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.

Answer 2

The time it will take for all of the water to drain out of that barrel is 35.5 hours approx.

What is directly proportional and inversely proportional relationship?

Let there are two variables p and q

Then, p and q are said to be directly proportional to each other if

$p = kq$

where k is some constant number called constant of proportionality.

This directly proportional relationship between p and q is written as

$p \propto q$  where that middle sign is the sign of proportionality.

In a directly proportional relationship, increasing one variable will increase another.

Now let m and n are two variables.

Then m and n are said to be inversely proportional to each other if

$m = \dfrac{c}{n} \\\\ \text{or} \\\\ n = \dfrac{c}{m}$

(both are equal)

where c is a constant number called constant of proportionality.

This inversely proportional relationship is denoted by

$m \propto \dfrac{1}{n} \\\\ \text{or} \\\\n \propto \dfrac{1}{m}$

As visible, increasing one variable will decrease the other variable if both are inversely proportional.

For the considered case, let we take three variables as:

• t = time passed since start (in hours)
• $x$ = depth of water at time 't'
• $y$ = amount of water(in terms of depth) leaking per hour, in liters

Now, when  t = 0, x = 36 inches.

and at t = 1, x = 34 inches.

So depth of water is function of time passed. Let x = f(t)

Also, y is negative rate of change of x with respect to t(since depth is decreasing, and draining is measuring decrement rate, thus, negative of increment rate), or

$y = -\dfrac{dx}{dt}$

We're given 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"

That means, $y \propto \sqrt{x}$

Let the constant of proportionality be k, then,

$y = \sqrt{x}$

Since we've $y = -\dfrac{dx}{dt}$, therefore,

$-\dfrac{dx}{dt} = k\sqrt{x}\\\\-\dfrac{dx}{\sqrt{x}} = kdt\\\\\text{Integrating both the sides, we get}\\\\\int -\dfrac{dx}{\sqrt{x}} = \int kdt\\\\-2\sqrt{x} = kt + C$

where C is integration constant.

Since at t = 0 hours passed, x = 36 inches, and at t = 1 hour passed, x = 34 inches, we get two equations as:

$-2\sqrt{x} = kt + C\\-2\sqrt{36} = -12 = C\\-2\sqrt{34} = k + C$

Putting value of C from first equation in second, we get:

$k = -2\sqrt{34} + C \\k = -2\sqrt{34} - 12$

Therefore, the relationship between depth of water and time we get is:

$-2\sqrt{x} = (-2\sqrt{34} - 12)t -12\\\sqrt{x} = (\sqrt{34} - 6)t + 6$

When the whole barrel gets empty, the depth of water becomes 0. The time for it is calculated using above equation as:

$\sqrt{x} = (\sqrt{34} - 6)t + 6\\0 = (\sqrt{34} - 6)t + 6\\t = \dfrac{6}{6 - \sqrt{34}} \approx 35.5$(in hours)

Thus, the time it will take for all of the water to drain out of that barrel is 35.5 hours approx.

Learn more about differential equations here:

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