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quiz3soln

# quiz3soln - y so y 4 y 2 4 z 2 = 16 Now we’ll parametrize...

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MATH 231 / Spring 2008 February 6, 2008 Name: Key 1. (4 points) Suppose u and v are diﬀerentiable vector functions and f is a real-valued function. Then (a) d dt [ u · v ] = u · v 0 + u 0 · v (b) d dt [ u ( f ( t ))] = u 0 ( f ( t )) f 0 ( t ) 2. (6 points) Find a vector function that represents the curve of intersection of the parabolic cylinder y = x 2 and the top half of the ellipsoid x 2 + 4 y 2 + 4 z 2 = 16. Solution: We want to ﬁnd the intersection of the two shapes. To do this, we will “set them equal” and solve. Since y = x 2 , we can replace x 2 in the second equation with
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Unformatted text preview: y , so y + 4 y 2 + 4 z 2 = 16. Now, we’ll parametrize this set of equations. Let x = t . Then we ﬁnd that y = t 2 and 16 = t 2 +4 t 4 +4 z 2 . Solving this equation for z , we get z = ± q 4-t 2 4-t 4 . However, since we only need consider the top half of the ellipsoid, we don’t need to worry about negative values of z . Thus, we ﬁnd that ~ r ( t ) = * t,t 2 , r 4-t 2 4-t 4 + 1...
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