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Asked Nov 9, 2019
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Find an equation of the tangent plane to the given parametric surface at the specified point. 

x = u + v,  y = 3u^2,  z = u - v;  (2,3,0)

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Then, u 1 and v = 1 The normal vector at (2,3,0) is (-6, -2,-6) The equation of the tangent plane at the point (2,3,0) is -6 (x-2)+2(y-3)-6 (z-0) 0 —6х +12+2у-6— 62 %3D 0 -6х +2 у-62 %3D -6 3х— у+3z %3D 3

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Math

Calculus

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