Answer to Question #335070 in Calculus for Johannes Steven

Question

Find the length of cardioid r = 1+sin$\theta$

Answer 1

This is the equation of acardioid. The polar equation is expressed in terms of a periodic function, so we have to know its period to get the exact length (and not just a portion or some excess of it).

$r=1+\sin\theta$ has period T=2π (the function can be obtained as a vertical translation of the sine function in the plane of coordinates (θ,r)).The length of a periodic polar curve can be computed by integrating the arc length on a complete period of the function, i.e. on an interval I of length T=2π:

$L=\int_Ids$

where $ds=\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta$

So we have to compute the derivative:

$\frac{dr}{d\theta}=\frac{d}{d\theta}(1+\sin{\theta})=cos\theta$

and this implies

$ds=\sqrt{(1+sin\theta)^2+cos\theta^2}d\theta=\sqrt{1+2sin\theta+sin^2\theta+cos^2\theta}d\theta$

By the Pythagorean trigonometric identity cos

$cos^2\theta+sin^2\theta=1$

$ds=\sqrt{2+2sin\theta}d\theta=\sqrt{2(1+sin\theta)}d\theta$

Integrating that square root is not easy... we'd better take advantage of trigonometric identities. We have $\sqrt{1+sin\theta}$

 (multiplied by a numerical coefficient) and we would like to get rid of that square root. The half angle formula for cosine states that

 $cos^2(\frac{\alpha}{2})=\frac{1+cos\alpha}{2}$.

We can use this with sine too, because if

$\alpha=\theta-\frac{\pi}{2}$

 then

 $cos\alpha=cos(\theta-\frac{\pi}{2})=sin\theta$.

So we substitute in the formula and get $cos^2\frac{\theta-\frac{\pi}{2}}{2}=\frac{1+cos (\theta-\frac{ \pi}{2})}{2}$

and then

$2cos^2 {(\frac{\theta}{4}-\frac{\pi}{4})}=1+sin\theta$

$ds=\sqrt{4 cos^2 {(\frac{\theta}{4}-\frac{\pi}{4})}}d\theta= 2| cos {(\frac{\theta}{4}-\frac{\pi}{4})} |d\theta$

Now we can integrate this on an interval of length 2π. To do that, we would like to get rid of the absolute value.

We know that

$cos {(\frac{x}{2}-\frac{\pi}{4})} \eqslantgtr0 \iff -\frac{\pi}{2}+2\pi k \eqslantless \frac{x}{2}-\frac{\pi}{4}\eqslantless \frac{\pi}{2}+2\pi k$

for k any integer value, so we get that −

$-\frac{\pi}{2}\pi+4\pi k \eqslantless x\eqslantless \frac{3}{2}\pi+4\pi k$

In the end, we can integrate the arc length on 

$I=[-\frac{\pi}{2}; \frac{3}{2}\pi]$, since this is an interval of length 2π. We can remove the absolute value, since $cos (\frac{x}{2}-\frac{\pi}{2}) \eqslantgtr 0$ on I. $\int_{-\frac{\pi}{2}}^\frac{3\pi}{2} 2cos (\frac{\theta}{2}-\frac{\pi}{4})d\theta=4sin\frac{\pi}{2}-4sin\frac{-\pi}{2}=4+4=8$