Q. 26 - 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 $y_k=0,1,2,3........n$ .

$f\left(x\right)=\cos x , \left[a,b\right]=\left[0,\pi \right] , n=3$ 

Step-by-Step Solution

Answer

The arc length is $2\sqrt{3601}+\sqrt{14401}$ .

Step 1. Given information .

Consider the given function $f\left(x\right)=\cos x$ .

Step 2. Formula used .

The formula to find arc length 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 arc length .

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

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

The arc length of $$\begin{gathered} f\left(x\right)=\underset{n\to \infty }{\lim }\sum _{k=1}^3\sqrt{1+{\left(\frac{\delta y_1}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_2}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_3}{\delta x}\right)}^2}·\delta x \\ =\sqrt{1+{\left(\frac{-1}{60}\right)}^2}·60+\sqrt{1+{\left(\frac{-1}{60}\right)}^2}·60+\sqrt{1+{\left(\frac{-1}{120}\right)}^2}·60 \\ =2\sqrt{3601}+\sqrt{14401} \end{gathered}$$