Question
Fish enter a lake at a rate modeled by the function E given by \(E(t)=20+15 sin(\frac{\pi t}{6}).\) Fish leave the lake at a rate modeled by the function L given by \(L(t)=4+2^{0.1t^{2}}.\) Both E(t) and L (t) are measured in fish per hour, and t is measured in hours since midnight (t = 0).
(a) How many fish enter the lake over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5) ? Give your answer to the nearest whole number.
(b) What is the average number of fish that leave the lake per hour over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5) ?
(c) At what time t, for 0 ≤ t ≤ 8, is the greatest number of fish in the lake? Justify your answer.
(d) Is the rate of change in the number of fish in the lake increasing or decreasing at 5 A.M. (t = 5) ? Explain your reasoning.
Answer/Explanation
Ans:
(a) \(\int_{0}^{5}E(t)dt=153.457690\)
To the nearest whole number, 153 fish enter the lake from midnight to 5 A.M.
(b) \(\frac{1}{5-0}\int_{0}^{5}L(t)dt=6.059038\)
The average number of fish that leave the lake per hour from midnight to 5 A.M. is 6.059 fish per hour.
(c) The rate of change in the number of fish in the lake at time t is given by E(t)-L(t) .
E(t)-L(t) =0⇒ t = 6.20356
E(t) – L(t) > 0 for 0 ≤ t < 6.20356, and E(t) – L(t) < 0 for 6.20356 < t ≤ 8. Therefore the greatest number of fish in the lake is at time 6.204 (or 6.203).
— OR —
Let A (t) be the change in the number of fish in the lake from midnight to t hours after midnight.
\(A(t)=\int_{0}^{t}(E(s)-L(s))ds\)
A'(t) = E(t) – L(t) = 0 ⇒ t = C = 6.20356
t | A(t) |
0 C 8 | 0 135.01492 80.91998 |
Therefore the greatest number of fish in the lake is at time t = 6.204 (or 6.203).
(d) E'(5) – L'(5) = -10.7228 < 0
Because E ‘ (5) – L’ (5) < 0, the rate of change in the number of fish is decreasing at time t = 5.