Topic 4.3 Notes – Rates of Change in Applied Contexts Other Than Motion

What a derivative means in context

Suppose a function $f(t)$ models some quantity over time.

• $f(t)$ = the amount at time $t$
• $f'(t)$ = the instantaneous rate of change of that amount with respect to time

The derivative answers:

• How fast is the quantity changing right now?
• Is it increasing or decreasing?
• In what units?

Units always follow this structure

$$\text{Units of } f'(t) = \frac{\text{output units}}{\text{input units}}$$

Examples:

• If $P(t)$ is population (people) and $t$ is years
  → $P'(t)$ is people per year
• If $C(x)$ is cost (dollars) and $x$ is items produced
  → $C'(x)$ is dollars per item
• If $V(t)$ is volume (liters) and $t$ is minutes
  → $V'(t)$ is liters per minute

The structure never changes. The interpretation does.

What the sign tells you

• $f'(a) > 0$ → increasing at that moment
• $f'(a) < 0$ → decreasing
• $f'(a) = 0$ → momentarily not changing
• Larger magnitude → changing faster

On the AP exam, a number without units or interpretation will lose credit.

Finding an instantaneous rate of change

When a problem asks, “What is the instantaneous rate of change at ___?” that means take the derivative and evaluate it there.

Here’s the logic you should follow every time:

1. Identify what the function represents (and its units).
2. Recognize that instantaneous rate = derivative.
3. Compute $f'(x)$.
4. Plug in the requested value.
5. State your answer with units and meaning.

Quick example

Suppose a company’s profit (in thousands of dollars) is
$$P(x) = 5x^2 - 40x + 120$$ where $x$ is the number of products (in hundreds).

Differentiate:
$$P'(x) = 10x - 40$$

At $x = 6$:

$$P'(6) = 60 - 40 = 20$$

Interpretation:

At a production level of 600 items, profit is increasing at 20 thousand dollars per hundred items.

Notice how we translated the scaled units correctly. That kind of detail shows up in FRQs.

Common applied contexts

The calculus is the same every time. Only the story changes.

Marginal interpretation (important)

If $C(x)$ is cost, then $C'(x)$ approximates the additional cost of producing one more unit when production is at $x$.

On multiple choice, they love phrasing like:

• “Approximate cost of producing the 51st item.”

That means evaluate $C'(50)$, not $C'(51)$. Students miss that every year.

Derivatives from graphs

Sometimes they won’t give you a formula. You’ll see a graph.

For example, here’s a graph of $f(x) = x^2$ with tangent lines drawn at different points.

Tangent lines and derivative values for $f(x)=x^2$

What you’re reading from a graph like this:

• Slope of the tangent line = $f'(x)$
• Steeper tangent = larger magnitude derivative
• Horizontal tangent = derivative is 0
• Downward slope = negative derivative

Notice in the picture: slopes are negative for $x < 0$, zero at the minimum, and positive for $x > 0$. That sign change tells you how the function is behaving.

If the graph is curved but flattening out, the derivative is getting closer to zero.

You are always reading slope, not height.

Instantaneous vs average rate

Don’t mix these up.

Average rate on $[a,b]$:
$$\frac{f(b)-f(a)}{b-a}$$ This is slope of a secant line.

Instantaneous rate at $a$:
$$f'(a)$$ This is slope of the tangent line.

If the question says:

• “Between year 2 and 5” → average rate
• “At year 5” → derivative

That wording difference matters.

Common mistakes

• Plugging in before differentiating.
• Forgetting units.
• Ignoring the sign when interpreting.
• Evaluating the derivative at the wrong value in marginal problems.
• Describing what the function value means instead of what the derivative means.

If the question gives you $f'(3) = -8$, don’t talk about the amount at time 3. Talk about how it’s changing at time 3.