Answer:
The amount of oil was decreasing at 69300 barrels, yearly
Step-by-step explanation:
Given
[tex]Initial =1\ million[/tex]
[tex]6\ years\ later = 500,000[/tex]
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:
[tex]\frac{dA}{dt} = kA[/tex]
Where k represents the constant of proportionality
[tex]\frac{dA}{dt} = kA[/tex]
Multiply both sides by dt/A
[tex]\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}[/tex]
[tex]\frac{dA}{A} = k\ dt[/tex]
Integrate both sides
[tex]\int\ {\frac{dA}{A} = \int\ {k\ dt}[/tex]
[tex]ln\ A = kt + lnC[/tex]
Make A, the subject
[tex]A = Ce^{kt}[/tex]
[tex]t = 0\ when\ A =1\ million[/tex] i.e. At initial
So, we have:
[tex]A = Ce^{kt}[/tex]
[tex]1000000 = Ce^{k*0}[/tex]
[tex]1000000 = Ce^{0}[/tex]
[tex]1000000 = C*1[/tex]
[tex]1000000 = C[/tex]
[tex]C =1000000[/tex]
Substitute [tex]C =1000000[/tex] in [tex]A = Ce^{kt}[/tex]
[tex]A = 1000000e^{kt}[/tex]
To solve for k;
[tex]6\ years\ later = 500,000[/tex]
i.e.
[tex]t = 6\ A = 500000[/tex]
So:
[tex]500000= 1000000e^{k*6}[/tex]
Divide both sides by 1000000
[tex]0.5= e^{k*6}[/tex]
Take natural logarithm (ln) of both sides
[tex]ln(0.5) = ln(e^{k*6})[/tex]
[tex]ln(0.5) = k*6[/tex]
Solve for k
[tex]k = \frac{ln(0.5)}{6}[/tex]
[tex]k = \frac{-0.693}{6}[/tex]
[tex]k = -0.1155[/tex]
Recall that:
[tex]\frac{dA}{dt} = kA[/tex]
Where
[tex]\frac{dA}{dt}[/tex] = Rate
So, when
[tex]A = 600000[/tex]
The rate is:
[tex]\frac{dA}{dt} = -0.1155 * 600000[/tex]
[tex]\frac{dA}{dt} = -69300[/tex]
Hence, the amount of oil was decreasing at 69300 barrels, yearly
