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.
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Expert's answer
2021-01-21T00:05:48-0500
r=2−sinx1 (1)
r2dtdx=4 (2)
(i)
rdtdx=r4=4(2−sinx)
(ii)
dtdr=dxdrdtdx=
=(2−sinx)2(−1)(−cosx)∗4(2−sinx)2=4cosx
−1⩽−sinx⩽1
1⩽2−sinx⩽3
1⩾2−sinx1⩾31
31⩽r⩽1
(iii)
v−speed(velocity)
v=vr2+vx2=(dtdr)2+(rdtdx)2=
=16cos2x+16(2−sinx)2=
=16+16∗4−64sinx=80−64sinx
v(x=0)=80=45
(iv)
a−acceleration
ax=r(dt2d2x)2+2dtdrdtdx=
=2−sinx1dtd(4(2−sinx)2)+2∗4cosx∗4(2−sinx)2=
=2−sinx4∗2(2−sinx)(−cosx)dtdx+32cosx(2−sinx)2=
=(−8)cosx∗4(2−sinx)2+32cosx(2−sinx)2=0
ar=dt2d2r−r(dtdx)2=
=dtd(4cosx)−2−sinx1(4(2−sinx)2)2=
=−4sinxdtdx−16(2−sinx)3=
=−4sinx∗4(2−sinx)2−16(2−sinx)3=
=−16(2−sinx)2(sinx+2−sinx)=
=−32(2−sinx)2
F=ma , so the force acting on P is directed towards the pole
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