Answer to Question #156185 in Differential Geometry | Topology for Ulrich

Question

A particle P moves on the curve with polar equation r = 1/ (2 - sinx) . Given that at any instant t, during the motion, r^2 (dx/dt) = 4,
(i) write an expression for r(dx/dt) in terms of x.
(ii) Show that dr/dt = 4cosx and 1/3 <=r<=1.
(iii) Find the speed of P when x = 0.
(iv) Prove that the force acting on P is directed towards the pole.

Answer 1

1. As $r^2\frac{dx}{dt}=4$ and $r=\frac{1}{2-\sin x}$ , $r\frac{dx}{dt} = \frac{4}{r} = 4\cdot(2-\sin x)=8-4\sin x$.
2. As $r=\frac{1}{2-\sin x}$, we will apply the chain rule to find $\frac{dr}{dt} = -\frac{1}{(2-\sin x)^2}\cdot(-\cos x)\cdot \frac{dx}{dt}$. Now we remark that $\frac{1}{(2-\sin x)^2} = r^2$ and as $r^2 \frac{dx}{dt} =4$, $\frac{dr}{dt} = 4\cos x$ . Also, as $-1\leq \sin x \leq 1$ we obtain $\frac{1}{3}\leq r\leq 1$ by applying the former inequality directly to the expression of $r$.
3. Speed expression in polar coordinates is $\frac{d(r\vec{e}_r)}{dt} = \frac{dr}{dt} \vec{e}_r + r\frac{dx}{dt} \vec{e}_x$ (where $x$ means angle variable, not a cartesian $x$). But now, using the previous results, we find that $\vec{v}(x) = 4\cos (x) \vec{e}_r + (8-4\sin x)\vec{e}_x$. Thus $\vec{v}(0) = 4\vec{e}_r+8\vec{e}_x$, $||\vec{v}(0)||=4\sqrt{5}$.
4. By Newton's second law, $\vec{F} = m\frac{d^2}{dt^2}\vec{r} =m\frac{d\vec{v}}{dt}$. Now using the chain rule (and the fact that $\frac{d\vec{e}_r}{dx} = \vec{e}_x, \frac{d\vec{e}_x}{dx}=-\vec{e}_r$) we find $\vec{F} = (-4\sin(x)\frac{dx}{dt}-8\frac{dx}{dt}+4\sin(x)\frac{dx}{dt})\vec{e}_r +(4\cos(x)\frac{dx}{dt} -4\cos(x)\frac{dx}{dt}) \vec{e}_x$. We see that the component along $\vec{e}_x$ is zero and thus the force is directed towards the pole.