Question

Water is leaking out of a large barrel at a rate proportional to the square rooot of the depth of the water at that time. If the water level starts at 36 inches and drops to 34 inches in 1 hour. How long will it take for all of the water to drain out of that barrel?

Answer 1

Answer:

It will take about 35.49 hours for the water to leak out of the barrel.

Step-by-step explanation:

Let $y(t)$ be the depth of water in the barrel at time $t$,  where $y$ is measured in inches and $t$ in hours.

We know that water is leaking out of a large barrel at a rate proportional to the square root of the depth of the water at that time. We then have that

                                                 $\frac{dy}{dt}=-k\sqrt{y}$

where $k$ is a constant of proportionality.

Separation of variables is a common method for solving differential equations. To solve the above differential equation you must:

Multiply by $\frac{1}{\sqrt{y}}$

$\frac{1}{\sqrt{y}}\frac{dy}{dt}=-k$

Multiply by $dt$

$\frac{1}{\sqrt{y}}\cdot dy=-k\cdot dt$

Take integral

$\int \frac{1}{\sqrt{y}}\cdot dy=\int-k\cdot dt$

Integrate

$2\sqrt{y}=-kt+C$

Isolate $y$

$y(t)=(\frac{C}{2} -\frac{k}{2}t)^2$

We know that the water level starts at 36, this means $y(0)=36$. We use this information to find the value of $C$.

$36=(\frac{C}{2} -\frac{k}{2}(0))^2\\C=12$

$y(t)=(\frac{12}{2} -\frac{k}{2}t)^2\\\\y(t)=(6 -\frac{k}{2}t)^2$

At t = 1, y = 34

$34=(6 -\frac{k}{2}(1))^2\\k=12-2\sqrt{34}$

So our formula for the depth of water in the barrel is

$y(t)=(6 -\frac{12-2\sqrt{34}}{2}t)^2\\\\y(t)=\left(6-\left(6-\sqrt{34}\right)t\right)^2$

To find the time, $t$, at which all the water leaks out of the barrel, we solve the equation

$\left(6-\left(6-\sqrt{34}\right)t\right)^2=0\\\\t=3\left(6+\sqrt{34}\right)\approx 35.49$

Thus, it will take about 35.49 hours for the water to leak out of the barrel.