# Ap Calculus BC 2002 Form B Question 2?

## Apr 30, 2018

a) The derivative, $P ' \left(t\right)$, has value $P ' \left(9\right) = 1 - 3 {e}^{- 0.2 \sqrt{9}} \approx - 0.646$, so the amount of pollutant is decreasing at $t = 9$

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

$0 = 1 - 3 {e}^{- 0.2 \sqrt{t}}$

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

$P \left(30.174\right) = P \left(0\right) + {\int}_{0}^{30.174} P ' \left(t\right) \mathrm{dt}$

$P \left(30.174\right) = 50 - 14.89566 = 35.104$ gallons

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

d) The slope of this line will be $P ' \left(0\right) = 1 - 3 {e}^{- 0.2 \left(0\right)} = - 2$. Thus the equation will be

$y - 50 = - 2 \left(x - 0\right)$

$y = - 2 x + 50$

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!