Question

Find the arc length of the graph of \(y=\sqrt{16-x^{2}}\) over the interval [0,4]

Short answer

The arc length of the graph of the function \(y=\sqrt{16-x^{2}}\) over the interval [0,4] is \(2\pi\).

Step-by-step solution

Differentiate the function

The first step is to obtain the derivative of the function \(f(x)=\sqrt{16-x^{2}}\). Applying the chain rule, we obtain \(f'(x)=-\frac{x}{\sqrt{16-x^{2}}}\).

Insert the derivative into the arc length formula

To apply the arc length formula, square the derivative and add 1: \(1 + [f'(x)]^{2}=1+\left(\frac{x^{2}}{16-x^{2}}\right)= \frac{16}{16-x^{2}}\).

Integrate

Now, we should integrate this from 0 to 4 to find the arc length: \(L=\int_{0}^{4} \sqrt{\frac{16}{16-x^{2}}} dx\). Make a substitution \(x=4sin(\theta)\), \(dx=4cos(\theta)d\theta\). The integral becomes \(L= \int_{0}^{\frac{\pi}{2}} 4 d\theta = 2\pi\).