Question

The number of gallons, P(t), of a pollutant in a lake changes at the rate P'(t) = 1-3e^(-.2sqrt(x)) gallons per day, where t is measured in days. There are 50 gallons of the pollutant in the lake at time t = 0. The lake is considered to be safe when it contains 40 gallons or less of pollutant.
a.) Is the amount of pollutant increasing at time t = 9 ? Why or why not?
b.) For what value of t will the number of gallons of pollutant be at its minimum? Justify your answer.
c.) Is the lake safe when the number of gallons of pollutant is at its minimum? Justify your answer.
d.) An investigator..

Answer 1

d.) An investigator uses the tangent line approximation to P(t) at t = 0 as a model for the amount of pollutant in the lake. At what time t does this model predict that the lake becomes safe?

Answer 2

@UnkleRhaukus

Answer 3

a) solve P'(9) if positive, then it is increasing, and vice versa
b) set P'(t) = 0 and solve for t

Answer 4

@epicfail for b.) how do i solve for t with the equation P'(t) = 1- 3e^(-.2sqrt(t))

Answer 5

well P'(t) = 0 =1-3e^(-2sqrtt)

Answer 6

^-.2

Answer 7

so 1=3e^(-2sqrtt)

Answer 8

you forgot the decimal xD

Answer 9

1/3=e^-2sqrtt

Answer 10

ln 1/3= -.2sqrtt

Answer 11

5ln1/3=-sqrtt

Answer 12

ln(1/3^5) squared = t

Answer 13

divide through by 0.2

Answer 14

nvm :)

Answer 15

when you take 0.2 to the other side, 0.2=1/5, so effectively you multiply through by 5

Answer 16

ok :)

Answer 17

why did you put down 3^5? @epicfail

Answer 18

@phi

Answer 19

@satellite73

Answer 20

@shamim

Answer 21

5ln(1/3) = ln((1/3)^5) =ln((1^5)/(3^5)) = ln(1/3^5)

Answer 22

ok ty :)

Answer 23

btw if we weren't going to simply .2 to 1/5 can u show me how we do the problem by just using .2 instead of 1/5?

Answer 24

@epicfail

Answer 25

exactly the same, ln(...)/0.2=5ln(...)

Answer 26

okay thank you lol