Chapter 6.4, Problem 21E

a.

To determine

To graph: the curve $y=x\sqrt[3]{4-x},0\le x\le 4.$

Answer to Problem 21E

Explanation of Solution

Given information: Given curve is $y=x\sqrt[3]{4-x},0\le x\le 4.$

Calculation:

Graph of the given curve $y=x\sqrt[3]{4-x},0\le x\le 4.$ is shown below:

  

b.

To determine

To compute: the lengths of inscribed polygons with n = 1, 2 and 4 sides.

Answer to Problem 21E

The polygon with side one, n =1 length = 4.

The polygon with side two, n =2 length $\approx 6.43$ .

The polygon with side four, n =4 length $\approx 7.5$ .

Explanation of Solution

Given information: Given curve is $y=x\sqrt[3]{4-x},0\le x\le 4.$

Calculation:

The polygon with side is just the line segment joining (0, f (0 )) and (4, f (4)) = (4, 0). So its length is 4.

The polygon with two sides joins (0, 0), (2, f (2))=(2, $2\sqrt[3]{2}$ ) and (0,4).

So its length is $\sqrt{{(2-0)}^2+{(2\sqrt[3]{2}-0)}^2}+\sqrt{{(4-2)}^2+{(0-2\sqrt[3]{2})}^2}\approx 6.43.$

The polygon with four sides joins the points (0,0) , (1, $\sqrt[3]{3}$ ) , (2, $2\sqrt[3]{2}$ ) , (3,3) , and (4,0). So its length is:

  $\sqrt{1+{(\sqrt[3]{3})}^2}+\sqrt{1+{(2\sqrt[3]{2}-\sqrt[3]{3})}^2}+\sqrt{1+{(3-2\sqrt[3]{2})}^2}+\sqrt{1+9}\approx 7.5$ .

  

c.

To determine

To set up: an integral for the length of the curve.

Answer to Problem 21E

  $L={\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{1+{\left[\frac{12-4x}{3{(4-x)}^{\frac{2}{3}}}\right]}^2}}dx$ .

Explanation of Solution

Given information: Given curve is $y=x\sqrt[3]{4-x},0\le x\le 4.$

Formula Used:

Arc Length Formula:

If f ‘( x ) is continues on [a ,b] then the length of the curve y = f ( x ) , $a\le x\le b$ , is

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left[f\text{'}(x)\right]}^2}}dx$

Calculation:

To find the length of the given curve $y=x\sqrt[3]{4-x},0\le x\le 4.$

Compute f ‘( x ):

  $\begin{array}{c}f(x)=x\sqrt[3]{4-x}\\ f\text{'}(x)=\frac{d[x\sqrt[3]{4-x}]}{dx}\\ \text{Use}\text{product}\text{rule:}\end{array}$

  $f\text{'}(x)=\frac{dx}{dx}.{(4-x)}^{\frac{1}{3}}+x.\frac{d{(4-x)}^{\frac{1}{3}}}{dx}$

  $\text{Using}\text{chain}\text{rule}:$

  $f\text{'}(x)={(4-x)}^{\frac{1}{3}}+x.\frac{d{(4-x)}^{\frac{1}{3}}}{d(4-x)}\times \frac{d(4-x)}{dx}$

  $f\text{'}(x)={(4-x)}^{\frac{1}{3}}+x.\frac{1}{3}.{(4-x)}^{-\frac{2}{3}}\times (-1)$

  $f\text{'}(x)={(4-x)}^{\frac{1}{3}}-\frac{x}{3{(4-x)}^{\frac{2}{3}}}.$

  $f\text{'}(x)=\frac{4-x}{{(4-x)}^{\frac{2}{3}}}-\frac{x}{3{(4-x)}^{\frac{2}{3}}}.$

  $f\text{'}(x)=\frac{1}{{(4-x)}^{\frac{2}{3}}}((4-x)-\frac{x}{3}).$

  $f\text{'}(x)=\frac{1}{{(4-x)}^{\frac{2}{3}}}(\frac{12-4x}{3}).$

  $f\text{'}(x)=\frac{12-4x}{3{(4-x)}^{\frac{2}{3}}}.$

  $\text{Therefore:}$

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{[f\text{'}(x)]}^2}}dx={\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{1+{\left[\frac{12-4x}{3{(4-x)}^{\frac{2}{3}}}\right]}^2}}dx.$

d.

To determine

To find: the length of the curve to four decimal places using calculator and compare with the approximation in part (b).

Answer to Problem 21E

  $L\approx 7.79879$ .

Explanation of Solution

Given information: Given curve is $y=x\sqrt[3]{4-x},0\le x\le 4.$

Calculation:

From part (c),

  $L={\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{1+{\left[\frac{12-4x}{3{(4-x)}^{\frac{2}{3}}}\right]}^2}}dx$

The problem specifically asks to use a calculator.

This Integral approximates to 7.79879.

In part ( b) , approximated the length of the curve using polygons with 1 ,2 and 4 sides.

All these approximations were underestimates.

  $L\approx 7.79879$ .