Chain Rule For Function of Severable Variables

[jjstuart79]

Homework Statement

I'm trying to follow my textbook 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?

Homework Equations

distance S = √(x2 - x1)2 + (y2 - y1)2

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.

 

[SammyS]

Homework Statement

I'm trying to follow my textbook 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?

Homework Equations

distance S = √((x2 - x1)² + (y2 - y1)²)

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.

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.

 

[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?

 

[jedishrfu]

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?

 

[SammyS]

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?

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.

 

[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.

 

[SammyS]

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.

I'm using Latex.

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

It takes a bit of getting used to.

 

[jjstuart79]

Thank you. I will give it a try.

 

[SammyS]

Here is a Link to a LaTeX Guide by one of the PF Mentors, Rebbelly98.

https://www.physicsforums.com/showthread.php?t=546968