Chapter 8.4, Problem 38E

To determine

To calculate the starting point of lane 2 for which the lengths of two lanes will be equal.

Answer to Problem 38E

Therefore the starting point of lane 2 is $\left(-19.909,8.409\right)$

Explanation of Solution

Given information:

The equation for lane 1is $y=\sqrt{100-0.2x^2}$

The equation for lane 2 is $y=\sqrt{150-0.2x^2}$

The starting point for the lane 1 is at the negative x -intercepts $\left(-\sqrt{500},0\right)$ and the finishing point for both lanes are positive x -intercepts.

Formula used:

If a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$ , $a<b,c<d$ , then the length of the curve

  $\begin{array}{l}L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\text{ if }y\text{ is smooth function of }x\text{ on }\left[a,b\right]\\ L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy}\text{ if }x\text{ is smooth function of }y\text{ on }\left[c,d\right]\end{array}$

Calculation :

Consider the starting point of lane 2 be $s_2$

Arrange the equation of lane 1 in standard form

  $\begin{array}{l}y=\sqrt{100-0.2x^2}\\ y^2=100-0.2x^2\\ y^2+0.2x^2=100\\ \frac{y^2}{100}+\frac{x^2}{500}=1\\ \frac{y^2}{{\left(\sqrt{100}\right)}^2}+\frac{x^2}{{\left(\sqrt{500}\right)}^2}=1\end{array}$

Therefore end point of lane 2 is $\sqrt{500}$ and the starting point is $-\sqrt{500}$

Arrange the equation of lane 2 in standard form

  $\begin{array}{l}y=\sqrt{150-0.2x^2}\\ y^2=150-0.2x^2\\ y^2+0.2x^2=150\\ \frac{y^2}{150}+\frac{x^2}{750}=1\\ \frac{y^2}{{\left(\sqrt{150}\right)}^2}+\frac{x^2}{{\left(\sqrt{750}\right)}^2}=1\end{array}$

Therefore end point of lane 2 is $\sqrt{750}$

Equation for the length of lane 1

  ${\displaystyle \underset{-\sqrt{500}}{\overset{\sqrt{500}}{\int }}\sqrt{1+{\left(\frac{d}{dx}\sqrt{100-0.2x^2}\right)}^2}dx}$

Equation for the length of lane 2

  ${\displaystyle \underset{s_2}{\overset{\sqrt{750}}{\int }}\sqrt{1+{\left(\frac{d}{dx}\sqrt{150-0.2x^2}\right)}^2}dx}$

Since these two lengths are equal

  $\begin{array}{l}{\displaystyle \underset{-\sqrt{500}}{\overset{\sqrt{500}}{\int }}\sqrt{1+{\left(\frac{d}{dx}\sqrt{100-0.2x^2}\right)}^2}dx}={\displaystyle \underset{s_2}{\overset{\sqrt{750}}{\int }}\sqrt{1+{\left(\frac{d}{dx}\sqrt{150-0.2x^2}\right)}^2}dx}\\ {\displaystyle \underset{-\sqrt{500}}{\overset{\sqrt{500}}{\int }}\sqrt{1+{\left(\frac{-0.2x}{\sqrt{100-0.2x^2}}\right)}^2}dx}={\displaystyle \underset{s_2}{\overset{\sqrt{750}}{\int }}\sqrt{1+{\left(\frac{-0.2x}{\sqrt{150-0.2x^2}}\right)}^2}dx}\\ {\displaystyle \underset{-\sqrt{500}}{\overset{\sqrt{500}}{\int }}\sqrt{1+\left(\frac{0.04x}{100-0.2x^2}\right)}dx}={\displaystyle \underset{s_2}{\overset{\sqrt{750}}{\int }}\sqrt{1+\left(\frac{0.04x}{150-0.2x^2}\right)}dx}\\ s_2=-19.909\text{ use NINT}\end{array}$

The value of abscissa at $s_2$ is $-19.909$

Put the value in the equation of lane 2 to get the value of ordinate of $s_2$

  $\begin{array}{l}y=\sqrt{150-0.2{\left(-19.909\right)}^2}\\ y=\sqrt{150-0.2{\left(-19.909\right)}^2}\\ y=8.409\end{array}$

Therefore the starting point of lane 2 is $\left(-19.909,8.409\right)$