A. Rate of Change · Limits and Rates of Change

Estimating Instantaneous Rates from Average Rates

A1.3 A1.1About 65 minutes

Learning goals

I can approximate an instantaneous rate of change at a point using average rates over intervals that contain the point.
I can explain why a centred interval usually gives a better estimate than a preceding or following interval.
I can estimate instantaneous rates from an equation, a table of data or a graph.
I can explain how the error in an estimate depends on the width of the interval.

Words to know

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

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

centred interval
An interval [a − h, a + h] with x = a in the middle.

estimate
An approximate value. Good estimates come with a sense of how close they are.

error
The difference between an estimate and the true value.

graphing technology
Software or a calculator that graphs functions, such as Desmos or GeoGebra (both free online).

Two points from one

An instantaneous rate is a slope at one point, but a slope needs two points. In the MHF4U lesson Approximating Instantaneous Rates by Shrinking the Interval (unit Rates of Change), the fix was to pick a second point very close to the first, using preceding, following and centred intervals. The average rate over a tiny interval containing the point is almost the instantaneous rate.

This lesson sharpens that idea. You will compare the three kinds of intervals again, see exactly why the centred one is usually much better, measure how the error shrinks with h, and estimate rates when you only have data (no equation), which is how scientists usually meet rates of change.

Three intervals around x = a

Pick a small positive number h.

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

For a curve that bends upward (like y = 2ˣ), the following secant is too steep and the preceding secant is not steep enough. The true tangent slope is trapped between them. The centred secant balances the two errors and is usually much closer.

Near P(1, 2) on y = 2ˣ, the following secant is too steep and the preceding secant too shallow. The centred secant is almost parallel to the tangent.

Worked example

Example 1: Squeezing a rate from an equation

Estimate the instantaneous rate of change of f(x) = √x at x = 4 using preceding, following and centred intervals with h = 1, 0.1 and 0.01.

Following, h = 0.1: (√4.1 − 2) ÷ 0.1 ≈ (2.024 846 − 2) ÷ 0.1 ≈ 0.248 46.
Preceding, h = 0.1: (2 − √3.9) ÷ 0.1 ≈ (2 − 1.974 842) ÷ 0.1 ≈ 0.251 58.
Centred, h = 0.1: (√4.1 − √3.9) ÷ 0.2 ≈ 0.250 02.
Full results (rounded to 5 decimal places): h = 1 gives following 0.23607, preceding 0.26795, centred 0.25201. h = 0.1 gives 0.24846, 0.25158, 0.25002. h = 0.01 gives 0.24984, 0.25016, 0.25000.
This curve bends downward, so this time the following estimates are too low and the preceding ones too high. The true value is trapped between them, and the centred values settle fastest.

The instantaneous rate at x = 4 is about 0.25. (In the lesson on first principles you will prove it is exactly 1/4.)

t (s)	0	10	20	30	40
altitude (m)	0	42	95	160	238

Altitude of a hot-air balloon, recorded every 10 s.

Worked example

Example 2: A rate from data only

A hot-air balloon rises from a field. Its altitude is recorded every 10 s (table above). Estimate how fast the balloon is rising at t = 20 s.

Without an equation, h cannot be smaller than the data spacing, 10 s.
Preceding [10, 20]: (95 − 42) ÷ 10 = 5.3 m/s.
Following [20, 30]: (160 − 95) ÷ 10 = 6.5 m/s.
Centred [10, 30]: (160 − 42) ÷ 20 = 5.9 m/s. This is also the average of 5.3 and 6.5.

The balloon is rising at about 5.9 m/s at t = 20 s. The rate is increasing, so the true value should be between 5.3 and 6.5 m/s.

Secant slopes for y = 2ˣ at x = 1, plotted against h. The following and preceding estimates approach 1.386 from above and below; the centred estimate is close for every h shown.

Worked example

Example 3: How small is small enough?

For f(x) = x³ at x = 2, the following slope simplifies to 12 + 6h + h² and the centred slope to 12 + h². (You will learn to simplify like this in later lessons.) How small must h be for each estimate to be within 0.01 of the true rate, 12?

Following error: 6h + h². With h = 0.001: 0.006 + 0.000 001 ≈ 0.006 < 0.01 ✓. With h = 0.01: 0.06 + 0.0001 > 0.01 ✗. So you need h about 0.001 (more precisely, h < about 0.0017).
Centred error: h². With h = 0.1: 0.01, right at the edge. Any h a little below 0.1 works, for example h = 0.09 gives 0.0081.
Halving h roughly halves the following error, but it divides the centred error by 4.

The following estimate needs h ≈ 0.001; the centred estimate is good enough with h just under 0.1. Centred intervals are much more efficient.

With graphing technology

In Desmos or GeoGebra, define f, a point P(a, f(a)) and a slider h. Plot Q(a + h, f(a + h)) and the line through P and Q, and display its slope. Drag h toward 0 from both sides and watch the slope settle. Most tools can also draw the exact tangent so you can check. A scientific calculator works too: just type the slope formula for each h.

Just for fun

Asking for the slope at a single point is like asking how fast a photograph of a car is going. One photo cannot say. Two photos taken a split second apart can: that is the whole idea of shrinking the interval.

Hands-on activity

Rate of a melting ice cube

You need: an ice cube, a small bowl, a kitchen scale that reads in grams, a timer and paper.

Put the bowl on the scale and press tare (zero). Add the ice cube and record its mass. (It may already be melting slightly; that is fine.)
Every 5 minutes for 40 minutes, pour the meltwater into the sink (or a cup) with a quick tilt and weigh the ice again. Record time and mass.
Make a table of mass against time.
Choose a time in the middle, such as t = 20 min. Estimate the instantaneous rate of melting there using a preceding, a following and a centred interval.
Which estimate do you trust most? Is the ice melting faster at the start or the end?

What to look for: all rates are negative (the mass is decreasing), in grams per minute. Your centred estimate should be between the preceding and following estimates. Melting often slows as the cube shrinks, because a smaller cube has less surface touching warm air. Wipe up any drips, and dry the scale when you finish.

Your turn

For f(x) = x² at x = 5, find the following and preceding slopes with h = 0.1, and the centred slope.

A car's distance (m) is recorded: t = 0, 1, 2, 3, 4 s gives d = 0, 2.1, 8.3, 18.6, 33.2. Estimate the velocity at t = 2 s with a centred interval.

Estimate the instantaneous rate of change of f(x) = 1/x at x = 2 with a centred interval and h = 0.1.

Estimate the instantaneous rate of change of f(x) = x³ at x = 1 with a following interval and h = 0.01.

For y = 2ˣ, which estimate is too high: preceding or following? Why?

For the hot-air balloon (Example 2), estimate the rate of climb at t = 30 s.

Why must the interval contain the point x = a?

More practice: warm-up

Write the preceding interval for a = 3 and h = 0.1.

Write the centred interval for a = 3 and h = 0.1.

For f(x) = 3x + 1, what does every estimate give at x = 2, and why?

For f(x) = x² at x = 1, find the following slope with h = 0.1.

For f(x) = x² at x = 1, find the preceding slope with h = 0.1.

More practice: on your own

Estimate the instantaneous rate of change of y = sin x (x in radians) at x = 0 using a centred interval with h = 0.01.

A car's value is V(t) = 30 000(0.85)ᵗ dollars after t years. Estimate the rate of change at t = 2 years with a centred interval, h = 0.01.

A mug cools: t = 0, 5, 10, 15, 20 min gives T = 85, 68, 56, 47, 41 °C. Estimate the rate of cooling at t = 10 min.

A toy rocket's height (m) is: t = 0, 1, 2, 3, 4 s gives 0, 25.1, 40.4, 45.9, 41.6. Estimate its velocity at t = 2 s and at t = 3 s.

For f(x) = x³ at x = 2, compute the following slope with h = 0.1 and compare it with 12 + 6h + h².

Challenge

Show that for f(x) = x², the centred slope at x = a equals 2a for every h > 0.

For f(x) = x³, show the centred slope at x = a is 3a² + h². How small must h be for an error under 0.0001?

Show that the centred slope is the average of the preceding and following slopes.

Check your understanding

Quiz

Which interval is centred at x = 3 with h = 0.2?

For a curve that bends upward near x = a, the following secant slope is usually:

A table gives f(1) = 4, f(2) = 9 and f(3) = 16. What is the centred estimate of the instantaneous rate at x = 2?

With only a data table spaced every 5 minutes, what is the smallest centred interval around t = 10 min you can use?

For f(x) = x³ at x = 2, the following slope is 12 + 6h + h². What is it when h = 0.1?

Why do we let h get smaller when estimating an instantaneous rate?

For parents and teachers

Big idea (A1.3): an instantaneous rate can be approximated by average rates over smaller and smaller intervals that contain the point. Centred intervals usually work best. This builds directly on the MHF4U unit Rates of Change.

How to help: Use the ice-cube activity, or any data your teen collects (a phone's battery percentage every 10 minutes, the temperature outdoors each hour). Ask for preceding, following and centred estimates at one time, and ask which one they trust most and why.

Common difficulties:

Dividing by h instead of 2h for a centred interval.
Rounding f(a + h) and f(a) too early: with small h, keep at least 6 to 8 digits until the end.
Using an interval that does not contain the point.
Expecting more accuracy from data than the spacing allows.