For the rate of change of the number of mosquitoes on Tropical Island at time t days is modeled by mosquitoes per day. There are 1000 mosquitoes on Tropical Island at time t = 0. (a) Show that the number of mosquitoes is increasing at time t = 6. (b) At time t = 6, is the number of mosquitoes increasing at an increasing rate, or is the number of mosquitoes increasing at a decreasing rate? Give a reason for your answer. (c) According to the model, how many mosquitoes will be on the island at time t = 31? Round your answer to the nearest whole number. (d) To the nearest whole number, what is the maximum number of mosquitoes for Show the analysis that leads to your conclusion.

(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  =  

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

     = -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 + = 1000 + 39.3569 = 1039.3569

Which gives 1039 to the nearest whole number