Chapter 7.4, Problem 38E

To determine

To Find: The starting point of the lane 2 so that the lengths of two lanes are equal.

Answer to Problem 38E

The starting point of the Lane 2 is $\left(-19.909,8.410\right)$ .

Explanation of Solution

Given information:

The figure is

  

Also, the equations of the two lanes are, $y=\sqrt{100-0.2x^2}$ and $y=\sqrt{150-0.2x^2}$ and the starting point of the lane 1 is given as $\left(-\sqrt{500},0\right)$ .

Formula used:

The formula to find the length of the curve from a to b is ${\displaystyle {\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}$ .

Calculation:

Consider lane 1 as $y_1=\sqrt{100-0.2x^2}$ .

Differentiating both sides with respect to x ,

  $\frac{d{y}_1}{dx}=-\frac{0.2x^2}{\sqrt{100-0.2x^2}}$

Now, $y_1=0$ implies,

  $\sqrt{100-0.2x^2}=0$

Square both the sides:

  $\begin{array}{c}100-0.2x^2=0\\ 0.2x^2=100\\ x^2=\frac{100}{0.2}\\ x\approx \pm 22.3607\end{array}$

Now, ${\displaystyle {\int }_{-22.3607}^{22.3607}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}$ fails to be evaluated by NINT ( The nearest integer function ) So, the track’s symmetry is to be used and back way from the upper limit,

Find,

  $\begin{array}{c}{\displaystyle {\int }_{-22.3607}^{22.3607}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}={\displaystyle {\int }_{-22.3607}^{22.3607}\sqrt{1+{\left(\frac{-0.2x}{\sqrt{100-0.2x^2}}\right)}^2}dx}\\ ={\displaystyle {\int }_{-22.3607}^{22.3607}\sqrt{1+\left(\frac{0.4x^2}{100-0.2x^2}\right)}dx}\\ ={\displaystyle {\int }_{-22.3607}^{22.3607}\sqrt{\left(\frac{100+0.2x^2}{100-0.2x^2}\right)}dx}\\ \approx 52.548\end{array}$

Next, approximate the last stretch at each end to be a vertical line and to get the total length of track 1,

  $\begin{array}{c}0.2\sqrt{100-0.2{\left(22.3607\right)}^2}\approx 2\left(0.078\right)\\ \approx 0.156\end{array}$

Get the final length of the line 1 to be 52.704.

In the similar way, find the length of the right half of track 2 which is 32.2734.

Now, let $Y_1=52.702$ and $Y_2=32.274+NINT\left(\sqrt{1+{\left(\frac{-0.2t}{150-0.2t^2}\right)}^2}\right)$ .

Plot these in graph and find out the starting point to be $\left(-19.909,8.410\right)$ .