MHB 11.e.28 Find the length of the cardroid

[karush]

find the length of the cardroid
\begin{align*}\displaystyle
L&=\sqrt{r^2+\left( \frac{dr}{d\theta}\right)^2}\\
r&=8(1-\sin {\theta})\therefore
\frac{dr}{d\theta}&=-8\cos\left(\theta\right)\\
r^2+\left( \frac{dr}{d\theta}\right)^2
&=(8(1-\sin {\theta}))^2+((-8)\cos(\theta))^2\\
&=64\left[1-2\sin\left({\theta}\right)
+\sin^2\left({\theta}\right)+\cos^2\left({\theta}\right)\right]\\
&=64\left[2-2\sin\left({\theta}\right)\right]
=128\left[1-\sin\left({\theta}\right)\right]\\
&=128\left[2\sin^2\left({\frac{\pi}{4}
-\frac{\theta}{2}}\right)\right]\\
L&=16\sqrt{\sin^2\left({\frac{\pi}{4}
-\frac{\theta}{2}}\right)}\\
&= 16\sin\left(\frac{\pi}{4}\right)=16\left(\frac{\sqrt{2}}{2}\right)=8\sqrt{2}
\end{align*}

 

[Greg]

Re: 11.e.28 find the length of the cardroid

Hmm...

Arc length, $s$, is given by

$$s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$

 

[karush]

Re: 11.e.28 find the length of the cardroid

$\text{find the length of the cardroid-}\\$
$\begin{align*}\displaystyle
S&=\int_{0}^{2\pi} \sqrt{r^2+\left( \frac{dr}{d\theta}\right)^2} \, d\theta\\
r&=8(1-\sin {\theta})\therefore
\frac{dr}{d\theta}=-8\cos\left(\theta\right) \\
&=\int_{0}^{2\pi}
\sqrt{(8(1-\sin {\theta}))^2+((-8)\cos(\theta))^2} \, d\theta \\
&=\int_{0}^{2\pi}\sqrt{
64\left[1-2\sin\left({\theta}\right)
+\sin^2\left({\theta}\right)+\cos^2\left({\theta}
\right)\right]}\, d\theta \\
&=\int_{0}^{2\pi}\sqrt{
64\left[2-2\sin\left({\theta}\right)\right]}\, d\theta
\end{align*} $

so far?