Q:

Taylors Series + differential style exam question

Hi guys! Is there anyone who can help me out with part (b) and (c) of this question (press link above)
Been stuck on it for days, just not sure how to even get started with either parts and so any help is really appreciated!
Thanks

A:

For a complete revolution we have, $$T = (\frac{2\pi} {\omega})$$ and $$\omega = (\frac{L}{mr^2})$$
Now from part a,$$L^2 = mr_0^2Kr_0^n = Kmr_0^{n+2}$$
or $$(\omega mr^2)^2 = Kmr_0^{n+2}$$
or$$\omega^2 = \frac{Kmr_0^{n+2}}{(mr^2)^2} = \frac{Kmr_0^{n+2}}{m^2r^4} = \frac{Kr_0^{n-2}}{m}$$
Taking the reciprocal and square root of both sides,$$1/\omega = \sqrt\frac{m}{Kr_0^{n-2}}$$
As, $$T = 2\pi/\omega$$
$$T= 2\pi\sqrt\frac{m}{Kr_0^{n-2}}$$
Now coming to part(c),by Taylor's expansion,
$$f(x_0 + \Delta x) = f(x_0) + 1/1! \Delta xf'(x_0) +1/2! (\Delta x)^2 f''(x_0) + ...$$
Here, $$ f(r) = -Kr^n$$ and hence, $$ -K(r_0 + \Delta r)^n = -Kr_0^n + 1.(-K\Delta x).nr_0^{n-1}$$ as $$\frac{dx^n}{dx} = nx^{n-1}$$As $$\Delta r<<r_0,$$ $$-K(r)^n \approx -Kr_0^n -K\Delta x.nr_0^{n-1}$$And for the last part$$f(r) = L^2/mr^2$$ Working similarly $$ \frac{d}{dr}(\frac{1}{r^2}) = \frac{-2}{r^3} and \frac{d^2}{dr^2}(\frac{1}{r^2}) = \frac{-2.-3}{r^4}$$
With a bit of approximation, you may arrive at the last equation. Thank you.

