Q. 29 - Calculus

Question

Approximate the arc length of f (x) on [a, b], using the approximation $\sum _{k=1}^n\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x$ with the given value of n. In each problem list the values of $\delta y_k$ for $k=0,1,2,3....n$ .

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

Step-by-Step Solution

Answer

The arc length is $4\sqrt{3}+\sqrt{1}$ .

Step 1. Given information .

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

Step 2. Formula used .

The formula used to find arc length of $f\left(x\right)=\sum _{k=1}^3\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x$ .

Step 3. Find the arc length .

$f\left(x\right)=\sum _{k=1}^3\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x$

$$\begin{gathered} \delta x=\frac{b-a}{n}=\frac{6}{3}=2 \\ {x}_0=a+k·\delta x=-3 , x_1=-3+2=-1 \\ {x}_2=-3+4=1 , x_3=-3+6=3 \\ \delta y_k=f\left(x_k\right)-f\left(x_{k-1}\right) \\ \delta y_1=f\left(x_1\right)-f\left(x_0\right)=\sqrt{8} , \delta y_2=f\left(x_2\right)-f\left(x_1\right)=0 \\ \delta y_3=f\left(x_3\right)-f\left(x_2\right)=\sqrt{8} \end{gathered}$$

$$\begin{gathered} Arc length =\sum _{k=1}^3\sqrt{1+{\left(\frac{\delta y_1}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_2}{\delta x}\right)}^2·}\delta x+\sqrt{1+{\left(\frac{\delta y_3}{\delta x}\right)}^2}·\delta x \\ =\sqrt{1+{\left(\frac{\sqrt{8}}{2}\right)}^2}·2+\sqrt{1+0}+\sqrt{1+{\left(\frac{\sqrt{8}}{2}\right)}^2}·2 \\ =2\sqrt{3}+\sqrt{1}+2\sqrt{3} \\ =4\sqrt{3}+\sqrt{1} \end{gathered}$$