Answer to Question #156186 in Calculus for Ulrich

Question #156186
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
2021-01-21T00:05:48-0500

r=12sinxr=\frac{1}{2-\sin{x}} (1)

r2dxdt=4r^2\frac{dx}{dt}=4 (2)


(i)

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


(ii)

drdt=drdxdxdt=\frac{dr}{dt}=\frac{dr}{dx}\frac{dx}{dt}=

=(1)(cosx)(2sinx)24(2sinx)2=4cosx=\frac{(-1)(-\cos{x})}{(2-\sin{x})^2}*4(2-\sin{x})^2=4\cos{x}


1sinx1-1\leqslant-\sin{x}\leqslant1

12sinx31\leqslant2-\sin{x}\leqslant3

112sinx131\geqslant\frac{1}{2-\sin{x}}\geqslant\frac{1}{3}

13r1\frac{1}{3}\leqslant r \leqslant 1


(iii)

vspeed(velocity)v - speed(velocity)

v=vr2+vx2=(drdt)2+(rdxdt)2=v=\sqrt{v_r^2+v_x^2}=\sqrt{(\frac{dr}{dt})^2+(r\frac{dx}{dt})^2}=

=16cos2x+16(2sinx)2==\sqrt{16\cos^2{x}+16(2-\sin{x})^2}=

=16+16464sinx=8064sinx=\sqrt{16+16*4-64\sin{x}}=\sqrt{80-64\sin{x}}

v(x=0)=80=45v(x=0)=\sqrt{80}=4\sqrt{5}


(iv)

aaccelerationa- acceleration

ax=r(d2xdt2)2+2drdtdxdt=a_x=r(\frac{d^2x}{dt^2})^2+2\frac{dr}{dt}\frac{dx}{dt}=

=12sinxddt(4(2sinx)2)+24cosx4(2sinx)2==\frac{1}{2-\sin{x}}\frac{d}{dt}(4(2-\sin{x})^2)+2*4\cos{x}*4(2-\sin{x})^2=

=42(2sinx)2sinx(cosx)dxdt+32cosx(2sinx)2==\frac{4*2(2-\sin{x})}{2-\sin{x}}(-\cos{x})\frac{dx}{dt}+32\cos{x}(2-\sin{x})^2=

=(8)cosx4(2sinx)2+32cosx(2sinx)2=0=(-8)\cos{x}*4(2-\sin{x})^2+32\cos{x}(2-\sin{x})^2=0


ar=d2rdt2r(dxdt)2=a_r=\frac{d^2r}{dt^2}-r(\frac{dx}{dt})^2=

=ddt(4cosx)12sinx(4(2sinx)2)2==\frac{d}{dt}(4\cos{x})-\frac{1}{2-\sin{x}}(4(2-\sin{x})^2)^2=

=4sinxdxdt16(2sinx)3==-4\sin{x}\frac{dx}{dt}-16(2-\sin{x})^3=

=4sinx4(2sinx)216(2sinx)3==-4\sin{x}*4(2-\sin{x})^2-16(2-\sin{x})^3=

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

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


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


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