Question

Referring to Applications of Differentiation (PDF) Quesion 2A-11, in problem set 3. The solution sheet says the answer is \[\frac{c}{v_0^k}(1-k\frac{\Delta v}{v_0}+\frac{k(k+1)}{2}(\frac{\Delta v}{v_0})^2)\]

What I get by using quadratic approx is very similar but with extra \[v_0\]'s. Does anyone have a comment on this?

\[\frac{c}{v_0^k}(1-k\frac{v_0+ \Delta v}{v_0}+\frac{k(k+1)}{2}(\frac{v_0+\Delta v}{v_0})^2)\]

Answer 1

Actually, if you follow the quadratic approx formula, we should replace x with (v_0 + Delta v) and should end up with my solution. The prof's solution seems to replace x with just Delta v. That is my question. how can you do this?

Answer 2

There are errors in the solutions to some of the problems, but this is not one of them. Here’s the explanation.

First, understand that we’re looking to use the quadratic approximation for (1 + x)^r near 0. This is a little tricky, because we’re working with (vo + deltav)^-k, so we don’t have a 1 up front in the parentheses. To remedy that situation, we divide by v0 inside the parens and multiply by vo^-k outside the parens to get an equivalent expression that has something that looks like (1+x)^r. This is what you see at the end of the first line in the solution that was provided (together with the constant c). Here’s what we have at this point:\[cv _{0}^{-k}\left( 1+\frac{ \Delta v }{ v _{0} } \right)^{-k}\]Everything to the left of the parens in this expression is a constant. We don’t replace x with vo + deltav as you suggest because the result (ignoring the constant multiplier) would be (1 + vo + deltav)^-k, and we don’t have an approximation formula for anything like that. Our formula is for (1 + x)^r, and what we have is (1 + deltav/vo)^-k, so the thing we have to substitute for x is deltav/vo. And of course we substitute -k for r.

Answer 3

I see now that's how the solution has been derived. I failed to make use of the formula (1+x)^r and actually did it the hard way. Following is how I did it and also where I wronged.
\[p=cv^{-k}\]
\[\frac{dp}{dv}=-ckv^{-k-1}\]
\[\frac{dp^2}{d^2v}=ck(k+1)v^{-k-2}\]
\[f(x)\approx f(x_0) + f'(x_0) (x-x_0) + \frac{1}{2}f''(x_0)(x-x_0)^2\]
\[v=v_0+\Delta v, so \]
\[f(v)\approx \frac{c}{v_0^k} +\frac{-ck}{v_0^{k+1}}(v_0+\Delta v - v_0) + \frac{1}{2}\frac{ck(k+1)}{2v_0^{k+2}}(v_0+\Delta v - v_0)^2\]
Simplifying the equation gives the exact same answer. It is not-so-good hard way but answer is the same. What I wronged was in substituting (x-x_0), which should give Delta v. Instead somehow got v_0+Delta v.

Thanks again for the explanation.