = 2yλ, 2(z −2) = 2zλ P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.7 839 Thus, x = 2 1 −λ , y = 1 1 −λ , z = 2 1 −λ . Using the fact that x 2 +y 2 +z 2 = 1, we have _ 2 1 −λ _ 2 + _ 1 1 −λ _ 2 + _ 2 1 −λ _ 2 = 1 =⇒ λ = −2, 4 At λ = −2, (x, y, z) = (2/3, 1/3, 2/3) and f(2/3, 1/3, 2/3) = 4 At λ = 4, (x, y, z) = (−2/3, −1/3, −2/3) and f(−2/3, −1/3, −2/3) = 16 Thus, (2/3, 1/3, 2/3) is the closest point and (−2/3, −1/3, −2/3) is the furthest point. 18. f(x, y, z) = sin xsin y sin z, g(x, y, z) = x +y +z −π ∇f = cos xsin y sin zi + sin xcos y sin zj + sin xsin y cos zk, ∇g = i +j +k ∇f = λ∇g =⇒ cos xsin y sin z = λ = sin xcos y sin z = sin z sin y cos z =⇒ cos x = cos y = cos z =⇒ x = y = z = π 3 Maximum of sin xsin y sin z is: 3 √ 3 8 19. f(x, y, z) = 3x −2y +z, g(x, y, z) = x 2 +y 2 +z 2 −14 ∇f = 3 i −2 j +k, ∇g = 2xi + 2y j + 2z k. ∇f = λ∇g =⇒ 3 = 2xλ, −2 = 2yλ, 1 = 2zλ. Thus, x = 3 2λ , y = − 1 λ , z = 1 2λ . Using the fact that x 2 +y 2 +z 2 = 14, we have _ 3 2λ _ 2 + _ − 1 λ _ 2 + _ 1 2λ _ 2 = 14 =⇒ λ = ± 1 2 . At λ = 1 2 , (x, y, z) = (3, −2, 1) and f(3, −2, 1) = 14 At λ = − 1 2 , (x, y, z) = (−3, 2, −1) and f(−3, 2, −1) = −14 Thus, the maximum value of f on the sphere is 14. 20. f(x, y, z) = xyz, g(x, y, z) = x 2 +y 2 +z −4 ∇f = yzi +xzj +xyk, ∇g = 2xi + 2yj +k ∇f = λ∇g =⇒ yz = 2λx, xz = 2λy, xy = λ =⇒ x 2 = y 2 = z 2 From g(x, y, z) = 0 we get 4x 2 = 4 =⇒ x = 1, y = 1, z = 2. Maximum volume is 2 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 840 SECTION 16.7 21. It’s easier to work with the square of the distance; the minimum certainly exists. f(x, y, z) = x 2 +y 2 +z 2 , g(x, y, z) = Ax +By +Cz +D ∇f = 2xi + 2yj + 2zk, ∇g = Ai +Bj +Ck. ∇f = λ∇g =⇒ 2x = Aλ, 2y = Bλ, 2z = Cλ. Substituting these equations in g(x, y, z) = 0, we have 1 2 λ _ A 2 +B 2 +C 2 _ +D = 0, λ = −2D A 2 +B 2 +C 2 . Thus, in turn, x = −DA A 2 +B 2 +C 2 , y = −DB A 2 +B 2 +C 2 , z = −DC A 2 +B 2 +C 2 so the minimum value of _ x 2 +y 2 +z 2 is |D| _ A 2 +B 2 +C 2 _ −1/2 . 22. f(x, y, z) = xyz, g(x, y, z) = 2xy + 2xz + 2yz −6a 2 ∇f = yzi +xzj +xyk, ∇g = 2(y +z)i + 2(x +z)j + 2(x +y)k ∇f = λ∇g =⇒ yz = 2λ(y +z), xz = 2λ(x +z), xy = 2λ(x +y) =⇒ x = y = z From g(x, y, z) = 0 we get 6x 2 = 6a 2 =⇒ x = y = z = a Maximum volume is a 3 . 23. area A = 1 2 ax + 1 2 by + 1 2 cz. The geometry suggests that x 2 +y 2 +z 2 has a minimum. f(x, y, z) = x 2 +y 2 +z 2 , g(x, y, z) = ax +by +cz −2A ∇f = 2xi + 2yj + 2zk, ∇g = ai +bj +ck. ∇f = λ∇g =⇒ 2x = aλ, 2y = bλ, 2z = cλ. Solving these equations for x, y, z and substituting the results in g(x, y, z) = 0, we have a 2 λ 2 + b 2 λ 2 + c 2 λ 2 −2A = 0, λ = 4A a 2 +b 2 +c 2 and thus x = 2aA a 2 +b 2 +c 2 , y = 2bA a 2 +b 2 +c 2 , z = 2cA a 2 +b 2 +c 2 . The minimum is 4A 2 (a 2 +b 2 +c 2 ) −1 . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.7 841 24. Use figure 16.7.6 and write the side condition as x +y +z = 2π. For (a) maximize f(x, y, z) = 8R 3 sin 1 2 xsin 1 2 y sin 1 2 z. For (b) maximize f(x, y, z) = 4R 2 (sin 2 1 2 x + sin 2 1 2 y + sin 2 1 2 z). Each maximum occurs with x = y = z = 2π 3 . This gives an equilateral triangle. 25. Since the curve is asymptotic to the line y = x as x →−∞ and as x →∞, the maximum exists. The distance between the point (x, y) and the line y −x = 0 is given by |y −x| √ 1 + 1 = 1 2 √ 2 |y −x|. (see Section 1.4) Since the points on the curve are below the line y = x, we can replace |y −x| by x −y. To simplify the work we drop the constant factor 1 2 √ 2. f(x, y) = x −y, g(x, y) = x 3 −y 3 −1 ∇f = i −j, ∇g = 3x 2 i −3y 2 j. We use the cross-product equation (16.7.4): 1 _ −3y 2 _ − _ 3x 2 _ (−1) = 0, 3x 2 −3y 2 = 0, x = −y (x = y). Now g(x, y) = 0 gives us x 3 −(−x) 3 −1 = 0, 2x 3 = 1, x = 2 −1/3 . The point is _ 2 −1/3 , −2 −1/3 _ . 26. Let r, s, t be the intercepts. We wish to minimize the volume V = 1 6 rst [volume of pyramid = 1 3 base ×height] subject to the side condition a r + b s + c t = 1. The minimum occurs when all the intercepts are: r = 3a, s = 3b, t = 3c 27. It suffices to show that the square of the area is a maximum when a = b = c. f(a, b, c) = s(s −a)(s −b)(s −c), g(a, b, c) = a +b +c −2s ∇f = −s(s −b)(s −c)i −s(s −a)(s −c) j −s(s −a)(s −b)k, ∇g = i +j +k. (Here i, j, k are the unit vectors in the directions of increasing a, b, c.) ∇f = λ∇g =⇒ −s(s −b)(s −c) = −s(s −a)(s −c) = −s(s −a)(s −b) = λ. Thus, s −b = s −a = s −c so that a = b = c. This gives us the maximum, as no minimum exists. [The area can be made arbitrarily small by taking a close to s.] 28. f(x, y, z) = 8xyz, g(x, y, z) = a 2 −x 2 −y 2 −z 2 , x > 0, y > 0, z > 0. ∇f = 8yzi + 8xzj + 8xyk, ∇g = −2xi −2y j −2z k ∇f = λ∇g =⇒ 8yz = −2λx, 8xz = −2λy, 8xy = −2λz =⇒ x = y = z The rectangular box of maximum volume inscribed in the sphere is a cube. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 842 SECTION 16.7 29. (a) f(x, y) = (xy) 1/2 , g(x, y) = x +y −k, (x, y ≥ 0, k a nonnegative constant) ∇f = y 1/2 2x 1/2 i + x 1/2 2y 1/2 j, ∇g = i +j. ∇f = λ∇g =⇒ y 1/2 2x 1/2 = λ = x 1/2 2y 1/2 =⇒ x = y = k 2 . Thus, the maximum value of f is: f(k/2, k/2) = k 2 . (b) For all x, y (x, y ≥ 0) we have (xy) 1/2 = f(x, y) ≤ f(k/2, k/2) = k 2 = x +y 2 . 30. (a) The maximum occurs when x = y = z = k 3 , where (xyz) 1/3 = k 3 . (b) If x +y +z = k, then, by (a), (xyz) 1/3 ≤ k 3 = x +y +z 3 . 31. Simply extend the arguments used in Exercises 29 and 30. 32. T(x, y, z) = xy 2 z, g(x, y, z) = x 2 +y 2 +z 2 −1 ∇T = y 2 zi + 2xyzj +xy 2 k, ∇g = 2xi + 2yj + 2zk ∇T = λ∇g =⇒ y 2 z = 2λx, 2xyz = 2λy, xy 2 = 2λz =⇒ x 2 = y 2 2 = z 2 From g(x, y, z) = 0 we get 4x 2 = 1 =⇒ x = ± 1 2 , y = ± 1 √ 2 , z = ± 1 2 Maximum 1 8 at ( 1 2 , ± 1 √ 2 , 1 2 ), (− 1 2 , ± 1 √ 2 , − 1 2 ) Minimum − 1 8 at ( 1 2 , ± 1 √ 2 , − 1 2 ), (− 1 2 , ± 1 √ 2 , 1 2 ) 33. S(r, h) = 2πr 2 + 2πrh, g(r, h) = πr 2 h −V, (V constant) ∇S = (4πr + 2πh) i + 2πr j, ∇g = 2πrhi +πr 2 j. ∇S = λ∇g =⇒ 4πr + 2πh = 2πrhλ, 2πr = πr 2 λ =⇒ r = 2 λ , h = 4 λ . Now πr 2 h = V, =⇒ λ = 3 _ 16π V =⇒ r = 3 _ V 2π , h = 3 _ 4V π . To minimize the surface area, take r = 3 _ V 2π , and h = 3 _ 4V π . 34. f(x, y, z) = xyz, g(x, y, z) = x +y +z −18 ∇f = yzi +xzj +xyk, ∇g = i +j +k ∇f = λ∇g =⇒ yz = xz = xy =⇒ x = y = z =⇒ x = y = z = 6 35. Same as Exercise 13. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.7 843 36. f(x, y, z) = xyz 2 , g(x, y, z) = x +y +z −30 ∇f = yz 2 i +xz 2 j + 2xyzk, ∇g = i +j +k ∇f = λ∇g =⇒ yz 2 = λ = xz 2 = 2xyz =⇒ x = y = z 2 =⇒ x = y = 15 2 , z = 15 37. Let x, y, z denote the length, width and height of the box. We want to maximize the volume V of the box given that the surface area S is constant. That is: maximize V (x, y, z) = xyz subject to S(x, y, z) = 2xy + 2xz + 2yz = S constant Let g(x, y, z) = 2xy + 2xz + 2yz −S. Then ∇V = yz i +xz j +xy k, ∇g = (2y + 2z) i + (2x + 2z) j + (2x + 2y) k ∇V = λ∇g and the side condition yield the system of equations: yz = λ(2y + 2z) xz = λ(2x + 2z) xy = λ(2x + 2y) xy + 2xz + 2yz = S. Multiply the first equation by x, the second by y and subtract. This gives 0 = 2λz(x −y) =⇒ x = y since z = 0 =⇒ V = 0. Multiply the second equation by y, the third by z and subtract. This gives 0 = 2λx(y −z) =⇒ y = z since x = 0 =⇒ V = 0. Thus the closed rectangular box of maximum volume is a cube. The cube has side length x = _ S/6. 38. Let x, y, z denote the length, width and height of the box. We want to maximize the volume V of the box given that the surface area S is constant. That is: maximize V (x, y, z) = xyz subject to S(x, y, z) = 2xy + 2xz + 2yz = S constant Let g(x, y, z) = xy + 2xz + 2yz −S. Then ∇V = yz i +xz j +xy k, ∇g = (y + 2z) i + (x + 2z) j + (2x + 2y) k ∇V = λ∇g and the side condition yield the system of equations: yz = λ(y + 2z) xz = λ(x + 2z) xy = λ(2x + 2y) xy + 2xz + 2yz = S. Multiplying the first equation by x, the second by y and subtracting, we get 0 = 2λz(x −y) =⇒ x = y since z = 0 =⇒ V = 0. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 844 SECTION 16.7 Now put y = x in the third equation. This gives x 2 = 4λx =⇒ x(x −4λ) = 0 =⇒ x = 4λ since x = 0 =⇒ V = 0. Thus, x = y = 4λ. Substituting x = 4λ in the second equation gives z = 2λ. Finally, substituting these values for x, y, z in the fourth equation, we get 48λ 2 = S =⇒ λ 2 = S 48 =⇒ λ = 1 4 _ S 3 To maximize the volume, take x = y = _ S 3 and z = 1 2 _ S 3 . 39. S(r, h) = 4πr 2 + 2πrh, g(r, h) = 4 3 πr 3 +πr 2 h −10, 000 ∇S = (8πr + 2πh)i + 2πrj, ∇g = (4πr 2 + 2πrh)i +πr 2 j (Here i, j are the unit vectors in the directions of increasing r and h.) ∇S = λ∇g =⇒ 2π(4r +h) = 2πrλ(2r +h), 2πr = λπr 2 =⇒ h = 0 Maximum volume for sphere of radius r = 3 _ 7500/π meters. 40. (a) f(x, y, l) = xyl, g(x, y, l) = 2x + 2y +l −108, ∇f = yl i +xl j +xy k, ∇g = 2 i + 2 j + k. ∇f = λ∇g =⇒ yl = 2λ, xl = 2λ, xy = λ =⇒ y = x and l = 2x. Now 2x + 2y +l = 108, =⇒ x = 18 and l = 36. To maximize the volume, take x = y = 18 in. and l = 36 in. (b) f(r, l) = πr 2 l, g(r, l) = 2πr +l −108, ∇f = 2πrl i +πr 2 j, ∇g = 2π i +j. ∇f = λ∇g =⇒ 2πrl = 2πλ, πr 2 = λ, l = πr. Now 2πr +l = 108, =⇒ r = 36 π and l = 36. To maximize the volume, take r = 36/π in. and l = 36 in. 41. f(x, y, z) = 8xyz, g(x, y, z) = 4x 2 + 9y 2 + 36z 2 −36. ∇f(x, y, z) = 8yzi + 8xzj + 8xyk, ∇g(x, y, z) = 8xi + 18yj + 72zk. ∇f = λ∇g gives yz = λx, 4xz = 9λy, xy = 9λz. 4 xyz λ = 4x 2 , 4 xyz λ = 9y 2 , 4 xyz λ = 36z 2 . Also notice 4x 2 + 9y 2 + 36z 2 −36 = 0 We have 12 xyz λ = 36 =⇒ x = √ 3, y = 2 √ 3 , z = 1 √ 3 . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.7 845 Thus, V = 8xyz = 8 · √ 3 · 2 √ 3 · 1 √ 3 = 16 √ 3 . 42. Let x, y, z denote the length, width and height of the crate, and let C be the cost. Then C(x, y, z) = 30xy + 20xz + 20yz subject to g(x, y, z) = xyz −96 = 0 ∇C = (30y + 20z) i + (30x + 20z) j + (20x + 20y) k, ∇g = yz i +xz j +xy k ∇C = λ∇g implies 30y + 20z = λyz 30x + 20z = λxz 20x + 20y = λxy xyz = 96 Multiplying the first equation by x, the second by y and subtracting, we get 20z(x −y) = 0 =⇒ x = y since z = 0 Now put y = x in the third equation. This gives 40x = λx 2 =⇒ x(λx −40) = 0 =⇒ x = 40 λ since x = 0 Thus, x = y = 40/λ. Substituting x = 40/λ in the second equation gives z = 60/λ. Finally, substituting these values for x, y, z in the fourth equation, we get 40 λ 40 λ 60 λ = 96 =⇒ 96λ 3 = 96, 000 =⇒ λ 3 = 1000 =⇒ λ = 10 To minimize the cost, take x = y = 4 meters and z = 6 meters. 43. To simplify notation we set x = Q 1 , y = Q 2 , z = Q 3 . f(x, y, z) = 2x + 8y + 24z, g(x, y, z) = x 2 + 2y 2 + 4z 2 −4, 500, 000, 000 ∇f = 2i + 8j + 24k, ∇g = 2xi + 4yj + 8zk. ∇f = λ∇g =⇒ 2 = 2λx, 8 = 4λy, 24 = 8λz. Since λ = 0 here, we solve the equations for x, y, z: x = 1 λ , y = 2 λ , z = 3 λ , and substitute these results in g(x, y, z) = 0 to obtain 1 λ 2 + 2 _ 4 λ 2 _ + 4 _ 9 λ 2 _ −45 ×10 8 = 0, 45 λ 2 = 45 ×10 8 , λ = ±10 −4 . Since x, y, z are non-negative, λ = 10 −4 and x = 10 4 = Q 1 , y = 2 ×10 4 = Q 2 , z = 3 ×10 4 = Q 3 . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 846 SECTION 16.7 44. f(x, y, z) = 8xyz, g(x, y, z) = x 2 a 2 + y 2 b 2 + z 2 c 2 −1. We take a, b, c, x, y, z > 0 ∇f(x, y, z) = 8yzi + 8xzj + 8xyk; ∇g(x, y, z) = 2x a 2 i + 2y b 2 j + 2z c 2 k. ∇f = λ∇g and the side condition yield the system of equations: 8yz = 2xλ a 2 8xz = 2yλ b 2 8xy = 2zλ c 2 x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 Multiply the first equation by x, the second by y and subtract. This gives 0 = 2λx 2 a 2 − 2λy 2 b 2 =⇒ y = b a x since λ = 0 =⇒ V = 0. Multiply the second equation by y, the third by z and subtract. This gives 0 = 2λy 2 b 2 − 2λz 2 c 2 =⇒ z = c b y = c a x. Substituting these results into the side condition, we get: 3x 2 a 2 = 1 =⇒ x = a √ 3 =⇒ y = b √ 3 and z = c √ 3 . The volume of the largest rectangular box is: V = 8 _ a √ 3 __ b √ 3 __ c √ 3 _ = 8 √ 3 9 abc. PROJECT 16.7 1. f(x, y, z) = xy +z 2 , g(x, y, z) = x 2 +y 2 +z 2 −4, h(x, y, z) = y −x ∇f = y i +xj + 2z k, ∇g = 2xi + 2y j + 2z k, ∇h = −i +j. ∇f = λ∇g +μ∇h =⇒ y = 2λx −μ, x = 2λy −μ, 2z = 2λz 2z = 2λz =⇒ λ = 0 or z = 1. λ = 0 =⇒ y = −x which contradicts y = x. z = 1 =⇒ x 2 +y 2 = 3, which, with y = x implies x = ± _ 3/2; _ ± _ 3/2, ± _ 3/2 _ P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.7 847 Adding the first two equations gives x +y = 2λ(x +y) =⇒ (x +y)[2λ −1] = 0 =⇒ λ = 1 c or x = y = 0. x = y = 0 =⇒ z = ±2; (0, 0, ±2). λ = 1 2 =⇒ z = 0 and y = x =⇒ 2x 4 = 4; x = ± √ 2; (± √ 2, ± √ 2, 0). f _ ± _ 3/2, ± _ 3/2, 1 _ = 5 2 ; f(0, 0, ±2) = 4; f(± √ 2, ± √ 2, 0) = 2. The maximum value of f is 4; the minimum value is 2. 2. D(x, y, z) = x 2 +y 2 +z 2 , g(x, y, z) = x + 2y + 3z, h(x, y, z) = 2x + 3y +z −4 ∇D = 2xi + 2yj + 2zk, ∇g = i + 2j + 3k, ∇h = 2i + 3j +k ∇D = λ∇g +μ∇h =⇒ 2x = λ + 2μ, 2y = 2λ + 3μ, 2z = 3λ +μ =⇒ z = 5y −7x Then g(x, y, z) = 0 and h(x, y, z) = 0 give x = 68 75 , y = 16 15 , z = − 76 75 Closest point _ 68 75 , 16 15 , − 76 75 _ 3. f(x, y, z) = x 2 +y 2 +z 2 , g(x, y, z) = x +y −z + 1, h(x, y, z) = x 2 +y 2 −z 2 ∇f = 2xi + 2y j + 2z k, ∇g = i +j −k, ∇h = 2xi + 2y j −2z k. ∇f = λ∇g +μ∇h =⇒ 2x = λ + 2xμ, 2y = λ + 2yμ, 2z = −λ −2zμ Multiplying the first equation by y, the second equation by x and subtracting, yields λ(y −x) = 0. Now λ = 0 =⇒ μ = 1 =⇒ x = y = z = 0. This is impossible since x +y −z = −1. Therefore, we must have y = x =⇒ z = ± √ 2 x. Substituting y = x, z = √ 2 x into the equation x +y −z + 1 = 0, we get x = −1 − √ 2 2 =⇒ y = −1 − √ 2 2 , z = −1 − √ 2 Substituting y = x, z = − √ 2 x into the equation x +y −z + 1 = 0, we get x = −1 + √ 2 2 =⇒ y = −1 + √ 2 2 , z = −1 + √ 2 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 848 SECTION 16.8 Since f _ −1 − √ 2 2 , −1 − √ 2 2 , −1 − √ 2 _ = 6 + 4 √ 2 and f _ −1 + √ 2 2 , −1 + √ 2 2 , −1 + √ 2 _ = 6 −4 √ 2, it follows that _ −1 + √ 2 2 , −1 + √ 2 2 , −1 + √ 2 _ is closest to the origin and _ −1 − √ 2 2 , −1 − √ 2 2 , −1 − √ 2 _ is furthest from the origin. SECTION 16.8 1. df = _ 3x 2 y −2xy 2 _ Δx + _ x 3 −2x 2 y _ Δy 2. df = ∂f ∂x Δx + ∂f ∂y Δy + ∂f ∂z Δz = (y +z)Δx + (x +z)Δy + (x +y)Δz 3. df = (cos y +y sin x) Δx −(xsin y + cos x) Δy 4. df = 2xye 2z Δx +x 2 e 2z Δy + 2x 2 ye 2z Δz 5. df = Δx −(tan z) Δy − _ y sec 2 z _ Δz 6. df = _ x −y x +y + ln(x +y) _ Δx + _ x −y x +y −ln(x +y) _ Δy 7. df = y(y 2 +z 2 −x 2 ) (x 2 +y 2 +z 2 ) 2 Δx + x(x 2 +z 2 −y 2 ) (x 2 +y 2 +z 2 ) 2 Δy − 2xyz (x 2 +y 2 +z 2 ) 2 Δz 8. df = _ 2x x 2 +y 2 +e xy (1 +xy) _ Δx + _ 2y x 2 +y 2 +x 2 e xy _ Δy 9. df = [cos(x +y) + cos(x −y)] Δx + [cos(x +y) −cos(x −y)] Δy 10. df = ln _ 1 +y 1 −y _ Δx + 2x 1 −y 2 Δy 11. df = _ y 2 ze xz + ln z _ Δx + 2ye xz Δy + _ xy 2 e xz + x z _ Δz 12. df = y(1 −2x 2 )e −(x 2 +y 2 ) Δx +x(1 −2y 2 )e −(x 2 +y 2 ) Δy P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.8 849 13. Δu = _ (x + Δx) 2 −3(x + Δx)(y + Δy) + 2(y + Δy) 2 ¸ − _ x 2 −3xy + 2y 2 _ = _ (1.7) 2 −3(1.7)(−2.8) + 2(−2.8) 2 ¸ − _ 2 2 −3(2)(−3) + 2(−3) 2 _ = (2.89 + 14.28 + 15.68) −40 = −7.15 du = (2x −3y) Δx + (−3x + 4y) Δy = (4 + 9)(−0.3) + (−6 −12)(0.2) = −7.50 14. du = _ √ x −y + x +y 2 √ x −y _ Δx + _ √ x −y − x +y 2 √ x −y _ Δy = 1 15. Δu = _ (x + Δx) 2 (z + Δz) −2(y + Δy)(z + Δz) 2 + 3(x + Δx)(y + Δy)(z + Δz) ¸ − _ x 2 z −2yz 2 + 3xyz _ = _ (2.1) 2 (2.8) −2(1.3)(2.8) 2 + 3(2.1)(1.3)(2.8) ¸ − _ (2) 2 3 −2(1)(3) 2 + 3(2)(1)(3) ¸ = 2.896 du = (2xz + 3yz) Δx + _ −2z 2 + 3xz _ Δy + _ x 2 −4yz + 3xy _ Δz = [2(2)(3) + 3(1)(3)](0.1) + [−2(3) 2 + 3(2)(3)](0.3) + [2 2 −4(1)(3) + 3(2)(1)](−0.2) = 2.5 16. du = y 3 +yz 2 (x 2 +y 2 +z 2 ) 3/2 Δx + x 3 +xz 2 (x 2 +y 2 +z 2 ) 3/2 Δy − xyz (x 2 +y 2 +z 2 ) 3/2 Δz = 77 4(14) 3/2 17. f(x, y) = x 1/2 y 1/4 ; x = 121, y = 16, Δx = 4, Δy = 1 f(x + Δx, y + Δy) ∼ = f(x, y) +df = x 1/2 y 1/4 + 1 2 x −1/2 y 1/4 Δx + 1 4 x 1/2 y −3/4 Δy √ 125 4 √ 17 ∼ = √ 121 4 √ 16 + 1 2 (121) −1/2 (16) 1/4 (4) + 1 4 (121) 1/2 (16) −3/4 (1) = 11(2) + 1 2 _ 1 11 _ (2)(4) + 1 4 (11) _ 1 8 _ = 22 + 4 11 + 11 32 = 22 249 352 ∼ = 22.71 18. f(x, y) = (1 − √ x)(1 + √ y), x = 9, y = 25, Δx = 1, Δy = −1 df = − 1 + √ y 2 √ x Δx + 1 − √ x 2 √ y Δy = − 4 5 f(10, 24) ∼ = f(9, 25) − 4 5 = −12 4 5 19. f(x, y) = sin xcos y; x = π, y = π 4 , Δx = − π 7 , Δy = − π 20 df = cos xcos y Δx −sin xsin y Δy f(x + Δx, y + Δy) ∼ = f(x, y) +df sin 6 7 π cos 1 5 π ∼ = sin π cos π 4 + _ cos π cos π 4 __ − π 7 _ − _ sin π sin π 4 __ − π 20 _ = 0 + _ 1 2 √ 2 _ _ π 7 _ + 0 = π √ 2 14 ∼ = 0.32 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 850 SECTION 16.8 20. f(x, y) = √ xtan y, x = 9, y = π 4 , Δx = −1, Δy = 1 16 π df = 1 2 √ x tan yΔx + √ xsec 2 yΔy = − 1 6 + 3π 8 f _ 8, 5 16 π _ ∼ = f _ 9, π 4 _ − 1 6 + 3π 8 = 17 6 + 3 8 π ∼ = 4.01 21. f(2.9, 0.01) ∼ = f(3, 0) +df, where df is to be evaluated at x = 3, y = 0, Δx = −0.1, Δy = 0.01. df = _ 2xe xy +x 2 ye xy _ Δx +x 3 e xy Δy = _ 2(3)e 0 + (3) 2 (0)e 0 ¸ (−0.1) + 3 3 e 0 (0.01) = −0.33 Thus, f(2.9, .01) ∼ = 3 2 e 0 −0.33 = 8.67. 22. x = 2, y = 3, z = 3, Δx = 0.12, Δy = −0.08, Δz = 0.02 df = 2xy cos πzΔx +x 2 cos πzΔy −πx 2 y sin πzΔz = −12(0.12) + 4(0.08) = −1.12 f(2.12, 2.92, 3.02) ∼ = f(2, 3, 3) −1.12 = −13.12 23. f(2.94, 1.1, 0.92) ∼ = f(3, 1, 1) +df, where df is to be evaluated at x = 3, y = 1, z = 1, Δx = −0.06, Δy = 0.1, Δz = −0.08 df = tan −1 yz Δx + xz 1 +y 2 z 2 Δy + xy 1 +y 2 z 2 Δz = π 4 (−0.06) + (1.5)(0.1) + (1.5)(−0.08) ∼ = −0.0171 Thus, f(2.94, 1.1, 0.92) ∼ = 3 4 π −0.0171 ∼ = 2.3391 24. x = 3, y = 4, Δx = 0.06, Δy = −0.12 df = x _ x 2 +y 2 Δx + y _ x 2 +y 2 Δy = 3 5 (0.06) + 4 5 (−0.12) = −0.06 f(3.06, 3.88) ∼ = f(3, 4) −0.06 = 4.94 25. df = ∂z ∂x Δx + ∂z ∂y Δy = 2y (x +y) 2 Δx − 2x (x +y) 2 Δy With x = 4, y = 2, Δx = 0.1, Δy = 0.1, we get df = 4 36 (0.1) − 8 36 (0.1) = − 1 90 . The exact change is 4.1 −2.1 4.1 + 2.1 − 4 −2 4 + 2 = 2 6.2 − 1 3 = − 1 93 . 26. V (r, h) = πr 2 h, r = 8, h = 12, Δr = −0.3, Δh = 0.2 dV = 2πrhΔr +πr 2 Δh = 192π(−0.3) + 64π(0.2) = −44.8π decreases by approximately 44.8π cubic inches. 27. S = 2πr 2 + 2πrh; r = 8, h = 12, Δr = −0.3, Δh = 0.2 dS = ∂S ∂r Δr + ∂S ∂h Δh = (4πr + 2πh) Δr + (2πr) Δh = 56π(−0.3) + 16π(0.2) = −13.6π. The area decreases about 13.6π in. 2 . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.8 851 28. dT = 2xcos πzΔx −2y sin πzΔy −(πx 2 sin πz +πy 2 cos πz)Δz = 4(0.1) −4π(0.2) = 2 5 − 4 5 π T decreases by about 4 5 π − 2 5 ∼ = 2.11 29. S(9.98, 5.88, 4.08) ∼ = S(10, 6, 4) +dS = 248 +dS, where dS = (2w + 2h) Δl + (2l + 2h) Δw + (2l + 2w) Δh = 20(−0.02) + 28(−0.12) + 32(0.08) = −1.20 Thus, S(9.98, 5.88, 4.08) ∼ = 248 −1.20 = 246.80. 30. f(r, h) = πr 2 h 3 , r = 7, h = 10, Δr = 0.2, Δh = 0.15 df = 2πrh 3 Δr + πr 2 3 Δh = 140 3 π(0.2) + 49 3 π(0.15) = π 3 (35.35) f(7.2, 10.15) ∼ = f(7, 10) + π 3 (35.35) = 525.35 π 3 ∼ = 550.15 31. (a) dV = yz Δx +xz Δy +xy Δz = (8)(6)(0.02) + (12)(6)(−0.05) + (12)(8)(0.03) = 0.24 (b) ΔV = (12.02)(7.95)(6.03) −(12)(8)(6) = 0.22077 32. (a) S(x, y, z) = 2(xy +xz +yz), x = 12, y = 8, z = 6, Δx = 0.02, Δy = −0.05, Δz = 0.03 dS = 2(y +z)Δx + 2(x +z)Δy + 2(x +y)Δz = 28(0.02) + 36(−0.05) + 40(0.03) = −0.04 (b) ΔS = S(12.02, 7.95, 6.03) −S(12, 8, 6) = −0.0438 33. T(P) −T(Q) ∼ = dT = (−2x + 2yz) Δx + (−2y + 2xz) Δy + (−2z + 2xy) Δz Letting x = 1, y = 3, z = 4, Δx = 0.15, Δy = −0.10, Δz = 0.10, we have dT = (22)(0.15) + (2)(−0.10) + (−2)(0.10) = 2.9 34. Amount of paint is increase in volume. f(x, y, z) = xyz, x = 48 in, y = 24 in, z = 36 in, Δx = Δy = Δz = 2 16 in. Δf ∼ = df = yzΔx +xzΔy +xyΔz = 3774( 2 16 ) = 468 The amount of paint is approximately 468 cubic inches. 35. (a) πr 2 h = π(r + Δr) 2 (h + Δh) =⇒ Δh = r 2 h (r + Δr) 2 −h = − (2r + Δr)h (r + Δr) 2 Δr. df = (2πrh) Δr +πr 2 Δh, df = 0 =⇒ Δh = −2h r Δr. (b) 2πr 2 + 2πrh = 2π(r + Δr) 2 + 2π(r + Δr)(h + Δh). Solving for Δh, Δh = r 2 +rh −(r + Δr) 2 r + Δr −h = − 2r +h + Δr r + Δr Δr. df = (4πr + 2πh) Δr + 2πr Δh, df = 0 =⇒ Δh = − _ 2r +h r _ Δr. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 852 SECTION 16.9 36. The area is given by A = 1 2 x 2 tan θ. (a) The change in area is approximated by: dA = xtan θ Δx + 1 2 x 2 sec 2 θ Δθ = 3Δx + 25 2 Δθ. (b) The actual change in area is 1 2 (x + Δx) 2 tan(θ + Δθ) − 1 2 x 2 tan θ = 1 2 [4 + Δx] 2 tan[arctan (3/4) + Δθ] −6. (c) The area is more sensitive to a change in θ. 37. (a) A = 1 2 x 2 sin θ; ΔA ∼ = dA = xsin θΔx + x 2 2 cos θΔθ (b) The area is more sensitive to changes in θ if x > 2 tan θ, otherwise it is more sensitive to changes in x. 38. (a) dV ∼ = yzΔx +xzΔy +xyΔz, x = 60 in, y = 36 in, z = 42 in Maximum possible error = 6192( 1 12 ) = 516 cubic inches. (b) dS ∼ = 2(y +z)Δx + 2(x +z)Δy + 2(x +y)Δz Maximum possible error = 552( 1 12 ) = 46 square inches 39. s = A A−W ; A = 9, W = 5, ΔA = ±0.01, ΔW = ±0.02 ds = ∂s ∂A ΔA+ ∂s ∂W ΔW = −W (A−W) 2 ΔA+ A (A−W) 2 ΔW = − 5 16 (±0.01) + 9 16 (±0.02) ∼ = ±0.014 The maximum possible error in the value of s is 0.014 lbs; 2.23 ≤ s + Δs ≤ 2.27 40. Assuming A > W, s is more sensitive to change in A. SECTION 16.9 1. ∂f ∂x = xy 2 , f(x, y) = 1 2 x 2 y 2 +φ(y), ∂f ∂y = x 2 y +φ (y) = x 2 y. Thus, φ (y) = 0, φ(y) = C, and f(x, y) = 1 2 x 2 y 2 +C. 2. ∂f ∂x = x, ∂f ∂y = y =⇒ f(x, y) = 1 2 (x 2 +y 2 ) +C 3. ∂f ∂x = y, f(x, y) = xy +φ(y), ∂f ∂y = x +φ (y) = x. Thus, φ (y) = 0, φ(y) = C, and f(x, y) = xy +C. 4. ∂f ∂x = x 2 +y =⇒ f(x, y) = x 3 3 +xy +φ(y); ∂f ∂y = x +φ (y) = y 3 +x =⇒ f(x, y) = 1 3 x 3 + 1 4 y 4 +xy +C 5. No; ∂ ∂y _ y 3 +x _ = 3y 2 whereas ∂ ∂x _ x 2 +y _ = 2x. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.9 853 6. ∂f ∂x = y 2 e x −y =⇒ f(x, y) = y 2 e x −xy +φ(y); ∂f ∂y = 2ye x −x +φ (y) = 2ye x −x =⇒ f(x, y) = y 2 e x −xy +C 7. ∂f ∂x = cos x −y sin x, f(x, y) = sin x +y cos x +φ(y), ∂f ∂y = cos x +φ (y) = cos x. Thus, φ (y) = 0, φ(y) = C, and f(x, y) = sin x +y cos x +C. 8. ∂f ∂x = 1 +e y =⇒ f(x, y) = x +xe y +φ(y); ∂f ∂y = xe y +φ (y) = xe y +y 2 =⇒ f(x, y) = x +xe y + y 3 3 +C 9. ∂f ∂x = e x cos y 2 , f(x, y) = e x cos y 2 +φ(y), ∂f ∂y = −2ye x sin y 2 +φ (y) = −2ye x sin y 2 . Thus, φ (y) = 0, φ(y) = C, and f(x, y) = e x cos y 2 +C. 10. ∂ 2 f ∂y∂x = −e x sin y, ∂ 2 f ∂x∂y = e x sin y = ∂ 2 f ∂y∂x ; not a gradient. 11. ∂f ∂y = xe x −e −y , f(x, y) = xye x +e −y +φ(x), ∂f ∂x = ye x +xye x +φ (x) = ye x (1 +x). Thus, φ (x) = 0, φ(x) = C, and f(x, y) = xye x +e −y +C. 12. ∂f ∂x = e x + 2xy =⇒ f(x, y) = e x +x 2 y +φ(y); ∂f ∂y = x 2 +φ (y) = x 2 + sin y =⇒ f(x, y) = e x +x 2 y −cos y +C 13. No; ∂ ∂y _ xe xy +x 2 _ = x 2 e xy whereas ∂ ∂x (ye xy −2y) = y 2 e xy 14. ∂f ∂y = xsin x + 2y + 1 =⇒ f(x, y) = xy sin x +y 2 +y +φ(x) ∂f ∂x = y sin x +xy cos x +φ (x) = y sin x +xy cos x =⇒ f(x, y) = xy sin x +y 2 +y +C 15. ∂f ∂x = 1 +y 2 +xy 2 , f(x, y) = x +xy 2 + 1 2 x 2 y 2 +φ(y), ∂f ∂y = 2xy +x 2 y +φ (y) = x 2 y +y + 2xy + 1. Thus, φ (y) = y + 1, φ(y) = 1 2 y 2 +y +C and f(x, y) = x +xy 2 + 1 2 x 2 y 2 + 1 2 y 2 +y +C. 16. ∂f ∂x = 2 ln 3y + 1 x =⇒ f(x, y) = 2xln 3y + ln |x| +φ(y); ∂f ∂y = 2x y +φ (y) = 2x y +y 2 f(x, y) = 2xln 3y + ln |x| + y 3 3 +C 17. ∂f ∂x = x _ x 2 +y 2 , f(x, y) = _ x 2 +y 2 +φ(y), ∂f ∂y = y _ x 2 +y 2 +φ (y) = y _ x 2 +y 2 . Thus, φ (y) = 0, φ(y) = C, and f(x, y) = _ x 2 +y 2 +C. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 854 SECTION 16.9 18. ∂f ∂x = xtan y + sec 2 x =⇒ f(x, y) = x 2 2 tan y + tan x +φ(y); ∂f ∂y = x 2 2 sec 2 y +φ (y) = x 2 2 sec 2 y +πy; =⇒ f(x, y) = x 2 2 tan y + tan x + π 2 y 2 +C. 19. ∂f ∂x = x 2 sin −1 y, f(x, y) = 1 3 x 3 sin −1 y +φ(y), ∂f ∂y = x 3 3 _ 1 −y 2 +φ (y) = x 3 3 _ 1 −y 2 −ln y. Thus, φ (y) = −ln y, =⇒ φ(y) = y −y ln y +C, and f(x, y) = 1 3 x 3 sin −1 y +y −y ln y +C. 20. ∂f ∂x = tan −1 y √ 1 −x 2 + x y =⇒ f = sin −1 xtan −1 y + x 2 2y +φ(y); ∂f ∂y = sin −1 x 1 +y 2 − x 2 2y 2 +φ (y) = sin −1 x 1 +y 2 − x 2 2y 2 + 1 =⇒ f(x, y) = sin −1 xtan −1 y + x 2 2y +y +C. 21. (a) Yes (b) Yes (c) No 22. (a) f(x, y) = (x −y)e −x 2 y +C (b) f(x, y) = sin(x +y) −cos(x −y) +C; f(π/3, π/4) = 6 =⇒ C = 6 f(x, y) = sin(x +y) −cos(x −y) + 6. 23. ∂f ∂x = f(x, y), ∂f/∂x f(x, y) = 1, ln f(x, y) = x +φ(y), ∂f/∂y f(x, y) = 0 +φ (y), ∂f ∂y = f(x, y). Thus, φ (y) = 1, φ(y) = y +K, and f(x, y) = e x+y+K = Ce x+y . 24. ∂f ∂x = e g(x,y) g x (x, y) =⇒ f(x, y) = e g(x,y) +φ(y); ∂f ∂y = e g(x,y) g y (x, y) +φ (y) = e g(x,y) g y (x, y) =⇒ f(x, y) = e g(x,y) +C. 25. (a) P = 2x, Q = z, R = y; ∂P ∂y = 0 = ∂Q ∂x , ∂P ∂z = 0 = ∂R ∂x , ∂Q ∂z = 1 = ∂R ∂y (b), (c), and (d) ∂f ∂x = 2x, f(x, y, z) = x 2 +g(y, z). ∂f ∂y = 0 + ∂g ∂y with ∂f ∂y = z =⇒ ∂g ∂y = z. Then, g(y, z) = yz +h(z) =⇒ f(x, y, z) = x 2 +yz +h(z), ∂f ∂z = 0 +y +h (z) and ∂f ∂z = y =⇒ h (z) = 0. Thus, h(z) = C and f(x, y, z) = x 2 +yz +C. 26. ∂f ∂x = yz =⇒ f(x, y, z) = xyz +g(y, z); ∂f ∂y = xz + ∂g ∂y = xz =⇒ f = xyz +h(z) ∂f ∂z = xy +h (z) = xy =⇒ f(x, y, z) = xyz +C P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.9 855 27. The function is a gradient by the test stated before Exercise 25. Take P = 2x +y, Q = 2y +x +z, R = y −2z. Then ∂P ∂y = 1 = ∂Q ∂x , ∂P ∂z = 0 = ∂R ∂x , ∂Q ∂z = 1 = ∂R ∂y . Next, we find f where ∇f = Pi +Qj +Rk. ∂f ∂x = 2x +y =⇒ f(x, y, z) = x 2 +xy +g(y, z). ∂f ∂y = x + ∂g ∂y with ∂f ∂y = 2y +x +z =⇒ ∂g ∂y = 2y +z. Then, g(y, z) = y 2 +yz +h(z), f(x, y, z) = x 2 +xy +y 2 +yz +h(z). ∂f ∂z = y +h (z) = y −2z =⇒ h (z) = −2z. Thus, h(z) = −z 2 +C and f(x, y, z) = x 2 +xy +y 2 +yz −z 2 +C. 28. ∂f ∂x = 2xsin 2y cos z =⇒ f(x, y, z) = x 2 sin 2y cos z +g(y, z); ∂f ∂y = 2x 2 cos 2y cos z + ∂g ∂y = 2x 2 cos 2y cos z =⇒ f(x, y, z) = x 2 sin 2y cos z +h(z) ∂f ∂z = −x 2 sin 2y sin z +h (z) = −x 2 sin 2y sin z =⇒ f(x, y, z) = x 2 sin 2y cos z +C 29. The function is a gradient by the test stated before Exercise 25. Take P = y 2 z 3 + 1, Q = 2xyz 3 +y, R = 3xy 2 z 2 + 1. Then ∂P ∂y = 2yz 3 = ∂Q ∂x , ∂P ∂z = 3y 2 z 2 = ∂R ∂x , ∂Q ∂z = 6xyz 2 = ∂R ∂y . Next, we find f where ∇f = Pi +Qj +Rk. ∂f ∂x = y 2 z 3 + 1, f(x, y, z) = xy 2 z 3 +x +g(y, z). ∂f ∂y = 2xyz 3 + ∂g ∂y with ∂f ∂y = 2xyz 3 +y =⇒ ∂g ∂y = y. Then, g(y, z) = 1 2 y 2 +h(z), f(x, y, z) = xy 2 z 3 +x + 1 2 y 2 +h(z). ∂f ∂z = 3xy 2 z 2 +h (z) = 3xy 2 z 2 + 1 =⇒ h (z) = 1. Thus, h(z) = z +C and f(x, y, z) = xy 2 z 3 +x + 1 2 y 2 +z +C. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 856 REVIEW EXERCISES 30. ∂f ∂x = y z −e z =⇒ f(x, y, z) = xy z −xe z +g(y, z) ∂f ∂y = x z + ∂g ∂y = x z + 1 =⇒ f(x, y, z) = xy z +y −xe z +h(z) ∂f ∂z = − xy z 2 −xe z +h (z) = −xe z − xy z 2 =⇒ f(x, y, z) = xy z −xe z +y +C 31. F(r) = ∇ _ GmM r _ 32. h(r) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∇ _ k n + 2 r n+2 _ , n = 2 ∇(k ln r) , n = −2. REVIEW EXERCISES 1. ∇f(x, y) = (4x −4y)i + (3y 2 −4x)j 2. ∇f(x, y) = y 3 −x 2 y (x 2 +y 2 ) 2 i + x 3 −xy 2 (x 2 +y 2 ) 2 j 3. ∇f(x, y) = (ye xy tan 2x + 2e xy sec 2 2x) i +xe xy tan 2xj 4. ∇f = 1 x 2 +y 2 +z 2 (xi +y j +z k) 5. ∇f(x, y) = 2xe −yz sec z i, −zx 2 e −yz sec zj −(x 2 ye −yz sec z −x 2 e −yz sec z tan z)k 6. ∇f(x, y) = ye −3z cos xy i +e −3z (xcos xy + sin y) j, −3e −3z (sin xy −cos y) k 7. ∇f(x, y) = (2x −2y) i −2xj, ∇f(1, −2) = 6 i −2 j; u a = 1 √ 5 i + 2 √ 5 j; f ua (1, −2) = ∇f(1, −2) · u a = 2 √ 5 . 8. ∇f(x, y) = (e xy +xye xy ) i +x 2 e xy j, ∇f(2, 0) = i + 4j; u a = 1 2 i + √ 3 2 j f ua (2, 0) = ∇f(2, 0) · u a = 1 2 + 2 √ 3. 9. ∇f(x, y, z) = (y 2 + 6xz) i + (2xy + 2z) j + (2y + 3x 2 ) k, ∇f(1, −2, 3) = 22 i + 2 j −k; u a = 1 3 i − 2 3 j + 2 3 k; f ua (1, −2, 3) = ∇f(1, −2, 3) · u a = 16 3 . 10. ∇f(x, y, z) = 2x x 2 +y 2 +z 2 i + 2y x 2 +y 2 +z 2 j + 2z x 2 +y 2 +z 2 k, ∇f(1, 2, 3) = 1 7 (i + 2 j + 3 k); u a = 1 √ 3 i − 1 √ 3 j + 1 √ 3 k; f ua (1, 2, 3) = ∇f(1, 2, 3) · u a = 2 7 √ 3 . P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 REVIEW EXERCISES 857 11. ∇f(x, y) = (6x −2y 2 ) i −4xy j, ∇f(3, −2) = 10 i + 24 j; a = (0, 0) −(3, −2) = (−3, 2) = −3 i + 2 j, u a = −3 √ 13 i + 2 √ 13 j; f ua (3, −2) = ∇f(3, −2) · u a = 18 √ 13 . 12. ∇f(x, y, z) = (y 2 z −3yz) i + (2xyz −3xz) j + (xy 2 −3xy) k, ∇f(1, −1, 2) = 8 i −10 j + 4 k; r (t) = i −π sin πt j + 2e t−1 k, a = r (1) = i + 2k, u a = 1 √ 5 i + 2 √ 5 k; f ua (1, −1, 2) = ∇f(1, −1, 2) · u a = 16 √ 5 . 13. ∇f(x, y, z) = 1 _ x 2 +y 2 +z 2 (xi +y j +z k), ∇f(3, −1, 4) = 1 √ 26 (3 i −j + 4 k); a = ±(4 i −3 j +k), u a = ± 1 √ 26 (4 i −3 j +k); f ua (3, −1, 4) = ∇f(3, −1, 4) · u a = ± 19 26 . 14. ∇f(x, y) = 2e 2x (cos y −sin y) i −e 2x (sin y + cos y) j, ∇f _ 1 2 , − 1 2 π _ = 2e i +e j; maximum directional derivative: ∇f _ 1 2 , − 1 2 π _ = e √ 5. 15. ∇f(x, y, z) = cos xyz(yz i +xz j +xy k), ∇f( 1 2 , 1 3 , π) = π √ 3 6 i + π √ 3 4 j + √ 3 12 k; minimum directional derivative: f u = −∇f( 1 2 , 1 3 , π) = − √ 39π 2 +3 12 16. Let r(t) = x(t) i +y(t) j be the path of the particle. ∇I(x, y) = −2xi −6y j. Then x (t) = −2x(t), y (t) = −6y(t) =⇒ x(t) = C 1 e −2t , y(t) = C 2 e −6t . r(0) = (4, 3) =⇒ C 1 = 4, C 2 = 3. Therefore the path of the particle is: r(t) = 4e −2t i + 3e −6t j, t ≥ 0, or, y = 3 64 x 3 , 0 < x ≤ 4 17. Let r(t) = x(t) i +y(t) j be the path of the particle. ∇T = −e −x cos y i −e −x sin y j. Then x (t) = −e −x(t) cos y(t), y (t) = −e −x(t) sin y(t) =⇒ y (t) x (t) = tan y(t) =⇒ dy dx = tan y The solution is sin y = Ce x . Since r(0) = 0, C = 0 and y = 0. The particle moves to the right the x-axis. 18. ∇z = 8xi + 2y j; r(t) = x(t) i +y(t) j. x (t) = −8x(t), y (t) = −2y(t) =⇒ x(t) = C 1 e −8t , y(t) = C 2 e −2t . (a) r(0) = (1, 1) =⇒ C 1 = 1, C 2 = 1; x = e −8t , y = e −2t or x = y 4 . (b) r(0) = (1, −2) =⇒ C 1 = 1, C 2 = −2; x = e −8t , y = −2e −2t or x = y 4 /16. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 858 REVIEW EXERCISES 19. ∇f(x, y) = e x arctan y i +e x 1 1 +y 2 j; ∇f(0, 1) = π 4 i + 1 2 j. u = ∇f(0, 1) ∇f(0, 1) = 1 √ 4 +π 2 (π i + 2 j); rate: ∇f(0, 1) = √ π 2 + 4 4 20. ∇f(x, y, z) = 1 (y +z) 2 [(y +z) i + (z −x) j −(x +y) k]; ∇f(−1, 1, 3) = 1 4 i + 1 4 j. u = ∇f(−1, 1, 3) ∇f(−1, 1, 3) = 1 2 √ 2 i + 1 2 √ 2 j; rate: ∇f(−1, 1, 3) = 1 4 √ 2 21. rate: df dt = ∇f · r = _ 4xi −9y 2 j _ · _ 1 2 t −1/2 i + 2e 2t j _ = 2 −18e 6t 22. f(r(t) = sin t 2 + cos t 2 , rate: f (r(t)) = 2t cos t 2 −2t sin t 2 23. rate: df dt = ∇f · r = __ 1 y + z x 2 _ i − x y 2 j − 1 x k _ · (cos t i −sin t j + sec 2 t k) = 1 −sin t cos 2 t 24. du dt = ∇u · r = 1 1 +x 2 y 2 (y i +xj) · (sec 2 t i + 2e 2t j) = e 2t 1 +e 4t tan 2 t (sec 2 t + 2 tan t) 25. du dt = ∇u · r = _ (3y 2 −2x) i + 6xy j ¸ · [(2t + 2) i + 3 j] = 104t 3 + 150t 2 −8t 26. u(r(t)) = 1 √ 1 +t 2 , du dt = −t (1 +t 2 ) 3/2 27. area A = 1 2 x(t)y(t) sin θ(t) dA dt = 0 = 1 2 y(t)x (t) sin θ(t) + 1 2 x(t)y (t) sin θ(t) + 1 2 θ (t)x(t)y(t) cos θ(t) = 0 At x = 4, y = 5, θ = π/3, dx dt = dy dt = 2, we have 5 dθ dt + 2 √ 3 + 5 √ 3 2 = 0 =⇒ dθ dt = − 9 √ 3 10 . 28. V = πr 2 h; dV dt = 2πrh dr dt +πr 2 dh dt Measure in centimeters: at r = 12, h = 1000, dr dt = 4, dh dt = 150, dV dt = 2π(12)(1000)(4) +π(144)(150) = 117, 600π cu.cm/yr ∼ = 0.37 cu m/yr. 29. ∂u ∂s = ∂u ∂x + ∂u ∂y ; ∂u ∂t = ∂u ∂x − ∂u ∂y ∂u ∂s ∂u ∂t = _ ∂u ∂x + ∂u ∂y _ _ ∂u ∂x − ∂u ∂y _ = _ ∂u ∂x _ 2 − _ ∂u ∂y _ 2 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 REVIEW EXERCISES 859 30. ∂u ∂s = u x e s cos t +u y e s sin t ∂ 2 u ∂s 2 = u xx e 2s cos 2 t +u xy e 2s sin t cos t +u x e s cos t +u y e s sin t +u yx e 2s cos t sin t +u yy e 2s sin 2 t ∂u ∂t = −u x e s sin t +u y e s cos t ∂ 2 u ∂t 2 = u xx e 2s sin 2 t −u xy e 2s sin t cos t −u x e s cos t −u y e s sin t −u yx e 2s cos t sin t +u yy e 2s cos 2 t ∂ 2 u ∂s 2 + ∂ 2 u ∂t 2 = e 2s (u xx +u yy ) =⇒ ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 = e −2s _ ∂ 2 u ∂s 2 + ∂ 2 u ∂t 2 _ 31. ∇f(x, y) = (3x 2 −6xy) i + (−3x 2 + 2y) j; ∇f(1, −1) = N = 9 i −5 j normal line: x = 1 + 9t, y = −1 −5t; tangent line: x = 1 + 5t, y = −1 + 9t 32. ∇f(x, y) = −πy sin πxy i −πxsin πxy j; ∇f(1/3, 2) = −π √ 3 i − π √ 3 6 j, take N = 6i+ j; normal line: x = 1/3 + 6t, y = 2 +t; tangent line: x = 1/3 +t, y = 2 −6t 33. Set f(x, y, z) = x 1/2 +y 1/2 −z ∇f(x, y, z) = 1 2 √ x i + 1 2 √ y j −k; ∇f(1, 1, 2) = 1 2 i + 1 2 j −k. Take N = i +j −2 k. tangent plane: (x −1) + (y −1) −2(z −2) = 0; normal line: x = 1 +t, y = 1 +t, z = 2 −2t 34. Set f(x, y, z) = x 2 +y 2 +z 2 . ∇f(x, y, z) = 2xi + 2y j + 2z k; ∇f(1, 2, −2) = 2 i + 4 j −4 k. Take N = i + 2 j −2 k. tangent plane: (x −1) + 2(y −2) −2(z + 2) = 0; normal line: x = 1 +t, y = 2 + 2t, z = −2 −2t 35. Set f(x, y, z) = z 3 +xyz −2. ∇f(x, y, z) = yz i +xz j + (3z 2 +xy) k; ∇f(1, 1, 1) = i +j + 4k. tangent plane: (x −1) + (y −1) + 4(z −1) = 0; normal line: x = 1 +t; y = 1 +t; z = 1 + 4t 36. Set f(x, y, z) = e 3x sin 3y −z. ∇f(x, y, z) = 3e 3x sin 3y i + 3e 3x cos 3y j −k; ∇f(0, π/6, 1) = 3i −k. tangent plane: 3(x −0) −(z −1) = 0 or 3x −z + 1 = 0; normal line: x = 3t, y = π/6, z = 1 −t 37. The point (2, 2, 1) is on each hyperboloid. Set f(x, y, z) = x 2 + 2y 2 −4z 2 , g(x, y, z) = 4x 2 −y 2 + 2z 2 . ∇f = 2xi + 4y j −8z k, ∇f(2, 2, 1) = (4, 8, −8); ∇g = 8xi −2y j + 4z k, ∇g(2, 2, 1) = (16, −4, 4). Since ∇f(2, 2, 1) · ∇g(2, 2, 1) = 0, the hyperboloids are mutually perpendicular at (2, 2, 1). P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 860 REVIEW EXERCISES 38. Set f(x, y, z) = x 2 +y 2 +z 2 . At each point (x 0 , y 0 , z 0 ), ∇f(x 0 , y 0 , z 0 ) = 2x 0 i + 2y 0 j + 2z 0 k. normal line to the sphere: x = x 0 +x 0 t, y = y 0 +y 0 t, z = z 0 +z 0 t. At t = −1, x = y = z = 0. 39. ∇f(x, y) = (2xy −2y) i + (x 2 −2x + 4y −15) j = 0 at (5, 0), (−3, 0), (1, 4). f xx = 2y, f xy = 2x −2, f yy = 4. point A B C D result (5, 0) 0 8 4 −64 saddle (−3, 0) 0 −8 4 −64 saddle (1, 4) 8 0 4 32 loc. min. f(1, 4) = −34 40. ∇f(x, y) = (6x −3y 2 ) i + (3y 2 + 6y −6xy) j = 0 at (0, 0), (2, 2), ( 1 2 , −1). f xx = 6, f xy = −6y, f yy = 6y −6x + 6. point A B C D result (0, 0) 6 0 6 36 loc. min. (2, 2) 6 −12 6 −108 saddle ( 1 2 , −1) 6 6 −3 −54 saddle f(0, 0) = 0 41. ∇f(x, y) = (3x 2 −18y) i + (3y 2 −18x) j = 0 at (0, 0), (6, 6). f xx = 6x, f xy = −18, f yy = 6y. point A B C D result (0, 0) 0 −18 0 −18 2 saddle (6, 6) 36 −18 36 > 0 loc. min. f(6, 6) = −216 42. ∇f(x, y) = (3x 2 −12x) i + (2y + 1) j = 0 at (0, − 1 2 ), (4, − 1 2 ). f xx = 6x −12, f xy = 0, f yy = 2. point A B C D result (0, − 1 2 ) −12 0 2 −24 saddle (4, − 1 2 ) 12 0 2 24 loc. min. f(4, − 1 2 ) = − 145 4 P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 REVIEW EXERCISES 861 43. ∇f(x, y) = (1 −2xy +y 2 ) i + (−1 −x 2 + 2xy) j = 0 at (1, 1), (−1, −1). f xx = −2y, f xy = −2x + 2y, f yy = 2x. point A B C D result (1, 1) −2 0 2 −4 saddle (−1, −1) 2 0 −2 −4 saddle 44. ∇f(x, y) = e −(x 2 +y 2 )/2 _ (y 2 −x 2 y 2 ) i + (2xy −xy 3 ) j ¸ = 0 at (±1, ± √ 2), (x, 0), x any real number. f xx = e −(x 2 +y 2 )/2 (−3xy 2 +x 3 y 2 ), f xy = e −(x 2 +y 2 )/2 (2y −y 3 −2x 2 y +x 2 y 3 ), f yy = e −(x 2 +y 2 )/2 (2x −5xy 2 +xy 4 ). point A B C D result (1, √ 2) −4e −3/2 0 −4e −3/2 16e −3 loc. max (1, − √ 2) −4e −3/2 0 −4e −3/2 16e −3 loc. max (−1, √ 2) 4e −3/2 0 4e −3/2 16e −3 loc. min (−1, − √ 2) 4e −3/2 0 4e −3/2 16e −3 loc. min local maxima: f(1, √ 2) = f(1, − √ 2) = 2e −3/2 ; local minima: f(−1, √ 2) = f(−1, − √ 2) = −2e −3/2 . At (x, 0), D = 0 and f(x, 0) ≡ 0. For x < 0, f(x, y) < f(x, 0) for all y = 0; for x > 0, f(x, y) > f(x, 0) for all y > 0; (0, 0) is a saddle point. Here is a graph of the surface. 45. ∇f = (2x −2) i + (2y + 2) j = 0 at (1, −1) in D; f(1, −1) = 0 Next we consider the boundary of D. We parametrize the circle by: C : r(t) = 2 cos t i + 2 sin t j, t ∈ [ 0, 2π ] The values of f on the boundary are given by the function F(t) = f(r(t)) = 6 −4 cos t + 4 sin t, t ∈ [ 0, 2π ] F (t) = 4 sin t + 4 cos t : F (t) = 0 =⇒ sin t = −cos t =⇒ t = 3 4 π, 7 4 π P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 862 REVIEW EXERCISES Evaluating F at the endpoints and critical numbers, we have: F(0) = F(2π) = f(2, 0) = 2; F _ 3 4 π _ = f _ − √ 2, √ 2 _ = 6 + 4 √ 2; F _ 7 4 π _ = f _ √ 2, − √ 2 _ = 6 −4 √ 2. f takes on its absolute maximum of 6 + 4 √ 2 at _ − √ 2, √ 2 _ ; f takes on its absolute minimum of 0 at (1, −1). 46. ∇f(x, y) = (4x −4) i + (2y −4) j = 0 at (1, 2) on the boundry of D; no critical points in D. Next we consider the boundary of D. We parametrize each side of the triangle: C 1 : r 1 (t) = t i + 2t j, t ∈ [ 0, 1 ] C 2 : r 2 (t) = (1 −t) i + 2 j, t ∈ [ 0, 1 ] C 3 : r 3 (t) = (2 −t) j, t ∈ [ 0, 2 ] 1 2 3 x 1 2 3 y Now, f 1 (t) = f(r 1 (t)) = 6t 2 −8t + 3, t ∈ [ 0, 1 ]; critical number: t = 2 3 f 2 (t) = f(r 2 (t)) = 2t 2 −3+, t ∈ [ 0, 1 ]; critical number f 3 (t) = f(r 3 (t)) = t 2 −1, t ∈ [ 0, 2 ]; critical number Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 1 (0) = f 3 (2) = f(0, 0) = 3; f 1 (2/3) = f(2/3, 4/3) = − 7 3 ; f 1 (1) = f 2 (0) = f(1, 2) = −3; f 2 (1) = f 3 (0) = f(0, 2) = −1. f takes on its absolute maximum of 3 at (0, 0) and its absolute minimum of −3 at (1, 2). 47. ∇f(x, y) = (8x −y) i + (−x + 2y + 1) j = 0 at (−1/15, −8/15) in D; f(−1/15, −8/15) = −4/15. On the boundary of D : x = cos t, y = 2 sin t. Set F(t) = f(cos t, 2 sin t) = 4 + 2 sin t −2 sin t cos t, 0 ≤ t ≤ 2π. Then F (t) = 2 cos t −4 cos 2 t + 2 = −2(2 cos t + 1)(cos t −1); F (t) = 0 =⇒ t = 2π 3 , 4π 3 . Evaluating F at the endpoints of the interval and at the critical points, we get F(0) = F(2π) = f(1, 0) = 4, F(2π/3) = f(−1/2, √ 3) = 4 + 3 √ 3 2 , F(4π/3) = f(−1/2, − √ 3) = 4 − 3 √ 3 2 > − 4 15 f takes on its absolute maximum of 2 at (0, 1); f takes on its absolute minimum of −4/15 at (−1/15, −8/15). P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 REVIEW EXERCISES 863 48. ∇f(x, y) = 4x 3 i + 6y 2 j = 0 at (0, 0) in D; f(0, 0) = 0. On the boundary of D : x = cos t, y = sin t. Set F(t) = f(cos t, sin t) = cos 4 t + 2 sin 3 t, 0 ≤ t ≤ 2π. Then F (t) = 4 cos 3 t sin t + 6 sin 2 t cos t = 2 sin t cos t(2 sin t −1)(sin t + 2); F (t) = 0 =⇒ t = π/6, π/2, 5π/6, π, 3π/2 Evaluating F at the endpoints of the interval and at the critical points, we get F(0) = F(2π) = f(1, 0) = 1, F(π/6) = f( √ 3/2, 1/2) = 13/16, F(π/2) = f(0, 1) = 2, F(5π/6) = f(− √ 3/2, 1/2) = 13/16, F(π) = f(−1, 0) = 1, F(3π/2) = f(0, −1) = −2. f takes on its absolute maximum of 2 at (0, 1); f takes on its absolute minimum of −2 at (0, −1). 49. Set f(x, y, z) = D 2 = (x −1) 2 + (y + 2) 2 + (z −3) 2 , g(x, y) = 3x + 2y −z −5. ∇f = 2(x −1) i + 2(y + 2) j + 2(z −3) k, ∇g = 3 i + 2 j −k. Set ∇f = λ∇g : 2(x −1) = 3λ =⇒ x = 3 2 λ + 1, 2(y + 2) = 2λ =⇒ y = λ −2, 2(z −3) = −λ =⇒ z = − 1 2 λ + 3. Substituting these values in 3x + 2y −z = 5 gives λ = 9 7 =⇒ x = 41 14 , y = − 5 7 , z = 33 14 . The point on the plane that is closest to (1, −2, 3) is (41/14, −5/7, 33/14). The distance from the point to the plane is 9 √ 14 . 50. Set f(x, y, z) = 3x −2y +z, g(x, y, z) = x 2 +y 2 +z 2 −14, ∇f = 3 i −2 j +k, ∇g = 2xi + 2y j + 2z k. Set ∇f = λ∇g : 3 = 2λx =⇒ x = 3/2λ, −2 = 2λy =⇒ y = −1/λ, 1 = 2λz =⇒ z = 1/2λ. Substituting these values in x 2 +y 2 +z 2 = 14 gives λ = ± 1 2 =⇒ x = 3, y = −2, z = 1 or x = −3, y = 2, z = −1. Evaluating f : f(3, −2, 1) = 14, f(−3, 2, −1) = −14. The maximum value of f on the sphere is 14. 51. Set f(x, y, z) = x +y −z, g(x, y, z) = x 2 +y 2 + 4z 2 −4, ∇f = i +j −k, ∇g = 2xi + 2y j + 8z k. Set ∇f = λ∇g : 1 = 2λx =⇒ x = 1/2λ, 1 = 2λy =⇒ y = 1/2λ, −1 = 8λz =⇒ z = −1/8λ. Substituting these values in x 2 +y 2 + 4z 2 = 4 gives λ = ± 3 8 =⇒ x = 4/3, y = 4/3, z = −1/3 or x = −4/3, y = −4/3, z = 1/3. Evaluating f : f( 4 3 , 4 3 , − 1 3 ) = 3, f(− 4 3 , − 4 3 , 1 3 ) = −3. The maximum value of f is 3, the minimum value is −3. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 864 REVIEW EXERCISES 52. Let the length, width and height be x, y, z respectively. Then the total cost is f(x, y, z) = 1 2 xy + 1 2 xz + 1 2 yz + 1 10 xy = 3 5 xy + 1 2 xz + 1 2 yz with the condition g(x, y, z) = xyz −16 = 0. Note first that xyz = 16 =⇒ x = 0, y = 0 z = 0. ∇f = _ 3 5 y + 1 2 z _ i + _ 3 5 x + 1 2 z _ j + _ 1 2 y + 1 2 x _ k, ∇g = yz i +xz j +xy k ∇f = λ∇g =⇒ 3 5 y + 1 2 z = λyz, 3 5 x + 1 2 z = λxz, 1 2 x + 1 2 y = λxy Multiply the first equation by x, the second equation by y and subtract. This gives: 1 2 (xz −yz) = 0 =⇒ z(x −y) = 0 =⇒ y = x Substituting y = x in the third equation yields x = λx 2 =⇒ x = 1 λ . Substituting x = y = 1 λ in the first equation yields z = 6 5λ . Finally, substituting these values for x, y and z into the equation xyz = 16, we get λ = 3 √ 3 2 3 √ 5 . Therefore, x = y = 2 3 √ 5 3 √ 3 = 10 3 √ 75 ∼ = 2.37 and z = 6 5 x = 12 3 √ 75 ∼ = 2.85. 53. df = (9x 2 −10xy 2 + 2) dx + (−10x 2 y −1) dy 54. df = (2xy sec 2 x 2 −2y 2 ) dx + (tan x 2 −4xy) dy 55. df = y 2 z +z 2 y (x +y +z) 2 dx + xz 2 +zx 2 (x +y +z) 2 dy + x 2 y +y 2 x (x +y +z) 2 dz 56. df = − z y 2 +xz dx + _ ze yz − 2y y 2 +xz _ dy + _ ye yz − x y 2 +xz _ dz 57. Set f(x, y, z) = e x _ y +z 3 . Then df = e x _ y +z 3 Δx + e x 2 1 _ y +z 3 Δy + e x 2 3z 2 _ y +z 3 Δz. With x = 0, y = 15, z = 1, Δx = 0.02, Δy = 0.2, Δz = 0.01, df = 4 Δx + 1 8 Δy + 3 8 Δz ∼ = 0.1088. Therefore, e 0.02 _ 15.2 + (1.01) 3 ∼ = e 0 √ 15 + 1 + 0.1088 = 4.1088. 58. Set f(x, y) = x 1/3 cos 2 y. Then df = 1 3 x −2/3 cos 2 y Δx −2x 1/3 cos y sin y Δy. With x = 64, y = 30 ◦ = π/6, Δx = 0.5, Δy = −2 ◦ = − π 90 , df = 1 64 Δx −2 √ 3 Δy ∼ = 0.1287. Therefore, (64.5) 1/3 cos 2 (28 ◦ ) ∼ = 64 1/3 cos 2 (30 ◦ ) + 0.1287 = 3.1287. P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 REVIEW EXERCISES 865 59. V = πr 2 h; r = 5 ft., h = 22 ft., Δr = 0.01 in. = 1 1200 ft., Δh = 0.01 = 1 1200 dV = 2πrhΔr +πr 2 Δh Using the values given above, dV = 2π(5)(22) 1 1200 +π(25) 1 1200 ∼ = 0.6414 cu. ft. ∼ = 1108.35 cu. in.; 1108.35 231 ∼ = 4.80. Approximately 4.80 gallons will be needed. 60. ∂P ∂y = 12x 2 y −8x = ∂Q ∂x ; the vector function is a gradient. ∂f ∂x = 6x 2 y 2 −8xy + 2x, f(x, y) = 2x 3 y 2 −4x 2 y +x 2 +φ(y), ∂f ∂y = 4x 3 y −4x 2 +φ (y) = 4x 3 y −4x 2 −8. Thus, φ (y) = −8, φ(y) = −8y +C, and f(x, y) = 2x 3 y 2 −4x 2 y +x 2 −8y +C. 61. ∂P ∂y = 2x −sin x = ∂Q ∂x ; the vector function is a gradient. ∂f ∂x = 2xy + 3 −y sin x, f(x, y) = x 2 y + 3x +y cos x +φ(y), ∂f ∂y = x 2 + cos x +φ (y) = x 2 + 2y + 1 + cos x. Thus, φ (y) = 2y + 1, φ(y) = y 2 +y +C, and f(x, y) = x 2 y + 3x +y cos x +y 2 +y +C. 62. ∂P ∂y = 2xy + 4y; ∂Q ∂x = −2xy + 2; ∂P ∂y = ∂Q ∂x ; the vector function is not a gradient. 63. ∂P ∂y = e y sin z = ∂Q ∂x , ∂P ∂z = e y cos z = ∂R ∂x , ∂Q ∂z = xe y cos z = ∂R ∂y ; the vector function is a gradient. f(x, y, z) = _ (e y sin z + 2x) dx = xe y sin z +x 2 +φ(y, z), f y = xe y sin z + ∂φ ∂y = xe y sin z −y 2 =⇒ ∂φ ∂y = −y 2 =⇒ φ = − 1 3 y 3 +ψ(z), f(x, y, z) = xe y sin z +x 2 − 1 3 y 3 +ψ(z), f z = xe y cos z +ψ (z) = xe y cos z =⇒ ψ (x) = 0 =⇒ ψ(x) = C Therefore f(x, y, z) = xe y sin z +x 2 − 1 3 y 3 +C.