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The surface given by $z = x^2 -y^2$ is cut by the plane given by $y = 3x$, producing a curve in the plane. Find the slope of this curve at the point $(1, 3, -8)$.

My answer is:

$$f(x, y, z) = x^2 - y^2 - z = 0$$ $$g(x, y, z) = y - 3x = 0$$

$\delta f = (2x, -2y, -1)$, evaluated at $(1, 3, -8) = (2, -6, -1)$ $\delta g = (-3, 1, 0)$

Take the cross product, we get $(1, 3, -16)$. I got stuck here. Any ideas?

The solution says -8sqrt(2/5)

gter
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1 Answers1

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the tangent line through $(1,3,-8)$ is given by $$l(t)=\left(1 + t ,3 + 3t , -8 - 16t\right)$$ that is $(1 , 3, -8) + (t , 3t , -16t)$ .

I think you had all the pieces.

Alan
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