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Math Help - Intersection of Surfaces

  1. #1
    Member billa's Avatar
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    Intersection of Surfaces

    "The surfaces z=f(x,y) and z=g(x,y) intersect in a curve C. Find an expression, in terms of the derivatives of f and g, for a vector tangent to C at a point P on C where x=a and y=b."

    Here is what I did.

    (1) let F(x,y) = f(x,y) - g(x,y) = 0
    (2) I found the gradient vector for F
    (3) I found a vector perpendicular to the gradient vector for F, this vector, which I called Txy must give the direction where F(x,y) doesn't change values, and this vector must be in the direction of the curve C.
    (4) I found the partial derivative of f(x,y) in the direction of Txy, I believe that this gives the rate of change of z along a direction (Txy) in the xy plane.

    (5) I said that the answer was...
    (i component of Txy)i +(j component of Txy)j + (|Txy| * partial derivative found in (4) ) k

    The book gives the same answer, except it does not include |Txy| in the k component, instead giving just the value of the partial derivative. I couldn't find any reason why this length should be 1, and I think that it needs to be there. Any help?
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  2. #2
    Member billa's Avatar
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    anyone?

    If am not making sense then say something
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  3. #3
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    Quote Originally Posted by billa View Post
    "The surfaces z=f(x,y) and z=g(x,y) intersect in a curve C. Find an expression, in terms of the derivatives of f and g, for a vector tangent to C at a point P on C where x=a and y=b."
    the vector you're looking for is simply \begin{vmatrix} i & j & k \\ f_x(P) & f_y(P) & -1 \\ g_x(P) & g_y(P) & -1 \end{vmatrix}. the reason is that a vector tangent to C at P must be perpendicular to the gradient of both surfaces at P and thus the cross product

    of the gradient of the surfaces at P wil give you a vector tangent to C at P.
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