Single-Variable-Calculus · Unit 2 · Video 1 · Interactive Practice

Twice the Average: The Derivative as a Rate of Change

IKey Formulas

Formula	Name	What it measures
$\dfrac{\Delta y}{\Delta x} = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1}$	Average rate of change	Slope of the chord across the interval
$\dfrac{dy}{dx} = \displaystyle\lim_{\Delta x \to 0} \dfrac{\Delta y}{\Delta x}$	Instantaneous rate of change	Slope of the tangent at a single instant
$h(t) = 80 - 5t^2$	Pumpkin drop, height in metres	$h(0) = 80$ and $h(4) = 0$ — a four-second fall
$\dfrac{dh}{dt} = -10t, \qquad \dfrac{d^2h}{dt^2} = -10$	Speed and acceleration	Each is the rate of change of the one above it

Key Insight: The pumpkin averages $-20$ m/s over the whole fall but lands with $h'(4) = -40$ m/s. An average taken over an interval can miss the moment that matters by a factor of two.

IIVisualization 1 — Average Rate of Change Over an Interval

Between any two instants of the fall, one number summarises the motion: the slope of the chord.

t₁ (s) 0 t₂ (s) 4

💡 On this parabola the average works out to $-5(t_1 + t_2)$ , so no interval inside the four-second fall can average the $-40$ m/s the pumpkin reaches at impact.

IIIVisualization 2 — Closing the Window on the Instant

As the averaging window closes on $t = 4$ , the chord slope runs from $-20$ all the way to $-40$ .

window Δt (s) 4.00

💡 Two honest footnotes on the model: the real building is a little taller than $80$ m, and $h = 80 - 5t^2$ ignores air resistance.

IVVisualization 3 — Height, Speed, Acceleration

Each graph's slope is the next graph's height, from height down to a constant acceleration.

t (s) 2.0

💡 Read the chain upward instead and it explains the formula: a constant acceleration of $-10$ m/s² forces the speed to be linear in $t$ , which forces the height to be quadratic.

VQuiz Questions

Problem 1 · Average Rate on an Interval

Given: the pumpkin's height is $h(t) = 80 - 5t^2$ metres — find the average rate of change of $h$ over $1 \le t \le 2$ .

✅ Correct!

h(1) = 75

and

h(2) = 60

, so the chord slope is

\frac{60 - 75}{2 - 1} = -15

m/s.

❌ Not quite. Those are the instantaneous speeds at the two endpoints:

h'(1) = -10

and

h'(2) = -20

. An average rate needs the chord,

\frac{h(2) - h(1)}{2 - 1}

, not a tangent.

❌ Right size, wrong sign. The numerator is final minus initial,

h(2) - h(1) = 60 - 75 = -15

, and the pumpkin is falling, so the rate is negative.

Show solution

Evaluate the height at both ends of the interval:

h(1) = 80 - 5(1)^2 = 75, \qquad h(2) = 80 - 5(2)^2 = 60

Now form the difference quotient over $\Delta t = 2 - 1 = 1$ :

\frac{\Delta h}{\Delta t} = \frac{h(2) - h(1)}{2 - 1} = \frac{60 - 75}{1} = -15 \ \mathrm{m/s}

As a check, the general average over $[t_1, t_2]$ for this height is

\frac{-5t_2^2 + 5t_1^2}{t_2 - t_1} = -5(t_1 + t_2),

and $-5(1 + 2) = -15$ . Notice it lies between $h'(1) = -10$ and $h'(2) = -20$ , as an average of a steadily increasing speed must.

Problem 2 · Watch the Order of Subtraction

Given: a student computes the pumpkin's average velocity over the whole fall as $\dfrac{80 - 0}{4 - 0} = +20$ m/s, but the correct value is $-20$ m/s. Which single step is wrong?

✅ Correct!

\frac{h(4) - h(0)}{4 - 0} = \frac{0 - 80}{4} = -20

m/s. Writing

80 - 0

on top reverses the change in height and flips the sign.

❌ That fixes the answer by accident. Reversing the bottom gives

\frac{80}{-4} = -20

, the right number — but the definition pins the order in both slots at once:

\frac{h(t_2) - h(t_1)}{t_2 - t_1}

, with the same

t_2

on top and bottom.

❌ Not quite.

h(4) = 80 - 5(4)^2 = 0

is correct, and

0 \le t \le 4

really is the whole fall. The error is in the order of subtraction.

Show solution

The average rate of change is defined as

\frac{\Delta h}{\Delta t} = \frac{h(t_2) - h(t_1)}{t_2 - t_1},

with the later value first in both the numerator and the denominator. Here $t_1 = 0$ , $t_2 = 4$ , $h(0) = 80$ and $h(4) = 0$ :

\frac{\Delta h}{\Delta t} = \frac{0 - 80}{4 - 0} = \frac{-80}{4} = -20 \ \mathrm{m/s}

The student wrote $h(t_1) - h(t_2)$ on top while keeping $t_2 - t_1$ underneath, which negates the whole quotient. The minus sign is not decoration: it records that the height is decreasing, so the pumpkin averages $20$ m/s downward.

Problem 3 · A Shorter Drop, End to End

Given: a stone is released from a $45$ m ledge and its height is $s(t) = 45 - 5t^2$ metres, with $t$ in seconds.

When does it reach the ground?

How fast is it moving at impact?

✅ Correct!

s(t) = 0

t = 3

, and

s'(t) = -10t

gives

s'(3) = -30

m/s — again exactly twice the

-15

m/s average over the fall.

❌ Check the landing time. Set

45 - 5t^2 = 0

, so

5t^2 = 45

t^2 = 9

and

t = 3

❌ Check the impact speed. Differentiate first —

s'(t) = -10t

— and only then substitute the landing time.

❌ That is the average, not the instant.

\frac{s(3) - s(0)}{3 - 0} = \frac{0 - 45}{3} = -15

m/s is the chord slope over the whole fall; the impact speed is the tangent slope at

t = 3

Show solution

Step 1 — landing time. The stone is on the ground when $s(t) = 0$ :

45 - 5t^2 = 0 \;\Longrightarrow\; t^2 = 9 \;\Longrightarrow\; t = 3 \ \mathrm{s}

(The root $t = -3$ is before the release.)

Step 2 — impact speed. Differentiate, using $\frac{d}{dt}(45) = 0$ and $\frac{d}{dt}t^2 = 2t$ :

\frac{ds}{dt} = -5 \cdot 2t = -10t \qquad\Longrightarrow\qquad s'(3) = -10(3) = -30 \ \mathrm{m/s}

Compare with the average.

\frac{\Delta s}{\Delta t} = \frac{0 - 45}{3 - 0} = -15 \ \mathrm{m/s}

The impact speed is twice the average again — that factor of two is not special to $80$ m, it is what a quadratic height with a start from rest always produces.

Problem 4 · From Charge to Current

Given: the charge on a plate is $q(t) = 3t^2 + \dfrac{1}{t}$ coulombs and the current is $i = \dfrac{dq}{dt}$ — find the current at $t = 1$ s.

✅ Correct!

\frac{d}{dt}(3t^2) = 6t

and

\frac{d}{dt}t^{-1} = -t^{-2}

, so

i(t) = 6t - \frac{1}{t^2}

and

i(1) = 6 - 1 = 5

❌ Check the sign of the second term. The power rule at

n = -1

gives

(-1)t^{-2} = -\frac{1}{t^2}

, so that term is subtracted:

6 - 1 = 5

, not

6 + 1

❌ The $\frac{1}{t}$ term is not constant. Its derivative is

-\frac{1}{t^2}

, which contributes

-1

t = 1

❌ Not quite. Differentiate term by term with the power rule:

3t^2 \to 6t

and

t^{-1} \to -t^{-2}

, then set

t = 1

Show solution

Current is the rate of change of charge, so the whole problem is one derivative. Write the second term as a power:

q(t) = 3t^2 + t^{-1}

The power rule $\frac{d}{dt}t^n = nt^{\,n-1}$ was proved for $n = 1, 2, 3, \dots$ , but it also holds at $n = -1$ — that is exactly the rule $\frac{d}{dt}\frac{1}{t} = -\frac{1}{t^2}$ obtained straight from the limit definition:

\frac{dq}{dt} = 3(2t) + (-1)t^{-2} = 6t - \frac{1}{t^2}

Evaluate at $t = 1$ :

i(1) = 6(1) - \frac{1}{1^2} = 6 - 1 = 5 \ \mathrm{A}

Same derivative, new reading: with $q$ in coulombs and $t$ in seconds, $dq/dt$ comes out in coulombs per second — amperes.

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