Q. 30 - 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 $k=0,1,2...n$ .

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

Step-by-Step Solution

Answer

The arc length is $2\sqrt{3}+\sqrt{1}+\sqrt{-50}+\sqrt{-2935}$ .

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 the arc of $f\left(x\right)=\sum _{k=1}^n\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x$ .

Step 3. Find the arc length .

$f\left(x\right)=\sum _{k=1}^6{\sqrt{1+\left(\frac{\delta y_k}{\delta x}\right)}}^2·\delta x$

$$\begin{gathered} \delta x=\frac{b-a}{n}=\frac{6}{6}=1 \\ {x}_0=a+k·\delta x=-3+0=-3 , x_1=-3+2=-1 \\ {x}_2=-3+4=1 , x_3=-3+6=3 \\ {x}_4=-3+8=5 , x_5=-3+10=7 \\ {x}_5=-3+12=9 \end{gathered}$$

$$\begin{gathered} \delta y_k=f\left(y_k\right)-f\left(y_{k-1}\right) \\ \delta y_1=f\left(x_1\right)-f\left(x_0\right)=\sqrt{8} , \delta y_2=f\left(x_2\right)-f\left(x_1\right)=0 \\ \delta y_3=f\left(x_3\right)-f\left(x_2\right)=\sqrt{8} , \delta y_4=f\left(x_3\right)-f\left(x_2\right)=\sqrt{-16} \\ \delta y_5=f\left(x_5\right)-f\left(x_4\right)=\sqrt{-40}-\sqrt{-4} \\ \delta y_6=\sqrt{-72}-\sqrt{-40} \end{gathered}$$

$$\begin{gathered} Arc length =\sum _{k=1}^6\sqrt{1+{\left(\frac{\delta y_1}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_2}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_3}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_4}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_5}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_6}{\delta x}\right)}^2}·\delta x \\ =2\sqrt{3}+\sqrt{1}+\sqrt{1-51}+\sqrt{1-2936} \\ =2\sqrt{3}+\sqrt{1}+\sqrt{-50}+\sqrt{-2935} \end{gathered}$$