Derivative of physics equation

[Bmanmcfly]

I just wanted to make sure I was tackling this the right way... Thanks for the help once again.

If k=1/2mv^2 is dk/dt= v(dv/dt)?
or is it dk/dt= dm/dt(v)(dv/dt)

It seems if its the latter that dm/dt=1, which is odd cause the problem that continues has a given mass, probably to trip me up.
Am i on the right track here?

Thanks again.

m/dt

 

[Deleted member 4993]

I just wanted to make sure I was tackling this the right way... Thanks for the help once again.

If k=1/2mv^2 is dk/dt= v(dv/dt)?

No

If mass is constant - then,

dk/dt = m * v * (dv/dt)


or is it dk/dt= dm/dt(v)(dv/dt)

No - it looks like you have forgotten how to differentiate a function!!

If mass is variable - e.g. in a rocket-ship - you need to use the product rule of differentiation

dk/dt = m * v * (dv/dt) + v²/2 * dm/dt

It seems if its the latter that dm/dt=1, which is odd cause the problem that continues has a given mass, probably to trip me up.
Am i on the right track here?

Thanks again.

m/dt

.

 

[Bmanmcfly]

Thanks. I knew I was doing something wrong, cause the answers I was getting were way too small.

 

[HallsofIvy]

I just wanted to make sure I was tackling this the right way... Thanks for the help once again.

If k=1/2mv^2 is dk/dt= v(dv/dt)?

As Subhotosh Khan said, you have forgotten the "m". Assuming m is constant, ((1/2)mv^2)'= (1/2)m (2v dv/dt)= mv dv/dt.

or is it dk/dt= dm/dt(v)(dv/dt)

If m is variable as well as v, then you would use the "product rule" (uv)'= u'v+ uv'.
Then dk/dt= (1/2)(dm/dt)(v^2)+ m dv/dt.

It seems if its the latter that dm/dt=1, which is odd cause the problem that continues has a given mass, probably to trip me up.
Am i on the right track here?

Thanks again.

m/dt