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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.

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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:

$L=\int_{\pi/6}^{\pi/3}\sqrt{cos^2(\theta)+sin^2(\theta)}d\theta=\theta|_{\pi/6}^{\pi/3}=\pi/6$.

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