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Math Help - Partial Derivatives Problem

  1. #1
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    Partial Derivatives Problem

    Hi all,

    This is the following homework problem I am having trouble with:

    Consider the surface: 25x^2+25y^2+1z^2=51

    P=(1,1,1) on this surface.

    So far, I have gotten:

    x=1+50t
    y=1+50t
    z=1+2t

    The equation to the plane P must be written as z=, so I wrote:

    z-1=50(x-1)+50(y-1)
    z=50(x-1)+50(y-1)+1

    Whenever I type this into the online homework, it displays "incorrect". Any suggestions?

    Thank you in advance!
    Last edited by newman611; November 27th 2010 at 06:33 PM. Reason: Image could not appear
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  2. #2
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    I'm probably not going to be able to help you with this, and I don't know if it's just me, but I can't see either images you attempted to put in your post. Thought i'd give you the heads up.
    edit: I can see it now. I'll think about it for a little while to see if i can be of assistance but odds are I won't do much good :/
    edit2: Alright, thought about it. Not sure what the question is asking? It just says consider the surface and the point on the surface... what do you need to find?
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  3. #3
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    Thanks, wombat!
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  4. #4
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    Haha no problem but i'm not really sure what you're thanking me for. Is what you typed in your original post the entire question or is there more?
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  5. #5
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    No, this is it...I said thanks because you didn't have to mention anything, lol.
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  6. #6
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    What exactly is the question? If you're looking for the tangent plane at (1,1,1), then you can use the formula f'_x(a,b,c)(x-a)+f'_y(a,b,c)(y-b)+f'_z(a,b,c)(z-c)=0 where f(x,y,z)=25x^2+25y^2+1z^2 and (a,b,c)=(1,1,1)
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  7. #7
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    Quote Originally Posted by newman611 View Post
    Hi all,

    This is the following homework problem I am having trouble with:

    Consider the surface: 25x^2+25y^2+1z^2=51

    P=(1,1,1) on this surface.

    So far, I have gotten:

    x=1+50t
    y=1+50t
    z=1+2t
    For what? These are parametric equations of a line- the line perpendicular to the surface at (1, 1, 1). They do tell you that a vector perpendicular to the surface, and so perpendicular to the tangent plane, is 50\vec{i}+ 50\vec{j}+ 2\vec{z} but you don't need the equation of the line to determine that.

    The equation to the plane P must be written as z=, so I wrote:

    z-1=50(x-1)+50(y-1)
    z=50(x-1)+50(y-1)+1
    Well, you don't say how you got that from what you did before so I can't make heads or tails of it. The equation of a plane through (x_0, y_0, z_0) with normal vector A\vec{i}+ B\vec{j}+ C\vec{k} is A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0.
    Here A= B= 50, C= 2, and x_0= y_0= z_0= 1 so the equation is 50(x- 1)+ 50(y- 1)+ 2(z- 1)= 0. Solve that for z.

    Whenever I type this into the online homework, it displays "incorrect". Any suggestions?

    Thank you in advance!
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