A new oil field has just begun production. The first oil removed is the easiest to get out, and so production falls as time goes on. The instantaneous rate at which oil can be extracted is 14% of the amount of oil remaining per year. Here, "instantaneous" refers to the fact that as soon as any oil is removed, the rate of production falls proportionally in the "next" instant. If the company continues to extract oil at that instantaneous rate, when will the amount of oil left in the field first be less than 22 percent of the original amount?
To find out when the amount of oil left in the field first becomes less than 22 percent of the original amount, we can set up a differential equation and solve it.
Let's suppose the original amount of oil in the field is denoted by "A" and the amount of oil left at any given time is denoted by "y." Since the rate at which oil can be extracted is 14% of the amount remaining per year, we can express this as:
dy/dx = -0.14y
Where dy/dx represents the rate of change of y with respect to x (time). This is a separable differential equation, which means we can separate the variables and solve for y.
First, let's divide both sides of the equation by y:
1/y dy = -0.14 dx
Next, let's integrate both sides of the equation:
∫1/y dy = ∫-0.14 dx
ln|y| = -0.14x + C
Here, C is the constant of integration. We can determine its value by using the initial condition that when x = 0 (at the start), y = A (original amount of oil):
ln|A| = -0.14(0) + C
ln|A| = C
Now we can substitute this value of C back into the equation:
ln|y| = -0.14x + ln|A|
To get rid of the absolute values, we can apply the exponentiation function on both sides:
y = e^(-0.14x + ln|A|)
y = e^(-0.14x) * e^ln|A|
y = A * e^(-0.14x)
We want to find the point where y is less than 22% of the original amount (0.22A). So:
0.22A = A * e^(-0.14x)
Dividing both sides by A:
0.22 = e^(-0.14x)
Now, we can solve for x by taking the natural logarithm (ln) of both sides:
ln(0.22) = -0.14x
x = ln(0.22) / -0.14
Using a calculator, we can find:
x ≈ 14.925
Therefore, the amount of oil left in the field will first be less than 22 percent of the original amount after approximately 14.925 years.
To solve this problem, we can use exponential decay to represent the decreasing amount of oil over time. Let's denote the original amount of oil in the field as A₀.
We know that the instantaneous rate at which oil can be extracted is 14% of the amount of oil remaining per year. This can be expressed mathematically as:
dA/dt = -0.14A
Where dA/dt represents the rate of change of the amount of oil with respect to time, and A represents the amount of oil remaining in the field.
To solve this differential equation, we can separate variables:
dA/A = -0.14 dt
Now, we can integrate both sides:
∫ (1/A) dA = -0.14 ∫ dt
ln(A) = -0.14t + C
Where C is the constant of integration.
Now, let's solve for C using the given information that at t=0, A equals the original amount of oil (A₀). We substitute these values:
ln(A₀) = -0.14(0) + C
ln(A₀) = C
Substituting this value back into our equation:
ln(A) = -0.14t + ln(A₀)
To find when the amount of oil left in the field is less than 22% (0.22) of the original amount (A₀), we can set A equal to 0.22A₀ and solve for t:
ln(0.22A₀) = -0.14t + ln(A₀)
-0.14t = ln(0.22A₀) - ln(A₀)
-0.14t = ln(0.22)
t = (ln(0.22)) / -0.14
Using a calculator or computational tool, we can find the value of t. Plugging in the value of ln(0.22) divided by -0.14 will give us the time at which the amount of oil left in the field first falls below 22% of the original amount.