Question

Approximate the arc length of f (x) on [a, b], using n line segments and the distance formula. Include a sketch of f (x) and the line segments .

$f\left(x\right)=\sqrt{9-x^2},\left[a,b\right]=\left[-3,3\right],n=3$

Short answer

The arc length is $13.996$.

The graph is as below:

Step-by-step solution

Step 1. Given information .

Consider the given function$f\left(x\right)=\sqrt{9-x^2}$.

Step 2. Formula used .

Arc length$=\underset{n\to \infty }{\lim }\sum _{k=1}^n\sqrt{1+{\left(\frac{\partial y_k}{\partial x}\right)}^2}·\partial x$

Step 3. Find the arc length .

$$\begin{gathered} f\left(x\right)=\sqrt{9-x^2},a=-3,b=3,n=3 \\ \partial x=\frac{b-a}{n}=\frac{3+3}{3}=2 \\ {x}_k=a+k\partial x \\ {x}_0=-3+0=-3,x_1=-3+1\times 2=-1 \\ {x}_2=-3+2\times 2=1,x_3=-3+3\times 2=3 \\ \partial y_1=f\left(x_1\right)-f\left(x_0\right)=\sqrt{8}-0=\sqrt{8}=2·828 \\ \partial y_2=f\left(x_2\right)-f\left(x_1\right)=\sqrt{8}-\sqrt{8}=0 \\ \partial y_3=f\left(x_3\right)-f\left(x_2\right)=0-\sqrt{8}=-2.828 \\ Arclength=\left[\sqrt{1+{\left(\frac{2.828}{2}\right)}^2}+\sqrt{1+0}+\sqrt{1+{\left(\frac{-2.828}{2}\right)}^2}\right]\times 2 \\ =\left[2\sqrt{1+{\left(1.414\right)}^2}+1\right]\times 2 \\ =\left(2\times 2.999+1\right)\times 2 \\ =13.996 \end{gathered}$$

Step 4. Sketch of the function fx=9-x2.

The sketch of the given function is shown below .