# Homework Help: Need help before class starts

1. Sep 24, 2010

### Fjolvar

I've been up for hours trying to figure out the proper notation to prove problem #4 (1.6.5) on the attached files. I'm not sure how to set this up. It is basically proving the product rule using the gradient of two functions. The rest of the problem is on the backside file, but help getting started on the first part would be enough. Any help would be greatly appreciated. Thanks!

#### Attached Files:

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• ###### HW 3 Back.pdf
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2. Sep 24, 2010

### Fjolvar

Any ideas?

3. Sep 24, 2010

### HallsofIvy

It looks to me like a pretty standard "just plug it in and calculate"!

You are asked to show that $\nabla(uv)= \nabla(u)v+ u\nabla(v)$ where u and v are scalar valued functions of variables x and y. This is, basically, just the "product rule" for gradients.

Okay,
$$\nabla(uv)= \frac{\partial uv}{\partial x}\vec{i}+ \frac{\partial uv}{\partial y}\vec{j}$$
Apply the product rule to both partial derivatives:
$$\nabla(uv)= \left(u\frac{\partial v}{\partial x}+ \frac{\partial u}{\partial x}v\right)\vec{i}+ \left(u\frac{\partial v}{\partial y}+ \frac{\partial u}{\partial y}v\right)\vec{j}$$
and regroup.

For part (a), if f(u, v)= 0 (where f is differentiable- that is NOT stated in the text but is necessary), then
$$\frac{\partial f}{\partial x}= \frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+ \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}= 0$$
and
$$\frac{\partial f}{\partial y}= \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+ \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}= 0$$
Think of those as two equations to solve for the "unknowns" $\partial f/\partial u$ and $\partial f/\partial v$. What condition on the "coefficients", $\partial u/\partial x$, $\partial u/\partial y$, $\partial v/\partial x$, and $\partial v/\partial y$ is necessary in order that that have a solution (think about the determinant of the matrix of coefficients- that's the point of (b)).

4. Sep 24, 2010

### Fjolvar

Thank you very much, I think I've got it from here.