Answer to Question #281580 in Calculus for Amit

Question

Determine the length of the curve 𝑥 = 𝑦^2 /2 for 0 ≤ 𝑥 ≤ 1/2 . Assume 𝑦 positive.

Answer 1

$L= \int \sqrt{1+(y')^2}dx\\ y=\sqrt{2x}\\ y'=\frac{1}{\sqrt{2x}}\\ L= \int_0^\frac{1}{2}\sqrt{1+\frac{1}{2x}}dx\\$

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

we apply linearity

$\frac{1}{\sqrt{2}}\int \sqrt{\frac{1}{x}+2}dx\\$

let u = $\sqrt{\frac{1}{x}+2}\\$ , dx = -2$\sqrt{\frac{1}{x}+2x^2}du, dx=-2\int \frac{u^2}{(u^2-2)}du\\$

we integrate $\int \frac{u^2}{(u^2-2)}du = \int (\frac{u^2-2}{(u^2-2)^2}+\frac{2}{(u^2-2)^2})du=\\$

$\int (\frac{1}{(u^2-2)}+\frac{2}{(u^2-2)^2})du\\$

we split and integrate differently.

$\int (\frac{1}{(u^2-2)}du \implies \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u+\sqrt{2})}{2^{\frac{3}{2}}}$

also

$\int\frac{2}{(u^2-2)^2}du \implies \int (\frac{1}{(u^2-2)}du \implies \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}} - \frac{1}{4(u+\sqrt{2})}- \frac{1}{4(u-\sqrt{2})}\\$

but substituting the solution of the integrations back, we get

$\frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u+\sqrt{2})}{2^{\frac{3}{2}}}+ \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}} - \frac{1}{4(u+\sqrt{2})}- \frac{1}{4(u-\sqrt{2})}$

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

$\frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u+\sqrt{2})}{2^{\frac{3}{2}}}+ \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}}- \frac{Ln(u-\sqrt{2})}{2^{\frac{3}{2}}} - \frac{1}{4(u+\sqrt{2})}- \frac{1}{4(u-\sqrt{2})}|_0^\frac{1}{2}$

We have.

$\frac{1}{\sqrt{2}}+\frac{Ln(\sqrt{2}+1)-Ln(\sqrt{2}-1)}{4}$