Question

A 100-liter tank initially full of water develops a leak at the bottom. Given that 10% of the water leaks out in the first 5 minutes, find the amount of water left in the tank 20 minutes after the leak develops if the water drains off at a rate that is proportional to the amount of water present.

Answer 1

Answer:

$A(t) = A_o e^{\frac{ln(0.9)}{5} t}$

And we want to find the amount remaining after t = 20 minutes so we just need to replace into the model and we got:

$A(20) = 100e^{\frac{ln(0.9)}{5} *20}= 65.61 L$

Step-by-step explanation:

For this case if we assume the proportional model we have the following differential equation:

$\frac{dA}{dt}= kA$

Where A represent the amount in the tank, t the time and k the proportional constant.

We can reorder the differential equation like this:

$\frac{dA}{A}= kdt$

If we integrate both sides we got:

$ln |A|= kt + C$

Using exponential on both sides we got:

$A = e^{kt} e^c = A_o e^{kt}$

And for this case  we have the following conditions given:

$A(0) = 100 L$

$A(5) = 100-10 = 90 L$

Using the first condition we have that:

$100 = A_o e^{0}, A_o = 100$

Using the second condition we have:

$90 =100e^{5k}$

And we can solve for k like this:

$ln (\frac{90}{100}) = 5k$

$k = \frac{ln(0.9)}{5}= -0.02107210313$

So then our model would be given by:

$A(t) = A_o e^{\frac{ln(0.9)}{5} t}$

And we want to find the amount remaining after t = 20 minutes so we just need to replace into the model and we got:

$A(20) = 100e^{\frac{ln(0.9)}{5} *20}= 65.61 L$