Question

400 gallons of pesticide is accidentally spilled into a lake and uniformly mixes with the water. The volume of the lake including the pesticide is 108 gallons. A river flows into the lake bringing 5,000 gallons of fresh water per minute, and the uniform mixture spills over the dam at the same rate. How long, in minutes, will it take to reduce the pesticide in the lake to a safe level of 1 part per million gallons? (Round your answer to the nearest minute.

Answer 1

Answer:

t=27725.8minutes

Explanation:

To determinate use the law the change of volume in determinate time is derivate so:

$\frac{dV}{dt}=flowin-flowout$

The flow in is zero and the flow out is modeling with V as the volume out in the volume of the lake in the reason of change.

$\frac{dV}{dt}=0-(\frac{V}{10x10^{8}})*5000\\\frac{dV}{dt}=-5x10^{-5}*V \\dV*V=-5x10^{-5}*dt$

Integrate to get the volume function

$\int\limits{\frac{1}{V}}\, dv=\int\limits{-5x10^{-5} }\, dt\\ ln(V)=-5x10^{-5}*t\\ln*e(V)=C*e^{-5x10^{-5}*t}$

The find the constant C in the time t=0 the volume is knowing so:

$t=0s\\V=400\\V=C*e^{0}\\C=400$

$V=400*e^{-5x10^{-5}*t}$

The volume is 1 part per million so:

$\frac{V}{10x^{8}}=\frac{1}{10x^{6}}\\ V=100$

$V=400*e^{-5x10^{-5}*t}\\100=400*e^{-5x10^{-5}*t}\\\frac{100}{400}=e^{-5x10^{-5}*t}\\ln*\frac{1}{4}=ln*e^{-5x10^{-5}*t}\\-ln(4)=-5x10^{-5}*t\\ t=\frac{-ln(4)}{-5x10^{-5}}=27725.88\\t=27725.8minutes$