Ap Calculus BC 2002 Form B Question 2?

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Answer 1 · 2018-04-30T00:38:35

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!