Problem simplifying v*dv/dt to get d/dt(V^2/2)

[JetteroHeller]

Homework Statement

we know dV/dt = a from the definition of acceleration
multiplying by sides by V, velocity yields
V dV/dt = aV
Left side can be reduced to d/dt(V^2/2)

so to be more clear, just looking at the left side
V dV/dt = d/dt( V^2/2)

I want to know what calculus topic that's addressed in, and how it works because I don't see it.

Homework Equations

The Attempt at a Solution

 

[Pengwuino]

It's just a chain rule. Let f(t) = v(t) and f'(t) = a(t). Then {d \over dt} {f(t)^2 \over 2} = f(t)f'(t)

All you're doing is assigning physical concepts (velocity, acceleration) to mathematical ideas (functions of a parameter, t)

 

[JetteroHeller]

Thank you very much pengwuino. I would never have thought of that.

 

[Xouthos]

I'm probably missing something obvious here. Can't find anything in the chain rule that explains this simplification. Can somebody elaborate this more?

 

[Muphrid]

You can explain it as either the chain rule or the product rule.

Chain rule: \frac{ds}{dt} = \frac{ds}{dv} \frac{dv}{dt}. Let s = v^2, which gives \frac{ds}{dt} = (2v)\frac{dv}{dt}.

Product rule: \frac{d}{dt} ab = \frac{da}{dt} b + a \frac{db}{dt}. Let a=b=v.

In general, this technique of manipulating an expression to turn it into a derivative of something else is useful because you can often use that to write some kind of conservation law. In this case, dealing with \frac{d}{dt} \frac{v^2}{2} was probably leading up to invoking conservation of energy (only a factor of mass is missing).

 

[iromkingman]

could you also think of it as: since you added a v to the right you must make the left match by first taking the anti derivative of v>>>(1/2)v^2?