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 6% 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 13 percent of the original amount?

Question

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 6% 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 13 percent of the original amount?

Answer 1

I have no idea. Sorry :(

Answer 2

dp/dt = -.06p

dp/p = -.06 dt
lnp = -.06t
p = c e^(-.06t)
where c is the starting value.

Answer 3

To solve this problem, we need to use exponential decay since the rate at which the oil is being extracted is proportional to the amount of oil remaining.

Let's assume that the initial amount of oil in the field is O_o (which we don't know), and at any given time t, the amount of oil remaining is O(t). The problem states that the instantaneous rate of extraction is 6% of the oil remaining per year. This can be mathematically represented as:

dO(t)/dt = -0.06 * O(t)

Here, the negative sign indicates that the amount of oil is decreasing with time.

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

1/O(t) dO(t) = -0.06 dt

Integrating both sides:

∫(1/O(t)) dO(t) = ∫(-0.06) dt

ln|O(t)| = -0.06t + C

Where C is the constant of integration.

Using the initial condition that at t = 0, the amount of oil remaining is O_o, we can substitute these values into the equation:

ln|O_o| = 0 + C
C = ln|O_o|

So, the equation becomes:

ln|O(t)| = -0.06t + ln|O_o|

Now, we need to find the value of t when the amount of oil remaining is less than 13% of the original amount. This can be expressed as:

O(t) < 0.13 * O_o

Taking the natural logarithm of both sides:

ln|O(t)| < ln|0.13 * O_o|

Since ln(x) is a monotonically increasing function, we can drop the logarithm and simplify:

O(t) < 0.13 * O_o

Now, substitute this inequality into our previous equation:

-0.06t + ln|O_o| < ln|0.13 * O_o|

Simplifying further:

-0.06t < ln|0.13 * O_o| - ln|O_o|

-0.06t < ln|0.13|

Divide both sides by -0.06 (remember to change the inequality sign since we are dividing by negative value):

t > ln|0.13| / 0.06

Using a calculator or approximation methods, we find:

t > 24.86

Therefore, the amount of oil left in the field will first be less than 13% of the original amount after approximately 24.86 years.