Question

A 100-gallon barrel, initially half-full of oil, develops a leak at the bottom. Let A(t) be the amount of oil in the barrel at time t. Suppose that the amount of oil is decreasing at a rate proportional to the product of the time elapsed and the amount of oil present in the barrel. The mathematical model is:___________.
(a) =kA, A(0) = 0
(b) A = ktA, A(O) = 50
(c) A tA, A(0) = 100
(d) A = két +A), A
(e) = 50 None of the above.

Answer 1

Answer:

the mathematical model is : $-\frac{1}{A}= kt - \frac{1}{50}$

Step-by-step explanation:

Given that:

Let A(t) to be the amount of oil in the barrel at time t.

However; Suppose that the amount of oil is decreasing at a rate proportional to the product of the time elapsed and the amount of oil present in the barrel.

Then,

$\frac{dA}{dt} \ \alpha \ A^2$

$\frac{dA}{dt}= KA^2$

Initially the 100 -gallon barrel is half-full of oil

So, A(0) = 100/2 = 50

$\frac{dA}{dt}= KA^2 \ \ \ \ \ \ :A(0)=50$

The variable is now being separated as:

$\frac{dA}{A^2}=kdl$

Taking integral of both sides; we have:

$\int\limits\frac{dA}{A^2}=\int\limits \ kdt$

$-\frac{1}{A}= kt +C$

However; since A(0) = 50; Then

t = 0  ; A =50 in the above equation

$-\frac{1}{50}= 0 +C$

$C = - \frac{1}{50}$

Thus; the mathematical model is : $-\frac{1}{A}= kt - \frac{1}{50}$