# Homework Help: Chain Rule For Function of Severable Variables

1. Dec 13, 2011

### jjstuart79

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

I'm trying to follow my text book on an application of the chain rule.

Two objects are traveling in elliptical paths given by the following parametric equation.

x1 = 4 cos t
x2 = 2 sin 2t
y1 = 2 sin t
y2 = 3 cos 2t

At what rate is the distance between the two objects changing when t = pi?

2. Relevant equations

distance S = √(x2 - x1)2 + (y2 - y1)2

3. The attempt at a solution

When t = pi
x1 = -4
y1 = 0
x2 = 0
y2 = 3

When t = pi the partial derivatives of s are as follows.

∂s/∂x1 = -(x2 - x1)/√(x2 - x1)2 + (y2 - y1)2 = -1/5(0 + 4) = -4/5

How come we're all of a sudden dividing -(x2 - x1) by S? I guess I'm not grasping the concept of the partial derivative of S with respect to x1 when x1 is an equation either. Would someone be able to elaborate on that? If you need me to explain more, please let me know.

Thanks in advance for any help.

2. Dec 13, 2011

### SammyS

Staff Emeritus
You need another set of grouping symbols.

Is this your solution or one that you copied?

It seems, by your question, that you don't know what's involved with taking the derivative.

3. Dec 13, 2011

### jjstuart79

That is the solution I copied from the text. I'm teaching myself this, so sometimes it helps to have someone be able to elaborate on the why part of what the text is showing.

To me I would think that since I'm taking the derivative with respect to x1, I take the derivative of x1 which is -4 sine t. So it would look like this?

-(2 sin 2t - -4 sine t)/√(2 sin 2t - -4 sine t)^2 + (3 cos 2t - 2 sin t)^2)

I still don't know why I have to divide -(x2 - x1) by S when taking the derivative?

Last edited: Dec 13, 2011
4. Dec 13, 2011

### Staff: Mentor

ok so you have two sets of parametric eqns all a function of t and you want the rate of change when t=pi

dist = (x1 - x2)^2 + (y1 - y2)^2

plug in your parametrics for the x1 ,x2, y1 and y2 and then differentiate with respect to t

the chain rule comes in when start to differentiate the terms with respect to t:

( x1 - x2 ) ^ 2 ==> ( 4 cos(t) - 2 sin(2t) ) ^ 2

differentiating to get: 2 ( 4 cos(t) - 2 sin(2t) ) ( - 4 sin(t) - 4 cos(2t) ) for the first term

and: ( 2 sin(t) - 3 cos(2t) ) ^ 2

differentiating to get: 2 ( 2 sin(t) - 3 cos(2t) ) ( 2 cos(t) + 6 sin(2t) ) for the second term

next we can plug in t=pi to get:

sin(t) = 0
sin(2t) = 0
cos(t) = -1
cos(2t) = 1

plugging these values in:

2 (-4)(-4) + 2(-3)(-2) = 2 * 16 + 2 * 6 = 44

does that look right?

Last edited: Dec 13, 2011
5. Dec 13, 2011

### SammyS

Staff Emeritus
What you have with $\displaystyle -\frac{2 \sin(2t) - -4 sin(t)}{\sqrt{2 \sin (2t) - -4 \sin (t))^2 + (3 \cos (2t) - 2 \sin (t))^2}}$ looks like an attempt to take the derivative of S with respect to t, although it's incorrect.

Let's look at a simpler example:
Suppose that $T(x,y) = \sqrt{x^2 + y^2}$

Then $\displaystyle \frac{\partial T}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$

Much like $\displaystyle \frac{d }{dx}\sqrt{x^2+25}=\frac{x}{\sqrt{x^2+25}}$
It's just using the chain rule along with the derivative of the square root function.

6. Dec 14, 2011

### jjstuart79

Thanks for the help. It makes more sense to me now. I was getting confused because I was applying the chain rule and not simplifying enough, I think. For instance using your example.

d/dx = √(x^2 + 25) = (x^2 + 25)^1/2 = 1/2(x^2 + 25)^-1/2 * (2x) = x(x^2 + 25)^-1/2 =

x/(x^2 + 25)^1/2

btw, how are you embedding your equations in the posts? The format is really nice.

Last edited: Dec 14, 2011
7. Dec 14, 2011

### SammyS

Staff Emeritus
I'm using Latex.

Use the Ʃ Icon just above the "Advanced" messaging window.

It takes a bit of getting used to.

8. Dec 14, 2011

### jjstuart79

Thank you. I will give it a try.

9. Dec 14, 2011

### SammyS

Staff Emeritus