Science:Math Exam Resources/Courses/MATH100/December 2019/Question 5

Critical points and intervals of increase and decrease

MATH100 December 2019

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Question 5
A $2$ metre tall woman (yes, she’s tall!) is running away from a $6$ metre tall lamp-post (with a light on top – it’s dark out). Her velocity t seconds after leaving the lamp-post is given by the equation $v(t)=4-\sin(2\pi t)$ metres per second. How quickly is the length of her shadow changing after $3$ seconds?

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
Draw a picture! Label the following quantities in your picture: the height of the lamp post, the height of the woman, the total length she has travelled, and the length of her shadow.

Hint 2
The first step is to notice the similar triangles in your picture. How can you use the information you know (height of the lamp post/height of the woman) to determine the ratio of these similar triangles. Can you think of an equation that relates the total length the woman has walked to the length of her shadow?

Hint 3

The ratio of our similar triangles is ${\frac {2}{6}}={\frac {1}{3}}$. Using this information, we know that $L_{S}$, the length of her shadow, and $L_{T}$, the total length, are related by the equation $L_{S}={\frac {1}{3}}L_{T}$. Now, we are asked how quickly the quantity $L_{S}$ is changing. "How quickly the quantity" changes is synonymous with the second derivative, or acceleration. How can you relate $L_{T}=L_{T}(t)$ as a function of time, to $v(t)$, the velocity of the woman? Remember, if $a(t)$ denotes acceleration, we have $v^{'}(t)=a(t)$.

Hint 4
From the previous hints, we know $L_{T}(t)={\frac {1}{3}}L_{S}(t)$. We also have $a_{T}(t)=v^{'}(t)=L^{''}(t)$. Also, if $a_{S}(t)$ denotes the acceleration of the shadow, that $a_{S}(t)=L_{S}^{''}(t)=1/3L_{T}^{''}(t)$. Combining this information, we have: $a_{S}(t)=1/3L_{S}^{''}(t)=1/3v^{'}(t)$. Now, we only need to evaluate this quantity when $t=3$.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 us label some variables before beginning. We let $L_{T}$ denote the total length (or distance) which the woman has walked from the lamp post, and $L_{S}$ denote the length of the shadow. Also, let $H_{L}$ and $H_{W}$ denote the height of the lamp post and woman, respectively. After drawing a picture, we notice two similar triangles. One of these triangles has short sides made by the lamp post ($H_{L}$), and the distance $L_{T}$. The second, smaller triangle has short sides made by the woman ($H_{W}$), and $L_{S}$, which is the length of her shadow. Using our knowledge of similar triangles, we know that: ${\frac {L_{S}}{L_{T}}}={\frac {H_{W}}{H_{L}}}={\frac {1}{3}}\Rightarrow L_{S}={\frac {1}{3}}L_{T}$. Now, from the problem statement, we know that $L_{S}=L_{S}(t)$ and $L_{T}=L_{T}(t)$ are both functions of time. Moreover, from introductory physics, we know that the derivative of position is velocity, and the second derivative of position is acceleration. Since the question asks us to determine "how quickly" $L_{S}(t)$ is changing, this corresponds to determining $L_{S}^{''}(t)$, or the acceleration of the woman, at time $t_{0}=3$. Now, the only piece of information we have not used so far is that the woman's velocity is given by the function $v(t)=4-\sin(2\pi t)$. From our above discussion, we know that: $L_{T}^{'}(t)=v(t)\Rightarrow L^{''}(t)=v^{'}(t)={\frac {d}{dt}}[4-\sin(2\pi t)]=-2\pi \cos(2\pi t).$ Also, we know that: $L_{S}(t)={\frac {1}{3}}L_{T}(t)\Rightarrow L_{S}^{''}(t)={\frac {1}{3}}L_{T}^{''}(t)$. Therefore, combining the two equations, we know that: $L_{S}^{''}(t)={\frac {1}{3}}L_{T}^{''}(t)=-{\frac {2\pi }{3}}\cos(2\pi t)$. Evaluating this expression at $t=3$, we see that: $L_{S}^{''}(t)=-{\frac {2\pi }{3}}\cos(6\pi )=-{\frac {2\pi }{3}}{\frac {m}{s^{2}}}$. Therefore, the acceleration of the woman's shadow at time $t=3$ is $-{\frac {2\pi }{3}}{\frac {m}{s^{2}}}$.