# Homework Help: Integration question

1. May 12, 2006

### Luminous Blob

I'm trying to follow a derivation in a book on marine acoustics for finding the amplitude along a sound ray, as a function of the arc length, s, of the ray.

The book gives the following equation:

$$2\frac{dA_0}{ds}+ \left[ \frac{c}{J}\frac{d}{ds}\left(\frac{J}{c}\right) \right]A_0 = 0$$

The book then says that by integrating the above equation, we get:

$$A_0(s)= A_0(0)\left| \frac{c(s)J(0)}{c(0)J(s)} \right|^{1/2}$$

The book doesn't give any intermediate steps, and I'm not really sure how the integration is actually done. I gather that the limits are from 0 to s, but I don't know how you deal with something like this where J, c and A_0 all seem to depend on s.

In these equations, c is the sound speed at a given arc length s along the ray, and J is the Jacobian determinant, given as:

$$J = r\left[\frac{dr}{ds}\frac{dz}{d\theta} - \frac{dz}{ds}\frac{dr}{d\theta}\right]$$

or alternatively:

$$J = r\left[\left(\frac{dz}{d\theta}\right)^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{1/2}$$

So...can anyone explain to me the steps involved in going from the first equation to the second? Any help would be appreciated.

Last edited: May 12, 2006
2. May 12, 2006

### J77

You can get it in the form...

$$\frac{2}{A_0}\frac{dA_0}{ds}+\frac{1}{J}\frac{dJ}{ds}+\frac{1}{C}\frac{dC}{ds}=0$$

where $$C(s)=1/c(s)$$

3. May 12, 2006

### Curious3141

As a follow up to J77's excellent hint, think of chain rule (this is like reversing chain rule).

4. May 12, 2006

### Luminous Blob

Aah, thanks guys! I see what's happening now.