4.3 Rates of Change in Applied Contexts other than Motion

TLDR

In AP Calculus, a derivative is just an instantaneous rate of change, and that idea works for way more than motion. If a function models volume, temperature, likes, or population over time, then its derivative gives how fast that quantity is changing, and the units come straight from the original function divided by the input variable.

Topic 4.3 sets you up for related rates in Topics 4.4 and 4.5, where the same skill of reading a context and interpreting a derivative gets more complex.

Derivatives Give Rates and Units in Any Context

• A derivative is an instantaneous rate of change with respect to the independent variable, no matter the context.
• The unit for $f'(x)$ is the unit of $f$ divided by the unit of $x$ (for example, liters per minute or likes per day).
• Phrases like "rate of change," "how fast," or "per minute" signal that you should differentiate.
• To find a rate at a specific moment, take the derivative and then evaluate it at that input value.
• Always interpret your final answer in context, using the right units and the right noun (not "velocity" when you mean likes per day).

Rates of Change in Applied Contexts Other Than Motion

To find the meaning of a rate of change or derivative in a non-motion context, focus on what the function models.

If a problem states that $f(x)$ gives the volume, in liters, of the water remaining in a tank $t$ minutes after the drain is opened, that means $f(x)$ models volume with respect to time in minutes.

For example, $f(15)$ gives the volume of water in the tank $15$ minutes after opening the drain. Since the derivative is the rate of change, $f'(x)$ gives the rate at which the volume is changing, in liters per minute. So if you are asked for the rate the volume is changing $15$ minutes after opening the drain, you find $f'(15)$.

Worked Example

Karen is pogo stick jumping. The following function gives her height above ground, in feet, $t$ seconds after jumping:

$H(t)=3\sin\left(\frac{t}{10}\right)+\frac{1}{2}$

What is the instantaneous rate of change of Karen's height after $10$ seconds?

Because the problem asks for the instantaneous rate of change at a specific point in time, take the derivative of the function and evaluate it at $t=10$. So you want $H'(10)$.

$H'(t)= \frac{3}{10}\cos\left(\frac{t}{10}\right)$

$H'(10)= \frac{3}{10}\cos(1)$

$\approx 0.162$

Since $H(t)$ models height above ground in feet with respect to time in seconds, the unit for the derivative is feet per second. The answer is about $0.162$ feet per second.

Turning Word Problems Into Rates With Units

Problem Solving:

1. Read carefully and identify what the function models and the units of both the output and the input.
2. Spot the trigger words. "Rate of change," "how fast," or any "per unit" phrasing means take a derivative.
3. Differentiate, then evaluate at the moment the problem asks about.
4. State your answer with the correct units, found by dividing the output unit by the input unit.
5. Interpret the result in context using the right noun for the quantity.

Common Trap: Using motion words like "velocity" or "speed" when the context is something else, such as likes per day or liters per minute. The structure is the same, but the language has to match the situation. Clear units and clear wording make your exam work easy to follow.

Practice Problems

Give these a try yourself.

Problems

Question 1

Thomas posted on Instagram, which rapidly gained likes over time. The following function gives the number of likes $t$ days after posting:

$L(t)=200\cdot e^{0.1t}$

What is the instantaneous rate of change of the number of likes $5$ days after the post was uploaded?

Question 2

Jen is filling up her car's gas tank. The following function gives the volume, in liters, of gas in the tank $t$ minutes after she starts pumping:

$G(t)=300+4t$

What is the instantaneous rate of change of the volume of gas $4$ minutes after she started filling up the tank?

Answers and Solutions

Question 1

The problem asks for the instantaneous rate of change at a specific point in time, so take the derivative and evaluate it at $t=5$ days.

$L'(t)= 200 \cdot 0.1 \cdot e^{0.1t}$

$L'(t)=20 \cdot e^{0.1t}$

$L'(5)= 20 \cdot e^{0.5}$

$\approx 32.97$

Since $L(t)$ models the number of likes with respect to time in days, the unit for the derivative is likes per day. The answer is about $32.97$ likes per day.

Question 2

The problem asks for the instantaneous rate of change at a specific point in time, so take the derivative and evaluate it at $t=4$ minutes.

$G'(t)=4$

$G'(4)=4$

Since $G(t)$ models the volume of gas in liters with respect to time in minutes, the unit for the derivative is liters per minute. The answer is $4$ liters per minute.

Common Misconceptions

• Misconception: Rates of change only apply to motion.
  Actually: The same derivative idea works for volume, temperature, population, concentration, and more.
• Misconception: You do not need to evaluate at the given moment.
  Actually: The derivative function gives a general rate, but you usually need its value at a specific input.
• Misconception: You can drop units or use the wrong ones.
  Actually: The unit of $f'(x)$ is always the output unit divided by the input unit.
• Misconception: The function value and its rate of change are the same.
  Actually: $f(15)$ tells you the amount at time $15$; $f'(15)$ tells you how fast it is changing at time $15$.
• Misconception: You should use motion vocabulary in non-motion contexts.
  Actually: Match your words to what the quantity actually represents.

Related AP Calculus Guides

• Unit 4 Review: Contextual Applications of Differentiation
• 4.1 Interpreting the Meaning of the Derivative in Context
• 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
• 4.4 Intro to Related Rates
• 4.6 Approximating Values of a Function Using Local Linearity and Linearization
• 4.5 Solving Related Rates Problems