# Hard Multivariable Calculus Question: Surfaces

The graph of a function z=f(x,y) can always be parametrized as (x,y,f(x,y)). Show that in this case the "old way" (using the derivative) and the "new way" (Parametrizing the surface) for finding the tangent plane at a point always agree.

I have no idea how to do this problem. I think he wants us to show that two different ways lead to the same tangent plane at a point. one is using the derivative, the second is doing it via surfaces. Can someone run me through the steps? Thanks in advance!

• Does the parametrized way not use the derivative? – Sonal_sqrt May 8 '18 at 4:27
• @PiyushDivyanakar i'm not sure. i think he wants us to use the gradient as one way and the T vectors as the second way – MathWannaBe May 8 '18 at 4:38

## 1 Answer

Suppose a surface is parametrized by a function of two real variables $\vec r(u,v)$. The vector $$\frac{\partial \vec r}{\partial u} \times \frac{\partial\vec r}{\partial v}$$ is normal to the tangent plane at any point. [see Stewart, p.416]

Using the parametrization given in the problem and the vector equation for a plane, you can show that you recover precisely the general form $$z-z_0=\frac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}(x-x_0)+\frac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}(y-y_0)$$ which is the equation of the tangent plane found the «old way». [see Stewart, p.160]

[Stewart, James. Calcul à plusieurs variables, Modulo, 2012]

• But, then how would we do it the new way using T vectors? – MathWannaBe May 8 '18 at 5:14
• I think the main idea in this question (in my understanding) is that if you have the surface z=f(x,y), you can substitute the parameters (u,v) by the coordinates (x,y); in that way everything you proved for parametrized surfaces now correspond to the initial setting with T vectors in terms of x and y. So going the other way you would only have to substitute (u,v) for (x,y) in the T vectors to get the tangent plane the «new way». – Guillaume Laplante-Anfossi May 8 '18 at 5:30