17.3 Ex 12 - 24

17.3 Ex 12 - 24 - S E C T I O N 17.3 Conservative Vector...

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SECTION 17.3 Conservative Vector Fields (ET Section 16.3) 1099 h 0 ( z ) = 0 h ( z ) = C Substituting in (2) we get ϕ ( x , y , z ) = x 2 yz + C Since only one potential function is needed, we choose the one corresponding to C = 0. That is, ( x , y , z ) = x 2 Using the Fundamental Theorem for Gradient Vectors we obtain: Z c F · d s = ( 3 , 2 , 1 ) ( 0 , 0 , 0 ) = 3 2 · 2 · 1 0 = 18 In Exercises 12–17, determine whether the vector feld is conservative and, iF so, fnd a potential Function. 12. F = h z , 1 , x i SOLUTION We check whether the vector ±eld F = h z , 1 , x i satis±es the cross partials condition: ± 1 y = y ( z ) = 0 ± 2 x = x ( 1 ) = 0 ± 1 y = ± 2 x ± 2 z = z ( 1 ) = 0 ± 3 y = y ( x ) = 0 ± 2 z = ± 3 y ± 3 x = x ( x ) = 1 ± 1 z = z ( z ) = 1 ± 3 x = ± 1 z F satis±es the cross partials condition everywhere. Hence, F is conservative. We ±nd a potential function ( x , y , z ) . Step 1. Use the condition x = ± 1 . is an antiderivative of ± 1 = z when y and z are ±xed, therefore: ( x , y , z ) = Z zdx = zx + g ( y , z ) (1) Step 2. Use the condition y = ± 2 . By (1) we have: y ( + g ( y , z )) = 1 g y ( y , z ) = 1 Integrating with respect to y , while holding z ±xed, gives: g ( y , z ) = Z 1 dy = y + h ( z ) We substitute in (1) to obtain: ( x , y , z ) = + y + h ( z ) (2) Step 3. Use the condition z = ± 3 .Using(2)weget: z ( + y + h ( z )) = x x + h 0 ( z ) = x h 0 ( z ) = 0 h ( z ) = c Substituting in (2) gives the following potential functions: ( x , y , z ) = + y + c . One of the potential functions is obtained by choosing c = 0: ( x , y , z ) = + y
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1100 CHAPTER 17 LINE AND SURFACE INTEGRALS (ET CHAPTER 16) 13. F = h 0 , x , y i SOLUTION Since F 1 y = y ( 0 ) = 0and F 2 x = x ( x ) = 1, we have F 1 y 6= F 2 x . Therefore F does not satisfy the cross-partial condition, hence F is not conservative. 14. F = - y 2 , 2 xy + e z , ye z ® We examine whether F satisFes the cross partials condition: F 1 y = y ³ y 2 ´ = 2 y F 2 x = x ( 2 + e z ) = 2 y F 1 y = F 2 x F 2 z = z ( 2 + e z ) = e z F 3 y = y ( z ) = e z F 2 z = F 3 y F 3 x = x ( z ) = 0 F 1 z = z ³ y 2 ´ = 0 F 3 x = F 1 z We see that F satisFes the cross partials condition everywhere, hence F is conservative. We Fnd a potential function for F . Step 1. Use the condition ϕ x = F 1 . is an antiderivative of F 1 = y 2 when y and z are Fxed. Hence: ( x , y , z ) = Z y 2 dx = y 2 x + g ( y , z ) (1) Step 2. Use the condition y = F 2 . By (1) we have: y ³ y 2 x + g ( y , z ) ´ = 2 + e z 2 yx + g y ( y , z ) = 2 + e z g y ( y , z ) = e z We integrate with respect to y , holding z Fxed: g ( y , z ) = Z e z dy = e z y + h ( z ) Substituting in (1) gives: ( x , y , z ) = y 2 x + e z y + h ( z ) (2) Step 3. Use the condition z = F 3 .By(2),weget: z ³ y 2 x + e z y + h ( z ) ´ = z e z y + h 0 ( z ) = z h 0 ( z ) = 0 Therefore h ( z ) = c . Substituting in (2) we get: ( x , y , z ) = y 2 x + e z y + c The potential function corresponding to c = 0is: ( x , y , z ) = y 2 x + e z y .
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This homework help was uploaded on 04/13/2008 for the course MATH 32B taught by Professor Rogawski during the Winter '08 term at UCLA.

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17.3 Ex 12 - 24 - S E C T I O N 17.3 Conservative Vector...

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