Q. 16 - Calculus

Question

In this exercise you will approximate the arc length of f (x) = $x^2$ on [0, 2] in two ways and compare your answers .

(a) By using four line segments and the distance formula.

(b) By using a right sum with four rectangles to approximate the area under the graph of the function $y=\sqrt{1+4x^2} , \left[a,b\right]=\left[0,2\right]$ .

(c) Why do the approximations you found in parts (a) and (b) both approximate the arc length of f (x) = $x^2$ on  [0, 2]? Which, if either, do you think might be a better approximation .

Step-by-Step Solution

Answer

(a) Arc length by using line segment is $2·1216$ .

(b) Arc length by using right Riemann sum is $5·967$ .

(c) Right sum is better than distance formula .

Step 1. Given information .

Consider the given function $f\left(x\right)=x^2$ .

Step 2. Formula used for part (a) .

$D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Step 3. (a) Find arc length .

Divide the given open interval $\left[0,2\right]$ into four sub interval $\left(0 , 0·5\right) , \left(0·5 , 1\right) , \left(1 , 1·5\right) , \left(1·5 , 2\right)$ .

Distance between $\left(0, 0·5\right) , \left(0·5 , 1\right)$

$$\begin{gathered} D={\sqrt{{\left(0·5-0\right)}^2+\left(1-0·5\right)}}^2 \\ =\frac{1}{\sqrt{2}} \end{gathered}$$

Distance between $\left(0·5 , 1\right) , \left(1, 1·5\right)$ is $\frac{1}{\sqrt{2}}$ .

Distance between $\left(1 , 1·5\right) , \left(1·5 , 2\right)$ is $\frac{1}{\sqrt{2}}$ .

$$\begin{gathered} Arc length =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ =0·7072+0·7072+0·7072 \\ =2·1216 \end{gathered}$$

Step 4. Formula used for part (b) .

Right sum $=\sum _{i=1}^n\delta x·f\left(x_i\right)$

Step 5. (b) Find the arc length .

$$\begin{gathered} S=\sum _{i=1}^4\delta x·f\left(x_i\right) \\ \delta x=\frac{b-a}{n}=\frac{2-0}{4}=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} =\delta x·f\left(0\right)+\delta x·f\left(0·5\right)+\delta x·f\left(1\right)+\delta x·\left(1·5\right)+\delta x·f\left(2\right) \\ =\frac{1}{2}\times 1+\frac{1}{2}\times \sqrt{2}+\frac{1}{2}\times \sqrt{5}+\frac{1}{2}\times \sqrt{10}+\frac{1}{2}\times \sqrt{17} \\ =\frac{1}{2}+\frac{\sqrt{2}}{2}+\frac{\sqrt{5}}{2}+\frac{\sqrt{10}}{2}+\frac{\sqrt{17}}{2} \\ =\frac{1+1·414+2·2360+1·414\times 2·2360+4·123}{2} \\ =\frac{11·934}{2} \\ =5·967 \end{gathered}$$

Step 6. (c) Which approximation is better .

The approximation in part (a) and part (b) both are better but the  approximation according to the Riemann right sum is more way better than the distance formula in Riemann sum the given function will be continuous and differentiable on open interval .