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CalcIVPSol2

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

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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 ) = ı + 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 C f ds = 2 1 2 t 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 ) = ı + 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 )) = ı + t 3 , so that C F · dr = 2 1 F ( x ( t ) , y ( t )) · r ( t ) dt = 2 1 ( 1 + 2 t 4 ) dt = t + 2 5 t 5 2 1 = 67 5 . 2. Let F ( x, y, z ) = 2 xzı + e z + ( x 2 + ye z ) k . (a) If F has a potential, find it. If not, say why not. We compute curl F = ı k ∂x ∂y ∂z 2 xz e z x 2 + ye z = ( e z - e z ) ı - (2 x - 2 x ) + (0 - 0) k = 0 . 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 C F · dr where C is the curve x ( t ) = cos 2 ( πt ) , y ( t ) = 1 + t 3 , z ( t ) = 1 4 t 2 , for 0 t 2 . 1

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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 ) C F · dr = f (1 , 9 , 1) - f (1 , 1 , 0) = 1 + 9 e - 1 = 9 e. 3. Let F ( x, y ) = ( yx + yx 2 ) ı + ( 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 C F · dr . The curve C is a simple, closed curve oriented clockwise. Thus we can use Green’s 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 Green’s theorem is ∂Q ∂x - ∂P ∂y = x 2 - ( x + x 2 ) = - x. Using Green’s theorem, with a minus sign because the curve is oriented clockwise, and polar coordinates to write the integral over the half-disk, we have C F · dr = - D ∂Q ∂x - ∂P ∂y dA = - π/ 2 - π/ 2 1 0 - r cos θ · r dr dθ = π/ 2 - π/ 2 cos θ dθ 1 0 r 2 dr = [sin θ ] π/ 2 - π/ 2 1 3 r 3 1 0 = (1 - ( - 1)) 1 3 - 0 = 2 3 .
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CalcIVPSol2 - Solutions to the 2nd Practice Midterm 1 Let C...

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