Answer to Question #180272 in Calculus for Prathamesh

Question

Determine the length of arc of curve 𝑦

2 = 2𝑥 (𝑓𝑜𝑟 0 ≤ 𝑥 ≤

1

2

) in the first 

quadrant.

Answer 1

Determine the length of arc of curve $𝑦^2 = 2𝑥$ (𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 12)

$y=f(x)=\sqrt{2x} \newline y'=\sqrt{2}\frac{1}{2}x^{-\frac{1}{2}} \newline L(y) = \int_0^{12}\sqrt{1+\frac{1}{2x}}dx=\frac{1}{2}(ln{5+2\sqrt{6}}+10\sqrt{6})=13.3937$

$\int_0^{12}\sqrt{1+\frac{1}{2x}}dx$

apply substitution $u=2x$

$=\int_0^{24}\frac{\sqrt{u+1}}{2\sqrt{u}}du$

take the constant out $\int{a f(x)}dx=a\int{f(x)}dx$

$=\frac{1}{2}\int_0^{24}\frac{\sqrt{u+1}}{\sqrt{u}}du$

apply integration by parts $u=\frac{\sqrt{u+1}}{\sqrt{u}}, v'=1$

$=\frac{1}{2}[\sqrt{\frac{u+1}{u}}u-\int-\frac{1}{2u^{\frac{1}{2}}\sqrt{u+1}}du]_0^{24} \newline \int-\frac{1}{2u^{\frac{1}{2}}\sqrt{u+1}}du=-ln|\sqrt{u+1}+\sqrt{u}| \newline =\frac{1}{2}[\sqrt{\frac{u+1}{u}}u-(-ln|\sqrt{u+1}+\sqrt{u}| )]_0^{24}$

Simplify

$=\frac{1}{2}[\sqrt{\frac{u+1}{u}}u+ln|\sqrt{u+1}+\sqrt{u}|]_0^{24}$

Compute the boundaries:

$ln(5+2\sqrt{6}) + 10\sqrt{6}\newline =\frac{1}{2}(ln(5+2\sqrt{6}) + 10\sqrt{6})=13.3937$