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, $\theta =0,\quad \theta '={\frac {1}{10}},$ We also get $x'=0$ 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 $y'=v$. Further, we obtain $x=30$ from the equation of the ellipse $x^{2}+{\dfrac {y^{2}}{4}}=900$ by substituting $y=0$ and noting that thus $x=\pm 30$. Then we differentiate $\tan(\theta )$ with respect to time to get

${\begin{aligned}\sec ^{2}(\theta )\theta '&={\frac {y'x-yx'}{x^{2}}}\\1\cdot \theta '&={\frac {y'x-yx'}{x^{2}}}\\\theta '&={\frac {v\cdot 30-0}{30^{2}}}\\&={\frac {v}{30}}\end{aligned}}$

Then $v=30\theta '=3m/s$

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