What is the arclength of #f(x)=sqrt(4-x^2) # in the interval #[-2,2]#?

Question

Charles Anctil · Accepted Answer

The arc length is #2pi# units.

This is a semi circle of radius #2#. We simply find the perimeter of the full circle of radius #2# and divide it by #2# to get our arc length. #P = 2(2)pi = 4pi# #P/2 = (4pi)/2 = 2pi# Therefore the arc length is #2pi# units. Hopefully this helps!

Emilia Aguilar · Answer

undefined To find the arc length of the function $f(x) = \sqrt{4 - x^2}$ on the interval $[-2, 2]$, we use the arc length formula:

$$ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx $$

In this case, $f(x) = \sqrt{4 - x^2}$, so $f'(x) = -\frac{x}{\sqrt{4 - x^2}}$.

Plugging these into the formula and evaluating the integral on $[-2, 2]$, we get:

$$ L = \int_{-2}^{2} \sqrt{1 + \left(-\frac{x}{\sqrt{4 - x^2}}\right)^2} \, dx $$

$$ L = \int_{-2}^{2} \sqrt{1 + \frac{x^2}{4 - x^2}} \, dx $$

$$ L = \int_{-2}^{2} \sqrt{\frac{4 - x^2 + x^2}{4 - x^2}} \, dx $$

$$ L = \int_{-2}^{2} \sqrt{\frac{4}{4 - x^2}} \, dx $$

Now, perform a trigonometric substitution by letting $x = 2 \sin \theta$. This leads to $dx = 2 \cos \theta \, d\theta$.

Substitute $x$ and $dx$ in terms of $\theta$ into the integral:

$$ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{4}{4 - (2\sin\theta)^2}} \cdot 2\cos\theta \, d\theta $$

$$ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{4}{4 - 4\sin^2\theta}} \cdot 2\cos\theta \, d\theta $$

$$ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{4}{4(1 - \sin^2\theta)}} \cdot 2\cos\theta \, d\theta $$

$$ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2\cdot 2\cos\theta}{\sqrt{1 - \sin^2\theta}} \, d\theta $$

$$ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4\, d\theta $$

$$ L = 4\theta \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} $$

$$ L = 4\left(\frac{\pi}{2} + \frac{\pi}{2}\right) $$

$$ L = 4\pi $$

So, the arc length of $f(x) = \sqrt{4 - x^2}$ on the interval $[-2, 2]$ is $4\pi$.