Science:Math Exam Resources/Courses/MATH102/December 2014/Question B 07/Solution 1

We draw a diagram as shown:

Graphs

Suppose the finish line is located on the positive side of the $x$-axis. Assume the angle between the line pass through the runner and the origin and the positive x-axis is $\theta$, then we have $\tan (\theta )=\frac{y}{x}$, where $(x,y)$ is the runner’s position. When the runner crosses the finish line, $\theta =0,{\theta }^{\prime }=\frac{1}{10},$ We also get $x^{\prime }=0$ as the elliptical track is perpendicular to the $x$-axis at the finish point so there is no change in $x$. Thus, at the finish point, the velocity of the runner can be described solely as the change in the y-axis. Hence, we have $y^{\prime }=v$. Further, we obtain $x=30$ from the equation of the ellipse $x^2+\frac{y^2}{4}=900$ by substituting $y=0$ and noting that thus $x=\pm 30$. Then we differentiate $\tan (\theta )$ with respect to time to get

$\begin{array}{rl}{\sec }^2(\theta ){\theta }^{\prime }& =\frac{y^{\prime }x-y{x}^{\prime }}{x^2}\\ 1\cdot {\theta }^{\prime }& =\frac{y^{\prime }x-y{x}^{\prime }}{x^2}\\ {\theta }^{\prime }& =\frac{v\cdot 30-0}{30^2}\\ & =\frac{v}{30}\end{array}$

Then $v=30{\theta }^{\prime }=3m/s$

Note: The third line of the equation above is where the observation that $x^{\prime }=0$ is applied into the question as indicated by the substitution of $y^{\prime }$ with v, the variable of $x$ with 30, and the rate of $x^{\prime }$ with 0.