Answer to Question #284164 in Calculus for Mao

Question #284164

Determine the length of the curve π‘₯ = 𝑦^2 /2 for 0 ≀ π‘₯ ≀ 1/2 . Assume 𝑦 positive.




1
Expert's answer
2022-01-10T17:50:37-0500

L=∫1+(yβ€²)2dxy=2xyβ€²=12xL=∫0121+12xdxL= \int \sqrt{1+(y')^2}dx\\ y=\sqrt{2x}\\ y'=\frac{1}{\sqrt{2x}}\\ L= \int_0^\frac{1}{2}\sqrt{1+\frac{1}{2x}}dx\\

This can also be written as L=∫01212x+1dxL= \int_0^\frac{1}{2}\sqrt{\frac{1}{2x}+1}dx\\

we apply linearity

12∫1x+2dx\frac{1}{\sqrt{2}}\int \sqrt{\frac{1}{x}+2}dx\\

let u = 1x+2\sqrt{\frac{1}{x}+2}\\ , dx = -21x+2x2du,dx=βˆ’2∫u2(u2βˆ’2)du\sqrt{\frac{1}{x}+2x^2}du, dx=-2\int \frac{u^2}{(u^2-2)}du\\

we integrate βˆ«u2(u2βˆ’2)du=∫(u2βˆ’2(u2βˆ’2)2+2(u2βˆ’2)2)du=\int \frac{u^2}{(u^2-2)}du = \int (\frac{u^2-2}{(u^2-2)^2}+\frac{2}{(u^2-2)^2})du=\\

∫(1(u2βˆ’2)+2(u2βˆ’2)2)du\int (\frac{1}{(u^2-2)}+\frac{2}{(u^2-2)^2})du\\

we split and integrate differently.

∫(1(u2βˆ’2)duβ€…β€ŠβŸΉβ€…β€ŠLn(uβˆ’2)232βˆ’Ln(u+2)232\int (\frac{1}{(u^2-2)}du \implies \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u+\sqrt{2})}{2^{\frac{3}{2}}}

also

∫2(u2βˆ’2)2duβ€…β€ŠβŸΉβ€…β€Šβˆ«(1(u2βˆ’2)duβ€…β€ŠβŸΉβ€…β€ŠLn(uβˆ’2)232βˆ’Ln(uβˆ’2)232βˆ’14(u+2)βˆ’14(uβˆ’2)\int\frac{2}{(u^2-2)^2}du \implies \int (\frac{1}{(u^2-2)}du \implies \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}} - \frac{1}{4(u+\sqrt{2})}- \frac{1}{4(u-\sqrt{2})}\\

but substituting the solution of the integrations back, we get

Ln(uβˆ’2)232βˆ’Ln(u+2)232+Ln(uβˆ’2)232βˆ’Ln(uβˆ’2)232βˆ’14(u+2)βˆ’14(uβˆ’2)\frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u+\sqrt{2})}{2^{\frac{3}{2}}}+ \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}} - \frac{1}{4(u+\sqrt{2})}- \frac{1}{4(u-\sqrt{2})}

recall, u= 1x+2\sqrt{\frac{1}{x}+2}\\ then we substitute u in terms of x back and introduce the lower and upper limits

Ln(uβˆ’2)232βˆ’Ln(u+2)232+Ln(uβˆ’2)232βˆ’Ln(uβˆ’2)232βˆ’14(u+2)βˆ’14(uβˆ’2)∣012\frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u+\sqrt{2})}{2^{\frac{3}{2}}}+ \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}} - \frac{1}{4(u+\sqrt{2})}- \frac{1}{4(u-\sqrt{2})}|_0^\frac{1}{2}

We have.

12+Ln(2+1)βˆ’Ln(2βˆ’1)4\frac{1}{\sqrt{2}}+\frac{Ln(\sqrt{2}+1)-Ln(\sqrt{2}-1)}{4}

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