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Find the length of the curve specified by equation $y=\frac{1}{3}\left(x^{2}+2\right)^{\frac{3}{2}}$ where x changes form 0 to 1.

(1) $y=\frac{1}{3}\left(x^{2}+2\right)^{3 / 2}$

(2) $y^{\prime}=\frac{d y}{d x}=\frac{1}{3} \cdot \frac{3}{2}\left(x^{2}+2\right)^{\frac{1}{2}} \cdot 2 x$

$=x\left(x^{2}+2\right)^{\frac{1}{2}}=x \sqrt{x^{2}+2}$

(3) $\left(y^{\prime}\right)^{2}=\left(x \sqrt{x^{2}+2}\right)^{2}=x^{2}\left(x^{2}+2\right)=x^{4}+2 x^{2}$

(4) $1+\left(y^{\prime}\right)^{2}=x^{4}+2 x^{2}+1=\left(x^{2}+1\right)\left(x^{2}+1\right)=\left(x^{2}+1\right)^{2}$

(5) $\sqrt{1+\left(y^{\prime}\right)^{2}}=\sqrt{\left(x^{2}+1\right)^{2}}=x^{2}+1$

(6) Arc length $=L=\int_{0}^{1} \sqrt{1+(y^{\prime})^{2}} d x=\int_{0}^{1}\left(x^{2}+1\right) d x$

$L=\left[\frac{x^{3}}{3}+x\right]_{0}^{1}=\left(\frac{(1)^{3}}{3}+1\right)-(0)=\frac{1}{3}+1=\frac{4}{3}$

Find the length of the acr of the parabola $y^{2}=x$ from (0 , 0) to (1 , 1)

(1) $x=y^{2}$

(2) $x^{\prime}=2 y$

(3) $(x^{\prime})^{2}=(2 y)^{2}=4 y^{2}$

(4) $1+\left(x^{\prime}\right)^{2}=1+4 y^{2}$

(5) $\sqrt{1+\left(x^{\prime}\right)}^{2}=\sqrt{1+4 y^{2}}$

(6) Arc length $=L=\int_{0}^{1} \sqrt{1+\left(x^{\prime}\right)^{2}} dy=\int_{0}^{1} \sqrt{1+4 y^{2}} d y$

$y=\frac{1}{2} \tan \theta \Rightarrow d y=\frac{1}{2} \sec ^{2} \theta d \theta \; \& \; \sqrt{1+4 y^{2}}=\sec \theta$

$y=0 \Rightarrow 0=\frac{1}{2} \tan \theta \Rightarrow 0={tan}^{-1}(0)=0$

$y=1 \Rightarrow 1=\frac{1}{2} \tan \theta \Rightarrow \theta=\tan ^{-1}(2)=\alpha$

$\alpha=\tan ^{-1}(2)$

$\tan \alpha=2$

arc length $=L=\int_{0}^{1} \sqrt{1+4 y^{2}} d y=\int_{0}^{\alpha} \sec \theta \cdot \frac{1}{2} \sec ^{2} \theta d{\theta}$

$L=\frac{1}{2} \int_{0}^{\alpha} \sec ^{3} \theta d \theta=\int_{0}^{\alpha} \sec \theta \cdot \sec ^{2} \theta d \theta$

$u=\sec \theta \Rightarrow d u= \sec \theta \tan \theta d \theta$

$v=\tan \theta \Rightarrow d v=\sec ^{2} \theta d \theta$

arc length $=\frac{1}{2} \int_{0}^{\alpha} \sec ^{3} \theta d \theta=\frac{1}{2}\left[u \cdot\left.v\right|_{0} ^{\alpha}-\int_{0}^{\alpha} v \cdot d u\right]$

$L=\frac{1}{2}\left[\sec \theta \tan \left.\theta\right|_{0} ^{\alpha}-\int_{0}^{\alpha} \sec \theta \tan ^{2} \theta d \theta\right]$

$\tan ^{2} \theta=\sec ^{2} \theta-1$

$L=\frac{1}{2}\left[\sec \theta \tan \left.\theta\right|_{0}^{\alpha}-\int_{0}^{\alpha} \sec \theta\left(\sec ^{2} \theta-1\right) d \theta\right]$

$=\frac{1}{2}\left[\sec \theta \tan \theta |_{0}^{\alpha}-\int_{0}^{\alpha}\left(\sec ^{3} \theta-\sec \theta\right) d \theta\right]$

arc length $=L=\frac{1}{2} \int_{0}^{\alpha} \sec ^{3} \theta d \theta$

$=\frac{1}{2}[\left.{sec \theta \tan \theta} \right|_{0} ^{\alpha}-\int_{0}^{\alpha} (\sec ^{3} \theta-\sec \theta ) d \theta ]$

$\frac{1}{2} \int_{0}^{\alpha} \sec ^{3} \theta d \theta$

$=\frac{1}{2}\left[\sec \theta \tan \left.\theta\right|_{0} ^{\alpha}-\int_{0}^{\alpha} \sec ^{3} \theta d \theta+\int_{0}^{\alpha} \sec \theta d \theta\right]$

$\frac{1}{2} \int_{0}^{\alpha} \sec ^{3} \theta d \theta$

$=\frac{1}{2} \sec \theta \tan\theta \left.\right|_{0} ^{\alpha}-\frac{1}{2} \int_{0}^{\alpha} \sec ^{3} \theta d \theta+\frac{1}{2} \int_{0}^{\alpha} \sec \theta d \theta$

$\frac{1}{2} \int_{0}^{\alpha} \sec ^{3} \theta d \theta+\frac{1}{2} \int_{0}^{\alpha} \sec ^{3} \theta d \theta$

$=\frac{1}{2} \sec \theta \tan \left.\theta\right|_{0}^{\alpha}+\frac{1}{2} \int_{0}^{\alpha} \sec \theta d \theta$

$\int_{0}^{\alpha} \sec ^{3} \theta d \theta=\frac{1}{2} \sec \theta \tan \left.\theta \right|_{0}^{\alpha}+\frac{1}{2} \int_{0}^{\alpha} \sec \theta d \theta$

$\int_{0}^{\alpha} \sec ^{3} \theta d \theta$

$=\frac{1}{2} \sec \alpha \tan \alpha-0+\frac{1}{2}[\ln |{sec \theta}+ \tan \theta| ]_{0}^{\alpha}$

$\frac{1}{2} \int_{1}^{\alpha} \sec ^{3} \theta=\frac{1}{4} \sec \alpha \tan \alpha+\frac{1}{4}[\ln |\sec \alpha+\tan \alpha |-0]$

$=\frac{1}{4}[\sec \alpha \tan \alpha+\ln |\sec x+\tan \alpha|]$

$=L=\frac{1}{2} \int_{0}^{\alpha} \sec ^{2} \theta d \theta$

$=\frac{1}{4}[\sec \alpha \tan \alpha+\ln |\sec x+\tan \alpha|]$

$\alpha=\tan ^{-1}(2) \rightarrow \tan (\alpha)=2$

$\sec ^{2} \alpha=1+\tan ^{2}(\alpha)=1+(2)^{2}=1+4=5$

$\sec \alpha=\sqrt{5}$

arc length $=\frac{1}{4}[2 \cdot \sqrt{5}+\ln |\sqrt{5}+2|]$