contestada

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.

Respuesta :

MrRoyal

Answer:

127.53 liters left after 10 minutes

Step-by-step explanation:

Let

[tex]A \to Amount[/tex]

[tex]t \to time[/tex]

Given

[tex]A(0) = 200[/tex] --- initial

[tex]A(5) = 200 * (1 - 20\%) = 160[/tex] --- the amount left, after 5 minutes

Required

[tex]A(10)[/tex] --- amount left after 5 minutes

To do this, we make use of:

[tex]A(t) = A(0) * e^{kt}[/tex]

[tex]A(5) = 160[/tex] implies that:

[tex]160 = 200 * e^{k*5}[/tex]

Divide both sides by 200

[tex]0.80 = e^{k*5}[/tex]

Take natural logarithm of both sides

[tex]\ln(0.80) = \ln(e^{k*5})[/tex]

[tex]\ln(0.80) = \ln(e^{5k})[/tex]

[tex]\ln(0.80) = 5k\ln(e)[/tex]

So, we have:

[tex]-0.223 = 5k[/tex]

Divide by 5

[tex]k = -0.045[/tex]

So, the function is:

[tex]A(t) = A(0) * e^{kt}[/tex]

[tex]A(t) = 200 * e^{-0.045t}[/tex]

The amount after 10 minutes is:

[tex]A(10) = 200 * e^{-0.045*10}[/tex]

[tex]A(10) = 200 * e^{-0.45}[/tex]

[tex]A(10) = 127.53[/tex]

Otras preguntas