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

1. Sep 11, 2010

### JetteroHeller

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 12, 2010

### 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)

3. Sep 12, 2010

### JetteroHeller

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

4. Jul 17, 2012

### 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?

5. Jul 17, 2012

### 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).

6. Mar 4, 2014

### 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?