The plane $y = −2$ intersects the surface

$$ z = x^3 − y \sin(x + y) $$

in a curve.

What is the slope of the tangent line to this intersection curve at the point $(2, −2, 8)$?

I've tried substituting in $y = -2$ and then finding the derivative of that but I don't think it's right but I'm not too sure what else to do.

  • $\begingroup$ What you did is correct, substitute $y=-2$ since that it where the surface will intersect the plane. Then evaluate the derivative at the x and z coordinate provided. $\endgroup$ – TSF Nov 19 '15 at 19:30
  • $\begingroup$ Sorry @TonyS.F. I don't really understand what you mean by the second part? I differentiate with respect to x then z and then do I substitute in the x and z co-ordinates given? $\endgroup$ – Luke McNeil Nov 19 '15 at 19:42
  • $\begingroup$ mathforum.org/library/drmath/view/52062.html $\endgroup$ – TSF Nov 20 '15 at 2:50

\begin{gather} z=x^3-y\sin(x+y)\\ z=x^3+2\sin(x-2)\\ \frac{dz}{dx}=3x^2+2\cos(x-2)=14 \end{gather}


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