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Math Help - [SOLVED] Tangent plane

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Tangent plane

    At what angle does the tangent plane to the surface S: z=2x^3-x^2y-y^2-2y+11 in the point (2,3,0) meet the xy-plane?

    F=2x^3-x^2y-y^2-2y+11 - z=0

    grad F(x,y,z) = (6x^2-2xy,-x^2-2y-2,-1)
    grad F(2,3,0) = (12,-12,-1)

    The normal vector to the xy-plane, n, is (0,0,1).


    grad F \cdot n = |17||1|cos\theta
    grad F \cdot n = -1

    So we have cos\theta = \frac{-1}{17} \implies \theta = arccos \frac{-1}{17}

    The correct answer is \theta = arccos \frac{1}{17} though, which is what you get if you select (0,0,-1) as the normal vector. So are both answers correct? Why is it that you choose (0,0,-1) as the normal vector?
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  2. #2
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    Quote Originally Posted by Spec View Post
    At what angle does the tangent plane to the surface S: z=2x^3-x^2y-y^2-2y+11 in the point (2,3,0) meet the xy-plane?

    F=2x^3-x^2y-y^2-2y+11 - z=0

    grad F(x,y,z) = (6x^2-2xy,-x^2-2y-2,-1)
    grad F(2,3,0) = (12,-12,-1)

    The normal vector to the xy-plane, n, is (0,0,1).


    grad F \cdot n = |17||1|cos\theta
    grad F \cdot n = -1

    So we have cos\theta = \frac{-1}{17} \implies \theta = arccos \frac{-1}{17}

    The correct answer is \theta = arccos \frac{1}{17} though, which is what you get if you select (0,0,-1) as the normal vector. So are both answers correct? Why is it that you choose (0,0,-1) as the normal vector?
    we usually take 0 \leq \theta \leq \frac{\pi}{2}. so if n_1, n_2 are the normal vectors of two planes, then \theta, the angle between the planes, is definied to be \theta = \arccos \left|\frac{n_1 \cdot n_2}{||n_1|| \cdot ||n_2||} \right|.

    what you've found is the supplement of \theta.
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