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 will first be less than 22% of the original amount, we can set up a differential equation and solve it.

Let's denote the amount of oil remaining at any given time as R(t), where t represents the number of years since production began.

According to the problem, the instantaneous rate at which oil can be extracted is 14% of the amount of oil remaining per year. This means that the rate of change of oil remaining, dR/dt, is equal to 14% of R(t).

We can express this relationship mathematically as:
dR/dt = -0.14R(t)

Now, we need to solve this differential equation to find R(t) when it reaches 22% of the original amount. To do this, we can separate variables:

1/R dR = -0.14 dt

Integrating both sides gives us:
ln|R| = -0.14t + C

Where C is the constant of integration.

Now, let's solve for R(t) when it becomes 22% of the original amount. Denote the original amount as R0, then we have:
R(t) = 0.22R0

Substituting this into the equation, we get:
ln|0.22R0| = -0.14t + C

To find the value of the constant, C, we can use the initial condition that at t=0, R(0) = R0:
ln|R0| = C

Substituting this into the equation, we have:
ln|0.22R0| = -0.14t + ln|R0|

Now, let's solve for t:
ln(0.22R0) - ln(R0) = -0.14t

Using the logarithm properties, we can simplify further:
ln(0.22) = -0.14t

Now, isolating t:
t = (ln(0.22)) / (-0.14)

Calculating the value:
t ≈ 8.741 years

Therefore, the amount of oil left in the field will first be less than 22% of the original amount after approximately 8.741 years of production.

To solve this problem, you can set up a differential equation to represent the rate of change of the oil remaining over time. Let's denote the amount of oil remaining at any point in time as y(t), where t is the time in years.

According to the given information, the instantaneous rate at which oil can be extracted is 14% of the amount of oil remaining per year. This can be written as:

dy/dt = -0.14y

The negative sign represents the fact that the amount of oil remaining decreases over time.

To solve this differential equation, we can separate variables and integrate both sides:

1/y dy = -0.14 dt

Integrating both sides gives:

ln|y| = -0.14t + C

where C is the constant of integration.

To find the value of the constant C, we can use the initial condition that at t = 0, the amount of oil remaining is 100% or 1. Substituting these values into the equation:

ln|1| = -0.14(0) + C
0 = C

So C = 0, and the equation becomes:

ln|y| = -0.14t

Next, we can solve for y when it becomes less than 22% or 0.22 of the original amount:

ln|y| = -0.14t
ln|0.22| = -0.14t

Using logarithm properties, we can rewrite this as:

-1.514 = -0.14t

Now, divide both sides by -0.14 to solve for t:

t = -1.514 / (-0.14)

t ≈ 10.815 years

Therefore, the amount of oil left in the field will be less than 22% of the original amount after approximately 10.815 years.