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

Question #156185
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.
1
Expert's answer
2021-01-20T12:54:27-0500
  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.

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