What is the arclength of #r=3/4theta # on #theta in [-pi,pi]#?

Question

Ezra Adams · Accepted Answer

#L=3/4pisqrt(pi^2+1)+3/4ln(pi+sqrt(pi^2+1))# units.

#r=3/4theta# #r^2=9/16theta^2# #r'=3/4# #(r')^2=9/16# Arclength is given by: #L=int_-pi^pisqrt(9/16theta^2+9/16)d theta# Simplify: #L=3/4int_-pi^pisqrt(theta^2+1)d theta# From symmetry: #L=3/2int_0^pisqrt(theta^2+1)d theta# Apply the substitution #theta=tanphi#: #L=3/2intsec^3phidphi# This is a known integral: #L=3/4[secphitanphi+ln|secphi+tanphi|]# Reverse the substitution: #L=3/4[thetasqrt(theta^2+1)+ln|theta+sqrt(theta^2+1)|]_0^pi# Insert the limits of integration: #L=3/4pisqrt(pi^2+1)+3/4ln(pi+sqrt(pi^2+1))#

Isaiah Allen · Answer

undefined To find the arc length of the polar curve $ r = \frac{3}{4}\theta $ on the interval $ \theta \in [-\pi, \pi] $, you can use the formula for arc length of a polar curve:

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

In this case, $ r = \frac{3}{4}\theta $. Differentiating $ r $ with respect to $ \theta $ yields:

$$ \frac{dr}{d\theta} = \frac{3}{4} $$

So, the formula for the arc length becomes:

$$ L = \int_{-\pi}^{\pi} \sqrt{\left(\frac{3}{4}\theta\right)^2 + \left(\frac{3}{4}\right)^2} \, d\theta $$

$$ L = \int_{-\pi}^{\pi} \sqrt{\frac{9}{16}\theta^2 + \frac{9}{16}} \, d\theta $$

$$ L = \int_{-\pi}^{\pi} \sqrt{\frac{9}{16}\left(\theta^2 + 1\right)} \, d\theta $$

$$ L = \frac{3}{4} \int_{-\pi}^{\pi} \sqrt{\theta^2 + 1} \, d\theta $$

This integral can be evaluated using trigonometric substitution. After integration, the result will be the arc length of the polar curve $ r = \frac{3}{4}\theta $ on the interval $ \theta \in [-\pi, \pi] $.