Question

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were million barrels of oil in the well; six years later barrels remain.At what rate was the amount of oil in the well decreasing when there were barrels remaining

Answer 1

The amount of oil was decreasing at 69300 barrels, yearly

Explanation:

Given

`Initial =1\ million`

`6\ years\ later = 500,000`

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

`(dA)/(dt) = kA`

Where k represents the constant of proportionality

`(dA)/(dt) = kA`

Multiply both sides by dt/A

`(dA)/(dt) * (dt)/(A) = kA * (dt)/(A)`

`(dA)/(A) = k\ dt`

Integrate both sides

`\int\ {(dA)/(A) = \int\ {k\ dt}`

`ln\ A = kt + lnC`

Make A, the subject

`A = Ce^(kt)`

`t = 0\ when\ A =1\ million` i.e. At initial

So, we have:

`A = Ce^(kt)`

`1000000 = Ce^(k*0)`

`1000000 = Ce^(0)`

`1000000 = C*1`

`1000000 = C`

`C =1000000`

Substitute
`C =1000000` in
`A = Ce^(kt)`

`A = 1000000e^(kt)`

To solve for k;

`6\ years\ later = 500,000`

i.e.

`t = 6\ A = 500000`

So:

`500000= 1000000e^(k*6)`

Divide both sides by 1000000

`0.5= e^(k*6)`

Take natural logarithm (ln) of both sides

`ln(0.5) = ln(e^(k*6))`

`ln(0.5) = k*6`

Solve for k

`k = (ln(0.5))/(6)`

`k = (-0.693)/(6)`

`k = -0.1155`

Recall that:

`(dA)/(dt) = kA`

Where

`(dA)/(dt)` = Rate

So, when

`A = 600000`

The rate is:

`(dA)/(dt) = -0.1155 * 600000`

`(dA)/(dt) = -69300`

Hence, the amount of oil was decreasing at 69300 barrels, yearly