Science:Math Exam Resources/Courses/MATH102/December 2015/Question 16

MATH102 December 2015

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Question 16
A squirrel sitting 6 m up in a tree is watching a coyote walk past the tree. The squirrel measures the angle formed between a vertical line directly below her and the line connecting her and the coyote and finds that it is changing at a rate of 1/12 radians per second when the coyote is 8 m away from the base of the tree. How fast is the coyote walking?

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Denote the distance of the coyote from the tree by $x$ and the angle between the tree and the line connecting the squirrel and the coyote by $\theta$. Then, find the relations between them. (It would be helpful to draw a diagram describing the statements in the question)

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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. Let $x=x(t)$ be the distance from the coyote to the tree and $\theta =\theta (t)$ be the angle between the tree and the line connecting the squirrel and the coyote. Then, since the height of the tree is 6, these variables are related to each other by $\tan \theta ={\frac {x}{6}}$. From the question, the rate of change of $\theta$ at $x=8$ is given by $\theta '|_{x=8}={\frac {1}{12}}{\text{rad/sec}}$. On the other hand, taking the derivative on both sides of the equation gives$\tan \theta ={\frac {x}{6}}$, we have ${\begin{aligned}&(\sec ^{2}\theta )\theta '=(\tan \theta )'=({\frac {x}{6}})'={\frac {1}{6}}x'\implies \\&x'=6(\sec ^{2}\theta |_{x=8})\theta '|_{x=8}=6\cdot (1+\tan ^{2}\theta |_{x=8})\cdot {\frac {1}{12}}=6\cdot \left(1+\left({\frac {4}{3}}\right)^{2}\right)\cdot {\frac {1}{12}}={\frac {25}{18}}\end{aligned}}$ Therefore, the coyote is walking at the speed $\color {blue}{\frac {25}{18}}m/sec$.