Emmy Dillon

2022-06-19

What is the arc length of $f(x)=\sqrt{4-{x}^{2}}$ on $x\in [-2,2]$?

Nia Molina

Beginner2022-06-20Added 21 answers

$\frac{dy}{dx}=\frac{-x}{\sqrt{4-{x}^{2}}}$

$(\frac{dy}{dx}{)}^{2}=\frac{{x}^{2}}{4-{x}^{2}}$

arc length $=\int \sqrt{1+\frac{{x}^{2}}{4-{x}^{2}}}dx=2\int \frac{1}{\sqrt{4-{x}^{2}}}dx$

$=\mathrm{arcsin}(\frac{x}{2})+C$

Finally, evaluate over [-2,2]

arc length $=2\pi $

$(\frac{dy}{dx}{)}^{2}=\frac{{x}^{2}}{4-{x}^{2}}$

arc length $=\int \sqrt{1+\frac{{x}^{2}}{4-{x}^{2}}}dx=2\int \frac{1}{\sqrt{4-{x}^{2}}}dx$

$=\mathrm{arcsin}(\frac{x}{2})+C$

Finally, evaluate over [-2,2]

arc length $=2\pi $

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