A contaminant is leaking into a lake at a rate of Rt 1600e0

A contaminant is leaking into a lake at a rate of R(t) = 1600e^0.06t gallons/h Enzymes have been added to the lake that neutralize the contaminant over time so that after t hours the fraction of the contaminant that remains is S(t) = e^-0.32t. If there are currently 10,000 gallons of the contaminant in the lake, how many gallons are present in the lake 18 hours from now?

Solution

contiminate added in dt time is R(t).dt

let contiminante reduced at O(t) rate .

then continant removed in dt time is O(t).dt

contiminant remaining S(t) = (R(t)-O(t))dt

O(t) = ds/dt -R(t) = f(t)                                   (f(t) is the function given in the question)

ds = (f(t) +R(t))dt

on integarting we will get contiminat remaining at some t time

int(ds) = int((e^-0.32t + 1600 e^0.06t)dt)

s(t) =     (-1/0.32)e^-0.32t +(1600/0.06)e^0.06t +C₀   (where C₀ is integartion constant)

s(t=0) = 10000 = (-1/0.32)+(1600/0.06) +C₀

hence C₀ = 10000 +100/32 - 160000/6 = -16663.5416667

now contiminant at time t is

s(t) = (160000/6)e^0.06t (-1/0.32)e^-0.32t -16663.5416667

so at t= -18hr is

s(t= -18) = (160000/6)e^-1.08 -(100/32) e^5.76 -16663.5416667