How do you solve this question The population of fish in a l

How do you solve this question?

The population of fish in a lake is observed to fluctuate during a six month period according to the formula P(t) = 80000 + 20000 sin t where t is time in months. Answer the questions below. Note that the units are provided outside the answer boxes. You don\'t have to include them in your answers. You may find it useful to graph this function on the domain 0 lessthanorequalto t lessthanorequalto 6 months. How fast is the population changing at the instant t = 4 months? fish/month, rounded to the nearest whole number. What is the smallest number of fish during the six month period? fish When does the population low point occur? Be accurate to 2 decimal places. months Shortly before the low point, the population was 70,000 fish. How fast was the population changing at that instant? fish/month, rounded to the nearest whole number. Shortly after the low point there is an instant when the population is changing at 10,000 fish/month. When is it? Be accurate to 2 decimal places. months If the population is changing at -11,000 fish/month, how many fish are there? Larger number = fish, rounded to the nearest fish. Smaller number = fish, rounded to the nearest fish.

Solution

P(t) = 80000 + 20000sint
dP/dt = 20000cost ; dP/dt = rate of change of population
so, 20000cost = -11000 => cost = -11/20
Now using identity sin²t + cos²t = 1 ; we get sint = (1-cos²t)^1/2 = +0.8352 & -0.8352
Using these two results in determining P(t) ; we get
P_large = 80000 + 20000*0.8352 = 96704
P_small = 80000 - 20000*0.8352 = 63296