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Math Help - equation of the tangent plane and normal

  1. #1
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    equation of the tangent plane and normal

    hello, i have task Z= y*cos*(x-y) in point T (π; π/2; ?) .....is this derivation all right?

    Zx = y*sin*(x-y)/T1= - π/2*sin*( π- π/2) = - π/2

    Zy = y*sin*(x-y) + cos*(x-y)/T1 = π/2 * sin*( π π/2) + cos*( π π/2) = π/2

    Z0 = π/2*cos*(2 π π) = π/2*cos* π/2 = π/2
    2


    Could anyone help me please?
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  2. #2
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  3. #3
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    Quote Originally Posted by snopy View Post
    hello, i have task:
    I have edited your post as I understand it, & re-done the mathematical statements in LaTeX.

    From the title of the original post, I understand that you need to:

    Find the equation of the tangent plane and normal to the surface:
    \ z= y\cos(x-y) at point T: (\pi,\  \pi/2,\ z_0)

    \displaystyle {{\partial z}\over{\partial x}} = -y\sin(x-y). Evaluated at point T, this is  - {{\pi}\over{2}}\sin( \pi- {{\pi}\over{2}}) = - {{\pi}\over{2}}

    \displaystyle {{\partial z}\over{\partial y}} = y\sin(x-y)+\cos(x-y). Evaluated at point T, this is {{\pi}\over{2}}\sin( \pi- {{\pi}\over{2}})+\cos(\pi- {{\pi}\over{2}})={{\pi}\over{2}}

    z_0={{\pi}\over{2}}\cos({{2\pi-\pi}\over{2}})={{\pi}\over{2}}\cos({{\pi}\over{2}}  )={{\pi}\over{2}}

    .....is this derivation all right?

    Could anyone help me please?

    \displaystyle z_0=0 because \displaystyle \cos(\pi/2)=0.

    I'll get back to look at the rest, when I get a chance.

    Good Luck!
    Last edited by SammyS; December 15th 2010 at 12:24 PM.
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  4. #4
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    thanks
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