# Intersection of Surfaces

• June 14th 2009, 01:18 AM
billa
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?
• June 15th 2009, 01:16 AM
billa
anyone?

If am not making sense then say something
• June 15th 2009, 03:25 AM
NonCommAlg
Quote:

Originally Posted by billa
"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.