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Math Help - Tangent plane and normal line..

  1. #1
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    Question Tangent plane and normal line..

    I got 2 questions that I am finding problems with, would someone help me please?

    (1) Fine equation of the tangent plane
    (2) Find equation of normal line to the surface
    represented by the equation:

    x-z = 4arctan(yz)

    at the points (1+Pi, 1, 1)

    Thanks.
    taurus
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  2. #2
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    Quote Originally Posted by taurus View Post
    I got 2 questions that I am finding problems with, would someone help me please?

    (1) Fine equation of the tangent plane
    (2) Find equation of normal line to the surface
    represented by the equation:

    x-z = 4arctan(yz)

    at the points (1+Pi, 1, 1)
    Let f(x,y,z) = x - z - 4\tan^{-1} yz so the surface is represented as f(x,y,z) = 0.
    The normal to the surface at (1+\pi , 1 , 1) is given by \nabla f(1+\pi , 1 , 1) = a\bold{i}+b\bold{j}+c\bold{k}. (Whatever a,b,c are, you need to compute that).

    For (1) to find the tangent plane you use the fact that is passes through the point (1+\pi , 1,1) and its normal vector is a\bold{i}+b\bold{j}+c\bold{k}.

    For (2) to find the normal line procede as above. The line contains (1+\pi,1,1) and is aligned with vector a\bold{i}+b\bold{j}+c\bold{k}.
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  3. #3
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    Am still a bit confused with my equation how its achieved?

    Could someone show me some steps?

    thanks
    Last edited by taurus; May 14th 2008 at 08:18 AM.
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  4. #4
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    Question

    if i have f(x,y,z) = x - z - 4arctan(yz) = 0

    So i need to differentiate for fx, fy and fz.

    so is the following right:
    fx = dx

    but whats fy and fz?

    Thanks
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  5. #5
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    Quote Originally Posted by taurus View Post
    if i have f(x,y,z) = x - z - 4arctan(yz) = 0
    If f(x,y,z) = x - z - 4\tan^{-1} (yz).

    Then, \frac{\partial f}{\partial x} = 1, \frac{\partial f}{\partial y} = -\frac{4z}{1+(yz)^2}, \frac{\partial f}{\partial z} = -1-\frac{4y}{1+(yz)^2}.
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