# MT2Sol - Solutions for the second midterm 1 Let C be the...

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Solutions for the second midterm 1. Let C be the helix given by x ( t ) = cos t, y ( t ) = sin t, z ( t ) = t for 0 t 2 π . (a) Integrate f ( x, y, z ) = x + 3 z 2 along C with respect to arc length. We have been given a parametrization; using it, we have ~ r 0 ( t ) = sin t~ ı + cos t~ + ~ k and | ~ r 0 ( t ) | = p sin 2 t + cos 2 t + 1 = 2 . Writing the function in terms of this parametrization gives f ( x ( t ) , y ( t ) , z ( t )) = cos t +3 t 2 . We have Z C f ds = Z 2 π 0 ( cos t + 3 t 2 ) 2 dt = 2 ± sin t + t 3 ² 2 π 0 = 8 2 π 3 . (b) Compute R C ~ F · d~ r where ~ F ( x, y, z ) = z 2 ~ ı + z~ +(2 xz + y ) ~ k . (Hint: ~ F is conservative.) Computing the integral directly doesn’t look promising, so instead we follow the hint and ﬁnd the potential function. Integrating with respect to x gives f = xz 2 + g ( y, z ). Diﬀer- entiating with respect to y gives ∂g/∂y = z and thus g ( y, z ) = yz + h ( z ). Diﬀerentiating with respect to z gives 2 xz + y + h 0 ( z ) = 2 xz + y , so that h ( z ) = K for some constant K . We conclude that the potential is f = xz 2 + yz + K for any constant K . Note that the curve C begins at (1 , 0 , 0) and ends at (1 , 0 , 2 π ). We have Z C ~ F · d~ r = f (1 , 0 , 2 π ) - f (1 , 0 , 0) = 4 π 2 - 0 = 4 π 2 . 2. Let C be the curve consisting of the following four line segments: from (1 , 0 , 0) to (1 , 1 , 0) , followed by (1 , 1 , 0) to (0 , 1 , 1) , followed by (0 , 1 , 1) to (0 , 0 , 1) , followed by (0 , 0 , 1) to (1 , 0 , 0) . In particular, C is a rectangle which lies in the plane

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MT2Sol - Solutions for the second midterm 1 Let C be the...

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