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

MATH102 December 2014

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Question B 07
A runner on an elliptical track described by the equation $x^{2}+{\dfrac {y^{2}}{4}}=900$ runs at a constant speed $v$ where $x$ and $y$ are measured in metres. Standing at the origin, you must rotate your head at 1/10 radians per second to watch her as she crosses the finish line located on the x axis. How fast is she running?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Set the angle between the line which connects the runner and the origin and the positive x-axis to be $\theta$ , then if the runner is at location $(x,y)$ , we have $\tan(\theta )={\frac {y}{x}}$ . We know $\theta '=1/10$ , so we can use implicit differentiation to relate $\theta '$ with $x'$ and $y'$ .

Hint 2
When the runner crosses the finish line, $x'=0$ as the elliptical track is perpendicular to the $x$ -axis at that point. So there is no $x$ component then $y'=v$ .

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.