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?

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 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?

Answer 1

.86=e^-k

k=-ln.86

.78=e^-kt
t= -ln.78 /k= ln.78 / ln.88

Answer 2

^ Answer above is not working

Answer 3

sine the rate of decrease is 14% of the amount present,

dp/dt = -.14p
dp/p = -.14 dt
lnp = -.14t + c
p = c e^-.14t

here, c is the original amount

when is only .22 left?

e^-.14t = .22
-.14t = ln.22
t = ln.22/-.14 = 10.8 years

Answer 4

To solve this problem, we'll use differential equations. Let's denote "t" as the number of years that have passed, and "Q(t)" as the amount of oil in the field at time "t", as a fraction of the original amount.

Given that the rate of oil extraction at any time is 14% of the amount of oil remaining, we can write this relationship as a differential equation:

dQ/dt = -0.14 * Q(t)

This equation tells us that the rate of change of the amount of oil in the field (dQ/dt) is proportional to the amount of oil remaining (Q(t)), with a proportionality constant of -0.14.

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

∫ (1/Q) dQ = ∫ (-0.14) dt

This simplifies to:

ln|Q| = -0.14t + C

Where "C" is the constant of integration. We can find the value of "C" by using the initial condition that at time t=0, the amount of oil left is the original amount, which is 1. Therefore:

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

Now, we can rewrite the equation as:

ln|Q| = -0.14t

To find the time when the amount of oil left is less than 22% of the original amount, we can substitute Q(t) with 0.22 (22% as a decimal):

ln|0.22| = -0.14t

Using the properties of logarithms, we can rewrite this equation as:

-0.14t = ln(0.22)

Finally, we can solve for "t" by dividing both sides by -0.14:

t = ln(0.22) / -0.14

Using a calculator, we find that t ≈ 5.606 years.

Therefore, the amount of oil left in the field will first be less than 22% of the original amount approximately 5.606 years after production begins.