#AP Pre-Calculus Study Guide: Rates of Change

Hey there, future AP Pre-Calc master! Let's dive into rates of change. This is a crucial topic, and we're going to make sure you're totally comfortable with it. Think of this as your go-to guide for a quick review before the big exam. Let's get started! 🚀

#1.2 Rates of Change

# 📐 Average Rates of Change: Speaking in Ratios

The average rate of change measures how much a function's output changes over a specific interval. It's all about ratios! Think of it as the slope of a line connecting two points on the function's graph. 📏

Key Concept

It's calculated as the change in output (y-values) divided by the change in input (x-values) over an interval [a, b]:

$\frac{f(b) - f(a)}{b - a}$

In simpler terms, it tells you how fast or slow a function is changing over a period.

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# 📈 Trends in Change

The rate of change at a specific point tells you how quickly the function's output is changing at that exact spot. It's like measuring the steepness of the curve at that point. 😁

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Image Courtesy of CK-12

To compare rates of change at different points, we use the average rate of change over small intervals around those points. This helps us understand where the function is changing more rapidly. 🤓

# ↕️ Positive and Negative Rates of Change

Quick Fact

Rates of change show how two quantities vary together. They can be positive or negative. 🔄

Positive Rate of Change: Both quantities increase or decrease together. Think of a car accelerating – as time increases, so does speed.
Negative Rate of Change: As one quantity increases, the other decreases. Like an object falling – as time increases, height decreases.

Common Mistake

Don't confuse the sign of the rate of change with the value of the function. A negative rate of change doesn't mean the function is negative, just that it's decreasing!

Exam Tip

Remember that the average rate of change is the slope of the secant line, while the instantaneous rate of change (which you'll see more in calculus) is the slope of the tangent line. This is a key concept for later topics.

Practice Question

Multiple Choice Questions

The function $f(x)$ is given by $f(x) = 2x^3 - 3x^2 + 5$ . What is the average rate of change of $f(x)$ over the interval $[-1, 2]$ ? (A) -1 (B) 1 (C) 3 (D) 6
A particle moves along a straight line such that its position at time $t$ is given by $s(t) = t^2 - 4t + 3$ . What is the average velocity of the particle over the time interval $[1, 3]$ ? (A) -2 (B) -1 (C) 0 (D) 1

Free Response Question

The temperature of a room, in degrees Fahrenheit, is modeled by the function $T(t) = 70 + 10\sin(\frac{\pi t}{12})$ , where $t$ is the time in hours since midnight.

(a) Find the average rate of change of the temperature between $t = 2$ and $t = 8$ hours. Show the work that leads to your answer. (2 points)

(b) At what time(s) in the interval $[0, 12]$ is the temperature increasing at a rate of 0 degrees per hour? (2 points)

(c) Estimate the instantaneous rate of change of the temperature at $t=6$ hours using the average rate of change over the interval $[5.9, 6.1]$ . (2 points)

Answer Key

Multiple Choice Answers

(C) 3
(B) -1

Free Response Question Scoring

(a) Average rate of change = $\frac{T(8) - T(2)}{8 - 2}$ (1 point) $\frac{(70 + 10\sin(\frac{8\pi}{12})) - (70 + 10\sin(\frac{2\pi}{12}))}{6} = \frac{10\sin(\frac{2\pi}{3}) - 10\sin(\frac{\pi}{6})}{6} = \frac{10(\frac{\sqrt{3}}{2}) - 10(\frac{1}{2})}{6} = \frac{5\sqrt{3} - 5}{6} \approx 0.60$ degrees per hour (1 point)

(b) The temperature is increasing at a rate of 0 degrees per hour when the derivative is zero. $T'(t) = 10\cos(\frac{\pi t}{12}) * \frac{\pi}{12} = 0$ $Rightarrow \cos(\frac{\pi t}{12}) = 0$ (1 point) $\frac{\pi t}{12} = \frac{\pi}{2}, \frac{3\pi}{2}$ $\Rightarrow t = 6, 18$ . In the interval $[0, 12]$ , the temperature is increasing at a rate of 0 degrees per hour at $t=6$ hours. (1 point)

(c) Average rate of change = $\frac{T(6.1) - T(5.9)}{6.1 - 5.9}$ (1 point) $\frac{(70 + 10\sin(\frac{6.1\pi}{12})) - (70 + 10\sin(\frac{5.9\pi}{12}))}{0.2} \approx \frac{10\sin(1.592) - 10\sin(1.540)}{0.2} \approx \frac{9.998 - 9.998}{0.2} \approx 0.00$ degrees per hour (1 point)

#Final Exam Focus

High-Priority Topics: Average rate of change calculations, understanding positive and negative rates of change, and interpreting rates of change from graphs.
Common Question Types: Multiple-choice questions testing your calculation of average rate of change, free-response questions that require you to interpret rates of change in a real-world context, and questions that combine rate of change with other concepts like function analysis.
Last-Minute Tips:
- Time Management: Don't spend too long on a single question. If you get stuck, move on and come back later.
- Common Pitfalls: Pay close attention to units, and remember that a negative rate of change indicates a decreasing function, not necessarily a negative function.
- Strategies: Always show your work, even for multiple-choice questions. This can help you catch errors and earn partial credit on free-response questions.