Science:Math Exam Resources/Courses/MATH102/December 2015/Question 06

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MATH102 December 2015

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Question 06
If we assume that the amount of food gained in a food patch during time $t$ is $f(t)=F_{max}{\frac {t^{2}}{k^{2}+t^{2}}}$ , then (a) The total amount of food in the patch is ${\frac {F_{max}}{k^{2}}}$ . (b) Getting food from the patch is initially a slow process but then it speeds up for a while. (c) It takes a time $t=k$ to get all the energy out of the patch. (d) It takes a time $t={\frac {k}{2}}$ to get all the energy out of the patch. (e) There is unlimited energy in the patch.

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
This is a Hill function. If you understand the behaviour of Hill functions, this problem can be easily solved. If not, think about how statements (a), (b), (c), (d), and (e) can be rephrased in terms of the asymptotes of $f(t)$ .

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Solution 1
This solution will use the theory of Hill functions. An introduction can be found in this video. You may also want to view the second solution presented below. This is a Hill function with horizontal asymptote $F_{\text{max}}$ , half-max $k$ , and Hill coefficient 2. For positive values of t, it follows that the function is strictly increasing toward a horizontal asymptote $F_{\text{max}}$ . (a) does not apply because f(t) approaches $F_{\text{max}}$ as $t\to \infty$ . (b) is reasonable. For values of t near zero, f(t) is close to $t^{2}$ , which is small near zero. This means that food is gathered slowly at times close to zero. (c) is not correct: k is the half-max, which means that half of the food will have been taken out of the patch at t = k. (d) is also not correct: k/2 is even less time than k, so less than half of the food will be recovered in this time. (e) is wrong; the maximum amount of food available is $F_{\text{max}}$ . answer: $\color {blue}(b)$

Solution 2
Solution 2: The statements (a), (b), (c), (d), and (e) can all be phrased in terms of $f'(t)$ and $\lim _{t\to \infty }f(t)$ : (a) states that $\lim _{t\to \infty }f(t)={\frac {F_{\text{max}}}{k^{2}}}$ . (b) states that $f'(t)$ is small for t close to zero, but larger for some larger values of t. (c) states that f(t) achieves a global maximum when t=k. (d) states that f(t) achieves a global maximum when t=k/2. (e) states that f(t) grows without bound. Let's focus on $\lim _{t\to \infty }f(t)$ first. We have that ${\begin{aligned}\lim _{t\to \infty }{\frac {F_{\text{max}}t^{2}}{t^{2}+k^{2}}}&=\lim _{t\to \infty }{\frac {F_{\text{max}}}{1+k^{2}/t^{2}}}\\&=F_{\text{max}}\end{aligned}}$ So f(t) neither approaches ${\frac {F_{\text{max}}}{k^{2}}}$ nor grows without bound, so neither (a) nor (e) is the correct answer. Next, we compute $f'(t)$ . This can be computed via the quotient rule: ${\begin{aligned}f'(t)&=F_{\text{max}}{\frac {2t(t^{2}+k^{2})-t^{2}(2t)}{(t^{2}+k^{2})^{2}}}&=F_{\text{max}}{\frac {2tk^{2}}{(t^{2}+k^{2})^{2}}}.\end{aligned}}$ This vanishes only when t is equal to 0. Therefore, f(t) has no local maxima or minima and both (c) and (d) are incorrect. If t is near zero, $f'(t)$ will be small. Furthermore, plugging in t = k gives a value of ${\frac {1}{2k}}$ for $f'(t)$ . This is significantly larger than 0. Therefore, the feeding starts out slow, but gets faster for a while (the feeding will then slow down again). Therefore, the answer is (b). answer: $\color {blue}(b)$