If water is lost at 20ft^3/day when W= 10,000 ft^3 then the leakage rate constant in the differential equation is 20/10,000 or 0.002. Suppose that water flows in at F= 200,000 ft^3/day. How much water would you predict in the lake after 100 days?\
W = Ce^(-kt)
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To find the amount of water in the lake after 100 days, we can use the differential equation:
W = Ce^(-kt)
Given that water flows in at a rate of F = 200,000 ft^3/day, we can substitute this value into the equation:
F = C * e^(-kt)
We know that the leakage rate constant k is 0.002, so we can substitute that as well:
200,000 = C * e^(-0.002t)
Now, we can solve for the constant C. Let's use the initial condition W = 10,000 ft^3 at t = 0:
10,000 = C * e^(-0.002 * 0)
10,000 = C * e^0
10,000 = C * 1
C = 10,000
Now, let's substitute the value of C back into the equation:
200,000 = 10,000 * e^(-0.002t)
To find the water in the lake after 100 days (t = 100), we can substitute t = 100 into the equation:
W = 10,000 * e^(-0.002 * 100)
W = 10,000 * e^(-0.2)
Using a calculator, we can find the value of e^(-0.2) to be approximately 0.8187:
W = 10,000 * 0.8187
W ≈ 8,187 ft^3
Therefore, we would predict that there would be approximately 8,187 ft^3 of water in the lake after 100 days.
To find the amount of water in the lake after 100 days, we can use the given differential equation:
W = Ce^(-kt)
We are given the leakage rate constant, k, which is 0.002. This means that water is lost at a rate of 0.002 ft^3/day for every 1 ft^3 of water in the lake.
We are also given the inflow rate of water, F, which is 200,000 ft^3/day.
To find the constant C, we can use the initial condition where water is lost at 20 ft^3/day when the volume is 10,000 ft^3. Plugging these values into the differential equation, we have:
20 = Ce^(-0.002 * 10,000)
Solving this equation for C, we find:
C = 20 / e^(-0.002 * 10,000)
Now, we can use this value of C to find the amount of water in the lake after 100 days. Plugging in the values into the equation, we have:
W = (20 / e^(-0.002 * 10,000)) * e^(-0.002 * 100)
Simplifying this expression gives us the predicted amount of water in the lake after 100 days.