What is the arclength of #r=costheta-3sin2theta # on #theta in [-pi/4,pi]#?

Question

Emma Aldridge · Accepted Answer

#approx 16.7916#

From #r(theta)=cos(theta)-3sin(3theta)# we get by differentiating

#r'(theta)=-sin(theta)-6cos(2theta)# so we get

#int_(-pi/4)^pisqrt((cos(theta)-3*sin(2theta))^2+(-sin(theta)-6cos(2theta))^2)d theta#

By a numerical method we obtain

#16.7916#

Charles Arocha · Answer

undefined To find the arc length of the polar curve $ r = \cos(\theta) - 3\sin(2\theta) $ on the interval $ \left[-\frac{\pi}{4}, \pi\right] $, we use the formula for arc length in polar coordinates:

$$ L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta $$

where $ r $ is the function defining the polar curve, $ \alpha $ and $ \beta $ are the starting and ending angles respectively.

First, we find $ \frac{dr}{d\theta} $:

$$ \frac{dr}{d\theta} = -\sin(\theta) - 6\cos(2\theta) $$

Now, we plug $ r $ and $ \frac{dr}{d\theta} $ into the arc length formula and integrate over the given interval:

$$ L = \int_{-\frac{\pi}{4}}^{\pi} \sqrt{(\cos(\theta) - 3\sin(2\theta))^2 + (-\sin(\theta) - 6\cos(2\theta))^2} \, d\theta $$

Solving this integral will give us the arc length of the polar curve over the interval $ \left[-\frac{\pi}{4}, \pi\right] $.