Homework 3, number 7

This is a very tough one, and I planned on going over it in class on Wednesday. I'll sketch out the approach below.

If power is constant (which we're basically told it is for the default car), then the work done is W=P*T where T is the time over which the power is being supplied. This work done must be equal to the car's change in kinetic energy. If the car starts from rest, the work just equals the final kinetic energy, 

W =PT = (1/2)mv^2

This relates P, T, and velocity, but we don't know velocity. We do know the track length though. What we want to do is solve that for v, and integrate it to get x. Since the length of the track (x) is fixed, that will let us relate power and time by themselves.

v = sqrt(2PT/m) 

x = (integral) v dt = sqrt(4PT^3 / 3m)

The question now is what happens if we vary P by some little amount dP, what happens with T? By how much does it decrease dT? The distance x is a function of the variables P and T. It is fixed, so any change in P will have to be accompanied by a change in T to keep it constant. 

The question we're really asking then is for the function x(P,T) to remain constant, what must the rates of change in P and T be? We'd need to know the slope along the "P axis" (so dx/dP) and multiply by the tiny change in P (let's call that change DP instead of dp to keep the change straight from the derivative). We'd also need to know the same along the "T axis". Basically, the change in any function is slope times displacement for each axis, all added together. This is the same way we propagate experimental uncertainties, by the way, something I hope we will cover soon.

If the function were f(x,y), we'd approximate a small change in f due to small changes DX and DY in x and y as

Df = f(x+dx) - f(x) = (df/dx)*DX + (df/dy)*DY

In the simpler case, if you just have y(x), all this says is DY = (dy/dx)*DX. Back to the problem at hand, if we have x(P,T), 

Dx = (dx/dP)*DP + (dx/dT)*DT

Since the track length is fixed, we know Dx = 0. Thus,

(dx/dP)*DP = - (dx/dT)*DT

or   DT = -DP*(dx/dP)/(dx/dT)

Given the function above, take the derivatives with respect to P and T, divide them, and that times the change in power gives you the corresponding change in time.

I'll plan on going over this on Wednesday too.