# Finding tangent plane at a point to the surface

Z is a surface satisfying $$e^{xz} \sin (y) =yz$$

The tangent plane at point $$(0, \pi/2, 2/\pi)$$

I know that the tangent plane will be given by $$z - c = f(a, b) + f_x(a,b) (x-a) + f_y(a,b)(y-b)$$

-

## 1 Answer

Consider the function $g: \mathbb{R}^3 \to \mathbb{R}$ given by $g(x,y,z) = e^{xz}\sin y-yz$. Clearly, the surface $Z$ is given by the set of $(x,y,z) \in \mathbb{R}^3$ such that $g(x,y,z) = 0$. Consider the gradient vector:

$$\nabla g = e^{xz}(z\sin y,\cos y,x\sin y) \, .$$

The surface is non-singular provided $\nabla g \neq (0,0,0).$ At the point $(x,y,z) = (0,\pi/2,2/\pi)$ we have

$$(\nabla g)(0,\pi/2,2/\pi) = (-2/\pi,0,0) \neq (0,0,0).$$

Thus, the surface $Z$ is non-singular sufficiently close to that point. Moreover, the gradient vector is normal to the tangent plane at that point. Thus, the tangent plane passes through $(0,\pi/2,2/\pi)$ and is perpendicular to the vector $(-2/\pi,0,0) \propto (1,0,0).$ The equation of the tangent plane is thus:

$$x = 0 \, .$$

-