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The function provided in the problem is:

\[
P(t) = -20 \cos(300\pi t) + 100
\]

where \( P(t) \) represents the blood pressure in millimeters of mercury (mm Hg) at time \( t \) in seconds.

We are asked to find the rate of change of blood pressure at \( t = 3 \) seconds. To do this, we need to calculate the derivative of \( P(t) \), which will give us the rate of change of blood pressure with respect to time, denoted as \( \frac{dP}{dt} \).

### Step 1: Differentiate \( P(t) \)

We will differentiate \( P(t) = -20 \cos(300\pi t) + 100 \) with respect to \( t \):

\[
\frac{dP}{dt} = \frac{d}{dt} \left[ -20 \cos(300\pi t) + 100 \right]
\]

The constant 100 disappears after differentiation, so we focus on differentiating \( -20 \cos(300\pi t) \).

Using the chain rule, the derivative of \( \cos(k t) \) is \( -\sin(k t) \cdot k \), where \( k = 300\pi \) in this case:

\[
\frac{d}{dt} \left[ -20 \cos(300\pi t) \right] = -20 \cdot (-\sin(300\pi t)) \cdot (300\pi)
\]

Simplifying this gives:

\[
\frac{dP}{dt} = 6000\pi \sin(300\pi t)
\]

### Step 2: Evaluate the derivative at \( t = 3 \)

Now, we substitute \( t = 3 \) into the derivative:

\[
\frac{dP}{dt} \bigg|_{t=3} = 6000\pi \sin(300\pi \cdot 3)
\]

Simplifying the argument of the sine function:

\[
300\pi \cdot 3 = 900\pi
\]

Since \( \sin(900\pi) = 0 \) (because sine of any multiple of \( \pi \) is zero), we have:

\[
\frac{dP}{dt} \bigg|_{t=3} = 6000\pi \cdot 0 = 0
\]

### Conclusion:

The rate of change of blood pressure at \( t = 3 \) seconds is:

\[
\boxed{0 \, \text{mm Hg/s}}
\]

This means that at \( t = 3 \) seconds, the blood pressure is not changing.

Would you like further details or have any questions?

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Here are some related questions to expand your understanding:

1. What does the value of the derivative at a particular point represent?
2. How would the graph of \( P(t) \) look over time based on this equation?
3. What is the maximum and minimum blood pressure according to this model?
4. How would the rate of change differ if the frequency in the cosine function was altered?
5. Why is the rate of change zero at this specific time \( t = 3 \)?

**Tip**: When working with trigonometric functions and their derivatives, always remember to apply the chain rule correctly, especially when the argument involves a constant multiplied by \( t \).

For a person at rest, the function P(t) = -20 cos(300πt) + 100 models blood pressure, in millimeters of mercury (mm Hg), at time t seconds. What is the rate of change in blood pressure at 3 seconds?

Learn how to find the rate of change in blood pressure using the derivative of P(t) = -20 cos(300πt) + 100. This problem covers key concepts such as differentiation, trigonometric functions, and the chain rule. Suitable for high school students.

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