Question

A contaminant is leaking into a lake at a rate of R(t) = 1400e0.06t gallons/h. Enzymes have been added to the lake that neutralize the contaminant over time so that after t hours the fraction 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? (Round your answer to the nearest whole number.) gal

Answer 1

Answer:

Step-by-step explanation:

Rate of leakage, R(t) = 1400 e^0.06t gallons/h

fraction remains , S(t) = e^(-0.32t)

initial contaminant = 1000 gallon

gallons contaminant present after t hour is S(t) R(t)

G(t) = S(t) R(t)

$G(t) = 1400 e^{0.06t}\times e^{-0.32t}$

$G(t) = 1400 e^{- 0.26t}$

Put t = 18 hours

$G(t) = 1400 e^{- 0.26\times 18}$

Taking log on both the sides

ln G = ln 1400 - 0.26 x 18

ln G = 7.244 - 4.68

ln G = 2.564

G = 13 gallons