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

Question

Christopher Agresta · Accepted Answer

arc length #=intsqrt [1+(dy/dx)^2]dx#

#dy/dx = (-x)/[sqrt(4-x^2)]#

#(dy/dx)^2=x^2/(4-x^2)#

arc length #=int sqrt[1+x^2/(4-x^2)]dx=2int1/sqrt(4-x^2)dx#

#=arcsin(x/2)+C#

Finally, evaluate over [-2,2]

arc length #=2pi#

hope that helped

Samantha Adkison · Answer

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

$$ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx $$

where $ a $ and $ b $ are the endpoints of the interval, and $ \frac{dy}{dx} $ is the derivative of $ f(x) $ with respect to $ x $.

First, we find $ \frac{dy}{dx} $ by differentiating $ f(x) $ with respect to $ x $:

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

Now we plug $ \frac{dy}{dx} $ into the formula for arc length:

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

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

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

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

Now, to solve this integral, we can use trigonometric substitution. Let $ x = 2\sin(\theta) $. Then, $ dx = 2\cos(\theta) \, d\theta $.

When $ x = -2 $, $ \theta = -\frac{\pi}{2} $.

When $ x = 2 $, $ \theta = \frac{\pi}{2} $.

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

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

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

$$ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 8 \, d\theta $$

$$ = 8\theta \Bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} $$

$$ = 8\left(\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right) $$

$$ = 8\pi $$

So, the arc length of $ f(x) = \sqrt{4 - x^2} $ on $ x $ in the interval $[-2, 2]$ is $ 8\pi $.