Question

A contaminant is leaking into a lake at a rate of R(t) = 1700e^0.06t gal/h. Enzymes that neutralize the contaminant have been added to the lake over time so that after t hours the fraction of 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 will be present in the lake 17 hours from now? (Round your answer to the nearest integer.)

Answer 1

Answer:

16,460 gallons

Step-by-step explanation:

This is a differential equation problem, we have a constant flow of contaminant into the lake, but also we know that only a fraction of that quantity of contaminant remains because of the enzymes. For that reason, the differential equation of contaminant's flow into the lake would be:

$\frac{dQ}{dt} =1700exp(0.06t)*exp(-0.32t)\\\frac{dQ}{dt} =1700exp(-0.26t)$

Then, we have to integrate in order to find the equation for Q(t), as the quantity of contaminant in the lake, in function of time.

$\int\limits^0_t {dQ}=\int\limits^0_t {1700exp(-0.26t)dt}\\Q(t)=\frac{1700}{-0.26} exp(-0.26t)+C$

Now, we use the given conditions to replace them in the equation, in order to solve for $C$

$t_{0} =0\\Q_{0}=10,000\\Q_{0}=-6538exp(-0.26*0)+C\\C=10,000+6538=16538$

Then, we reorganize the equation and we replace t for 17 hours, in order to determine the quantity of contaminant at that time:

$Q_{t} =-6538exp(-0.26t)+16538\\Q_{17} =-6538exp(-0.26*17)+16538\\Q_{17} =16460 gallons$