# Homework Help: Partial Differentiation Help with Chain Rule

1. Jun 21, 2009

In fluid mechanics velocity is given in the form

$$\textbf{V}=u\textbf{i}+v\textbf{j}+w\textbf{k}$$

1. The problem statement, all variables and given/known data

A two-dimensional velocity field is given by

$$\textbf{V}=(x^2-y^2+x)\textbf{i}+(-2xy-y)\textbf{j}$$

At $(x_o,y_o)$ compute the accelerations $a_x\text{ and }a_y$

I am having trouble with the books definition of $a_x\text{ and }a_y$

Now I can see that the w term is zero. They define ax to be

$$a_x=\frac{\partial{u}}{\partial{t}}=u\frac{\partial{u}}{\partial{x}}+v\frac{\partial{u}}{\partial{y}}$$

I know that the chain rule has been used to arrive at this, but I seem to be getting lost along the way. So I am attempting to do it out here. So when we say

$\frac{\partial{u}}{\partial{t}}$ we are really saying

$$\frac{\partial{}}{\partial{t}}[u(x(t),y(t),z(t))]$$

right?

I am just confused as to how to evaluate this

2. Jun 21, 2009

### slider142

That is a bit strange, seeing as u is only a function of one variable, t. I don't see why the partial symbol is being used, as it is thus equivalent to the normal derivative du/dt. Since u(t) = u(x(t), y(t), z(t)) is a map from R into R, we have:
$$\frac{du}{dt} = \frac{\partial u}{\partial x}\frac{dx}{dt} + \frac{\partial u}{\partial y}\frac{dy}{dt} + \frac{\partial u}{\partial z}\frac{dz}{dt}$$
by the chain rule.
I do not see any way in which they can justify v being part of the partial derivative of u with respect to time.

Last edited: Jun 21, 2009
3. Jun 21, 2009

I guess I should have stated that u,v,w are ALL functions of x,y,z,t (or can be)

So u could be a function of x, y, & z each of which are functions of time....I can see where this is going, but I do not know how to get there. Unfortunately there are some "gaps" in my fundamental knowledge of calculus. They are mostly notational in nature.

I know that

$$\frac{\partial{}}{\partial{t}}[u(x(t),y(t),z(t))]$$

$$=\frac{\partial{u}}{\partial{x}}*\frac{\partial{x}}{\partial{t}}+\frac{\partial{u}}{\partial{y}}*\frac{\partial{y}}{\partial{t}}+\frac{\partial{u}}{\partial{z}}*\frac{\partial{z}}{\partial{t}}$$

but I am not sure how to get from that first line to the next.

Seriously, I am just confused by the commas. :rofl:

4. Jun 21, 2009

### slider142

Oh, never mind. they're using the fact that
$$u(t) = \frac{dx}{dt}$$
and
$$v(t) = \frac{dy}{dt}$$
which goes right into the chain rule we already posted.

Last edited: Jun 21, 2009
5. Jun 21, 2009

I was actually being serious about the whole "being confused by the commas" thing.

How do we actually "operate on" $[u(x(t),y(t),z(t))]$ ? That is, what are the mechanical steps to get from

$$\frac{\partial{}}{\partial{t}}[u(x(t),y(t),z(t))]$$

to

$$=\frac{\partial{u}}{\partial{x}}*\frac{\partial{x} }{\partial{t}}+\frac{\partial{u}}{\partial{y}}*\frac{\partial{y}}{\partial{t}}+\frac{\partial{u}}{\partial{z}}*\frac{\partial{z}}{\partial{t}}$$

6. Jun 21, 2009

### Fredrik

Staff Emeritus
There are no intermediate steps. That is the chain rule.

I like to write the single-variable version of it in the form

$$(f\circ g)'(x)=f'(g(x))g'(x)$$

and the many-variables version of it in the form

$$(f\circ g)^i{}_{,j}(x)=f^i{}_{,k}(g(x))g^k{}_{,j}(x)$$

because this form is fairly easy to remember. Repeated indices are summed over (Einstein's summation convention), and a ",j" denotes partial derivation with respect to variable number j. For example, $g^k{}_{,j}(x)$ is the partial derivative with respect to the jth variable of the kth component of the function g.

In your case, $g:\mathbb R\rightarrow\mathbb R^3,\ f:\mathbb R^3\rightarrow\mathbb R$, so the above reduces to

$$(f\circ g)'(x)=f_{,k}(g(x))g^k'(x)$$

Last edited: Jun 22, 2009
7. Jun 21, 2009

Wow. Engineers really get the short end of the stick when it comes to thoroughness in their maths. I have been thinking about reteaching myself Calculus from a more 'Pure' standpoint.

Do you have any suggestions on a good textbook for someone who already knows how to 'do' calculus problems (for the most part) but really wants to understand the 'hows and whys' of it all?

8. Jun 22, 2009

### Fredrik

Staff Emeritus
Theorems like the chain rule are proved in all the standard textbooks. I'm sure you can find recommendations in the science book forum. (The books I studied are in Swedish, so you probably wouldn't like them). If you want a book that's just about definitions and theorems, I can recommend "Principles of mathematical analysis" by Walter Rudin. I should warn you though. It's a pretty difficult subject.

9. Jun 22, 2009

### slider142

The commas are just there to denote a map from R into R3.
Spivak's "Calculus on Manifolds" is the definitive text on the topic. It is concise and easily followable by anyone who has taken single-variable analysis and has exposure to multivariable calculus. The proofs are either left on exercises The following two theorems are proven in the chapter on differentiation and can be used to differentiate any function using the chain rule (the notation is Euler's notation, $D_if^j$ refers to the partial derivative with respect to the ith variable of the jth component of f):
This theorem allows you to find the derivative of vector-valued functions in matrix form.
Theorem: If $f:R^n\rightarrow R^m$ is differentiable at a, then $D_jf^i(a)$ exists for $1\leq i\leq m, 1\leq j\leq n$ and f'(a) is the $m\times n$ matrix $(D_jf^i(a))$.
This theorem is the multivariable chain rule.
Theorem:If $g_1,\cdots,g_m:R^n\rightarrow R$ are continuously differentiable at a, and $f:R^m\rightarrow R$ is differentiable at $(g_1(a),\cdots,g_m(a))$, then $F:R^n\rightarrow R$ defined by $F(x) = f(g_1(x),\cdots,g_m(x))$ is also differentiable at a and
$$D_iF(a) = \sum_{j=1}^m D_jf(g_1(a),\cdots,g_m(a)) D_ig_j(a).$$

Spivak makes the proofs of these theorems very easy by using very good definitions and leading the reader in the right directions with the exercises so that the reader can usually prove the theorem themselves without Spivak's proof. Ie., Stokes's Theorem, a crown jewel of vector calculus, simply falls out of the definitions by rote algebra.

Last edited: Jun 22, 2009
10. Jun 22, 2009