Question

For the cardioids r = a(1+cosθ) , prove that (radius of curvature)^2/r is constant

Answer 1

Curvature for a polar curve defined by \(r(\theta)\) is
\[\kappa (\theta) = \frac{\bigg|2 f'(\theta)^2 + f(\theta)^2 - f(\theta)f''(\theta)\bigg|}{\bigg[f'(\theta)^2 + f(\theta)^2 \bigg]^{3/2}}\]

Answer 2

Whoops, those \(f\)s should be \(r\)s.

First compute your derivatives:
\[r(\theta)=a(1+\cos\theta)\\[1ex]
\quad\implies r'(\theta)=-a\sin\theta\\[1ex]
\quad\implies r''(\theta)=-a\cos\theta\]

Answer 3

I am aware of that I have also proceeded with the same... but im struggling to bring it to a constant value.

Answer 4

That is the radius of curvature has a formula[\[\rho=\frac{ (1+y _{1}^{2})^{3/2} }{ y _{2^{2}}}\]

Answer 5

Ah okay. You have
\[\begin{align*}
\kappa(\theta)&=\frac{\bigg|2(-a\sin\theta)^2+(a(1+\cos\theta))^2-a(1+\cos\theta)(-a\cos\theta)\bigg|}{\bigg[(-a\sin\theta)^2+(a(1+\cos\theta))^2\bigg]^{3/2}}\\[1ex]
&=\frac{\bigg|2a^2\sin^2\theta+a^2+2a^2\cos\theta+a^2\cos^2\theta+a^2\cos\theta+a^2\cos^2\theta\bigg|}{\bigg[a^2\sin^2\theta+a^2+2a^2\cos\theta+a^2\cos^2\theta\bigg]^{3/2}}
\end{align*}\]Every instance of \(a^2\sin^2\theta+a^2\cos^2\theta\) reduces to \(a^2\), so you can simplify this to
\[\begin{align*}
\kappa(\theta)&=\frac{\bigg|3a^2+3a^2\cos\theta\bigg|}{\bigg(2a^2+2a^2\cos\theta\bigg)^{3/2}}\\[1ex]
&=\frac{3a^2\bigg|1+\cos\theta\bigg|}{(2a^2)^{3/2}\bigg(1+\cos\theta\bigg)^{3/2}}\\[1ex]
&=\frac{3}{\sqrt8|a|(1+\cos\theta)^{1/2}}
\end{align*}\]The radius of curvature is given by
\[\rho(\theta)=\frac{1}{\kappa(\theta)}\]You need to show that \(\dfrac{\rho(\theta)^2}{r}\) is constant. Can you continue from here?

Answer 6

Yep, I got there too just now :)
I'll complete the problem. Thanks a ton :)

Answer 7

Man, how do you type so many equations so fast. I need to get a hang of that. I'll try to attach a pic :)

Answer 8

Lots of practice and a penchant for having things look nice :)

Answer 9

\[\frac{ \rho(\theta)^2 }{ r}= \frac{ 8 a}{ 9 }\]

Answer 10

Looks good!

Answer 11

Thanks a lot :)

Answer 12

You're welcome, happy to help