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Benny
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Calculus question
« on: Nov 18th, 2008, 2:37pm » |
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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?
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towr
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Re: Calculus question
« Reply #1 on: Nov 18th, 2008, 3:02pm » |
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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)
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Benny
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Re: Calculus question
« Reply #2 on: Nov 19th, 2008, 9:26am » |
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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).
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towr
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Re: Calculus question
« Reply #3 on: Nov 19th, 2008, 10:50am » |
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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.htm l I get
z(x) = ( ex y(x) dx + c)/ ex + A e-x+B
then z'(x) = y(x) - (ex y(x) dx + c)/ex - A e-x+B
so z(x)+z'(x) = y(x)
[/edit]
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| « Last Edit: Nov 19th, 2008, 11:21am by towr » |
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Benny
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Re: Calculus question
« Reply #4 on: Nov 19th, 2008, 4:35pm » |
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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?
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towr
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Re: Calculus question
« Reply #5 on: Nov 20th, 2008, 12:38am » |
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on Nov 19th, 2008, 4:35pm, BenVitale wrote:Towr,
I don't understand your solution. |
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I'm solving y = z + z' for z.
Quote:| I'm trying to find dy/dz' when y = f(z,z') |
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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? |
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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)
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| « Last Edit: Nov 20th, 2008, 12:39am by towr » |
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Benny
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Re: Calculus question
« Reply #6 on: Nov 20th, 2008, 10:49am » |
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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.
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Benny
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Re: Calculus question
« Reply #7 on: Nov 20th, 2008, 6:54pm » |
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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)
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towr
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Re: Calculus question
« Reply #8 on: Nov 21st, 2008, 12:04am » |
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Why would z be sin(x)?
If you know that to start with, you can solve the problem much more easily.
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Benny
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Re: Calculus question
« Reply #9 on: Nov 21st, 2008, 10:21am » |
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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.
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