# How to solve this problem using diff equations?

1. Feb 28, 2013

### supermiedos

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

When a light ray goes trough a transparent surface, its intensity rate of change decreases proportionaly to I(t), where t is the thickness. On a specific kind of water, intensity 3 feet from below the surface is 25% its original intensity. What is the intensity from 15 ft below the surface? (sorry for my english lol)

2. Relevant equations

dI/dt = kI, I(3) = 0.25I_0

3. The attempt at a solution
My attempt is here: http://img69.imageshack.us/img69/4237/zill2.png [Broken]

The correct answer is I = 0.0009765625 I_0, but I don't know how to get there!!! Any hint please?

Last edited by a moderator: May 6, 2017
2. Feb 28, 2013

### pasmith

You know that $I(t) = I_0 e^{-kt}$ for some $k > 0$. The first step is to find $k$ from the condition $I(3) = 0.25I_0$, so that
$$I_0 e^{-3k} = 0.25 I_0$$
Then you can find $I(15) = I_0 e^{-15k}$.

Last edited by a moderator: May 6, 2017
3. Feb 28, 2013

### supermiedos

Ah of course!!! Because I(0) = I_0, and thus, c = I_0. The rest is easy. Thank you!

4. Mar 2, 2013

### Coelum

Supermiedos,
there is a much simpler approach: since the light intensity decreases by a factor of 4 every time it goes through a layer of 3 meters of water, the light intensity after 5 layers is just $(1/4)^5=1/1024=0.0009765625$. No need for an ODE

5. Mar 3, 2013

### supermiedos

Hehe, that's actually right. Thank you :D