MCV4U Average and Instantaneous Rates of Change: Notes and Practice

A car covers a 450 km highway drive in 5 hours. Its average speed is $\frac{450}{5} = 90$ km/h, but the speedometer spends very little of the trip at exactly 90: the car slows for traffic, speeds up to pass, and stops for gas. The trip average and the speedometer reading answer two different questions. One asks how fast the distance changed over the whole interval; the other asks how fast it is changing at one instant. Calculus starts with the second question, and this lesson shows how to answer it using nothing more than slopes.

A rate of change compares how much one quantity changes with how much another changes. It always has units of (output units) per (input units): metres per second, litres per minute, dollars per item, degrees per kilometre. A positive rate means the output is increasing; a negative rate means it is decreasing.

Representation	Example: water draining from a tank
Words	The tank drains quickly at first, then more and more slowly.
Numbers	A table: $V = 500$ L at $t = 0$ min, $420$ L at $2$ min, $352$ L at $4$ min, $296$ L at $6$ min, $252$ L at $8$ min.
Graph	A curve of volume against time that falls steeply at first and then flattens; its steepness is the rate.
Algebra	An equation for $V (t)$ , from which rates are computed exactly.

The same rate of change can be described in words, in a table, on a graph or with an equation.

A car's lit instrument panel at night, with the speedometer needle partway up a scale marked in km/h. — PhotoA speedometer shows how fast the car is going at one instant, not its average over the whole trip. That instantaneous reading is what this lesson learns to calculate.Nakonana, CC BY-SA 4.0, via Wikimedia Commons

Average rate of change: the slope of a secant

average rate of change from x = a to x = b = \frac{Δ y}{Δ x} = \frac{f ( b ) - f ( a )}{b - a}

This is the slope of the secant line joining the points (a, f(a)) and (b, f(b)) on the graph.

A secant is a line through two points on a curve. Its slope is exactly the average rate of change between those points. For a stone dropped from a cliff, the distance fallen is about $d (t) = 4.9 t^{2}$ metres after $t$ seconds. Between $t = 1$ and $t = 3$ the stone falls from $4.9$ m to $44.1$ m, so its average velocity is $\frac{44.1 - 4.9}{3 - 1} = 19.6$ m/s. Over the shorter interval from $t = 2$ to $t = 3$ it is $\frac{44.1 - 19.6}{1} = 24.5$ m/s. Different intervals give different averages because the stone keeps speeding up.

VideoSal Khan finds the slope of a secant line between two points on a curve and reads it as the average rate of change over that interval, the same idea this section uses for the falling stone.Watch on YouTube

Instantaneous rate of change: the slope of a tangent

How fast is the stone moving at exactly $t = 2$ ? One point does not give a slope (you need a rise AND a run), so instead take secants from $t = 2$ to a second point $t = 2 + h$ and let the second point slide closer. On a smooth graph (no breaks, corners or other sharp points) the secants settle onto one line, the tangent at that point. The slope of the tangent is the instantaneous rate of change.

Fig 1On the curve y = 0.5x² + 1, secants from x = 1 to x = 4, 2.5 and 1.5 (h = 3, 1.5 and 0.5) have slopes 2.5, 1.75 and 1.25, closing in on the tangent slope of 1 at x = 1.

h	Interval	Secant slope (average velocity, m/s)
$1$	$2 \leq t \leq 3$	$24.5$
$0.1$	$2 \leq t \leq 2.1$	$20.09$
$0.01$	$2 \leq t \leq 2.01$	$19.649$
$0.001$	$2 \leq t \leq 2.001$	$19.6049$
$- 0.001$	$1.999 \leq t \leq 2$	$19.5951$
$- 0.01$	$1.99 \leq t \leq 2$	$19.551$
$- 0.1$	$1.9 \leq t \leq 2$	$19.11$
$- 1$	$1 \leq t \leq 2$	$14.7$

Secant slopes of d(t) = 4.9t² through the point t = 2. Positive h uses a later second point, negative h an earlier one.

From both sides, the secant slopes close in on $19.6$ . So the stone's instantaneous velocity at $t = 2$ s is $19.6$ m/s. Notice that every secant slope differs from $19.6$ by $4.9 h$ : as $h$ shrinks, the error shrinks with it. The next two lessons make this exact with the idea of a limit.

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Pick one of four functions (x², x³ − 3x, x³/4 or 2 sin x), set the point a and the interval width h, and watch the secant slope approach the exact tangent slope as h shrinks toward 0.

VideoA short walkthrough of estimating an instantaneous rate of change by computing secant slopes and sliding the second point closer, exactly the shrinking-interval method in this section.Watch on YouTube

When there is no tangent: a corner

The method needs a smooth graph. Take $f (x) = ∣ x - 2∣ + 1$ , whose V-shaped graph has a sharp corner at $P (2, 1)$ . Every secant from $P$ to a point on the right has slope $1$ , and every secant from $P$ to a point on the left has slope $- 1$ , no matter how small $h$ is. The secants never settle on a single line, so there is no tangent at $P$ and the instantaneous rate of change at $P$ cannot be determined. Away from the corner the graph is a straight line and the rate is easy: $- 1$ to the left, $1$ to the right.

Average versus instantaneous

An average rate of change belongs to an interval and is the slope of a secant:

\frac{f ( b ) - f ( a )}{b - a}

. An instantaneous rate of change belongs to a single point and is the slope of the tangent there. You estimate it with secants through the point from both sides, using smaller and smaller intervals, and it only exists where the graph is smooth.

Average and Instantaneous Rates of Change

Key things to remember

Average rate of change: the slope of a secant

Instantaneous rate of change: the slope of a tangent

When there is no tangent: a corner

Worked examples

Where marks get lost

Flashcards 7

Practice