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

$r=\frac{1}{2-\sin{x}}$ (1)

$r^2\frac{dx}{dt}=4$ (2)

(i)

$r\frac{dx}{dt}=\frac{4}{r}=4(2-\sin{x})$

(ii)

$\frac{dr}{dt}=\frac{dr}{dx}\frac{dx}{dt}=$

$=\frac{(-1)(-\cos{x})}{(2-\sin{x})^2}*4(2-\sin{x})^2=4\cos{x}$

$-1\leqslant-\sin{x}\leqslant1$

$1\leqslant2-\sin{x}\leqslant3$

$1\geqslant\frac{1}{2-\sin{x}}\geqslant\frac{1}{3}$

$\frac{1}{3}\leqslant r \leqslant 1$

(iii)

$v - speed(velocity)$

$v=\sqrt{v_r^2+v_x^2}=\sqrt{(\frac{dr}{dt})^2+(r\frac{dx}{dt})^2}=$

$=\sqrt{16\cos^2{x}+16(2-\sin{x})^2}=$

$=\sqrt{16+16*4-64\sin{x}}=\sqrt{80-64\sin{x}}$

$v(x=0)=\sqrt{80}=4\sqrt{5}$

(iv)

$a- acceleration$

$a_x=r(\frac{d^2x}{dt^2})^2+2\frac{dr}{dt}\frac{dx}{dt}=$

$=\frac{1}{2-\sin{x}}\frac{d}{dt}(4(2-\sin{x})^2)+2*4\cos{x}*4(2-\sin{x})^2=$

$=\frac{4*2(2-\sin{x})}{2-\sin{x}}(-\cos{x})\frac{dx}{dt}+32\cos{x}(2-\sin{x})^2=$

$=(-8)\cos{x}*4(2-\sin{x})^2+32\cos{x}(2-\sin{x})^2=0$

$a_r=\frac{d^2r}{dt^2}-r(\frac{dx}{dt})^2=$

$=\frac{d}{dt}(4\cos{x})-\frac{1}{2-\sin{x}}(4(2-\sin{x})^2)^2=$

$=-4\sin{x}\frac{dx}{dt}-16(2-\sin{x})^3=$

$=-4\sin{x}*4(2-\sin{x})^2-16(2-\sin{x})^3=$

$=-16(2-\sin{x})^2(\sin{x}+2-\sin{x})=$

$=-32(2-\sin{x})^2$

$F=ma$ , so the force acting on P is directed towards the pole