What is the arclength of the polar curve #f(theta) = 2thetasin(5theta)-thetacot2theta # over #theta in [pi/12,pi/3] #?

Question

What is the arclength of the polar curve #f(theta) = 2thetasin(5theta)-thetacot2theta  # over #theta in [pi/12,pi/3]  #?

Christopher Agresta · Accepted Answer

#approx 2.73636# From #r(theta)=2theta*sin(5theta)-thetacot(2theta)# we get by differentiating with respect to #theta#: #r'(theta)=10thetacos(5theta)-cot(2theta)+2thetacsc^2(2theta)+2sin(5theta)# and our integral is given by #int_(pi/12)^(pi/3)sqrt((2theta*sin(5theta)-thetacot(2theta))^2+(10thetacos(5theta)-cot(2theta)+2thetacsc^2(2theta)+2sin(5theta))^2)d theta#

Elijah Abram · Answer

undefined To find the arc length of the polar curve $ f(\theta) = 2\theta \sin(5\theta) - \theta \cot(2\theta) $ over $ \theta $ in $ \left[\frac{\pi}{12}, \frac{\pi}{3}\right] $, we use the formula for arc length of a polar curve:

$$ L = \int_{\theta_1}^{\theta_2} \sqrt{[r(\theta)]^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta $$

where $ r(\theta) $ is the polar curve function and $ \frac{dr}{d\theta} $ is its derivative with respect to $ \theta $.

Here, $ r(\theta) = 2\theta \sin(5\theta) - \theta \cot(2\theta) $.

Taking the derivative:

$$ \frac{dr}{d\theta} = 2\sin(5\theta) + 10\theta \cos(5\theta) - \frac{\cot(2\theta)}{2} + \theta\csc^2(2\theta) $$

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

$$ L = \int_{\frac{\pi}{12}}^{\frac{\pi}{3}} \sqrt{\left(2\theta \sin(5\theta) - \theta \cot(2\theta)\right)^2 + \left(2\sin(5\theta) + 10\theta \cos(5\theta) - \frac{\cot(2\theta)}{2} + \theta\csc^2(2\theta)\right)^2} \, d\theta $$

This integral can be evaluated numerically using computational tools like calculators or software.