Science:Math Exam Resources/Courses/MATH110/April 2019/Question 09

MATH110 April 2019

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Question 09
A jogger runs around an elliptical track, so that her position is described by the equation ${\frac {x^{2}}{4}}+y^{2}=1,$ where $(x,y)$ are the jogger’s coordinates. A camera is located on the ground at the point of coordinates (1, 0). Find the coordinates of the jogger when she is closest to the camera. Justify your answer.

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
If the jogger is closest to the camera, then her distance from it is minimal. What is the function that is being minimized?

Hint 2
We need to minimize the function whose input is the position of the jogger and whose output is the distance from the camera. Recall that the distance between two points $(a,b)$ and $(x,y)$ is $d((a,b),(x,y))={\sqrt {(a-x)^{2}+(b-y)^{2}}}$ and remember that the jogger is staying on the ellipse, so her coordinates $(x,y)$ satisfy the equation of the ellipse.

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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. If $(x,y)$ are the coordinates of the jogger, then her distance from the camera is $d=d((1,0),(x,y))={\sqrt {(x-1)^{2}+(y-0)^{2}}}.$ Since she is staying on the ellipse, her coordinates are related by the equation ${\frac {x^{2}}{4}}+y^{2}=1,$ so $y^{2}=1-{\frac {x^{2}}{4}}$ and the distance from the camera is ${\begin{aligned}d&={\sqrt {(x-1)^{2}+1-{\frac {x^{2}}{4}}}}\\&={\sqrt {x^{2}-2x+1+1-{\frac {x^{2}}{4}}}}\\&={\sqrt {{\frac {3x^{2}}{4}}-2x+2}}\end{aligned}}$ Now this is a function of a single variable. To minimize it, first compute its derivative (by using the chain rule): ${\begin{aligned}d'(x)&={\frac {1}{2{\sqrt {{\frac {3x^{2}}{4}}-2x+2}}}}\cdot \left({\frac {6x}{4}}-2\right)\\&={\frac {{\frac {3x}{4}}-1}{\sqrt {{\frac {3x^{2}}{4}}-2x+2}}}\end{aligned}}$ If $x$ is a minimal point, then $d'(x)=0$, but this can happen only if the numerator, ${\frac {3x}{4}}-1$, equals 0, so $x={\frac {4}{3}}$. If this is the first coordinate of the jogger, the second is found by isolating $y$ in the ellipse equation: ${\frac {1}{4}}\cdot \left({\frac {4}{3}}\right)^{2}+y^{2}=1\Rightarrow y^{2}={\frac {5}{9}}\Rightarrow y=\pm {\sqrt {\frac {5}{9}}}.$ By drawing the ellipse and the position of the camera, we can see that these are the points where the function $d$ attains its minima (as opposed to maxima or saddle points). So the coordinates of the jogger when she is closest to the camera are $\color {blue}\left({\frac {4}{3}},-{\frac {\sqrt {5}}{3}}\right),\ \left({\frac {4}{3}},{\frac {\sqrt {5}}{3}}\right).$