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Math Help - Find the equation for the tangent plane to the surface. Answer doesn't seem right.

  1. #1
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    Find the equation for the tangent plane to the surface. Answer doesn't seem right.

    z=ln(3x^{2}+6y^{2}+1), P(0,0,0)

    Ok, so I found \frac{\partial f}{\partial x} and \frac{\partial f}{\partial y}

    \frac{\partial f}{\partial x}=\frac{6x}{3x^{2}+6y^{2}+1}

    \frac{\partial f}{\partial y}=\frac{12y}{3x^{2}+6y^{2}+1}

    So then,

    f_{x}(0,0)=\frac{0}{1}=0

    f_{y}(0,0)=\frac{0}{1}=0

    And I have a normal vector n=(0,0,-1) at point (0,0,0)

    So then,

    0(x-0)+0(y-0)-(z-0)=0

    =>z=0

    Did I do this correctly? Thanks for looking.
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  2. #2
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    Quote Originally Posted by downthesun01 View Post
    z=ln(3x^{2}+6y^{2}+1), P(0,0,0)
    You should have looked at  F(x,y,z)=\ln(3x^{2}+6y^{2}+1)-z.

    Where is your F_z?
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  3. #3
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    f_{z}=-1

    f_{z}(0,0,0)=-1

    n=0i+0j-k at point (0,0,0)

    0(x-0)+0(y-0)-(z-0)=0

    z=0

    Still getting z=0

    And it still seems like an odd answer for some reason.
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  4. #4
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    Quote Originally Posted by downthesun01 View Post
    f_{z}=-1

    f_{z}(0,0,0)=-1

    n=0i+0j-k at point (0,0,0)

    0(x-0)+0(y-0)-(z-0)=0

    z=0

    Still getting z=0

    And it still seems like an odd answer for some reason.
    1. Which reasons?

    2. I've attached the graph of z and the tangent plane. And it looks as if your calcualtions are OK. So again: which reasons?
    Attached Thumbnails Attached Thumbnails Find the equation for the tangent plane to the surface. Answer doesn't seem right.-tangplane.png  
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  5. #5
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    Honestly, z=0 just seems kind of boring. I know, not a very good reason. Thanks for the graph though. It makes it really easy to see that the answer is correct. Much appreciated.
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