Question

Prove that for cardiod r=a(1+cosθ)
ρ^2/r is constant

Answer 1

$\rho =\frac{(r^2+(r')^2)^\frac{3}{2}}{r^2+2(r')^2-rr''}$

$r'=-a\sin\theta; r''=-a\cos \theta$

$\rho=\frac{(a^2(1+\cos \theta)^2+a^2 \sin^2\theta)^\frac{3}{2}}{a^2(1+\cos \theta)^2+2a^2\sin^2\theta+a^2(1+\cos \theta)\cos \theta}=$

$=\frac{(a^2+2a^2\cos \theta+a^2\cos^2\theta+a^2 \sin^2\theta)^\frac{3}{2}}{a^2+2a^2\cos \theta+a^2\cos^2\theta+2a^2\sin^2\theta+a^2\cos \theta+a^2 \cos^2 \theta}=$

$=\frac{(a^2+2a^2\cos \theta+a^2)^\frac{3}{2}}{a^2+2a^2\cos \theta+2a^2+a^2\cos \theta}=$ $\frac{(2a^2+2a^2\cos \theta)^\frac{3}{2}}{3a^2+3a^2\cos \theta}=\frac{2^{\frac{3}{2}}a^3(1+\cos \theta)^\frac{3}{2}}{3a^2+3a^2\cos \theta}=$

$=\frac{2^{\frac{3}{2}}a^3(1+\cos \theta )^\frac{1}{2}}{3a^2}=\frac{2^{\frac{3}{2}}a}{3}(1+\cos \theta)^{\frac{1}{2}}$

$\frac{\rho^2}{r}=\frac{2^3a^2}{3^2}\frac{(1+\cos \theta)}{a(1+\cos \theta)}=\frac{8a}{9}\, is \, constant$