Answer :
Answer:
(a) R(t) = 5√6 × cos(6/5) = 4.43796 > 0 Therefore the number of mosquitoes are increasing
(b) The number of mosquitoes is increasing at a decreasing rate
(c) At time t =31 the number of mosquitoes on the island is 964.34 mosquitoes
(d) The maximum number of mosquitoes is 1039 mosquitoes to the nearest whole number
Step-by-step explanation:
(a) To solve the question, we note that
R(t) = 5√t × cos(t/5)
When t = 6, R(t) = 5√6 × cos(6/5) = 4.43796 > 0 Therefore the Therefore the rate of change is increasing because the change rate at t = 6 seconds is + 4.43796
(b) To check whether the number of mosquitoes is increasing at a decreasing, we find the derivative of the expression for the rate of change as follows.
dR(t)/dt = [tex]\frac{5\cdot cos(\frac{t}{5}) }{2\sqrt{t} } - \sqrt{t}\cdot sin (\frac{t}{5} )[/tex]
At t = 6, dR(t)/dt = -1.9132
As the derivative of R(t) is negative, the curve is tipping downwards and the rate of increase is becoming lesser or decreasing.
(c) At time t =31, the number of mosquitoes on the island is given as
The area under the slope of the rate of increase curve from 0 to 31 or
[tex]\int\limits^{31}_0 {5 \cdot\sqrt{t} } \times cos \frac{t}{5} , dt[/tex] = -35.6648
However at time t = 0 there are 1000 mosquitoes, which gives a balance of
1000 - 35.6648 = 964.34 mosquitoes
(d) From the equation for the rate of change of the number of mosquitoes, it is seen that
R(t) = 0 when t = 0
R(t) > 0 for 0 < t < 2.5π
R(t) < 0 for 2.5π < t < 7.5π
R(t) > 0 for 7.5π < t < 31
As seen from the above, the maximum number of mosquitoes occurs at either t = 2.5 π or t = 31
Therefore, the maximum number of mosquitoes is given by;
1000 + [tex]\int\limits^{2.5\cdot \pi }_0 {5 \cdot\sqrt{t} } \times cos \frac{t}{5} , dt[/tex] = 1000 + 39.3569 = 1039.3569
Which gives 1039 to the nearest whole number