Navigation: Wiki Home > Skills > Average vs. Instantaneous Rate of Change
When someone says "the temperature is changing at 3 degrees per hour," do they mean an average over the whole day, or the rate right now? The answer matters. The distinction between average and instantaneous rates of change is at the heart of what derivatives measure.
Think about driving: your trip computer shows "45 mph average" but your speedometer shows "52 mph right now." Both describe how fast you're going, but they answer different questions. The average tells you about the whole journey; the instantaneous tells you about this exact moment.
Calculus gives us a way to compute the instantaneous rate from the average, by taking limits.
| Property | Value |
|---|---|
| Concept | Rates of Change |
| Chapter | 2.7 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
| Type | Formula | What It Measures |
|---|---|---|
| Average rate | $\displaystyle \frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$ | Change over an interval |
| Instantaneous rate | $\displaystyle \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ | Change at a single point |
If $y = f(x)$, then the average rate of change of $y$ with respect to $x$ over the interval $[x_1, x_2]$ is:
$$\boxed{\text{Average rate} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{\Delta y}{\Delta x}}$$
The instantaneous rate of change at $x = x_1$ is:
$$\boxed{\text{Instantaneous rate} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx}}$$
This is the derivative, and it equals the slope of the tangent line at $x_1$.
y
│
│ ·Q
│ ·╱
│ ·╱
│ ·P╱
│ · ╱ curve y = f(x)
│ ╱
│────────────────── x
x₁ x₂
Secant line PQ: slope = average rate of change
As Q → P: secant line → tangent line
average rate → instantaneous rateThe average rate of change is the slope of the secant line between two points.
The instantaneous rate of change is the slope of the tangent line at one point.
As the interval shrinks:
$$[x_1, x_2] \to [x_1, x_1 + 0.1] \to [x_1, x_1 + 0.01] \to [x_1, x_1 + 0.001] \to \cdots$$
The secant slopes approach the tangent slope:
$$\frac{f(x_2) - f(x_1)}{x_2 - x_1} \to \frac{f(x_1 + 0.1) - f(x_1)}{0.1} \to \frac{f(x_1 + 0.01) - f(x_1)}{0.01} \to \cdots \to f'(x_1)$$
The rate of change $\frac{dy}{dx}$ always has units:
$$\text{units of rate} = \frac{\text{units of } y}{\text{units of } x}$$
| Quantity $y$ | Quantity $x$ | Rate $dy/dx$ |
|---|---|---|
| meters | seconds | m/s (velocity) |
| dollars | items | $/item (marginal cost) |
| kilograms | meters | kg/m (linear density) |
| population | years | people/year (growth rate) |
For each statement, identify whether it describes an average or instantaneous rate of change:
(a) "The car was traveling at 70 mph when the officer pulled it over."
(b) "The flight took 3 hours and covered 1500 miles."
(c) "At noon, the temperature was rising at 2°F per hour."
(d) "Between 2020 and 2025, the population grew by 5000 people per year."
The population of a bacterial colony is given by $P(t) = 200 + 50t + 10t^2$ cells, where $t$ is in hours.
(a) Find the average rate of change from $t = 1$ to $t = 4$ hours.
(b) What are the units of your answer?
Let $f(x) = x^2 + 3x$.
(a) Compute the average rate of change from $x = 2$ to $x = 2 + h$ (leave your answer in terms of $h$).
(b) Find the instantaneous rate of change at $x = 2$ by taking the limit as $h \to 0$.
The temperature $T$ (in °C) of a cooling cup of coffee is recorded:
| $t$ (min) | 0 | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|---|
| $T$ (°C) | 90 | 75 | 63 | 54 | 47 | 42 |
(a) Find the average rate of cooling from $t = 5$ to $t = 15$ minutes.
(b) Estimate the instantaneous rate of cooling at $t = 10$ minutes using data from both sides.
(c) Is the coffee cooling faster at $t = 5$ or at $t = 20$? Explain.
For $f(x) = x^2$:
(a) Find the average rate of change from $x = a$ to $x = b$.
(b) The Mean Value Theorem says there exists some $c$ in $(a, b)$ where the instantaneous rate equals the average rate. Find this value of $c$ in terms of $a$ and $b$.
(c) What is special about where $c$ is located in the interval $[a, b]$?
The graph shows $y = f(x)$ with points $P$ and $Q$ marked.
y
│ Q
│ ·
│ ·
│ ·
│ P
│ ·
└──────── xAs $Q$ moves closer to $P$ along the curve, what happens to the secant line $PQ$?
(A) It becomes vertical (B) It approaches the tangent line at $P$ (C) Its slope approaches zero (D) It rotates away from the curve
Which statement best describes what $\displaystyle\lim_{h \to 0} \frac{f(3+h) - f(3)}{h} = 5$ means?
(A) $f(3) = 5$
(B) The average rate of change from $x = 3$ to $x = 8$ is 5
(C) The tangent line at $x = 3$ has slope 5
(D) The function increases by 5 units at $x = 3$
The Zoom Lens:
Imagine looking at a curvy road on a map. Zoomed out, you see the whole winding path: that is like average rate of change over a long interval.
Now zoom in closer and closer to one point. The curve looks straighter and straighter. When you are zoomed in infinitely close, the curve looks like a straight line. That straight line is the tangent, and its slope is the instantaneous rate of change.
The derivative is what you get when you "zoom in all the way."
Looking back:
Looking ahead:
The Big Picture: This skill is the conceptual bridge between algebra (difference quotients) and calculus (derivatives). The derivative does not just measure slope. It measures how fast things change. Physics, economics, biology, and chemistry all use this same idea.
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|---|---|---|
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Last updated: 2026-01-22