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Need help with chain rule for relating ds/dt to dx/dt and dy/dt 
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#1
Jun312, 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. 


#2
Jun312, 05:58 PM

P: 294

You are taking the derivative with respect to t.
So d/dt of 3x^{2} = 6x * dx/dt, not 6x. Maybe this helps figure out the whole derivative? 


#3
Jun312, 06:03 PM

Admin
P: 21,827

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 s^{2} = 3x^{2} + 6y^{2}, and differentiate each term with respect to t. 


#4
Jun312, 06:16 PM

P: 10

Need help with chain rule for relating ds/dt to dx/dt and dy/dt
Thank you for the links. 


#5
Jun312, 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. 


#6
Jun312, 08:19 PM

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Thanks
PF Gold
P: 39,348

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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