Need help with chain rule for relating ds/dt to dx/dt and dy/dt


by rectifryer
Tags: chain, ds or dt, dx or dt, dy or dt, relating, rule
rectifryer
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#1
Jun3-12, 05:46 PM
P: 10
1. The problem statement, all variables and given/known data

s=[itex]\sqrt{(3x^2)+(6y^2)}[/itex]

2. Relevant equations
None


3. The attempt at a solution
[itex]\stackrel{ds}{dt}[/itex]=[itex]\stackrel{d}{dt}[/itex][itex]\sqrt{(3x^2)+(6y^2)}[/itex]

[itex]\stackrel{3x}{\sqrt{(3x^2)+(6y^2)}}[/itex]

The problem with that is its only d/dx if y is a set number. I don't know how to differentiate the entire thing properly. I have been hacking at this for 8 hours. I feel like mental jello.
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Villyer
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#2
Jun3-12, 05:58 PM
P: 294
You are taking the derivative with respect to t.

So d/dt of 3x2 = 6x * dx/dt, not 6x.


Maybe this helps figure out the whole derivative?
Astronuc
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#3
Jun3-12, 06:03 PM
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Well the relevant equation under 2. Relevant equations would be an expression of the chain rule.

d/dt(f(g(t)) = f'(g(t))*g'(t)

http://archives.math.utk.edu/visual....e.4/index.html
http://www.math.ucdavis.edu/~kouba/C...ChainRule.html
http://mathworld.wolfram.com/ChainRule.html

Let g(t) = g(x(t),y(t)) and f = √

One could also write the original equations as s2 = 3x2 + 6y2, and differentiate each term with respect to t.

rectifryer
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#4
Jun3-12, 06:16 PM
P: 10

Need help with chain rule for relating ds/dt to dx/dt and dy/dt


One could also write the original equations as s2 = 3x2 + 6y2, and differentiate each term with respect to t.
That doesn't really seem like it would get me anywhere. I know I am wrong, but why would that work?

Thank you for the links.
rectifryer
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#5
Jun3-12, 07:37 PM
P: 10
I have narrowed down my question, specifically to the area I have highlighted on this picture (bear in mind, I can't post pics under 10 posts):

http:// i.imgur .com /62erw.png

Where did all the dx/dt and dy/dt come from on the right side? I don't understand that step. I know how to do this when thinking about it in function form, but it confuses me to think about it in fraction form, which is whats required to answer.
HallsofIvy
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#6
Jun3-12, 08:19 PM
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PF Gold
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If s is a function of two variables, x and y, which are themselves functions of t. The "chain rule" says
[tex]\frac{ds}{dt}= \frac{\partial s}{\partial x}\frac{dx}{dt}+ \frac{\partial s}{\partial y}\frac{dy}{dt}[/tex]

Here, [itex]s(x,y)= \sqrt{3x^2+ 6y^2}= (3x^2+ 6y^2)^{1/2}[/itex]
What are [itex]\partial s/\partial x[/itex] and [itex]\partial s/\partial y[/itex]?


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