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:

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

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

Answer 2

The rate of change of volume of pesticide in the lake relative to the

volume of water in the lake pesticide gives change in concentration.

Response:

• The time it will take fir the volume of pesticide in the lake to reduce to a safe level of 1 parts per million is 27,725.9 minutes

Which method can be used to determine the time the water in the lake will be safe?

Given:

Volume of water in the lake + The pesticide = 10⁸ gallons

Volume of water that flows into the lake = 5,000 gallons

Volume of water and pesticide that flows into the dam = 5,000 gallons

Concentration of pesticide that flows into the dam is therefore;

$C = \mathbf{\dfrac{V}{10^8} \times 5,000}$

Rate at which the volume of pesticide, V changes in the lake is therefore;

$\dfrac{dV}{dt} = \mathbf{0 - \dfrac{V}{10^8} \times 5,000}$

Which gives;

$\dfrac{dV}{V} = \mathbf{-\dfrac{V}{10^8} \times 5,000 \times dt}$

• $\displaystyle \int\limits {\dfrac{dV}{V} = \mathbf{\int\limits -\dfrac{1}{10^8} \times 5,000 \, dt}}$

Which gives;

㏑V = -5 × 10⁻⁵·t + c

$V = e^{-5 \times 10^{-5} \cdot t + c} = \mathbf{C \cdot e^{-5 \times 10^{-5} \cdot t}}$

$V = C \cdot e^{-5 \times 10^{-5} \cdot t}$

From the given constraints, we have;

At t = 0, V = 400, which gives;

$400 = \mathbf{C \cdot e^{-5 \times 10^{-5} \times 0}}= C$

C = 400

When the volume is 1 part per million, we have;

$\dfrac{1}{1,000,000} = \mathbf{ \dfrac{V}{10^8}x^{2} }$

Which gives;

$V = \dfrac{1}{1,000,000} \times 10^8= \mathbf{100}$

When the volume is 1 part per million, V = 100

Therefore;

$100 = \mathbf{400 \times e^{-5 \times 10^{-5} \cdot t}}$

$e^{-5 \times 10^{-5} \cdot t} = \dfrac{100}{400} = \dfrac{1}{4}$

${-5 \times 10^{-5} \cdot t} = \mathbf{ln \left(\dfrac{1}{4} \right)}$

• $The \ time \ in \ minutes \ to \ reach \ safe \ levels, \, t = \dfrac{ ln \left(\dfrac{1}{4} \right)}{-5 \times 10^5 } \approx \underline{27,725.9 \ minutes}$

Learn more about rate of change of a function here:

https://brainly.com/question/10507415