CalcIVPSol2 - Solutions to the 2nd Practice Midterm 1. Let...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to the 2nd Practice Midterm 1. Let C be the curve in the plane given by the graph of y = x 2 from the point (1 , 1) to the point (2 , 4) . (a) Integrate f ( x, y ) = 2 y/x along C (with respect to arc length). Since C is a graph, we can parametrize it by x = t , y = t 2 for t [1 , 2]. Then we see that ~ r ( t ) = ~ + 2 t~ and the arc-length element is ds = 1 + 4 t 2 dt . Further, we have f ( x ( t ) , y ( t )) = 2 t , so that Z C f ds = Z 2 1 2 t p 1 + 4 t 2 dt = 1 6 ( 1 + 4 t 2 ) 3 / 2 2 1 = 1 6 17 3 / 2- 5 3 / 2 . (b) Let ~ F ( x, y ) = ~ + xy~ . Integrate ~ F along C . We use the same parametrization for C as in the last part. Now we have ~ F ( x ( t ) , y ( t )) = ~ + t 3 ~ , so that Z C ~ F d~ r = Z 2 1 ~ F ( x ( t ) , y ( t )) ~ r ( t ) dt = Z 2 1 ( 1 + 2 t 4 ) dt = t + 2 5 t 5 2 1 = 67 5 . 2. Let ~ F ( x, y, z ) = 2 xz~ + e z ~ + ( x 2 + ye z ) ~ k . (a) If ~ F has a potential, find it. If not, say why not. We compute curl ~ F = ~ ~ ~ k x y z 2 xz e z x 2 + ye z = ( e z- e z ) ~ - (2 x- 2 x ) ~ + (0- 0) ~ k = ~ . Since ~ F is defined everywhere on R 3 , which is simply-connected, this means that ~ F is conservative. To find its potential, we integrate with respect to x to find f ( x, y, z ) = x 2 z + g ( y, z ). Differentiating with respect to y gives g y ( y, z ) = e z , and thus g ( y, z ) = ye z + h ( z ). Differentiating with respect to z gives x 2 + ye z + h ( z ) = x 2 + ye z , so that h ( z ) is constant. Putting this together, we see that the potential is f ( x, y, z ) = x 2 z + ye z + K for any constant K . (b) Compute R C ~ F d~ r where C is the curve x ( t ) = cos 2 ( t ) , y ( t ) = 1 + t 3 , z ( t ) = 1 4 t 2 , for t 2 . 1 The fundamental theorem for line integrals tells us that, because f is a potential for ~ F , the integral of ~ F along any curve is equal to f at the the end point minus f at the starting point. The above curve goes from (1 , 1 , 0) to (1 , 9 , 1), and thus (we set K = 0 in our definition of f ) Z C ~ F d~ r = f (1 , 9 , 1)- f (1 , 1 , 0) = 1 + 9 e- 1 = 9 e. 3. Let ~ F ( x, y ) = ( yx + yx 2 ) ~ + ( 1 3 x 3 ) ~ . Let C be the curve consisting of the line segment from (0 ,- 1) to (0 , 1) followed by the semi-circle from (0 , 1) through (1 , 0) to (0 ,- 1) . Compute R C ~ F d~ r . The curve C is a simple, closed curve oriented clockwise. Thus we can use Greens theorem to evaluate the integral. The region D bounded by C is the right half of the unit disk, that is the set of points where x 2 + y 2 1 with x 0. The integrand for Greens theorem is Q x- P y = x 2- ( x + x 2 ) =- x....
View Full Document

Page1 / 6

CalcIVPSol2 - Solutions to the 2nd Practice Midterm 1. Let...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online