Question

$$ \begin{array}{l}{\text { Modeling Running Tracks Two lanes of a running track }} \\ {\text { are modeled by the semiellipses as shown. The equation for }} \\\ {\text { lane } 1 \text { is } y=\sqrt{100-0.2 x^{2}} \text { , and the equation for lane } 2} \\ {\text { is } y=\sqrt{150-0.2 x^{2}} . \text { The starting point for lane } 1 \text { is at the }}\end{array} $$ $$ \begin{array}{l}{\text { negative } x \text { -intercept }(-\sqrt{500}, 0) . \text { The finish points for both lanes }} \\ {\text { are the positive } x \text { -intercepts. Where should the starting point be }} \\ {\text { placed on lane } 2 \text { so that the two lane lengths will be equal }} \\ {\text { (running clockwise)? }}\end{array} $$

Short answer

The starting point on lane 2 which would make the lengths of the two lanes equal when running clockwise should be \((\sqrt{500}, -5\sqrt{15})\)

Step-by-step solution

Understanding the Problem

Understand that we need to find the starting point on lane 2 that would make the two lane lengths equal. Given the equations of two semi-ellipses, with lane 1 starting at \(-\sqrt{500}, 0\), we need to find a symmetrical starting point on lane 2.

Setting Up the Equations

The equation for lane 1 is \(y = \sqrt{100 - 0.2x^{2}}\), which we can convert to parametric form as \((x,y) = (10\cos{t}, 10\sin{t})\) for \(-\pi /2\leq t \leq \pi /2\). Similarly, the equation for lane 2 would be \((x,y) = (15\cos{t}, 15\sin{t})\) for \( -\pi /2 \leq t \leq \pi /2 \).

Equalizing the Lengths of the Lanes

The length of an ellipse's arc can be obtained using the formula \(L = \int_{a}^{b} \sqrt{r_x^{2}\sin^{2}{t} + r_y^{2}\cos^{2}{t}}dt\). For both lanes to have the same length, their corresponding parameters \(t1\) and \(t2\) must satisfy the equation: \(\int_{-\pi /2}^{t1} \sqrt{10^{2}\sin^{2}{t} + 10^{2}\cos^{2}{t}}dt = \int_{t2}^{\pi /2} \sqrt{15^{2}\sin^{2}{t} + 15^{2}\cos^{2}{t}}dt\). Simplifying implies \(t1 = -t2\), i.e. the starting points on the lanes are symmetrical with respect to the x-axis.

Calculate the Starting Point Coordinates

Substitute \(-t2\) for \(t1\) in the parametric equation for lane 2 to obtain the starting point coordinates. The starting point on lane 2 is \( (15\cos{-t2}, 15\sin{-t2}) = (15\cos{t2}, -15\sin{t2}) \). Since the x-coordinates of the starting points are symmetric with respect to the origin, the x-coordinate of the starting point on lane 2 is equivalent to the x-coordinate of the starting point on lane 1, i.e. \(15\cos{t2} = \sqrt{500}\).

Solve for the y-coordinate

Solving the equation for \(t2\), we get \(15\cos{t2} = \sqrt{500}\) .Thus, \( t2= \arccos{(\sqrt{500} / 15)}\). Using this in our y-coordinate gives \(y= -15 \sin{t2}\), which evaluates to \(y=-5\sqrt{15} \) after substituting \(t2\) and simplifying.

Final Answer

Therefore, the coordinates of the starting point on lane 2 should be \((\sqrt{500}, -5\sqrt{15})\).