# Sound Wave Equation Deduction

1. Oct 8, 2014

### PeSoberbo

I am studying the sound wave equation deducted by Feynman in his lectures. In section 47-3:

P0 + Pe = f(d0 + de) = f(d0) + de f'(d0)

Where f'(d0) stands for the derivative of f(d) evaluated at d=d0. Also, de is very small.

I do not understand the second step of the equality. Can anyone help me?
Link to the lectures: http://www.feynmanlectures.caltech.edu/I_47.html#Ch47-S1

2. Oct 8, 2014

### Staff: Mentor

The last expression in your equation is only an approximation to the second expression, so '=' should really be $\approx$. The approximation is called a linear approximation that uses a value on the tangent line at (d0, f(d0)) instead of the value at (d0 + de, f(d0 + de)).

The underlying concept is this:
$$\frac{f(d_0 + d_e) - f(d_0)}{d_e} \approx f'(d_0)$$
Multiply both sides by de and then add f(d0 to both sides to get the relationship you show.

3. Oct 13, 2014

### PeSoberbo

Thank you Mark!