Question

Ap Calculus BC 2002 Form B Question 2?

Answer 1

a) The derivative, `P'(t)`, has value `P'(9) = 1 - 3e^(-0.2sqrt(9)) ~~-0.646`, so the amount of pollutant is decreasing at `t = 9`

b) This is your run of the mill critical number problem.

`0 = 1 - 3e^(-0.2sqrt(t))`

Solve using a calculator to get

`t = 30.174`

Since the amount of pollutant is decreasing at `t = 9`, and increasing at `t = 40`, this is a minimum.

c) Recall that

`P(30.174) = P(0) + int_0^30.174 P'(t)dt`

`P(30.174) = 50 - 14.89566 = 35.104` gallons

Since this is less than 40 gallons, the lake is safe at that point.

d) The slope of this line will be `P'(0) = 1 - 3e^(-0.2(0)) = -2`. Thus the equation will be

`y - 50 = -2(x - 0)`

`y = -2x + 50`

We need this to be less than `40`, so we write the inequality:

`40 ≥ 50 - 2x`

`2x ≥ 10`

`x ≥ 5`

Thus, after 5 days the lake should be safe.

Hopefully this helps!