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richyw Group Title

Find an equation for the tangent plane to the surface given by\[z=\ln{\left(1+x^2+y^2\right)}\]

  • 2 years ago
  • 2 years ago

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  1. richyw Group Title
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    Sorry, at the point (0,2,ln5). why can't I just say\[z_x=\frac{2y}{\left(1+x^2+y^2\right)}\]\[z_y=\frac{2x}{\left(1+x^2+y^2\right)}\]And then that \[z-ln(5)=z_x(0,2)(x-0)+z_y(0,2)(y-2)\]

    • 2 years ago
  2. richyw Group Title
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    so \[z=\frac{4}{5}x+\ln(5)\]

    • 2 years ago
  3. richyw Group Title
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    I don't see where I am going wrong....

    • 2 years ago
  4. richyw Group Title
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    nevermind. I see where I went wrong haha.

    • 2 years ago
  5. TuringTest Group Title
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    is it that you don't have the coefficient for z, since you didn't turn it into a function? I want to know because your way is different than mine.

    • 2 years ago
  6. TuringTest Group Title
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    er, I mean didn't turn it into a 3-variable function...

    • 2 years ago
  7. richyw Group Title
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    my notation was sloppy. All I do to find the plane tangent to \(z=f(x,y)\) at the point \(P\left(a,b,f(a,b)\right)\) is use the formula \[z-f(a,b)=f_x(a,b)(x-a)+f_y(a,b)(y-b)\]

    • 2 years ago
  8. richyw Group Title
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    I just accidentally put f_y where f_x should have been...

    • 2 years ago
  9. TuringTest Group Title
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    ah, gotchya thanks

    • 2 years ago
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