An insect is moving along the curve $r=|cos\theta|$ such that $\theta =\frac{\pi t}{6}$, where $t$ is time measured in seconds. What is the distance travelled by the insect in the time interval between $t=1$ and $t=2$ ?
My attempt: The arc length is given by the formula : $\int_{a}^{b} \sqrt{1+f'(x)^2}$.
So here in the given region, we have $r=|\cos\theta|=cos\theta$.
Now, we have $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sqrt{1+sin^2(x)}$.
This integration is difficult to carry out. I am unable to proceed further.
Any help is appreciated. Thanks in advance.
Best Answer
Recall that the formula for the length of the arc in polar coordinates is:
$L=\int_{\alpha}^{\beta}\sqrt{r^2+\left(\frac{dr}{dt}\right)^2}d\theta$.
Replacing we have that:
$L=\int_{\pi/6}^{\pi/3}\sqrt{cos^2(\theta)+sin^2(\theta)}d\theta=\theta|_{\pi/6}^{\pi/3}=\pi/6$.