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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 ) = ~ ı + 2 t~ and the arclength 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 ) = ~ 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 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~ ı + 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 = ~ . Since ~ F is defined everywhere on R 3 , which is simplyconnected, 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 ( πt ) , 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 ) ~ ı + ( 1 3 x 3 ) ~ . Let C be the curve consisting of the line segment from (0 , 1) to (0 , 1) followed by the semicircle 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 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....
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 Spring '07
 NEEL
 Calculus, Arc Length, Vector Calculus, Cos, STOKE

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