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April 23rd, 2010, 08:52 AM   #1
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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.
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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.
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April 23rd, 2010, 09:51 AM   #3
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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)
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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.
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April 24th, 2010, 06:54 PM   #5
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Re: Chain rule

assuming s denotes position.

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