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 April 23rd, 2010, 08:52 AM #1 Member   Joined: Apr 2010 Posts: 91 Thanks: 0 Chain rule A particle moves along a straight line with displacement s(t), velocity v(t), and acceleration a(t). Show that a(t) = v(t) * dv/ds Explain the difference between the meanings of the derivatives of dv/dt and dv/ds For the second part of the question I said that dv / dt is the rate of change of velocity with respect to time and dv / ds is the rate of change of velocity with respect to distance, but I am not sure of the first part where you have to prove the formula.
April 23rd, 2010, 08:59 AM   #2
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Re: Chain rule

Quote:
 Originally Posted by TsAmE A particle moves along a straight line with displacement s(t), velocity v(t), and acceleration a(t). Show that a(t) = v(t) * dv/ds Explain the difference between the meanings of the derivatives of dv/dt and dv/ds For the second part of the question I said that dv / dt is the rate of change of velocity with respect to time and dv / ds is the rate of change of velocity with respect to distance, but I am not sure of the first part where you have to prove the formula.
It is widely known that v(t) = ds/dt
So if you substitute it back into the equation:
a(t) = ds/dt * dv/ds
order doesn't matter in multiplication so
a(t) = dv/ds * ds/dt
= dv/dt since ds "cancels out"
And thats the chain rule for how to differentiate velocity wrt time. You need to know velocity wrt displacement first and then displacement wrt time.

 April 23rd, 2010, 09:51 AM #3 Member   Joined: Apr 2010 Posts: 91 Thanks: 0 Re: Chain rule Oh ok that makes sense, but how can a(t) = v(t) * (dv/ds) be the chain rule form? It looks nothing like the form f'(g(x)) * g'(x)
April 23rd, 2010, 01:23 PM   #4
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Re: Chain rule

Quote:
 Originally Posted by TsAmE Oh ok that makes sense, but how can a(t) = v(t) * (dv/ds) be the chain rule form? It looks nothing like the form f'(g(x)) * g'(x)
f=v
g=s
x=t
so f'(g(x)) pretty much means df/dg and g'(x) means dg/dx.
Hope this helps.

 April 24th, 2010, 06:54 PM #5 Senior Member   Joined: Jan 2009 Posts: 345 Thanks: 3 Re: Chain rule assuming s denotes position. $a(t)= v(t) \frac{dv}{ds}\Leftrightarrow v(t)dv(\frac{1}{ds}) \Rightarrow v(t)[\frac{ds}{dt}(\frac{1}{ds})] \Rightarrow \frac{v(t)}{dt} = a(t)$

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# a particle moves along a straight line with displacement s(t), velocity v(t), and acceleration a(t) explain the difference between the meanings of the derivativees

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