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        riddles >> hard >> Calculus question
        
(Message started by: BenVitale on Nov 18^th, 2008, 2:37pm)

Title: Calculus question
Post by BenVitale on Nov 18^th, 2008, 2:37pm

Has anyone ever seen a problem where one variable is the derivative of another?

Such as y = f(z,z') where z and z' are functions of x

I realize that the Chain Rule is a very powerful, but I've never seen how to solve it.

We can use the dz/dz' = dz/dx*dx/dz' trick.

Can anyone provide examples where you differentiate with respect to non-t or non-x variables?

Title: Re: Calculus question
Post by towr on Nov 18^th, 2008, 3:02pm

They're called differential equations. You encounter them a lot in physics.
For example for a mass-spring system,
-you have the position of the mass
-which changes depending on the velocity of the mass (time-derivitive of position)
-which changes depending on the acceleration (time-derivitive of velocity)

Title: Re: Calculus question
Post by BenVitale on Nov 19^th, 2008, 9:26am

Right. I've spoken too fast.

From the Chain Rule, we have

dy/dz' = df/dz * dz/dz' + df/dz' * dz'/dz'
dy/dz' = df/dz * dz/dz' + df/dz'

dz/dz' = dz/dx * dx/dz' = dz/dx / (dz'/dx)

...................................

Let's take for example, y = sin(x) + cos(x)
We know that: (sin x)'= cos x and (cos x)'= - sin x
Let z = sin(x)
Then, y = z + z'

dy/dz' = d(z+z')/dz * dz/dz' + d(z+z')/dz'
...... = 1 * dz/dz' + 1
...... = 1 + dz/dz'
...... = 1 + dz/dx / (dz'/dx)
...... = 1 + dsin(x)/dx / (dcox(x)/dx)
...... = 1 + cos(x) / (-sin(x))
...... = 1 - cos(x) / sin(x)
...... = 1 - cot(x)

Still leaving z = sin(x) and z' = cos(x).

Title: Re: Calculus question
Post by towr on Nov 19^th, 2008, 10:50am

If you have y = z + z'
Then you can add A * e^-x+B to any z(x) you start with. So instead of z(x) = sin(x), you'll have z(x) = sin(x) + A * e^-x+B

I'm still not quite sure what you're trying to do though. And it's been a long time since I did differential equations.

[edit]
Using some pointers from http://mathworld.wolfram.com/First-OrderOrdinaryDifferentialEquation.html I get
z(x) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^x y(x) dx + c)/ e^x + A e^-x+B

then z'(x) = y(x) - (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^x y(x) dx + c)/e^x - A e^-x+B
so z(x)+z'(x) = y(x)
[/edit]

Title: Re: Calculus question
Post by BenVitale on Nov 19^th, 2008, 4:35pm

Towr,
I don't understand your solution.

I'm trying to find dy/dz' when y = f(z,z')

y = sin(x) + cos(x) is f(z,z')
because z = sin(x) and z'= cos(x)

And I'm using the Chain rule to find dy/dz'

The Chain Rule does the trick here, doesn't it?

Title: Re: Calculus question
Post by towr on Nov 20^th, 2008, 12:38am

on 11/19/08 at 16:35:50, BenVitale wrote:

Towr,
I don't understand your solution.

I'm solving y = z + z' for z.

Quote:

I'm trying to find dy/dz' when y = f(z,z')

Why would you want to ever do a thing like that?

Quote:

y = sin(x) + cos(x) is f(z,z')
because z = sin(x) and z'= cos(x)

And I'm using the Chain rule to find dy/dz'

The Chain Rule does the trick here, doesn't it?

why not just take dy/dx = cos(x) - sin(x) and dz'/dx = - sin(x) and divide the two to get
dy/dz' = 1 - cos(x)/sin(x) = 1 - cot(x)

Title: Re: Calculus question
Post by BenVitale on Nov 20^th, 2008, 10:49am

That works too. Your solution is shorter.

I started with the idea of using the Chain Rule. It is such a powerful and seductive method. I wanted to use this method with a function where one variable is the derivative of another variable.

Title: Re: Calculus question
Post by BenVitale on Nov 20^th, 2008, 6:54pm

How about if I have y = z³ + (z') ²

The Chain Rule gives us:

dy/dz' = d(z³ + (z')²)/dz * dz/dz' + d(z³ + (z') ²)/dz'
dy/dz' = 3z² * dz/dz' + 2z'
dy/dz' = 3*sin²(x) * (-cot(x)) + 2*cos(x)
dy/dz'= cosx (2 - 3sinx)

Title: Re: Calculus question
Post by towr on Nov 21^st, 2008, 12:04am

Why would z be sin(x)?
If you know that to start with, you can solve the problem much more easily.

Title: Re: Calculus question
Post by BenVitale on Nov 21^st, 2008, 10:21am

Sorry for not explaining it clearly.
The second example/question is actually a follow-up to the first one.

z is still sin(x)

and y = z³ + (z')²

Instead of writing y = [3 sin x - sin 3x]/4 + cos² (x)
I wrote: y = z³ + (z')²

I know that I could avoid all that and use an easier way to solve this. I'm exploring the Chain Rule tool, and I would like to check with the readers, the mods and tell me what you do think about this.