Lesson 2 of 7Rates of Change & Combining Functions

Unit 6 · Lesson 2 of 7

Instantaneous Rate of Change

The instantaneous rate of change is the slope of the tangent line at a single point — estimate it by shrinking the interval in a difference quotient until the answers stop moving.

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Key things to remember

1Instantaneous rate of change at $x = a$ is the slope of the tangent line to $y = f (x)$ at $(a, f (a))$ .
2The difference quotient $\frac{f ( a + h ) - f ( a )}{h}$ is the slope of the secant from $a$ to $a + h$ .
3Shrinking $h$ makes the secant slope approach the tangent slope; $h$ may become tiny but never exactly $0$ .
4Centred difference estimate: $\frac{f ( a + h ) - f ( a - h )}{2 h}$ — usually more accurate than a one-sided estimate with the same $h$ .
5Algebraically: expand, cancel the common factor of $h$ , then discard the remaining terms containing $h$ because they vanish as $h$ becomes small.
6Units of an instantaneous rate are the same as for an average rate: output units per input unit.

Your car's speedometer does not report your average speed for the trip. It reports how fast you are going right now, at this instant. But "right now" is a problem for the formula from the last topic: a rate needs a change in input on the bottom, and at a single instant there is no interval at all — the denominator would be zero. The way around this is beautifully simple. Keep using intervals, but make them smaller and smaller, and watch where the answers are heading.

From secant to tangent

Fig 1As the second point slides toward the first, the secant line pivots and settles onto the tangent line.

Fix a point $(a, f (a))$ on a curve and pick a second point a little to the right. The secant line through them has a slope you can compute. Now slide the second point closer to the first. The secant pivots, and its slope changes — but not wildly. It homes in on one particular value: the slope of the tangent line, the line that just touches the curve at $(a, f (a))$ and runs in the same direction the curve is running at that exact spot. That tangent slope is the instantaneous rate of change of $f$ at $x = a$ .

The difference quotient

To slide the second point in a controlled way, give the gap a name. Let $h$ be the horizontal distance from $a$ to the second input, so the second point is $(a + h, f (a + h))$ . The slope of that secant is the difference quotient: the run is $(a + h) - a = h$ , so the $h$ lands neatly in the denominator.

slope of secant = \frac{f ( a + h ) - f ( a )}{( a + h ) - a} = \frac{f ( a + h ) - f ( a )}{h}

Fig 2The two points that build the difference quotient, with the run labelled h and the rise labelled as the difference of the outputs.

Why the interval can never shrink to nothing

Setting

h = 0

is not allowed. The denominator becomes

0

, and the numerator becomes

f (a) - f (a) = 0

as well, giving the meaningless form zero over zero. So

h

is allowed to become tiny — a millionth, a billionth — but never exactly zero. We find out what the quotient is heading toward without ever landing on it.

VideoAn MHF4U example that estimates an instantaneous rate of change with the difference quotient, the same method as this section.Watch on YouTube

A table of shrinking intervals

h	a + h	f(a + h)	Difference quotient
0.1	2.1	9.261	12.61
0.01	2.01	8.120601	12.0601
0.001	2.001	8.012006001	12.006001
0.0001	2.0001	8.00120006...	12.00060001

Difference quotients for the cube function at x = 2, with h shrinking by a factor of ten each row.

The most reliable numerical method is exactly this: choose several values of $h$ , each ten times smaller than the last, and compute the difference quotient for each. Here $f (x) = x^{3}$ and $a = 2$ , so $f (2) = 8$ , and the digits are settling — $12.61$ , then $12.06$ , then $12.006$ , then $12.0006$ . Each extra decimal place of accuracy costs one more row. The instantaneous rate of change of $x^{3}$ at $x = 2$ is $12$ . Keep shrinking $h$ until the leading digits of two consecutive rows agree to the precision you need, then report that value.

Loading interactive…

Drag the slider to shrink h and watch the secant slope converge on the tangent slope at the marked point.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Find the exact instantaneous rate of change of

f (x) = x^{2}

x = 3

by simplifying the difference quotient.

Example 2

A rocket's height in metres after

t

seconds is

h (t) = - 4.9 t^{2} + 25 t

. Find its exact instantaneous velocity at

t = 2

Example 3

A cyclist's distance is recorded as

t = 2

18

t = 4

27

t = 6

41

m. Estimate the instantaneous speed at

t = 4

s using a centred difference, and explain why it beats the forward estimate.

Example 4

Use a table of shrinking

h

to estimate the instantaneous rate of change of

f (x) = x^{3}

x = 2

, to three decimal places.

Where marks get lost

Substituting $h = 0$ into the difference quotient before cancelling, which produces the meaningless form $\frac{0}{0}$ .
Forgetting to expand $(a + h)^{2}$ properly and writing $a^{2} + h^{2}$ , which loses the crucial $2 ah$ middle term.
Cancelling $h$ from only one term of the numerator instead of factoring it out of the whole numerator first.
Using one large value of $h$ (like $h = 1$ ) and calling the result the instantaneous rate — that is still an average rate over a wide interval.
Dividing by $h$ instead of $2 h$ in a centred difference, which halves the answer.

Flashcards 7

Instantaneous rate of change

Tangent line

Difference quotient

Why can h never be 0?

+ 3 more

Practice

Which expression is the difference quotient for

f

x = a

Simplify the difference quotient for

f (x) = x^{2}

x = 5

and state the instantaneous rate of change there.

The average rate of change of

f

over the interval from

3

3 + h

7.4

when

h = 0.1

7.04

when

h = 0.01

, and

7.004

when

h = 0.001

. Estimate the instantaneous rate of change at

x = 3

Explain in one phrase why

h

can never be set exactly equal to

0

in the difference quotient.

A tank's volume is

V = 18

L at

t = 2

min,

V = 27

L at

t = 4

min, and

V = 41

L at

t = 6

min. Use a centred difference to estimate the instantaneous rate of change of volume at

t = 4

min.

L/min

h

gets smaller and smaller, the secant line through

(a, f (a))

and

(a + h, f (a + h))

does what?

Find the instantaneous rate of change of

f (x) = x^{2} + 2 x

x = 1

Order the steps for finding an exact instantaneous rate of change algebraically at x = a:

1Expand f(a + h) and subtract f(a), collecting like terms
2Let h become extremely small and discard every remaining term containing h
3Factor h out of the numerator and divide it out
4Write the difference quotient (f(a + h) − f(a)) / h

Which estimate of the instantaneous rate of change at

x = a

is usually the most accurate?

A rocket's height is

h (t) = - 4.9 t^{2} + 25 t

metres. Find its instantaneous velocity at

t = 2

m/s

Match each term to what it measures or means.

Secant line

Tangent line

The letter h

0/0

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