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Math Help - Tangent Planes

  1. #1
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    Tangent Planes

    Hello math help forums! This is my first post on these forums and I hope to come back here often. Go calc!

    Okay, so I just finished a problem on my homework, but for some reason the logic I used in completing this problem seems odd. I don't know why though.

    Here is the problem:

    Find a point on the surface of  z=8-3x^2-2y^2 at which the tangent plane is perpendicular to the line  x=2-3t, y=7+8t, z=5-t. Find the equation of the tangent plane at that point.

    First, I know that a plane in space is defined by normal vector to the plane and a point on that plane. Because the given line is perpendicular to the tangent plane, I used the direction numbers of the line to form a scalar equation for the tangent plane:

     -3(x-x_{0}) + 8(y-y_{0}) -1(z-z_{0})

    Where (x_{0}, y_{0}, z_{0}) is the point on the surface at which the tangent plane is perpendicular to the line.

    The equation of the tangent plane is also given by:

     z-z_{0} = f_{x}(x_{0}, y_{0})(x-x_{0}) + f_{y}(x_{0}, y_{0})(y-y_{0})

    f_{x}(x_{0}, y_{0}) = -6x_{0}

    f_{y}(x_{0}, y_{0}) = -4y_{0}

    Thus,

    -6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0}) + z_{0} - z = 0

    Since both equations equal 0, and because the point (x_{0}, y_{0}, z_{0}) is the same for each equation, I set the equations equal to each other. For some reason, I feel like I am doing something wrong at this point... I don't know why.

    -6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0}) + z_{0} - z = -3(x-x_{0}) + 8(y-y_{0}) -1(z-z_{0})

    The z terms cancel out, and I am left with:

    -6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0})= -3(x-x_{0}) + 8(y-y_{0})

    Thus,

    -6x_{0}=-3 \Rightarrow x_{0} = \frac{1}{2}

    and

    -4y_{0}=8 \Rightarrow y_{0} = -2

    I then plugged these points into the original equation for the surface to obtain  z_{0} = \frac{-3}{4}

    Does this logic seem right?

    Thank you for your help! It is greatly appreciated.

    Sincerely,

    Opiaboy
    Last edited by mr fantastic; October 6th 2009 at 04:11 AM. Reason: Fixed latex.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    It is kind of hard to follow because of the latex but you are not quite correct :

    You have for your normal

    N = 6x i + 4y j + k

    Further N is parallel to M = -3 i + 8 j - k

    Since N and M are parallel N = cM for some c

    her c = -1 so the z comp match

    -6x = -3 -4y = 8

    x= 1/2 y= -2

    To find z sub the x and y values into

    z=8-3x^2-2y^2 so z = 8 -3/4 -8 = -3/4
    so the pt is (1/2,-2, -3/4)

    Note for your point N = -3 i + 8j + k which is not parallel to M
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by opiaboy View Post
    *EDIT okay for some reason the Latex thing isn't working and my math symbols look like jank... gotta fix that later..
    Your LaTeX problem is that you are using [tex][\math] tags rather than [tex][/tex] tags.

    If you are using the advanced editor you will find a button on the tool bar with a \Sigma which will automatically put the tags around any selected text.

    CB
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