Finding the tangent plane of a point of a curve when using implicit differentiation

Question

I need to find the tangent plane of this surface:

$$(z-1)^3=\sin(y^2)e^{xz}$$

at the point $(0, \sqrt \pi ,1)$

I find $dz \over dx$ and $dz \over dy$

$${dz \over dx}={-e^{xz}\sin(y^2)z \over xe^{xz}\sin(y^2)-3(z-1)^2}$$

$${dz \over dy}={-2e^{xz}y \, \cos(y^2) \over xe^{xz}\sin(y^2)-3(z-1)^2}$$

I've got the equation for the tangent plane, but I'm not sure how to go from here?

$$z-z_0=f'_x(x_0, y_0)(x-x_0)+f'_y(x_0, y_0)(y-y_0)$$

Thanks!

Answer 1

Note that your solution is fine. You just need to evaluate your derivatives at the given point, so $x_0 = 0, y_0 = \sqrt{\pi}, z_0 = 1$.

Here is a good resource.

I will write you a solution (with hidden spoilers) so you can work it out and try to get it on your own. (This is a more general approach than what you have.)

First, let's think of the given surface as a level surface of a function in three variables.

Let $F: \mathbb{R}^3 \to \mathbb{R}$. What would be a good candidate for our function $F$? (There are more than one possible answers.)

$$F(x,y,z) = \sin(y^2)e^{xz} - (z-1)^3$$
With this $F$, your surface is the zero set $F(x,y,z) = 0$.
Also possible are $$F(x,y,z) = (z-1)^3 - \sin(y^2)e^{xz},$$
and the surface is the zero set $F(x,y,z) = 0.$
You can even do $$F(x,y,z) = \sin(y^2)e^{xz} - (z-1)^3 + 1,$$
and the surface is the level $F(x,y,z) = 1$.

Now, what we need to do is to find:

$F_x$, $F_y$, and $F_z$ at the point $(0,\sqrt{\pi},1)$

Then, the equation of the tangent plane according to the formula in the link is:

$0 = F_x(0,\sqrt{\pi},1)(x-0) + F_y(0,\sqrt{\pi},1)(y-\sqrt{\pi}) + F_z(0,\sqrt{\pi},1)(z-1)$

It remains to evaluate the partial derivatives at the given point.

Seeing that you don't really have problems taking derivatives, I'll leave that task to you.

Let me know if you need further clarification.