Derivative Help!

lily7

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Oct 17, 2021
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2
A particle of constant mass m moves along the x-axis. Its velocity v and position x satisfy the equation: 1/2m(v^2 - v0^2) = 1/2k(x0^2-x^2), where k, v0 and x0 are constants. Show that whenever v does not equal 0, mdv/dt=-kx.

I got to the point of doing
dv/dt(mv^2-mv0^2) = dv/dt(kx0^2-kx^2)

Is that right? How do I do it?
 

Jomo

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Dec 30, 2014
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11,180
m, k, v0 and x0 are constant.

mv^2 -mvo^2 = av^2 - b where a and b are constants. a=m and b = mv0^2.

I am sure you know how to compute the derivative of av^2 wrt t.

The rhs kx0^2-kx^2 = a - bx^2 where a= kx0^2 and b =k.

You need to show some work to get more help.
 

lily7

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Oct 17, 2021
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2
m, k, v0 and x0 are constant.

mv^2 -mvo^2 = av^2 - b where a and b are constants. a=m and b = mv0^2.

I am sure you know how to compute the derivative of av^2 wrt t.

The rhs kx0^2-kx^2 = a - bx^2 where a= kx0^2 and b =k.

You need to show some work to get more help.
Ok so then I get 2av=-2bx? And then divide both sides by 2 to get mv=-kx. But, how do I get from there to mdv/dt =-kx?
 

topsquark

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Aug 27, 2012
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1,267
[imath]\dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_0^2 = \dfrac{1}{2}kx_0^2 - \dfrac{1}{2}kx^2[/imath]

Take the time derivative of both sides:
[imath]\dfrac{1}{2}m \cdot 2 v \dfrac{dv}{dt} = - \dfrac{1}{2}k \cdot 2 x \dfrac{dx}{dt}[/imath]

Hint: What's dx/dt? Is there anything here that will cancel?

-Dan
 

Jomo

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Joined
Dec 30, 2014
Messages
11,180
Ok so then I get 2av=-2bx? And then divide both sides by 2 to get mv=-kx. But, how do I get from there to mdv/dt =-kx?
You did not take the derivative with respect to t.
You should have gotten 2avdv/dt=-2bxdx/dt
 
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