What is the arclength of the polar curve #f(theta) = 2thetasin(5theta)-cottheta # over #theta in [pi/12,pi/6] #?

Question

What is the arclength of the polar curve #f(theta) = 2thetasin(5theta)-cottheta  # over #theta in [pi/12,pi/6]  #?

William Abdalla · Accepted Answer

#approx 1.76588# We have given #r(theta)=2theta*sin(5theta)-cot(5theta)# then #r'(theta)=2sin(5theta)+10cos(5theta)-1/sin^2(theta)# and we get #int_(pi/12)^(pi/6)sqrt((2theta*sin(5theta)-cot(theta))^2+(2sin(5theta)+10cos(5theta)-1/sin(theta)^2)^2)d theta# I have found only a numerical value of that integral #approx 1.76588#

William Abdalla · Answer

undefined To find the arc length of the polar curve $ f(\theta) = 2\theta \sin(5\theta) - \cot(\theta) $ over the interval $ \theta \in [\frac{\pi}{12}, \frac{\pi}{6}] $, you can use the following formula:

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

Where $ r(\theta) = f(\theta) $ and $ \frac{dr}{d\theta} $ is the derivative of $ r(\theta) $ with respect to $ \theta $.

1. Find $ r(\theta) = f(\theta) $.
2. Compute $ \frac{dr}{d\theta} $.
3. Plug $ r(\theta) $ and $ \frac{dr}{d\theta} $ into the arc length formula.
4. Evaluate the integral over the given interval.

Once you have completed these steps, you'll have the arc length of the polar curve over the specified interval.