What is the arclength of the polar curve #f(theta) = sin4theta-cos3theta # over #theta in [pi/3,pi/2] #?

Question

What is the arclength of the polar curve #f(theta) = sin4theta-cos3theta  # over #theta in [pi/3,pi/2]  #?

Luke Alvarez · Accepted Answer

The arclength is approximately #0.46371#

The equation for arclength of a polar curve is:

#int_a^b sqrt(r^2 + ((dr)/(d theta))^2) d theta#

In this case,

#r^2 = (sin4theta-cos3theta)^2 #

#(dr)/(d theta) = d/(d theta) (sin4theta-cos3theta) #

#((dr)/(d theta))^2 = (4cos4theta + 3sin3theta)^2#

So, the arclength can be found by:

#int_(pi/3)^(pi/2) sqrt( (sin(4theta)-cos(3theta))^2 + (4cos4theta + 3sin3theta)^2) d theta#

#~~ 0.46371#

Final Answer

I am not entirely sure how to solve this integral manually; if anyone else has a valid method of integration, feel free to extend this answer, but as far as I can see, the only way to solve this problem is with a graphing calculator or similar interface.

Emily Ammerman · Answer

undefined The arc length of the polar curve $ f(\theta) = \sin(4\theta) - \cos(3\theta) $ over the interval $ \theta \in [\frac{\pi}{3}, \frac{\pi}{2}] $ can be calculated using the formula for arc length in polar coordinates:

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

where $ r = f(\theta) $ is the polar function and $ \alpha $ and $ \beta $ are the limits of integration. Compute $ \frac{dr}{d\theta} $ by differentiating $ r $ with respect to $ \theta $, and then evaluate the integral over the given interval.