Question

A 200-liter tank initially full of water develops a leak at the bottom. Given that 20% of the water leaks out in the first 5 minutes, find the amount of water left in the tank 10 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:

127.53 liters left after 10 minutes

Step-by-step explanation:

Let

$A \to Amount$

$t \to time$

Given

$A(0) = 200$ --- initial

$A(5) = 200 * (1 - 20\%) = 160$ --- the amount left, after 5 minutes

Required

$A(10)$ --- amount left after 5 minutes

To do this, we make use of:

$A(t) = A(0) * e^{kt}$

$A(5) = 160$ implies that:

$160 = 200 * e^{k*5}$

Divide both sides by 200

$0.80 = e^{k*5}$

Take natural logarithm of both sides

$\ln(0.80) = \ln(e^{k*5})$

$\ln(0.80) = \ln(e^{5k})$

$\ln(0.80) = 5k\ln(e)$

So, we have:

$-0.223 = 5k$

Divide by 5

$k = -0.045$

So, the function is:

$A(t) = A(0) * e^{kt}$

$A(t) = 200 * e^{-0.045t}$

The amount after 10 minutes is:

$A(10) = 200 * e^{-0.045*10}$

$A(10) = 200 * e^{-0.45}$

$A(10) = 127.53$