ch13 review

# ch13 review - MAT 267 1. d) 1. Complete Solutions to Review...

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MAT 267 Complete Solutions to Review Chapter 13 1. d) 1. b) 2. a) 3. e) 4. f) 5. c) 6. 2. (a) ∇× F = 0 , thus the vector field is conservative. (b) The potential function is f ( x , y , z ) = x 2 y + zy + e z + C (c) The initial point of the curve is P 0 = r 0 = 0,1,0 and the terminal point is P 1 = r / 2 =/ 2, 0,1 . Using the Fundamental Theorem of Line Integrals we have C F d r = f P 1 − f P 0 = e 1 3. ∇⋅ F = x 2 x y  y x 2 z  z y e z = 2 y 0 e z = 2 y e z 4. We can parametrize the curve using x = t , y = t 2 with 0 t 1. Then r t = < t ,t 2 > , r ' t = <1,2 t > and r ' t ∣= 1 4 t 2 . Thus C x ds = 0 1 t 1 4t 2 d t = 1 12 5 5 1 5. Parametrize the circle using r t = <2cos t , 2sin t > , −/ 2 t ≤/ 2. Then r ' t = < 2sin t , 2cos t > , r ' t ∣= 2 and C  x , y ds = −/ 2 / 2  r t ∣ r ' t ∣ dt = − / 2 / 2 2cos t 4sin 2 t 2 dt = 32 3 . 6. Let A(1,0,0) and B(0,1,1), then AB = < 1,1,1> and the line through A and B can be parametrized by r t = <1 t ,t ,t > , 0 t 1. Then Work = 0 1 <1 t ,t 1 t ,t 1 t > < 1,1 , 1> dt =− 1 6 . 7. C 1 can be parametrized by r t = < t ,t 2 > , 0 t 2. C 1 F d r = 0 2 F r t ⋅ r ' t dt = 0 2 < 2 t 3 t 2, 3 t > <1 , 2 t > dt = 0 2 2 t 9 t 2 dt = 28 For C 2 : The line from (0,0) to (1,1) has equation y = x and can be parametrized by r t = < t ,t > , 0 t 1. The line from (1,1) to (2,4) has equation y = 3 x – 2 and can be parametrized by r t = < t , 3 t 2> , 1 t 2. Then C 2 F d r = 0 1 <2 t 3 t , 3 t > <1,1 > dt 1 2 < 2 t 3 3 t 2 , 3 t > < 1,3> dt = 0 1 8 t dt 1 2 20 t 6 dt = 28. Since ∇× F = 0 the vector field is conservative and the line integral is independent of path. Note that C 1 and C 2 have the same initial and terminal point, thus the line integral has the same value for both paths. 8.

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## This note was uploaded on 08/25/2011 for the course CHE 113 taught by Professor Burrows during the Spring '11 term at ASU.

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ch13 review - MAT 267 1. d) 1. Complete Solutions to Review...

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