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April 13th, 2017, 04:02 AM   #1
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Differential equations

While deriving the equation for transverse acceleration, I came across following problem:

It is written that:
transverse acceleration=(dr/dt)*(ds/dt)+(dr/dt)*(ds/dt)+ r(d^2(s)/dt^2)
In the next step it is written that this is equal to:
Transverse acceleration = 1/r {(r^2)(d^2 (s))/(dt^2) +2r(ds/dt)}.......How??
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April 13th, 2017, 04:45 AM   #2
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Originally Posted by shashank dwivedi View Post
While deriving the equation for transverse acceleration, I came across following problem:

It is written that:
transverse acceleration=(dr/dt)*(ds/dt)+(dr/dt)*(ds/dt)+ r(d^2(s)/dt^2)
In the next step it is written that this is equal to:
Transverse acceleration = 1/r {(r^2)(d^2 (s))/(dt^2) +2r(ds/dt)}.......How??
Is this all there is to the problem statement? I notice that dr/dt has vanished.

-Dan
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April 13th, 2017, 07:27 AM   #3
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Well, yes I have the same doubt and after that they have written that:

Transverse acceleration= 1/r {d((r^2)ds/dt)/dt}
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April 13th, 2017, 08:41 AM   #4
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transverse acceleration=(dr/dt)*(ds/dt)+(dr/dt)*(ds/dt)+ r(d^2(s)/dt^2)
Hmmmmm.... Another weird thing. The first two terms are equal to each other. Why is that?

-Dan
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April 13th, 2017, 05:25 PM   #5
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transverse acceleration $= \dfrac{dr}{dt}*\dfrac{ds}{dt} + \dfrac{dr}{dt}*\dfrac{ds}{dt} + r \dfrac{d^2(s)}{dt^2}$

Transverse acceleration = $\dfrac{1}{r} * \dfrac{(r^2) (d^2 (s))}{dt^2} +2r \dfrac{ds}{dt}$

Transverse acceleration= $\dfrac{1}{r} \dfrac{\dfrac{d((r^2)ds}{dt}}{dt}$

This is the exact sequence of steps provided in your text?
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