# Tangent plane to the surface $\cos(x)\sin(y)e^z = 0$?

The surface is as in the title, $$\cos(x) \sin(y) e^z = 0$$ I'm looking for the tangent plane at the point $(\frac{\pi}{2},1,0)$

I know the equation of a tangent plane for $z = f(x,y)$ is $$z-z_0 = f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$ But in the surface given, I cannot isolate for $z$. What should I do?

Let $f : U\subset \mathbb{R}^3\to \mathbb{R}$ be given by $f(x,y,z)=\cos x\sin y e^z$ then we have that your surface is indeed a level set $M = f^{-1}(0)$. Then it's easy: remember that the gradient of a function is orthogonal to the level sets. Using this we have that
$$\nabla f(x,y,z)=(-\sin x\sin ye^z, \cos x\cos y e^z, \cos x\sin y e^{z})$$
So that at $(\pi/2,1,0)$ we have $\nabla f(\pi/2,1,0)=(-\sin 1,0,0)$, so that the normal is a multiple of the vector $e_1$, and hence since the magnitude of the normal vector doesn't matter, we can pick the normal vector to be $e_1$. Of course then, the tangent plane is just the $yz$ plane.
• Worth pointing out: this relies on an alternate characterization of planes, $\vec{p}\cdot\hat{n}=D$ where $\hat{n}$ is a normal to the plane. – Steven Stadnicki Oct 9 '13 at 2:03