The frequency F of a fire truck siren heard by a stationary observer is
F =
132,400
331土v
where v represents the velocity of the accelerating fire truck in meters per second (see figure). Find the rate of change of F with respect to v when

﻿﻿﻿﻿the fire truck is approaching at a velocity of 30 meters per second (use - v).

﻿﻿﻿﻿the fire truck is moving away at a velocity of 30 meters
per second (use + v).
F＝=
132,400
331 + v

To solve the problem, we'll differentiate the given frequency \( F \) with respect to the velocity \( v \) of the fire truck. The formula given for the frequency \( F \) is:

\[
F = \frac{132,400}{331 \pm v}
\]

Here, the sign \( \pm \) indicates two cases:

1. The fire truck is approaching, so use \( -v \) (approaching means a negative velocity).
2. The fire truck is moving away, so use \( +v \) (moving away means a positive velocity).

### 1. Fire Truck Approaching at 30 m/s

For the approaching case, the formula becomes:

\[
F = \frac{132,400}{331 - v}
\]

We need to find the rate of change of \( F \) with respect to \( v \), which means we need to find \( \frac{dF}{dv} \).

Differentiate \( F \) with respect to \( v \):

\[
\frac{dF}{dv} = \frac{d}{dv}\left(\frac{132,400}{331 - v}\right)
\]

Using the quotient rule or by recognizing that this is a simple derivative of a rational function, we get:

\[
\frac{dF}{dv} = 132,400 \times \frac{-1}{(331 - v)^2}
\]

Substitute \( v = 30 \) m/s:

\[
\frac{dF}{dv} = 132,400 \times \frac{-1}{(331 - 30)^2} = 132,400 \times \frac{-1}{301^2}
\]

\[
\frac{dF}{dv} = 132,400 \times \frac{-1}{90,601} \approx -1.461 \text{ Hz/m/s}
\]

So, the rate of change of \( F \) when the truck is approaching at 30 m/s is approximately \(-1.461\) Hz per meter per second.

### 2. Fire Truck Moving Away at 30 m/s

For the moving away case, the formula becomes:

\[
F = \frac{132,400}{331 + v}
\]

We again differentiate \( F \) with respect to \( v \):

\[
\frac{dF}{dv} = \frac{d}{dv}\left(\frac{132,400}{331 + v}\right)
\]

This gives:

\[
\frac{dF}{dv} = 132,400 \times \frac{-1}{(331 + v)^2}
\]

Substitute \( v = 30 \) m/s:

\[
\frac{dF}{dv} = 132,400 \times \frac{-1}{(331 + 30)^2} = 132,400 \times \frac{-1}{361^2}
\]

\[
\frac{dF}{dv} = 132,400 \times \frac{-1}{130,321} \approx -1.016 \text{ Hz/m/s}
\]

So, the rate of change of \( F \) when the truck is moving away at 30 m/s is approximately \(-1.016\) Hz per meter per second.

### Summary

- The rate of change of \( F \) when the truck is approaching at 30 m/s is approximately \(-1.461\) Hz/m/s.
- The rate of change of \( F \) when the truck is moving away at 30 m/s is approximately \(-1.016\) Hz/m/s.

Would you like more details on any part of the solution? Here are some related questions for further exploration:

1. How does the rate of change \( \frac{dF}{dv} \) behave as the truck's speed increases?
2. What happens to \( F \) when the fire truck is stationary (i.e., \( v = 0 \))?
3. Can you explain the significance of the Doppler effect in the context of this problem?
4. How would the formula change if the observer was also moving?
5. What if the truck were moving at the speed of sound? How would the calculation be different?

**Tip:** When dealing with rates of change, always check the units to ensure the calculations make physical sense.

Calculate the rate of change of the fire truck siren frequency when approaching or moving away at 30 m/s. Learn how to differentiate a rational function and apply it to real-world scenarios.

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