# 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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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!