# Rate of flow Question

How would you solve a general problem of a steady stream of leakage:

Water leaks at a rate

$$r(t)= 20 \sqrt{3} \sec^2 (2t) \, \frac{\text{gallons}}{\text{hour}}.$$

At time 0, there are 50 gallons of water.

So how should I find a function that represents the amount remaining at a certain time, say $\pi/6$ hours?

I understand to take the integral of $r(t)$, but how would you proceed then?

As a follow up question, how long will it take the sludge to leak completely? I got an answer of 2x = infinity...

• Hi! Welcome to MSE. I've editted your post for clarity without (hopefully) altering meaning. Best wishes. – Mark Fantini Feb 3 '14 at 0:32
• Thanks. This makes it better. – user235059 Feb 3 '14 at 3:40
• You are welcome. :) – Mark Fantini Feb 3 '14 at 9:29
• As a follow up question, how long will it take the sludge to leak completely? I got an answer of 2x = infinity... – user235059 Feb 5 '14 at 17:55

Hint: You are given the if $W(t)$ is the amount of water in the tank, your function $r(t)$ is $W'(t)$. So if you integrate $$\int W'(t) dt=\int r(t) dt$$ that will give you $W(t)$, the amount of water at time $t$. Use the fact that $W(0)=50$. Then the question is just asking what is $W(\frac{\pi}{6})$.

• Not quite: since $\ r(t) \$ is a volume rate of change, the integral represents the amount of water added to or lost from the volume in the tank. This is why we must include an "initial condition" for the actual amount of water in the tank in order to obtain a specific function for the tank's contents. – colormegone Feb 3 '14 at 0:42
• That's why I said to include $W(0)=50$! – Kyle L Feb 4 '14 at 7:23
• The rate function is a way of writing a differential equation, $\ \frac{dV}{dt} \ = \ r(t) \ .$ This does not give us exclusively a function for the rate of water volume change in the tank, but only the rate at which water is flowing. What I was taking issue with is that the integral can also describe the amount of water accumulating elsewhere, such as another tank, the outside environment, etc. Since the rate function given by OP is described as leakage, you need to write your integral as $\int - r(t) \ dt \$ in order to produce the function you call $\ W(t) \ .$ – colormegone Feb 4 '14 at 8:29

The "Net Change Theorem" tells us that the volume of water lost from the container is

$$\Delta V (T) \ = \ \int_0^{T} \ - r(t) \ \ dt \ ,$$

the negative sign being inserted since we are told this is "leakage", and thus a reduction of water volume in the container. What remains in the container at time $\ T \$ is then

$$V(T) \ = \ 50 \ + \ \Delta V(T) \ .$$

[We add the change in volume, which is a negative change for this problem.]