Derivative of kinetic energy with respect to position help

[Keshroom]

Homework Statement

Show dT/dx = ma

Homework Equations

T=1/2mv^2
F=ma

The Attempt at a Solution

dT/dx = d/dx(1/2mv^2)
= mv.dv/dx <--------------i believe you use the chain rule. But can someone explain exactly how to get to this step?
= mv. (dv/dt). (dt/dx)
= m(dx/dt) . a(dt/dx) <-------------can someone please expain how m(dx/dt) = m? and a(dt/dx) = just a?
= ma

 

[NascentOxygen]

dT/dx = d/dx(1/2mv^2)
= mv.dv/dx <--------------i believe you use the chain rule. But can someone explain exactly how to get to this step?
= mv. (dv/dt). (dt/dx)

In place of dv/dx it is valid to substitute (dv/dt)⋅(dt/dx) because the (dt) terms here would cancel like when simplifying fractions.

=m ⋅ (dx/dt) ⋅ (dv/dt) ⋅ (dt/dx)

One of these 'fractions' is the reciprocal of the other, so they cancel.

All done!

= m(dx/dt) . a(dt/dx) <-------------can someone please expain how m(dx/dt) = m? and a(dt/dx) = just a?
= ma

No need to.

 

[Keshroom]

In place of dv/dx it is valid to substitute (dv/dt)⋅(dt/dx) because the (dt) terms here would cancel like when simplifying fractions.

=m ⋅ (dx/dt) ⋅ (dv/dt) ⋅ (dt/dx)

One of these 'fractions' is the reciprocal of the other, so they cancel.

Yeah i understand that part, but how exactly do i use the chain rule to get to this step?
dT/dx = d/dx(1/2mv^2) to mv(dv/dx)

 

[NascentOxygen]

To differentiate something with respect to x:
you can instead differentiate it with respect to v
and then multiply the result by (dv/dx).

That's the application of the chain rule.

 

[Keshroom]

To differentiate something with respect to x:
you can instead differentiate it with respect to v
and then multiply the result by (dv/dx).

That's the application of the chain rule.

oooooo right. Makes so much sense now. Thanks heaps :)

 

[HallsofIvy]

I am really concerned when people talk about "canceling" parts of derivatives. It works, of course, but you should keep in mind that this is just a 'mnemonic'. What really is happening is that, going back to before the limit of the "difference quotient", where you really do have fractions, do the cancelling there, then taking the limit. You have to be careful that limits "respect" the fractions.

From $T= (1/2)mv^2$ we have $dT/dx= (1/2)m d(v^2)/dx$ (assuming that m is constant, of course) $= m v dv/dx$. Now, a slightly more "rigorous" argument would be that since x is itself a function of t, $dv/dt= (dv/dx)(dx/dt)$ (the chain rule). But, of course, $dx/dt= v$ so that says that $dv/dt= (dv/dx)(dx/dt)= (dv/dx)v$ and so $dv/dx= (dv/dt)/v$. Putting that into the above, $dT/dx= m v (dv/dt/v)= m dv/dt= ma$

 

[NascentOxygen]

I am really concerned when people talk about "canceling" parts of derivatives.

Hence the quotes.

It doesn't hurt to remind people, but the fact that it works is usually sufficient justification.

 

[SammyS]

I am really concerned when people talk about "canceling" parts of derivatives. ...

I wholeheartedly agree !

 

 

[Keshroom]

I am really concerned when people talk about "canceling" parts of derivatives. It works, of course, but you should keep in mind that this is just a 'mnemonic'. What really is happening is that, going back to before the limit of the "difference quotient", where you really do have fractions, do the cancelling there, then taking the limit. You have to be careful that limits "respect" the fractions.

From $T= (1/2)mv^2$ we have $dT/dx= (1/2)m d(v^2)/dx$ (assuming that m is constant, of course) $= m v dv/dx$. Now, a slightly more "rigorous" argument would be that since x is itself a function of t, $dv/dt= (dv/dx)(dx/dt)$ (the chain rule). But, of course, $dx/dt= v$ so that says that $dv/dt= (dv/dx)(dx/dt)= (dv/dx)v$ and so $dv/dx= (dv/dt)/v$. Putting that into the above, $dT/dx= m v (dv/dt/v)= m dv/dt= ma$

Thanks heaps HallsofIvy, this is very clear. :D