# A rate at which water flows out of a tank is equal to ` 3V^2 m^3/s` where V is the water left in the tank. How long does 300 m^3 take to flow out.

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### 1 Answer

Here you have an equation, where you're relating the flow rate to the volume contained inside the tank. We need to figure out how to set the equation up first.

`V` is clearly our volume, and our flow rate (`F`) is given as:

`F = 3V^2`

However, if you look at the units, you'll notice that the flow rate is the rate of change of the volume over time. In other words:

`F = (dV)/(dt)`

Well, this gives us a new equation, then! Keep in mind, since flow is out of the tank, flow rate will be negative.

`(dV)/(dt) = -3V^2`

Now, this is a first order nonlinear differential equation. However, unlike many second-order nonlinear D.E.'s, we can solve this one!

We'll start by dividing both sides by `V^2`:

`1/V^2 (dV)/(dt) =-3`

Now, we'll multiply by `dt` on both sides:

`(dV)/V^2 =- 3dt`

What have we done? we separated the variables! This means we can now just integrate both sides in order to find the function V with respect to t. Let's integrate:

`int(dV)/V^2 = int-3dt`

Now, if you remember your integrals for rational functions, the left side will become `-1/V + C_1` and the right side will become `-3t+C_2`, where `C_1` and `C_2` are just constants of integration.

Let's see where we're at, then, in our equation:

`-1/V + C_1 = -3t+C_2`

First, just to simplify things, let's just subtract `C_1` from both sides, and recognize that `C_2` as a random constant can have anything done to it and have it remain a random constant. This means we can just replace `C_2` (after subtracting `C_1`) with `C`:

`-1/V =-3t+C`

Now, we're ready to solve for V(t). Let's start by multiplying by -1:

`1/V = 3t+C`

Remember, C didn't become negative because it's just some random constant! Multiplying a random constant by -1 still gives you a random constant, so we're going to leave C alone in this case.

Now, we take the multiplicative inverse of both sides (we flip both sides):

`V =1/(3t+C)`

**And there you have it. We have solved for V(t).**

Now, we can figure out what C is supposed to be! We know that at time t = 0, V = 300 `m^3`. We can then enter these values into our equation:

`300 = 1/(3*0+C)`

Well, we can clearly get rid of that 3*0:

`300 = 1/C`

Well, now we can just flip both sides:

`1/300 = C`

And there you have it! We solved for our constant of integration, and, so, **we have our full V(t)**:

`V(t) = 1/(3t+1/300)`

Now, to solve for when V = 0 (when all water has flowed out of the tank).

`0 = 1/(3t+1/300)`

Well, I can already tell you this is going to have a problem. You can never bring the fraction on the right to 0 because it has a nonzero numerator! You can't flip (take the multiplicative inverse of) the fraction because 0 has no multiplicative inverse. You can't multiply both sides by the denominator because then the result is 0 = 1!

There is only one way we get that fraction to be 0. Here's how:

`lim_(t->oo) 1/(3t+1/300) = 0`

So, in order for V(t) to be zero, we must take the limit as t approaches infinity. Therefore, V(t) will never be zero, and the **tank will never be fully-drained!**

I hope that helps!

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