Question

Suppose water is added to a tank at 10 gal / min , but leakes out at the rate of 0.2 gal / min for each gallon in the tank . What is the smllest capacity the tank can have if the pocess is to continue indefinitely ?

Answer 1

Answer:

50 gallon

Step-by-step explanation:

We are given that

$\frac{dV_{in}}{dt}=10 gal/min$

Let V(in gallon) be the present volume of tank.

$\frac{dV_{out}}{dt}=0.2 V gal/min$

We have to find the smallest capacity  the tank can have if the process is to continue indefinitely.

According to question

$\frac{dV}{dt}$=Incoming volume-outgoing volume

$\frac{dV}{dt}=10-0.2 V=-(\frac{1}{5}V-10)=-(\frac{V-50}{5})$

$\frac{5dV}{V-50}=-dt$

Integrating on both sides

$\int \frac{5dV}{V-50}=-\int dt$

$5ln(V-50)=-t+ln C$

By using the formula

$\int \frac{dx}{x}=ln x+C$

$ln(V-50)=\frac{1}{5}(-t+ln C)$

$ln(V-50)=-\frac{1}{5}t+\frac{1}{5} ln C$

$ln(V-50)=-\frac{1}{5} t+lnC^{\frac{1}{5}}$

$ln(V-50)=-\frac{1}{5}t+ln C'$

Where $ln C'=ln C^{\frac{1}{5}}$

$ln(V-50)-ln C'=-\frac{1}{5} t$

$ln\frac{V-50}{C'}=-\frac{1}{5}$

Using the formula

$ln m-ln n=ln\frac{m}{n}$

$\frac{V-50}{C'}=e^{-\frac{1}{5}t}$

$V-50=C'e^{-\frac{1}{5}t}$

Substitute $t=\infty$

$V-50=0$

$V=50$

Hence, the smallest capacity the tank can have if the process is to continue indefinitely=50 gallon