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
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".
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".
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}".
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.
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments
Leave a comment