Separate amount, average rate, and instantaneous rate
Visitors to a website
After \(t\) days, a website has had \(W\) thousand visitors since a campaign began. The orange bar is the height of the curve at day 2. It shows an amount: by day 2 the site has had 10 thousand visitors.
A line joins the point at day 2 to the point at day \(b\), and its slope is the average rate over that interval. For \(b = 6\), it is 11 thousand visitors per day.
Move the second day toward day 2. The line rotates toward the orange tangent, and its slope approaches 7 thousand visitors per day, the rate at day 2.
Three readings of one curve
The curve shows any amount y against an input x, and the orange bar is the amount at x.
The slope of the line through two points is the change in y divided by the change in x, the average rate between them.
Move the second point toward the first. The line rotates toward the orange tangent, and its slope approaches the rate at x, which is the derivative.
The bar and the tangent stay in place because they belong to one input, while the line belongs to the whole interval.
Notation
The amount is the value of the function, a height on the graph, and its units have no “per”.
\[\cy{W(a)}\]The average rate over an interval is the change in the amount divided by the length of the interval.
\[\dfrac{W(b) - W(a)}{b - a}\]The rate at one input is the limit of the average rate as the interval length \(h = b - a\) shrinks to zero. This limit is the derivative.
\[\cy{W'(a)} = \lim_{h \to 0} \dfrac{W(a + h) - W(a)}{h}\]Both rates have the units of the amount for each unit of the input, here thousand visitors per day.
When it fails
On a long interval, the average rate can be far from the rate at either end, because the slope of the curve changes.
From day 2 to day 10, the average is 15 thousand visitors per day. The rate is 7 at day 2 and 23 at day 10, so the average describes the whole interval and equals neither rate.
Example 1: A phone battery
A phone’s charge is \(B = 30 + 40t - 5t^2\) percent after \(t\) hours. Find the charge at \(t = 1\), the average rate from \(t = 1\) to \(t = 3\), and the rate at \(t = 1\).
The charge at \(t = 1\) is an amount, the height of the curve.
\[\cy{B(1)} = 30 + 40 - 5 = 65\]The average rate is the change in charge divided by the length of the interval. Here \(B(3) = 105\).
\[\dfrac{B(3) - B(1)}{3 - 1} = \dfrac{105 - 65}{2} = 20\]The rate at \(t = 1\) is the derivative, the slope of the orange tangent.
\[\cy{B'(t)} = 40 - 10t, \quad B'(1) = 30\]The charge is 65 percent. Over the two hours it rises at 20 percent per hour on average, and at \(t = 1\) it rises at 30 percent per hour.
Example 2: Concert tickets
A concert has sold \(S = 20d + 3d^2\) tickets after \(d\) days. Find how fast tickets sell on day 4.
By day 4, the number of tickets sold is an amount.
\[S(4) = 80 + 48 = 128\]A tempting wrong move: The rate on day 4 is the total divided by the days.
\[\cr{\dfrac{S(4)}{4} = 32}\]The number 32 is the slope of the line from day 0 to day 4, so it is the average rate over four days. At day 4 the orange tangent is steeper, so the rate is larger.
The rate on day 4 is the derivative.
\[S'(d) = 20 + 6d\]Substitute \(d = 4\). On day 4, tickets sell at 44 tickets per day.
\[S'(4) = 44\]
Practice
A plant’s height is \(h = 12 + 6w - 0.5w^2\) centimeters after \(w\) weeks. Compare the average rate of growth from week 2 to week 6 with the rate at week 2.
Draw the line through the points at week 2 and week 6, and the tangent at week 2. Compare the slopes.
Solution
The average rate is the change in height, from 22 to 30, divided by the length of the interval.
\[\dfrac{30 - 22}{6 - 2} = 2\]The rate at week 2 is the derivative.
\[h'(w) = 6 - w, \quad h'(2) = 4\]The rate at week 2, 4 centimeters per week, is larger than the average, 2. The growth slows, so the tangent at week 2 is steeper than the line.
Questions
The graph shows the thousands of followers \(F\) of an account after \(d\) days. The line joins the points at day 2 and day 6. Question 3 does not use the graph.
What does the slope of the line measure?
- The average rate of change of F from day 2 to day 6
- The rate of change of F at day 2
- The number of followers on day 6
- The total change in F from day 2 to day 6
Answer
The average rate of change of F from day 2 to day 6
At day 6, the orange tangent is steeper than the line. What does this show?
- At day 6, F grows faster than its average rate over days 2 to 6
- At day 6, F grows more slowly than its average rate over days 2 to 6
- At day 6, F grows at the same rate as the average
- A rate at one day cannot be compared with an average
Answer
At day 6, F grows faster than its average rate over days 2 to 6
A colony has \(B = 100 + 6t^2\) cells \(t\) hours after the start. Which value is the rate at which the colony grows at \(t = 5\)?
- 60 cells per hour
- 30 cells per hour
- 250 cells
- 150 cells
Answer
60 cells per hour
Summary
An amount is a height on the curve, and its units have no “per”.
The average rate is the slope of the line over an interval, and the derivative is the slope of the tangent at one input.
When a question asks how fast a quantity changes at one input, use the derivative, not the amount or the average over an interval.
This is one of four skills in topic 4.3 of Which & Why: AP Calculus.