Amount of Change: Delta Notation and Net Change
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.1: “Defining the Derivative” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/3-1-defining-the-derivative |
| Supplementary | OpenStax Calculus Volume 1, Section 1.4: “The Tangent and Velocity Problems” |
| Supplementary link | https://openstax.org/books/calculus-volume-1/pages/1-4-the-tangent-and-velocity-problems |
Try This First
A savings account holds $\$500$ at the start of January. By the end of March, it holds $\$620$.
Before reading further:
- By how much did the account balance change over those three months?
- If you wanted to describe this change in two numbers, what two numbers would you write down?
- Is there enough information to know how much it changed each month? Why or why not?
Hold your answers. This lesson puts names and notation on what you just computed.
Key Idea: The Greek Letter Delta
The Greek letter $\Delta$ (delta) is a symbol for “the change in” a quantity. When a quantity $y$ changes from an old value $y_1$ to a new value $y_2$, the change is: $$\Delta y = y_2 - y_1 = \text{new value} - \text{old value}.$$
For a function $f$ with input $x$ moving from $a$ to $b$: $$\Delta x = b - a \qquad \text{(change in input)},$$ $$\Delta y = f(b) - f(a) \qquad \text{(change in output)}.$$
Net change is the same as $\Delta y$: it subtracts the starting value from the ending value and reports the difference, including its sign. A positive $\Delta y$ means the output increased. A negative $\Delta y$ means it decreased.
This is not yet a rate. It is an amount: how much did the output move?
Two Connected Representations
Numerically (from the savings account): $\Delta y = 620 - 500 = 120$ dollars. The input change is $\Delta x = 3 - 0 = 3$ months.
Graphically: on a graph of account balance vs time, $\Delta y$ is the vertical rise between two points on the graph, and $\Delta x$ is the horizontal run. The Greek letters $\Delta y$ and $\Delta x$ are the “rise” and “run” you have likely seen before.
Translation prompt: Locate two points on any graph. The vertical distance between them is $\Delta y$. The horizontal distance is $\Delta x$. Which is which, if $\Delta y < 0$? Draw the picture.
Worked Example
The position of an object along a track is given by $s(t) = t^2$ meters, where $t$ is time in seconds. Find $\Delta s$ as $t$ goes from $t = 1$ to $t = 4$.
Predict first: the position is increasing (since $s(t) = t^2$ grows as $t$ grows). The change should be positive, and probably more than 10 meters since the positions will be 1 and 16.
Compute: $$\Delta t = 4 - 1 = 3 \text{ seconds},$$ $$\Delta s = s(4) - s(1) = 16 - 1 = 15 \text{ meters}.$$
The object moved 15 meters in 3 seconds.
Compare with prediction: positive, and 15 is indeed more than 10. The ballpark estimate was right.
Now find $\Delta s$ as $t$ goes from $t = 1$ to $t = 2$, and as $t$ goes from $t = 2$ to $t = 4$.
From $t = 1$ to $t = 2$: $\Delta s = s(2) - s(1) = 4 - 1 = 3$ meters. From $t = 2$ to $t = 4$: $\Delta s = s(4) - s(2) = 16 - 4 = 12$ meters.
Notice: the first 1-second stretch gives $\Delta s = 3$ m; the 2-second stretch from $t=2$ to $t=4$ gives $\Delta s = 12$ m. The object traveled more in those 2 seconds than in the first 1 second, and also more per second. The amount of change depends on where in the interval you look. This variation in change-per-unit-input is exactly what makes non-linear functions interesting and what calculus is built to analyze.
“average” change means taking the arithmetic mean of the output values. The change $\Delta y = f(b) - f(a)$ is a subtraction, not an average. In the savings example, the change is $620 - 500 = 120$ dollars -- not $(620 + 500)/2 = 560$ dollars. Averaging the two balance values gives the balance at the midpoint of time (roughly), which answers a completely different question. “How much did the balance change?” and “what was the typical balance?” are distinct questions that require different computations. Later, “average rate of change” will divide $\Delta y$ by $\Delta x$, which is also different from an arithmetic mean. The word “average” has a specific meaning in each context.
doubling $\Delta x$ doubles $\Delta y$ for any function. For linear functions, where the output grows by a fixed amount per unit of input, doubling $\Delta x$ does double $\Delta y$. But for non-linear functions, it does not. In the position example: going from $t=1$ to $t=2$ (a $\Delta x$ of 1) gives $\Delta s = 3$ m. Doubling the interval to $t=1$ to $t=3$ (a $\Delta x$ of 2) gives $\Delta s = 9 - 1 = 8$ m, not 6. The relationship between $\Delta x$ and $\Delta y$ varies with position and is precisely what rates of change -- and eventually derivatives -- describe.
Predict-Then-Check Practice
Problem 1 (Low floor): For $f(x) = 3x + 2$, find $\Delta f$ as $x$ goes from 1 to 5. Predict the sign and rough size before computing.
Check your answer
Predict: $f$ is linear with positive slope, so $\Delta f$ is positive. The output should increase by roughly $3 \times 4 = 12$ (since slope is 3 and input increases by 4).
Compute: $\Delta x = 5 - 1 = 4$. $\Delta f = f(5) - f(1) = 17 - 5 = 12$.
The prediction was exact here because the function is linear (constant rate of change). For non-linear functions, the estimate would be approximate.
Problem 2: For $g(x) = x^2$, compare $\Delta g$ over $[0,1]$, over $[1,2]$, and over $[2,3]$. Each interval has $\Delta x = 1$. Do equal changes in input produce equal changes in output?
Check your answer
$\Delta g$ on $[0,1]$: $g(1) - g(0) = 1 - 0 = 1$. $\Delta g$ on $[1,2]$: $g(2) - g(1) = 4 - 1 = 3$. $\Delta g$ on $[2,3]$: $g(3) - g(2) = 9 - 4 = 5$.
The outputs change by 1, 3, 5 -- the changes are growing, even though each $\Delta x = 1$. Equal steps in the input produce unequal steps in the output for a quadratic function. This is the signature of non-constant rate of change.
Problem 3 (High ceiling): For $f(x) = x^2$, find $\Delta f$ over the interval $[a, a+h]$ for general values of $a$ and $h$. Write your answer in simplified form and explain what happens as $h$ gets smaller.
Check your answer
$\Delta f = f(a+h) - f(a) = (a+h)^2 - a^2 = a^2 + 2ah + h^2 - a^2 = 2ah + h^2$.
As $h$ gets smaller (shrinks toward 0), $\Delta f = 2ah + h^2$ gets smaller too. Specifically, the $h^2$ term shrinks much faster than the $2ah$ term. For small $h$, $\Delta f \approx 2ah$. Dividing by $h$ will give the average rate of change $2a + h$, which approaches $2a$ as $h \to 0$. This is the beginning of the derivative -- you will complete this calculation in the next node.
Ask Why
If two functions have the same value of $\Delta y$ over an interval, are they behaving the same on that interval? Why or why not?
Think of a function that goes up 10 and comes back down 7, ending at a net change of $+3$, versus a function that goes straight from the starting value to 3 above it. Both have $\Delta y = 3$, but the journeys are different. What does $\Delta y$ tell you, and what does it not tell you?