Find all the points (x0,y0,z0) on the surface z = (x^3)(y^2) at which the tangent plane is parallel to the plane 3x + 18y - z = 0. I'm not quite sure how to go about finding out how to do this.

Question

anonymous · Answer

Okay first of all, we want to get the normal to the plane.

anonymous · Answer

The normal vector to a surface is given by $$
\begin{bmatrix}
-f_x \
-f_y \
1
\end{bmatrix}
$$

anonymous · Answer

For the planes to be parallel, the normal vectors must be parallel.

anonymous · Answer

How do you know when two vectors are parallel?

anonymous · Answer

If $\exists c\quad \mathbf{a} = c\mathbf{b}$ then $\mathbf{a}$ is parallel to $\mathbf{b}$

anonymous · Answer

Also, the cross product will be 0, also

anonymous · Answer

I got (3,1/3,3) and (-3,-1/3,-3) after all of it. Thank you!

anonymous · Answer

Also, the dot product will be equal to the product of the magnitudes since $\cos\theta = 1$

anonymous · Answer

Umm, what exactly did you do?

anonymous · Answer

I found f(x)(x0,y0) and f(y)(x0,y0), replaced x with x0 and y with y0, and solved for the coefficients from the plane I'm trying to find the parallel of. Then I solved for z to get those points. According to the answer sheet, I'm correct; I just got snagged on the whole parallel concept

anonymous · Answer

Good thinking.  Glad I could help.