How did "v" come out of this derivative?

Dens

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






3. The attempt at a solution

I basically understood how to attack the problem and my answer was very close to the key. The only thing bothering me is that v on the top. I cannot see how that could come up anywhere.

I know that the "r" is r = r(t), so that I am only taking the derivative of $$\ln(1 + w/r)$$. I don't see where the v actually is

TSny

The flux is expressed as a function of r. You're taking the derivative with respect to t. So, recall the chain rule of calculus: first take the derivative with respect to r and then multiply by the derivative of r with respect to t.

Dens

Nope, implicit differentiation. I had forgotten about that

TSny

Here's the chain rule: If [itex]y=f[u][/itex] and [itex]u = g[x][/itex] are differentiable functions, then [itex]\frac{dy}{dx}[/itex]=[itex]\frac{dy}{du}[/itex][itex]\cdot[/itex][itex]\frac{du}{dx}[/itex]

You have the flux expressed as a function of r: [itex]\Phi[r][/itex] and r is some function of time: [itex]r[t][/itex].

So, the chain rule says [itex]\frac{d\Phi}{dt}[/itex] = [itex]\frac{d\Phi}{dr}[/itex][itex]\cdot[/itex][itex]\frac{dr}{dt}[/itex]

This get's you the answer and it shows why the speed v appears.

How do you get the answer using implicit differentiation?

Dens

No you are right, I forget the $\phi$ on the LHS. I was thinking something like

Something = r as a function of t ==> differentiate both sides wrt t ==> implicitly differentiate RHS

http://www.wolframalpha.com/input/?i=D[ln%281+%2B+w%2Fr%28t%29%29%2Ct]