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

MATH102 December 2014

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[hide] 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?

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[show] 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 $y'$ .

[show] 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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[show] 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.