The number of fish in a small bay is modeled by the function F defined by F(t) = 10(t^3 - 12t^2 + 45t +100), where t is measured in days

Question

Interval: 0 less than or equal to t which is less than or equal to 8

a) Using correct units, interpret the meaning of F'(4) = -30 in the context of the problem

b) Based on the model, what is the absolute minimum number of fish in the bay over the time interval 0 less than or equal to t less than or equal to 8? Justify your answer

c) For what values of t, 0 less than or equal to t less than or equal to 8, is the rate of change of the number of fish in the bay decreasing?

d) For 0 less than or equal to t less than or equal to 8, the number of pelicans flying near the bay is modeled by the differentiable function P, where P is a function of the number of fish in the bay. Based on the models, write an expression for the rate of change of the number of pelicans flying near the bay at time t=c

Answer 1

1) The direvative of the function gives the instant rate of change of the equation. In this case at f(4)=-30, on the 4th day the fish population is decreasing by 30 fish/day.

2) The absolute minimum is at t(0)=1000 fish

3) To find the negative rate of change we need to find the inflection points. To do this we set the derivative function to 0. so if f(t)=10(t^3-12*t^2+45t+100) then f'(t)=0=10*(3t^2-24t+45)=30*(t^2-8t+15). Factoring this we have (t-5)*(t-3)=0 so we have a min/max on day 3 & day 5. From this data we have known intervals (0-3), (3-5) & (5-8). We need to pick a number from each of those intervals and using the derivative function see if it is positive or negative.

0-3: +

3-5: -

5-8:+

From this we know between days 3 & 5, the population of the fish are decreasing.

4) The rate of change for the pelicans and the fish are direct correlation so the rate of change for the fish is equal to the rate of change for the pelican