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 .

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

Short answer

The arc length is$3$.

The graph is sketched as shown below:

Step-by-step solution

Step 1. Given information .

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

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 .

$$\begin{gathered} a=0,b=\mathrm{\pi },\mathrm{n}=3 \\ \partial \mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{\mathrm{\pi }}{3} \end{gathered}$$

$x_k=a+k\partial x$where k = 0,1,2,3,........n

$x_0=0,x_1=\frac{\mathrm{\pi }}{3},x_2=\frac{2\mathrm{\pi }}{3},x_3=\frac{3\mathrm{\pi }}{3}$

$$\begin{gathered} \partial y_1=f\left(x_1\right)-f\left(x_0\right)=\cos \frac{\mathrm{\pi }}{3}-\cos 0=\frac{1}{2}-1=-\frac{1}{2} \\ \partial y_2=f\left(x_2\right)-f\left(x_1\right)=\cos \frac{2\mathrm{\pi }}{3}-\cos \frac{\mathrm{\pi }}{3}=-\frac{1}{2}-\frac{1}{2}=-1 \\ \partial y_3=f\left(x_3\right)-f\left(x_2\right)=\cos \mathrm{\pi }-\cos \frac{2\mathrm{\pi }}{3}=-1+\frac{1}{2}=\frac{-1}{2} \end{gathered}$$

Substitute the values in given formula ,

Arc length $=\underset{n\to \infty }{\lim }\sum _{k=1}^n\sqrt{1+{\left(\frac{\partial y_k}{\partial x}\right)}^2}·\partial x$

$$\begin{gathered} =\left[\sqrt{1+{\left(-\frac{1/2}{\mathrm{\pi }/3}\right)}^2}+\sqrt{1+{\left(\frac{-1}{\mathrm{\pi }/3}\right)}^2}+\sqrt{1+{\left(\frac{-1/2}{\mathrm{\pi }/3}\right)}^2}\right]\times \frac{\mathrm{\pi }}{3} \\ =\frac{1}{6}\left[\sqrt{48}+\sqrt{18·85}+\sqrt{48}\right] \\ =\frac{1}{6}\left[2\sqrt{48}+\sqrt{18}\right] \\ =\frac{1}{6}\left[2\times 6·928+4·2426\right] \\ =3 \end{gathered}$$

Step 4. Sketch of the  function fx=cos x .

The sketch of the given function is shown below .