F flakine Junior Member Joined Aug 24, 2005 Messages 78 Jul 25, 2006 #1 Find Arc Length of Curve r=sin^2(θ/2) from θ=0 to θ=pi/2 √((sin^2(θ/2))^2 + 2(2sin(θ/2))(1/2)) I'm lost on how to pull this one together!
Find Arc Length of Curve r=sin^2(θ/2) from θ=0 to θ=pi/2 √((sin^2(θ/2))^2 + 2(2sin(θ/2))(1/2)) I'm lost on how to pull this one together!
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Jul 25, 2006 #2 Using the arc length formula for polar coordinates. \(\displaystyle \L\\\int{\sqrt{r^{2}+(\frac{dr}{d{\theta}})^{2}}}d{\theta}\) \(\displaystyle \frac{dr}{d{\theta}}=sin(\frac{\theta}{2})cos(\frac{\theta}{2})\) \(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}{\sqrt{(sin^{2}(\frac{\theta}{2}))^{2}+sin^{2}(\frac{\theta}{2})cos^{2}(\frac{\theta}{2})}}d{\theta}\) This looks monstrous, but it simplifies nicely. \(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}{\sqrt{(sin^{2}(\frac{\theta}{2}))(sin^{2}(\frac{\theta}{2})+cos^{2}(\frac{\theta}{2}))}}d{\theta}\) \(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}sin(\frac{\theta}{2})d{\theta}\)
Using the arc length formula for polar coordinates. \(\displaystyle \L\\\int{\sqrt{r^{2}+(\frac{dr}{d{\theta}})^{2}}}d{\theta}\) \(\displaystyle \frac{dr}{d{\theta}}=sin(\frac{\theta}{2})cos(\frac{\theta}{2})\) \(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}{\sqrt{(sin^{2}(\frac{\theta}{2}))^{2}+sin^{2}(\frac{\theta}{2})cos^{2}(\frac{\theta}{2})}}d{\theta}\) This looks monstrous, but it simplifies nicely. \(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}{\sqrt{(sin^{2}(\frac{\theta}{2}))(sin^{2}(\frac{\theta}{2})+cos^{2}(\frac{\theta}{2}))}}d{\theta}\) \(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}sin(\frac{\theta}{2})d{\theta}\)
