Question

find the length of the curve y=x^3/2 from (1,1) to (2,2√2)

Answer 1

$y=x^{3/2}\\\\x=r\ cos \theta\ \ y=r\ sin \theta,=>\ \ y^2=r^2\ sin^2\theta=x^3=r^3\ Cos^3\theta\\\\r=sec\theta\ tan^2\theta\\\\tan\theta=\frac{y}{x},\ when\ x=y=1,\ \theta=tan^{-1}1=\pi/4\\\\when\ x=2,\ y=2\sqrt2,\ \ \theta=tan^{-1}\sqrt2 \\\\arc\ length= \int\limits^{tan^{-1}\sqrt2}_{\pi/4} {r} \, d\theta =\int\limits^{tan^{-1}\sqrt2}_{\pi/4} {sec\theta\ tan^2\theta} \, d\theta$

$I= \int\limits^{}_{} {sec \theta(sec^2\theta\ -1)} \, d\theta\\\\ I=\int\limits^{}_{} {sec^3 \theta} \, d\theta- \int\limits^{}_{} {sec \theta} \, d\theta,\ \ \ --equation1\\\\\ I=\int\limits^{}_{} {(tan\theta)(sec\theta\ tan\theta)} \, d\theta\\\\ I = tan\theta\ sec\theta-\int\limits^{}_{} {(sec^2\theta)(sec\theta)} \, d\theta\\\\ I=tan\theta\ sec \theta\ - (I+\int\limits^{}_{} {sec \theta} \, d\theta)--using\ eq1\\\\ I=\frac{1}{2}tan\theta\ sec \theta-\frac{1}{2}Log|(sec\theta+tan\theta)|$

sec π/4=√2  ,   tan π/4=1  ,     sec (tan⁻¹√2) = √3 

Arc length =   ( [ I ] at θ = tan⁻¹ √2 )  -   ( [ I ] at  θ = π/4 )
  
               = [  1/2 * √2 * √3  - 1/2 * Log | √3+√2 | ]  -  [ 1/2 * 1 * √2 - 1/2 * Log | √2 + 1 | ]

               =  (√3 - 1)/√2  + 1/2 * Log [ (√2 + 1) / (√3+√2) ] 

               =  (√3 - 1)/√2  + 1/2 * Log (√6 + √3 - 2 - √2)