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 $y_k=0,1,2,3........n$.

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

Short answer

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

Step-by-step solution

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