Vector Concepts for Fluid Mechanics

Steven A.Jones BIEN 501, Physiological Modeling March 8, 00! Vector Concepts for Fluid Mechanics 1. A vector has "agnit#de and direction and can $e descri$ed as an arro% &ointing in the assigned direction %ith a length e'#al to its "agnit#de ()ig#re 1*. . +ectors can $e descri$ed in ,artesian coordinates as- k c j b i a v     + + = .. /he coe00icients a , b , and c are the x , y , and z co"&onents o0 the vector. 1. k j i    and , , are the"selves vectors. i  is a vector &ointing in the x direction, j  is a vector &ointing in the y direction, and k  is a vector &ointing in the z direction. See )ig#re . 5. /he "agnit#de o0 a vector is its length, %hich is given $y- 2 2 2 c b a v v + + = =  , %hich is si"&ly a state"ent o0 the Pythagorean theore". Problem 1: 2iven that the vector j  can $e re&resented as k j i j     0 1 0 + + = , sho%, 0ro" E'. , that the "agnit#de o0 j  is 1. Figure 1: A vector , re&resented $y the $old arro%. /he ti& o0 the arro% can $e reached 0ro" the origin $y "oving a distance in the direction, 0ollo%ed $y a distance in the direction, 0ollo%ed $y a distance in the direction. x y z v  b a c Figure 2: /he , and #nit vectors. /hese are tr#e vectors in the sense that they have $oth "agnit#de and direction. )or e3a"&le, the "agnit#de o0 is 1 and its direction is along the 3 a3is. x y z v  k  i  j  E'. 1 E'. Steven A. Jones BIEN 501, Physiological Modeling March 8, 00! Problem 2: Sho% that the e'#ation 0or the "agnit#de o0 a vector is si"&ly a state"ent o0 the Pythagorean theore" as 0ollo%s. )irst note that b and c are &er&endic#lar to one another $eca#se b is &arallel to the y a3is and c is &arallel to the z a3is (re0er to )ig#re 1*. 4se the Pythagorean theore" to o$tain d (in )ig#re .* in ter"s o0 b and c . No%, since b and c are $oth &er&endic#lar to the x a3is, d is also &er&endic#lar to the x a3is (i.e., it lies in a &lane that is &er&endic#lar to the x a3is. 4se this in0or"ation to %rite the length o0 the vector v  in ter"s o0 a and d . /hen $ac5 s#$stit#te 0or d to o$tain the length o0 v  in ter"s o0 a , b and c . !. /he dot &rod#ct $et%een t%o vectors, k c j b i a v     1 1 1 1 + + = and k c j b i a v     2 2 2 2 + + = is 2 1 2 1 2 1 2 1 c c b b a a v v + + = ⋅   Problem 3: Sho% that the dot &rod#ct o0 i  and j  is 6ero. Si"ilarly sho% that the dot &rod#ct o0 i  and k  is 6ero. (7int- %rite k j i i     0 0 1 + + = and k j i j     0 1 0 + + = and then #se E'. . 0or the dot &rod#ct. Problem 4: Sho% that the dot &rod#ct o0 k j i u     1 1 1 + + = and k j i j     5 . 0 5 . 0 1 + + − = is 6ero. 8. /he dot &rod#ct o0 t%o vectors is a scalar val#e. It has only "agnit#de, not direction. 8. /he val#e o0 the dot &rod#ct is- θ cos 2 1 v v   , %here θ is the angle $et%een the t%o vectors. Problem 5: Note that the vectors i  and j i v    1 1 + = are in directions that are 15° a&art 0ro" one another. ,alc#late their dot &rod#cts 0irst $y E'. . and then $y E'. 1 and sho% that they are e'#al. 9. /he dot &rod#ct o0 one vector %ith another is &hysically the &ro:ection o0 that vector on the other vector, "#lti&lied $y the length o0 the other vector. Figure 3: 2eo"etry to $e #sed to de"onstrate that the e'#ation 0or the "agnit#de o0 a vector is a si"&le state"ent o0 the Pythagorean theore". x y z v  b a c d E'. . E'. 1 Steven A. Jones BIEN 501, Physiological Modeling March 8, 00! 10. /he dot &rod#ct o0 a vector %ith itsel0 is the s'#are o0 the vector;s "agnit#de. /h#s, v v v v v v v v ⋅ = ≡ ⇒ = = ⋅ 2 2 . 11. A #nit vector is a vector %ith "agnit#de o0 one. 1. Any vector can $e nor"ali6ed to $eco"e a #nit vector $y si"&ly dividing $y its o%n "agnit#de. 2 2 2 c b a k c j b i a v v u + + + + = =       . 1.. <ecalling 9 a$ove, the dot &rod#ct o0 any vector %ith a #nit vector is the &ro:ection o0 that vector in the direction o0 the #nit vector. (/he &ro:ection is the shado% cast $y the vector $y a light so#rce that is in the &lane o0 the t%o vectors and that is &er&endic#lar to the #nit vector*. 11. /he cross &rod#ct o0 t%o vectors is a vector and can $e %ritten as- ( ) ( ) ( ) k a c b a j a c c a i b c c b c b a c b a k j i v v         2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 1 1 1 2 1 − + − − − = = × 11. /he "agnit#de o0 the cross=&rod#ct has the val#e o0 θ sin 2 1 2 1 v v v v     = × , %here θ is the angle $et%een the t%o vectors. 15. /he direction o0 the cross=&rod#ct is &er&endic#lar to $oth o0 the t%o original vectors, or, e'#ivalently, &er&endic#lar to the &lane in %hich the t%o original vectors lie. Problem 6: )ind the "agnit#de and direction o0 the cross &rod#ct o0 the t%o vectors given in Pro$le" 5. Problem : Sho% that the dot &rod#ct o0 the res#lt 0ro" Pro$le" ! %ith the t%o vectors given in Pro$le" 5 is 6ero. 7ence, sho% that the cross &rod#ct o0 these to vectors is &er&endic#lar to the original t%o vectors. /he a$ove conce&ts are ones that st#dents sho#ld 5no% 0ro" their statics co#rse. /he st#dents %ill need this "#ch revie%, and they %ill need to $e re"inded that they already 5no% these conce&ts 0ro" statics. Ne3t co"e so"e conce&ts that %ill $e ne% in the sense o0 #nderstanding vector a&&lications. In other %ords, %hile st#dents "ay have seen the gradient o&erator in their "ath class, they "ay not yet #nderstand %hat it "eans &hysically. Steven A. Jones BIEN 501, Physiological Modeling March 8, 00! 1!. >e0ine a vector k z j y i x r     + + ≡ . (/he o&erator ≡ "eans ?is de0ined as,@ as o&&osed to ?is e'#al to.@ /he di00erence $et%een ≡ and = is s#$tle, $#t o0ten i"&ortant*. Note that this vector "ay also $e re&resented $y ( ) z y x r , , ≡  . A 0#nction o0 the three s&atial coordinates ( ) z y x , , , i.e. ( ) z y x f , , , can then $e re&resented as ( ) r f . /he gradient o0 the 0#nction ( ) r f  is de0ined as k z f j y f i x f f    ∂ ∂ + ∂ ∂ + ∂ ∂ = ∇ . Problem !: I0 ( ) ( ) yz x r f sin =  , 0ind ( ) r f  ∇ . (Ans%er- ( ) ( ) ( ) ( ) k yz xy j yz xz i yz r f    cos cos sin + + = ∇ * 18. /he gradient is a vector 0#nction. In other %ords, 0or a given set o0 val#es o0 x , y , and z , the gradient can $e re&resented $y ( ) k c j b i a r f     + + = ∇ . Problem ": )or the ans%er to Pro$le" 8, %hat is the vector ( ) r f  ∇ at ( ) 4 1 , , 2 π = r  . (Ans%er- k j i    2 2 2 1 2 1 π + + *. 18. 7o%ever, the 0#nction itsel0 is not a vector. Problem 1#: Ahat is the val#e o0 the 0#nction ( ) r f  at ( ) 4 1 , , 2 π = r  . 19. /he gradient is a 0#nction o0 the s&atial coordinates. 0. Each ter" o0 the gradient is the rate o0 change o0 the 0#nction as one "oves in the given coordinate direction, ass#"ing that there is no change in &osition %ith res&ect to the other t%o coordinate directions. /o ill#strate the gradient, consider the sit#ation sho%n in )ig#re 1. A &a&er "ill &rod#ces a s#l0#ro#s odor that di"inishes in &#ngency %ith distance 0ro" the "ill. A &erson at &oint A e3&eriences a relatively large rate o0 red#ction o0 the odor as he "oves in the x direction. /he sa"e &erson e3&eriences a relatively s"all rate o0 red#ction o0 odor as he "oves in the y direction. /h#s, the x co"&onent o0 the gradient is large and the y y co"&onent o0 the gradient is s"all. x y Pa&er Mill A Figure 4: /he conce&t o0 gradient ill#strated $y the rate at %hich the odor 0ro" a &a&er "ill decreases as one "oves in a &artic#lar direction. Steven A. Jones BIEN 501, Physiological Modeling March 8, 00! Problem 11: Ass#"e in the &a&er "ill e3a"&le that the odor dro&s o00 %ith distance 0ro" the "ill according to the e'#ation 2 r A O = . 2ive the e'#ation 0or the gradient at any location ( ) y x, 0ro" the "ill, ass#"ing that the origin o0 the x and y a3es is at the center o0 the "ill. 7int- Bo# sho#ld %rite 2 2 y x r + = and then ta5e the derivatives %ith res&ect to x and y . Answer: ( ) 4 2 r j y i x A O   + − = ∇ . Problem 12: )or the sa"e &a&er "ill, %hat are the i  and j  co"&onents o0 the gradient 0o#nd in Pro$le" . at a &oint ( ) ( ) 0 , 2 , = y x C 7o% do they agree %ith the a$ove state"ent that the odor sho#ld dro& o00 "ore '#ic5ly along the x direction than along the y direction at this locationC 1. +elocity is a vector '#antity. It;s "agnit#de is the s&eed o0 an o$:ect. It;s direction is the direction in %hich the o$:ect is "oving. . /he gradient o&erator can $e descri$ed as a vector, k z j y i x    ∂ ∂ + ∂ ∂ + ∂ ∂ ≡ ∇ . /h#s, the gradient o0 a 0#nction is the gradient o&erator a&&lied to the 0#nction. .. /he divergence o0 a vector v  (s#ch as velocity* is the dot &rod#ct o0 the gradient o&erator %ith the vector. z v y v x v v v ∂ ∂ + ∂ ∂ + ∂ ∂ = ⋅ ∇ = 3 2 1 ) (   div 1. /he divergence is a scalar '#antity $eca#se it is a dot &rod#ct. 5. Dne "ay ta5e the divergence o0 the gradient o0 a 0#nction. /his scalar '#antity is 5no%n as the Ea&lacian o&erator. ( ) ( ) 2 2 2 2 2 2 2 ) ( z f y f x f k z f j y f i x f k z j y i x r f r f ∂ ∂ + ∂ ∂ + ∂ ∂ =         ∂ ∂ + ∂ ∂ + ∂ ∂ ⋅         ∂ ∂ + ∂ ∂ + ∂ ∂ = ∇ ≡ ∇ ⋅ ∇         !. Dne "ay also consider the divergence to $e an o&erator 2 ∇ de0ined as- 2 2 2 2 2 2 2 z y x k z j y i x k z j y i x ∂ ∂ + ∂ ∂ + ∂ ∂ =         ∂ ∂ + ∂ ∂ + ∂ ∂ ⋅         ∂ ∂ + ∂ ∂ + ∂ ∂ = ∇ ⋅ ∇ = ∇       8. /h#s, f z y x f         ∂ ∂ + ∂ ∂ + ∂ ∂ = ∇ 2 2 2 2 2 2 2 8. /he c#rl o0 a vector is the cross=&rod#ct o0 the gradient o&erator %ith the vector. Steven A. Jones BIEN 501, Physiological Modeling March 8, 00! k x v y v j x v z v i y v z v v v              ∂ ∂ − ∂ ∂ +       ∂ ∂ − ∂ ∂ −         ∂ ∂ − ∂ ∂ = × ∇ = 2 1 3 1 3 2 ) ( c#rl 9. /he c#rl o0 a vector is another vector. .0. /he c#rl o0 a vector can $e %ritten as- ( ) 3 2 1 v v v z y x k j i v v ∂ ∂ ∂ ∂ ∂ ∂ = × ∇ =      c#rl and this 0or" &rovides one %ith an easy %ay to re"e"$er %here the ?F? signs "#st go. Problem 13: Ahich o0 these are not legiti"ate vector o&erations. AhyC Ass#"e f is a scalar 0#nction. a. v  × ∇ $. f × ∇ c. f ∇ × ∇ d. ( ) v  ⋅ ∇ × ∇ e. ( ) v  ⋅ ∇ × ∇ 0. ( ) ( ) f f ∇ × ∇ g. ( ) f 2 ∇ ∇ h. ( ) f 2 ∇ ⋅ ∇ Problem 14: )or each legiti"ate o&eration in Pro$le" 5, state %hat 5ind o0 entity (vector or scalar* is the res#lt. .1. /he e'#ation ( ) 0 c r f = , %here 0 c is a constant, re&resents a s#r0ace in three= di"ensional s&ace. A vector nor"al to this s#r0ace is ( ) r f ∇ . .. ( ) r f ∇ is also the direction o0 "a3i"#" rate o0 change in ( ) r f . ... A c#rve in s&ace is the intersection o0 t%o s#r0aces. Problem 15: Ahat is the c#rve re&resented $y the intersection o0 the 0ollo%ing t%o s#r0aces- 25 2 2 2 = + + z y x G 0 2 2 = − + z y x C $olution: )irst notice that the 0irst c#rve is a s&here o0 radi#s 5 and the second c#rve is a &ara$ola o0 rotation. I0 %e s#$stit#te 2 2 y x z + = 0ro" the second c#rve into the 0irst c#rve, the res#lt is- 0 25 2 = − + z z , %hich has t%o sol#tions, 2 100 1 1 + ± − = z , one o0 %hich is negative and there0ore irrelevant. It 0ollo%s that the c#rve is in the &lane re&resented $y ( ) 2 1 101 − = z and on the c#rve in that &lane ( ) 2 1 101 2 2 − = + y x . Steven A. Jones BIEN 501, Physiological Modeling March 8, 00! .1. /here are several vector identities that sho#ld $e recogni6ed $y the st#dent (tho#gh not necessarily "e"ori6ed*. a. ( ) ( ) ( ) ( ) ( ) ( ) a b c c a b b c a b a c a c b c b a ⋅ × − = ⋅ × − = ⋅ × − = ⋅ × = ⋅ × = ⋅ × Proof of the first e%ualit&: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) a c b a k j i c b a k j i k j i b a ⋅ × = ⋅ = − + − − − = − + − − − = ⋅ × − + − − − = = × 3 2 1 3 2 1 3 2 1 2 1 2 1 3 3 1 1 2 3 3 2 3 1 2 2 1 2 1 3 3 1 1 2 3 3 2 1 2 2 1 1 3 3 1 2 3 3 2 3 2 1 3 2 1 c c c b b b a b c c b a c b c b a c b c b c b a b a c b a b a c b a b a b a b a b a b a b a b a b b b a a a /he other e'#alities 0ollo% si"ilarly. $. ( ) ( ) ( ) a c b b c a c b a ⋅ − ⋅ = × × Proof: ( ) b a × is a vector &er&endic#lar to $oth a and b. Any vector that is &er&endic#lar to b a × "#st there0ore $e in the &lane o0 $oth a and b. In &artic#lar ( ) c b a × × "#st $e in that &lane. Any vector in that &lane can $e descri$ed $y a linear co"$ination o0 a and b. /h#s, ( ) b a c b a n m + = × × . )#rther"ore, since c "#st $e &er&endic#lar to ( ) c b a × × , it 0ollo%s that ( ) ( ) 0 = ⋅ × × c c b a so that ( ) 0 = ⋅ + c b a n m , i.e. ( ) ( ) c b c a ⋅ − = ⋅ n m . /here0ore, ( ) b a c b a n m + = × × Steven A. Jones BIEN 501, Physiological Modeling March 8, 00!