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Greeley [361]
2 years ago
5

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

n 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
Mathematics
1 answer:
JulijaS [17]2 years ago
7 0

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step 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:

\frac{dA}{dt} = kA

Where k represents the constant of proportionality

\frac{dA}{dt} = kA

Multiply both sides by dt/A

\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}

\frac{dA}{A}  = k\ dt

Integrate both sides

\int\ {\frac{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 = \frac{ln(0.5)}{6}

k = \frac{-0.693}{6}

k = -0.1155

Recall that:

\frac{dA}{dt} = kA

Where

\frac{dA}{dt} = Rate

So, when

A = 600000

The rate is:

\frac{dA}{dt} = -0.1155 * 600000

\frac{dA}{dt} = -69300

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

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