# Homework Help: Modeling Growth and Decay Problem

1. Apr 1, 2014

### Burjam

1. The problem statement, all variables and given/known data

Suppose water leaks out of a barrel at a rate proportional to the square root of the depth of the water. If the level starts at 36 in. and drops to 35 in. after 1 hour, how long will it take for all the water to leak out of the barrel?

I have to choose and implement a differential equation solution method to determine a solution.

2. Relevant equations

N/A

3. The attempt at a solution

dL/dt = -√(L), where L is the water level and t is time in hours
∫dL/-√(L) = ∫1dt
-2√(L) = t + c
L = ((t+c)/2)^2
36 = (-c/2)^2
c = 12

So now that I've solved for the constant using the initial condition, I'm not sure what to do. I think I might've even set up this problem wrong to begin with because when I substitute 1 in for t, I don't get 35, which is what L should equal at that t.

Help.

2. Apr 1, 2014

### haruspex

It says proportional to, not equal to.

3. Apr 1, 2014

### Burjam

I'm not sure how to translate that to an equation. Proportional is a tricky word in this problem.

4. Apr 1, 2014

### Staff: Mentor

If, for example, x is proportional to y, then the ratio of these quantities is a constant. IOW, x/y = k.

Equivalently, x = ky, where k is the constant of proportionality.

5. Apr 1, 2014

### haruspex

To expand on that, this gives you another constant that needs to be adjusted to fit the data. This should resolve your problem of getting the wrong answer when you set t = 1.

6. Apr 2, 2014

### Burjam

Ok so (dp/dt)/20-p = k

then dp/dt = k(20-p)

The problem is there are no conditions I can use in this problem that allow me to solve for k. I have conditions for p, but none for dp/dt.

7. Apr 2, 2014

### haruspex

p? 20?

Your ODE was dL/dt = -√(L). Inserting the missing constant of proportionality we have dL/dt = -k√(L). Solve that.

8. Apr 2, 2014

### Burjam

I was thinking of the other problem I posted. I think this is the same problem with that one too.

9. Apr 2, 2014

### Burjam

Ok so solving for k.

The water level dropped 1 inch in an hour at L=35 so

1 = -k√35
k=-1/√35

Does that look right?

10. Apr 2, 2014

### haruspex

It might be near enough, but it's not exactly right. The rate is constantly changing. That 1 inch in an hour is the average rate over the first hour. It will be the rate at some L value between 35 and 36, but it won't be the rate at L=35.
Correct is to solve the integral first then plug in the numbers.