What is the arclength of #r=2sin(theta-(3pi)/16) + 9theta# on #theta in [(-5pi)/16,(9pi)/16]#?

Question

Scarlett Allison · Accepted Answer

Store #r=2sin(theta-(3pi)/16)+9theta# in your calculator. Find the derivative #(dr)/(d theta)=2cos(theta-(3pi)/16)+9# and store this separately. Plug these into the integral that gives the arc length for polar equations: #s=int_((-5pi)/16)^((9pi)/16)sqrt(r^2+((dr)/(d theta))^2)d theta=37.1478...#

William Abdalla · Answer

undefined To find the arc length of $ r = 2 \sin\left(\theta - \frac{3\pi}{16}\right) + 9\theta $ on the interval $ \left[-\frac{5\pi}{16}, \frac{9\pi}{16}\right] $, you can use the formula for polar arc length:

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

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

First, find $ \frac{dr}{d\theta} $ by taking the derivative of $ r $ with respect to $ \theta $, and then substitute it into the formula. After that, integrate the expression over the given interval to find the arc length.