Q. 24 - Calculus

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 .

$f\left(x\right)=\sqrt{9-x^2} , \left[a,b\right]=\left[-3,3\right] , n=6$

Step-by-Step Solution

Answer

The arc length is $4.625$ .

The graph is sketched as below:

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 arc length .

$$\begin{gathered} f\left(x\right)=\sqrt{9-x^2}, a=-3, b=3, n=6 \\ \partial x=\frac{b-a}{n}=\frac{3+3}{6}=1 \\ {x}_k=a+k\partial x \\ {x}_0=-3, x_1=-3+1=-2 \\ {x}_2=-3+2=-1, x_3=-3+3=0 \\ {x}_4=-3+4=1, x_5=-3+5=2 \\ {x}_6=-3+6=3 \\ \partial y_1=f\left(-2\right)-f\left(-3\right)=\sqrt{9-4}-\sqrt{9-9}=\sqrt{5}=2.236 \\ \partial y_2=f\left(-1\right)-f\left(-2\right)=\sqrt{9-1}-\sqrt{9-4}=\sqrt{8}-\sqrt{5}=2.828-2.236=0.592 \\ \partial y_3=f\left(0\right)-f\left(-1\right)=\sqrt{9}-\sqrt{8}=3-2.828=0.172 \\ \partial y_4=f\left(1\right)-f\left(0\right)=\sqrt{8}-\sqrt{9}=2.282-3=-0.172 \\ \partial y_5=f\left(2\right)-f\left(1\right)=\sqrt{5}-\sqrt{8}=-0.592 \\ \partial y_6=f\left(3\right)-f\left(2\right)=0-\sqrt{5}=-2.236 \\ Arc length=\left[\sqrt{1+{\left(2.236\right)}^2}+\sqrt{1+{\left(0.592\right)}^2}+\sqrt{1+{\left(0.172\right)}^2}+\sqrt{1+{\left(-0.172\right)}^2}+\sqrt{1+{\left(-0.592\right)}^2}+\sqrt{1+{\left(-2.236\right)}^2}\right]\times 1 \\ =\sqrt{5.999}+\sqrt{1.350}+\sqrt{1.0295}+\sqrt{1.0295}+\sqrt{1.350}+\sqrt{5.999} \\ =\left(2.4492+1.1618+1.0146\right)\times 1 \\ =4.625 \end{gathered}$$

 

Step 4. Sketch the graph of function f x = 9 - x 2 .

The sketch of the given function is shown below .