Math 1241, Spring 2014
Section 3.3
Rates of Change
Average vs. Instantaneous Rates
Average Speed
• The concept of speed (distance traveled
divided by time traveled) is a familiar instance
of a rate of change.
• Example: To drive the 15.5 miles from Clayton
State to Turner field, it takes 18 minutes.
•
15.5 𝑚𝑖𝑙𝑒𝑠
18 𝑚𝑖𝑛.
= 0.88 mi/ min = 52.7 mph
• Question: If you drove to Turner Field, would
your speedometer always read 52.7 mph?
Average Speed
• The 52.7 mph is an average speed. Your
speedometer measures something else.
• In pre-Calculus courses, you solve problems
assuming that speed (or some other rate of
change) is constant: it does not change.
• One of the main features of Calculus: we can
solve problems where speed (or some other
rate or change) is not necessarily constant.
Average Rate of Change
• For a function y = f(x), we define the average
rate of change from x = a to x = b as:
𝑓 𝑏 − 𝑓(𝑎)
𝑏−𝑎
• In the case of speed:
– We often use t instead of x (for obvious reasons).
– f(t) is the total distance we’ve travelled at time t.
– Numerator = Change/difference in distance
– Denominator = Change/difference in time
Average Rate of Change = Slope
• The blue curve is the
graph of y = f(x).
• The two blue dots show
the function’s values at
x = a and x = b.
• The red line is called a
secant line. Its slope
equals the average rate
of change of f(x) from a
to b.
Instantaneous Speed
• Your speedometer measures your speed “at a
given time.” What does this mean?
• Average speed: Change in distance divided by
change in time. We can’t do this “at a given
time,” because the change in time is zero (in
the denominator).
• Solution: Take the limit as change in time
approaches zero!
Instantaneous Rates of Change
• Take the limit of (average rate of change), as
the change in the independent (x) variable
approaches zero. There are two ways to do so:
𝑓 𝑏 − 𝑓(𝑎)
lim
𝑏→𝑎
𝑏−𝑎
𝑓 𝑎 + ℎ − 𝑓(𝑎)
lim
ℎ→0
ℎ
• Although these appear to be different
formulas, note that h = b – a (thus b = a + h).
Graphical Demonstration
• It’s somewhat difficult to do dynamic graphs in
Graph, so we’ll use the following link:
https://www.desmos.com/calculator/irip8pnpdf
• Left-click on one of the dots and hold down the
button. Drag your mouse to see how the secant line
(in red) changes.
• As the dots get closer together, the slope of the
secant line approaches the instantaneous rate of
change.
Tangent Lines
Consider what happens to the secant line as
𝑏 → 𝑎 (or as ℎ → 0).
• Any secant line contains the point (𝑎, 𝑓 𝑎 ).
• The slope of the secant line approaches the
instantaneous rate of change (at x = a).
The line through (𝑎, 𝑓 𝑎 ) with slope equal to
the instantaneous rate of change (at x = a) is
called the tangent line (at x = a).
• The tangent line of a circle is a special case.
Graphical Demonstration
• Using the link from earlier:
https://www.desmos.com/calculator/irip8pnpdf
• Change the function definition to sqrt(4-x^2).
This is the upper half of the circle centered at
(0,0) with radius 2.
• Drag the two points close together. The secant
line is very close to a tangent line of the circle.
Tangent Lines in Graph
• Fortunately, Graph will draw a tangent line.
1
𝑥
– Start by graphing a function. I’ll use 𝑥 + .
– Select Function -> Insert Tangent/Normal from the
menu (or press F2, or use the toolbar button).
– In the “x = “ field, type the x-value where you
want the tangent line (I’ll use x = 1). Use a
different color than the original function.
• Zoom in on the point where the function
touches the tangent line. What do you see?
An important note
• The instantaneous rate of change of the
function f(x) at x = a is a limit.
• To actually compute it, we need to know the
function value for x values closer and closer to
x = a. This would mean infinitely many values!
• We can avoid this if we have a formula for f(x)
that is valid near x = a. We’ll usually take this
approach.
Algebraic Example
Find the instantaneous rate of change of the
function 𝑓 𝑥 = 𝑥 2 at the point a = 1.
• Before computing the limit, use Graph to draw
the tangent line. What is the slope?
• We need to evaluate one of the following:
𝑥2 − 1 2
1 + ℎ 2 − (1)2
lim
lim
𝑥→1 𝑥 − 1
ℎ→0
ℎ
• Try both forms; which one is easier?