D. Characteristics of Functions · Rates of Change

Approximating Instantaneous Rates by Shrinking the Interval

D1.5 D1.6About 70 minutes

Learning goals

I can estimate an instantaneous rate of change using average rates over preceding, following and centred intervals.
I can make a table of average rates for smaller and smaller intervals and see what value they approach.
I can estimate an instantaneous rate from a table of data, without an equation.
I can explain why a centred interval usually gives a better estimate.

Words to know

preceding interval
An interval that ends at the point: [a − h, a].

following interval
An interval that starts at the point: [a, a + h].

centred interval
An interval with the point in the middle: [a − h, a + h].

h
A small positive number: the distance from the point a to the other end of the interval.

Sneaking up on a single point

To find a slope you need two points. To find an instantaneous rate you have only one. The trick is to use a second point that is very, very close to the first. Then the average rate over that tiny interval is almost the instantaneous rate.

In the last lesson, average velocities of the car over [4, 5], [4, 4.1] and [4, 4.01] were 11.25, 10.125 and 10.0125 m/s, closing in on 10 m/s. Today you will make this a reliable method that works from an equation or from a table of data.

Three ways to choose the interval

To estimate the instantaneous rate at x = a, pick a small number h and use one of these intervals:

Preceding: [a − h, a], rate = (f(a) − f(a − h)) ÷ h
Following: [a, a + h], rate = (f(a + h) − f(a)) ÷ h
Centred: [a − h, a + h], rate = (f(a + h) − f(a − h)) ÷ (2h)

Then make h smaller (1, 0.1, 0.01, 0.001, …) and watch what value the rates approach. That value is the instantaneous rate of change.

Around P(2, 21.9) on the ball's height graph, with h = 1: the preceding and following secants have very different slopes, but the centred secant (slope 0.4) is parallel to the tangent.

Worked example

Example 1: The ball at t = 2

The ball's height is h(t) = −4.9t² + 20t + 1.5. Estimate its velocity at t = 2 s using following, preceding and centred intervals with h = 1, 0.1, 0.01 and 0.001.

h(2) = 21.9. For h = 0.1: h(2.1) = 21.891 and h(1.9) = 21.811.
Following, h = 0.1: (21.891 − 21.9) ÷ 0.1 = −0.09. Preceding: (21.9 − 21.811) ÷ 0.1 = 0.89. Centred: (21.891 − 21.811) ÷ 0.2 = 0.4.
Do the same for the other values of h. The table below shows the results.
The following rates climb toward 0.4 and the preceding rates fall toward 0.4.

The velocity at t = 2 s is 0.4 m/s: the ball is still rising, but only just. It is very close to the top.

h	following [2, 2 + h]	preceding [2 − h, 2]	centred [2 − h, 2 + h]
1	−4.5	5.3	0.4
0.1	−0.09	0.89	0.4
0.01	0.351	0.449	0.4
0.001	0.3951	0.4049	0.4

Average velocities (m/s) for the ball near t = 2. For this quadratic, the centred rate is exactly 0.4 for every h.

The preceding rates (filled dots) come down toward 0.4 and the following rates (open dots) climb up toward it as h shrinks. (The h = 0.01 dots, 0.449 and 0.351, are the pair closest to 0.4.)

Why centred is usually better

On a curve, the following secant tends to err one way and the preceding secant the other way. The centred secant balances the two, so it is usually much closer to the tangent for the same h. For a quadratic, it is exactly right for every h (you will prove this in the Challenge).

Worked example

Example 2: Depreciation (exponential)

A car's value is V(t) = 28 000(0.85)^t dollars after t years. Estimate how fast it is losing value at t = 3 years, using centred intervals with h = 0.1, 0.01 and 0.001.

h = 0.1: (V(3.1) − V(2.9)) ÷ 0.2 ≈ −2794.72.
h = 0.01: (V(3.01) − V(2.99)) ÷ 0.02 ≈ −2794.60.
h = 0.001: (V(3.001) − V(2.999)) ÷ 0.002 ≈ −2794.59.
The estimates agree to the nearest dollar.

At t = 3 the car is losing value at about $2795 per year. (Compare: over the first 3 years the average loss was $3601.50 per year. The car loses value fastest when it is new.)

Worked example

Example 3: From data only

A mug of water cools on the counter. A thermometer gives these readings:

t (min): 0, 2, 4, 6, 8, 10
T (°C): 90.0, 76.7, 65.9, 57.2, 50.1, 44.4

Estimate the rate of cooling at t = 4 min.

There is no equation, so we cannot shrink h below the 2-minute spacing. Use the smallest centred interval available: [2, 6].
Rate ≈ (T(6) − T(2)) ÷ (6 − 2) = (57.2 − 76.7) ÷ 4 = −19.5 ÷ 4 = −4.875.
The following interval [4, 6] gives (57.2 − 65.9) ÷ 2 = −4.35 and the preceding interval [2, 4] gives (65.9 − 76.7) ÷ 2 = −5.4. The centred value is between them.

About −4.9 °C/min: at t = 4 min the water is cooling at roughly 4.9 degrees per minute. With real data, more frequent readings would give a better estimate.

A warning about calculators

Smaller h is better, but only up to a point. A calculator keeps about 10 to 15 digits. With h = 0.000 000 000 1, f(a + h) and f(a) agree in almost every digit, and the tiny difference left over can be mostly rounding error. In practice, h = 0.001 or 0.0001 is a good choice with a scientific calculator. If your estimates suddenly jump around as h gets tiny, rounding is the reason.

Just for fun

Shrinking the interval is the only time in math when being indecisive ("0.1? No, 0.01! No, 0.001!") is the correct method.

Hands-on activity

Calculator chase

You need: a scientific calculator (or a phone calculator turned sideways for the scientific keys), paper and a pencil.

Choose f(x) = 2ˣ and the point x = 3, where f(3) = 8.
Make a table with columns h, following rate (2^(3 + h) − 8) ÷ h and centred rate (2^(3 + h) − 2^(3 − h)) ÷ (2h).
Fill it in for h = 1, 0.1, 0.01, 0.001 and 0.0001. Use brackets around the exponent.
What value do the rates approach? Divide that value by f(3) = 8. Do you recognize the result?
Bonus: try h = 0.000 000 000 01. What goes wrong?

What to look for: the rates approach about 5.545. Dividing by 8 gives about 0.693. You will see this number again in the lesson Rates of Change of Exponential and Logarithmic Functions. With an extremely tiny h, many calculators give a rough or even 0 answer because of rounding.

Your turn

For f(x) = x², estimate the instantaneous rate of change at x = 3 using the following interval with h = 0.1.

Repeat with the preceding interval, h = 0.1.

Repeat with the centred interval, h = 0.1. What is your best estimate of the instantaneous rate?

Estimate the instantaneous rate of change of f(x) = 2ˣ at x = 1 with a centred interval, h = 0.001.

For the ball h(t) = −4.9t² + 20t + 1.5, find the centred rate at t = 1 with h = 0.01.

Use the cooling-water table in Example 3 to estimate the rate of cooling at t = 6 min.

Explain in a sentence or two why a centred interval usually gives a better estimate than a following interval with the same h.

More practice: warm-up

Write the preceding interval of length 0.1 that ends at x = 5.

Write the following interval of length 0.01 that starts at x = 2.

Write the centred interval with h = 0.5 around x = 3.

For f(x) = 3x + 1, find the following rate at x = 2 with h = 0.1.

For f(x) = x², find the following rate at x = 1 with h = 0.01.

More practice: on your own

For f(x) = x³, estimate the instantaneous rate at x = 2 using following intervals with h = 0.1 and h = 0.01. What value do they approach?

For the car V(t) = 28 000(0.85)^t, the centred estimate at t = 3 with h = 1 is about −2806.91. Why is it further from the true rate than the estimate with h = 0.01?

Estimate the instantaneous rate of change of f(x) = sin x at x = π/3 (radians) using a centred interval with h = 0.001.

Estimate the instantaneous rate of change of f(x) = 1/x at x = 2 using a centred interval with h = 0.01.

A town's population is P(t) = 15 000(1.03)^t after t years. Estimate the growth rate at t = 5 using a following interval with h = 0.001.

Challenge

For any quadratic f(x) = ax² + bx + c, show that the centred rate (f(x + h) − f(x − h)) ÷ (2h) equals 2ax + b for every h ≠ 0.

For f(x) = x², the following rate at x = 3 simplifies to 6 + h. How small must h be for the estimate to be within 0.001 of the true rate, 6?

A cart's position is recorded each second: t = 0, 1, 2, 3, 4 s and d = 0, 2.1, 8.3, 18.6, 33.0 m. Estimate its velocity at t = 2 and at t = 3, then estimate its acceleration (the rate of change of velocity) between those times.

Check your understanding

Quiz

What is the centred interval with h = 0.1 around x = 4?

For f(x) = x², what is the following rate at x = 3 with h = 0.01?

For the same small h, which interval usually gives the best estimate of an instantaneous rate?

For the ball h(t) = −4.9t² + 20t + 1.5, following rates at t = 2 are −4.5, −0.09, 0.351 and 0.3951 for h = 1, 0.1, 0.01, 0.001. What is the velocity at t = 2?

Water temperatures are 76.7 °C at t = 2 min and 57.2 °C at t = 6 min. What is the centred estimate of the rate of cooling at t = 4 min?

Why can't you just use h = 0 in (f(a + h) − f(a)) ÷ h?

For parents and teachers

Big ideas (D1.5, D1.6): you can estimate an instantaneous rate by computing average rates over smaller and smaller intervals and seeing what value they approach. Centred intervals usually work best. With data only, use the smallest centred interval available.

How to help: Sit with your teen as they fill in a table of h-values on a calculator. Ask: "What number are these getting close to? How do you know when to stop?"

Common difficulties:

Dividing by h instead of 2h for a centred interval.
Missing brackets on the calculator, such as typing 2^3.001 − 8 ÷ 0.001 instead of (2^3.001 − 8) ÷ 0.001.
Rounding f-values too early. With h = 0.001, you need many digits of f(a + h).
Thinking the last row of the table is the answer. It is an estimate; the instantaneous rate is the value the rows approach.

This idea, letting h approach 0, is the heart of the limit that starts Grade 12 Calculus and Vectors (MCV4U).