What is the arc length of the polar curve #f(theta) = 2costheta-theta # over #theta in [pi/8, pi/3] #?

Question

What is the arc length of the polar curve #f(theta) = 2costheta-theta  # over #theta in [pi/8, pi/3]  #?

William Abdalla · Accepted Answer

#(21pi)/8 -sqrt3 /2 -2 +1/sqrt2 +4 cos pi/8#

The given function is #r= 2cos theta - theta#

#(dr)/(d theta)# would be #-2sin theta -1#

Arc length formula is #int_a^b 1/2 r^2 d theta#

The required arc length would be #1/2 int_(pi/8) ^(pi/3) (4sin^2 theta + 4 sin theta +1) d theta#

=#1/2 int_(pi/8)^(pi/3) (2-2cos 2theta + 4sin theta +1)#

=# [3 theta - sin 2 theta- 4 cos theta]_(pi/8)^(pi/3)]#

=#3pi -sin( (2pi)/3)-4 cos (pi/3) -(3pi)/8 + sin (pi/4) +4 cos (pi/8)#

=#(21pi)/8 -sqrt3 /2 -2 +1/sqrt2 +4 cos pi/8#

Further calculations may be made as desired.

Charles Anctil · Answer

undefined To find the arc length of the polar curve $ f(\theta) = 2 \cos(\theta) - \theta $ over the interval $ \theta \in [\frac{\pi}{8}, \frac{\pi}{3}] $, you can 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 = f(\theta) $ is the polar function, and $ \frac{dr}{d\theta} $ is the derivative of $ r $ with respect to $ \theta $.

1. Find $ \frac{dr}{d\theta} $ by taking the derivative of $ f(\theta) $ with respect to $ \theta $:
   $$ \frac{dr}{d\theta} = \frac{d}{d\theta}(2\cos(\theta) - \theta) = -2\sin(\theta) - 1 $$

2. Substitute $ r = f(\theta) $ and $ \frac{dr}{d\theta} $ into the arc length formula.

3. Integrate the expression from $ \frac{\pi}{8} $ to $ \frac{\pi}{3} $ to find the arc length:

$$ L = \int_{\pi/8}^{\pi/3} \sqrt{(2\cos(\theta) - \theta)^2 + (-2\sin(\theta) - 1)^2} \, d\theta $$

4. Evaluate the integral to find the arc length $ L $.