🎯⭐ INTERACTIVE LESSON

Rates of Change

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Rates of Change - Complete Interactive Lesson

Part 1: Average Rate of Change

📈 Average Rate of Change

Part 1 of 7

What Is a Rate of Change?

A rate of change measures how fast one quantity changes relative to another.

$\text{Average Rate of Change} = \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}$

This is the slope of the secant line through $(a, f(a))$ and $(b, f(b))$ .

Familiar Examples

Context	Rate of Change
Distance/Time	Speed (mph)
Cost/Items	Price per item
Population/Year	Growth rate
Temperature/Hour	Cooling/heating rate

Connection to Slope

For a linear function $f(x) = mx + b$ :

The rate of change is constant = $m$
Every secant line has the same slope

For nonlinear functions, the rate of change varies depending on the interval.

Worked Examples

Example 1: Polynomial

$f(x) = x^2$ . Average rate of change on $[1, 4]$ :

$\frac{f (4) - f (1)}{4 - 1} = \frac{}{}$

Secant Lines

Drawing a Secant Line

The secant line through $(a, f(a))$ and $(b, f(b))$ has equation:

$y - f (a) = \frac{f (b) - f (a)}{b - a} (x - a)$

Average Rate of Change Quiz 🎯

Compute the AROC:

1) $f(x) = 3x + 2$ on $[1, 5]$ :

2) $f(x) = x^2 - 1$ on :

Rates Concepts 🔽

Exit Quiz ✅

Part 2: Secant Lines

🔬 The Difference Quotient

Part 2 of 7

From AROC to Instant Rate

The difference quotient uses a variable step size $h$ :

$\frac{f(x+h) - f(x)}{h}$

Part 3: Instantaneous Rate of Change

📐 Secant Lines to Tangent Lines

Part 3 of 7

The Visual Story

As the two points on a curve get closer together, the secant line rotates toward the tangent line:

Secant through $(a, f(a))$ and $(b, f(b))$ — wide interval
Move $b$ closer to — secant rotates

Part 4: Tangent Line Concept

⚡ Instantaneous Rate of Change

Part 4 of 7

AROC vs IROC

Feature	AROC	IROC
Formula	$\frac{f(b)-f(a)}{b-a}$

Part 5: Applications

🚗 Motion & Velocity Applications

Part 5 of 7

Position, Velocity, Acceleration

For a particle moving along a line with position $s(t)$ :

Quantity	Definition	Rate of
Position $s(t)$	Location at time $t$	—
Velocity

Part 6: Problem-Solving Workshop

📊 Real-World Rate Applications

Part 6 of 7

Rates of Change in Context

Rate of change applies to any quantity that varies:

Application	Function	Rate measures
Economics	Revenue $R(x)$	Marginal revenue
Biology	Population $P(t)$	Growth rate
Chemistry	Concentration

Part 7: Review & Applications

🏆 Rates of Change — Complete Synthesis

Part 7 of 7

Everything Connected

Rates of Change Master Map
│
├─ Average Rate (AROC)
│   ├─ Formula: [f(b)-f(a)]/(b-a)
│   ├─ Geometry: Secant line slope
│   └─ Physics: Average velocity
│
├─ Difference Quotient
│   ├─ Formula: [f(x+h)-f(x)]/h
│   ├─ Algebraic simplification
│   └─ Must cancel h from denominator
│
├─ Instantaneous Rate (IROC)
│   ├─ = lim(h→0) of DQ
│   ├─ Geometry: Tangent line slope
│   ├─ Physics: Instantaneous velocity
│   └─ THIS IS THE DERIVATIVE
│
└─ Applications
    ├─ Motion: position → velocity → acceleration
    ├─ Economics: cost → marginal cost
    ├─ Biology: population → growth rate
    └─ Always include units and context

Key Formulas Reference

The Core Three

$\text{AROC} = \frac{f(b)-f(a)}{b-a} \quad \text{(secant slope)}$

\frac{f ( 4 ) - f ( 1 )}{4 - 1} = \frac{16 - 1}{3} = \frac{15}{3} = 5

The secant line through $(1,1)$ and $(4,16)$ has slope 5.

Example 2: Square Root

$g(x) = \sqrt{x}$ . Average rate of change on $[4, 9]$ :

$\frac{\sqrt{9} - \sqrt{4}}{9 - 4} = \frac{3 - 2}{5} = \frac{1}{5} = 0.2$

Example 3: Word Problem

A ball's height is $h(t) = -16t^2 + 64t$ feet at time $t$ seconds.

Average velocity from $t = 1$ to $t = 3$ : $\frac{h(3) - h(1)}{3-1} = \frac{(-16(9)+192) - (-16+64)}{2} = \frac{48 - 48}{2} = 0 \text{ ft/s}$

The ball returns to the same height — zero average velocity!

y - f (a) = \frac{f ( b ) - f ( a )}{b - a} (x - a)

Decreasing Intervals

If $f(b) < f(a)$ when $b > a$ , the AROC is negative — the secant slopes downward.

Example: Finding a Secant Equation

$f(x) = x^3$ , points $(1,1)$ and $(2,8)$ :

Slope: $\frac{8-1}{2-1} = 7$
Equation: $y - 1 = 7(x - 1) \Rightarrow y = 7x - 6$

Multiple Intervals Show Changing Rates

On $[0,1]$ : AROC = $1$
On $[1,2]$ : AROC = $3$
On $[2,3]$ : AROC = $5$

The rate itself is increasing — the function curves upward faster and faster.

3) $f(x) = \frac{1}{x}$ on $[1, 4]$ :

This represents the AROC over the interval $[x, x+h]$ .

As $h \to 0$ , the secant line approaches the tangent line — giving the instantaneous rate of change.

$\text{IROC} = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$

This limit IS the derivative $f'(x)$ . But in precalculus, we focus on computing the difference quotient and understanding what happens as $h$ shrinks.

Worked Examples

Example 1: $f(x) = x^2$

$\frac{f(x+h) - f(x)}{h} = \frac{(x+h)^2 - x^2}{h}$

Expand: $(x+h)^2 = x^2 + 2xh + h^2$

$= \frac{x^2 + 2xh + h^2 - x^2}{h} = \frac{2xh + h^2}{h} = \frac{h(2x + h)}{h} = 2x + h$

As $h \to 0$ : difference quotient $\to 2x$ . So the slope at any point is $2x$ .

Example 2: $f(x) = 3x + 5$

$\frac{(3(x+h)+5) - (3x+5)}{h} = \frac{3h}{h} = 3$

Constant! The "derivative" of a linear function is its slope.

Example 3: $f(x) = 1/x$

$\frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \frac{-h}{hx(x+h)} = \frac{-1}{x(x+h)}$

As $h \to 0$ : $\frac{-1}{x^2}$ . The slope at $x$ is .

Simplification Strategies

Step-by-Step Process

Write $f(x+h)$ — replace every $x$ with $(x+h)$
Subtract $f(x)$
Expand all terms
Cancel — the $f(x)$ terms must vanish
Factor out $h$ from the numerator
Cancel the $h$ in numerator and denominator
Let $h \to 0$ (for the limit / IROC)

Common Expansion Patterns

$(x+h)^2 = x^2 + 2xh + h^2$

Key Insight

After simplifying, $h$ must cancel from the denominator. If it doesn't, you made an algebra error.

Difference Quotient Quiz 🎯

Simplify each difference quotient:

1) $f(x) = x^2 + 3x$ . Simplified DQ = $2x + h + ?$ (fill in the number)

2) $f(x) = 4x^2$ . Simplified DQ = $?x + 4h$ (fill the coefficient)

3) Limit as $h \to 0$ of DQ for $f(x) = x^3$ at $x = 2$ :

DQ Concepts 🔽

In the limit as

b \to a

— secant BECOMES the tangent

$\text{Tangent slope} = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h}$

The tangent line gives the best linear approximation to the curve at a point. It tells you:

The direction the curve is heading
The instantaneous rate of change
Whether the function is increasing or decreasing at that point

Finding Tangent Lines

Process

Compute $f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$ (the slope)
Use point-slope form: $y - f(a) = f'(a)(x - a)$

Example: Tangent to $f(x)=x^2$ at $x=3$

Slope: From the difference quotient, $f'(x) = 2x$ , so $f'(3) = 6$ .

Point: $(3, 9)$ .

Tangent: $y - 9 = 6(x - 3) \Rightarrow y = 6x - 9$

Example: Tangent to $f(x)=\sqrt{x}$ at $x=4$

DQ: $\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{\sqrt{x+h}+\sqrt{x}}$

As $h \to 0$ : slope $= \frac{1}{2\sqrt{x}}$ . At : slope .

Tangent: $y - 2 = \frac{1}{4}(x-4) \Rightarrow y = \frac{x}{4} + 1$

Linear Approximation Preview

Using the Tangent Line to Estimate

Near $x = a$ , the tangent line approximates $f$ :

$f(x) \approx f(a) + f'(a)(x - a)$

Example

Estimate $\sqrt{4.1}$ using tangent to $\sqrt{x}$ at :

$\sqrt{4.1} \approx 2 + \frac{1}{4}(4.1 - 4) = 2 + 0.025 = 2.025$

Actual: $\sqrt{4.1} = 2.02485...$ Error: $0.00015$ !

Secant Line Approximation (Less Accurate)

Using the secant through $(4,2)$ and $(9,3)$ :

$\text{slope} = \frac{3-2}{9-4} = 0.2$

Estimate: $2 + 0.2(0.1) = 2.02$ — less accurate than the tangent estimate.

This is why instantaneous rates beat average rates for local estimation.

Secant → Tangent Quiz 🎯

Find tangent line components:

1) $f(x) = x^2$ , at $x = 5$ . Tangent slope = ?

2) $f(x) = x^3$ , at $x = 1$ . Tangent slope (DQ limit: $3x^2$ ) = ?

3) Using tangent to $x^2$ at $x=3$ : estimate $f(3.1)$ ≈ ?

Tangent Concepts 🔽

Physical Interpretation

AROC of position = average velocity
IROC of position = instantaneous velocity (speedometer reading)
AROC of velocity = average acceleration
IROC of velocity = instantaneous acceleration

Computing IROC

Method 1: Difference Quotient Limit

For $f(x) = x^2 + 2x$ at $x = 3$ :

$\lim_{h \to 0}\frac{(3+h)^2 + 2(3+h) - (9+6)}{h}$

$= \lim_{h \to 0}\frac{9 + 6h + h^2 + 6 + 2h - 15}{h}$

$= \lim_{h \to 0}\frac{8h + h^2}{h} = \lim_{h \to 0}(8 + h) = 8$

Method 2: Shrinking Intervals

Approximate IROC at $x=3$ for $f(x)=x^2$ :

Interval	AROC
$[3, 4]$	$7$
$[3, 3.1]$	$6.1$

Pattern: AROC → $6$ as interval shrinks. So IROC at $x=3$ is $6$ .

Interpreting IROC

Sign of IROC

$f'(a) > 0$ : function is increasing at $a$
$f'(a) < 0$ : function is decreasing at $a$
$f'(a) = 0$ : function has a horizontal tangent (possible max/min)

Magnitude of IROC

$|f'(a)|$ is large: function is changing rapidly
$|f'(a)|$ is small: function is changing slowly
: momentarily not changing

Example: Population Growth

If $P(t) = 1000e^{0.05t}$ gives population at time $t$ :

$P'(0) = 50$ : growing at 50 organisms/year initially
$P'(10) \approx 82$ : growing faster later (exponential!)

The IROC itself is increasing — accelerating growth.

Find the IROC:

1) $f(x)=x^2-3x$ at $x=4$ (DQ simplifies to $2x+h-3$ ):

2) $f(x)=x^3$ at $x=1$ (DQ limit: $3x^2$ ):

3) Position $s(t) = t^2 + 5t$ . Instantaneous velocity at $t=3$ :

IROC Concepts 🔽

Positive vs Negative Velocity

$v(t) > 0$ : moving in the positive direction (right/up)
$v(t) < 0$ : moving in the negative direction (left/down)
$v(t) = 0$ : momentarily at rest (possible direction change)

Motion Example

Ball Thrown Upward

$s(t) = -16t^2 + 64t + 80$ feet, $t$ in seconds.

Velocity (DQ limit of $-16t^2 + 64t + 80$ gives): $v(t) = -32t + 64$

When is the ball at rest? $v(t)=0$ : $-32t+64=0 \Rightarrow t=2$ seconds

Maximum height: At $t=2$ : $s(2) = -16(4)+128+80 = 144$ feet

When does it hit ground? $s(t)=0$ : $-16t^2+64t+80=0$ seconds

Impact velocity: $v(5) = -32(5)+64 = -96$ ft/s (downward at 96 ft/s)

Average vs Instantaneous Velocity

Average velocity from $t=0$ to $t=5$ : $\frac{s(5)-s(0)}{5} = \frac{0-80}{5} = -16$ ft/s

Displacement vs Total Distance

Displacement

Change in position from $t=a$ to $t=b$ : $\text{Displacement} = s(b) - s(a)$

Can be positive, negative, or zero.

Total Distance Traveled

Sum of all |movement| regardless of direction. Must account for direction changes.

Example

A particle: $s(0)=2$ , moves right to $s(1)=7$ , then left to $s(3)=1$ .

Displacement: $s(3)-s(0) = 1-2 = -1$ (net: 1 unit left)
Total distance: $|7-2| + |1-7| = 5 + 6 = 11$ units

Key Insight

Average velocity = displacement / time (can be zero even if object moved!)

Average speed = total distance / time (always ≥ 0)

Motion Quiz 🎯

For $s(t) = -16t^2 + 48t$ :

1) Velocity function: $v(t) = -32t + ?$

2) Time when ball is at its highest (v=0): $t$ = ?

3) Maximum height: $s$ = ?

Motion Concepts 🔽

Marginal Analysis (Economics)

If $C(x)$ = total cost of producing $x$ items:

$\text{Marginal cost} \approx C'(x) = \text{IROC of cost}$

This is the cost of producing one more item. Similarly for revenue and profit.

Worked Examples

Example 1: Population Growth

$P(t) = 500e^{0.03t}$ (bacteria), $t$ in hours.

AROC from $t=0$ to $t=10$ : $\frac{P(10)-P(0)}{10} = \frac{500e^{0.3}-500}{10} = \frac{500(1.3499-1)}{10} = \frac{174.9}{10} \approx 17.5 \text{ bacteria/hr}$

Example 2: Cooling

A cup of coffee cools: $T(t) = 70 + 130e^{-0.05t}$ (°F).

$T(0) = 200°$ F (initial)
$T(10) = 70 + 130e^{-0.5} \approx 70 + 78.9 = 148.9°$ F

AROC: $\frac{148.9 - 200}{10} = -5.11°$ F/min (cooling at 5.1°/min average)

Example 3: Profit

$P(x) = -0.5x^2 + 100x - 500$ dollars for $x$ units.

AROC from $x=50$ to $x=60$ : $\frac{P(60)-P(50)}{10} = \frac{(4700)-(3250)}{10} = \frac{1450}{10} = 145 \text{ \$/unit}$

Interpreting Rates in Context

Units Matter!

Rate units = $\frac{\text{output units}}{\text{input units}}$

If $f$ measures...	And input is...	Rate units are...
Meters	Seconds	m/s
Dollars	Items	$/item
Bacteria	Hours	bacteria/hr
Gallons	Minutes	gal/min

Answering Rate Questions

Always include:

Value: the numerical rate
Units: output/input
Context: what it means practically

Good answer: "At $t=5$ minutes, the tank is draining at approximately 12 gallons per minute."

Bad answer: "The rate is 12." ❌ (no units, no context)

Related Rates Preview

If $A = \pi r^2$ and $r$ changes over time, then $A$ also changes. The rate $dA/dt$ depends on — this is in calculus.

Applications Quiz 🎯

Applied Rates:

1) Revenue $R(x) = 50x - 0.1x^2$ . AROC from $x=100$ to $x=200$ :

2) Tank drains: $V(t)= 1000 - 5t^2$ gallons. AROC from $t=0$ to $t=10$ :

3) If answer to (2) is your rate, what are its units? Enter "gal/min" or "min/gal":

Rate Contexts 🔽

$\text{DQ} = \frac{f(x+h)-f(x)}{h} \quad \text{(general secant)}$

$\text{IROC} = f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h} \quad \text{(tangent slope)}$

$f(x)$	DQ simplified	Limit ( $f'(x)$ )
$mx + b$	$m$	$m$
$x^2$	$2x + h$	$2x$
$x^3$	$3x^2 + 3xh + h^2$
$1/x$	$-1/[x(x+h)]$	$-1/x^2$
$\sqrt{x}$	$1/(\sqrt{x+h}+\sqrt{x})$

Tangent Line Formula

$y = f(a) + f'(a)(x - a)$

Bridge to Calculus

What Calculus Adds

In calculus, you'll learn shortcut rules so you don't need the limit process each time:

Power Rule: $\frac{d}{dx}[x^n] = nx^{n-1}$
Product Rule: $(fg)' = f'g + fg'$
Chain Rule: $[f(g(x))]' = f'(g(x)) \cdot g'(x)$

But the limit definition is where it all starts. Everything builds from here.

The Precalculus → Calculus Pipeline

✅ Functions & graphs (completed)
✅ Limits (computed and understood)
✅ Rates of change (AROC → IROC)
➡️ Next: Derivatives (formalized IROC)
➡️ Then: Integrals (reverse of derivatives)
➡️ Finally: FTC (connects derivatives & integrals)

You now have the conceptual foundation for ALL of calculus!

Master Rates Quiz 🎯

Mixed Practice:

1) AROC of $f(x) = x^2$ on $[1,5]$ :

2) IROC of $f(x) = x^2$ at $x = 3$ (use $f'(x)=2x$ ):

3) Tangent to $f(x)=x^2$ at $x=3$ : $y = 6x - ?$

(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3

\sqrt{x+h}

: rationalize with conjugate

∣ f^{'} (a) ∣ = 0

t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1)=0 \Rightarrow t=5

Instantaneous velocity at

t=1

v(1) = -32+64 = 32

ft/s (upward)

∣7 - 2∣ + ∣1 - 7∣ = 5 + 6 = 11

Rates of Change

Rates of Change - Complete Interactive Lesson

Part 1: Average Rate of Change

📈 Average Rate of Change

What Is a Rate of Change?

Familiar Examples

Connection to Slope

Worked Examples

Example 1: Polynomial

Secant Lines

Drawing a Secant Line

Part 2: Secant Lines

🔬 The Difference Quotient

From AROC to Instant Rate

Part 3: Instantaneous Rate of Change

📐 Secant Lines to Tangent Lines

The Visual Story

Part 4: Tangent Line Concept

⚡ Instantaneous Rate of Change

AROC vs IROC

Part 5: Applications

🚗 Motion & Velocity Applications

Position, Velocity, Acceleration

Part 6: Problem-Solving Workshop

📊 Real-World Rate Applications

Rates of Change in Context

Part 7: Review & Applications

🏆 Rates of Change — Complete Synthesis

Everything Connected

Key Formulas Reference

The Core Three

Example 2: Square Root

Example 3: Word Problem

Decreasing Intervals

Example: Finding a Secant Equation

Multiple Intervals Show Changing Rates

The Big Idea

Worked Examples

Example 1: f(x)=x2f(x) = x^2f(x)=x2

Example 2: f(x)=3x+5f(x) = 3x + 5f(x)=3x+5

Example 3: f(x)=1/xf(x) = 1/xf(x)=1/x

Simplification Strategies

Step-by-Step Process

Common Expansion Patterns

Key Insight

Why This Matters

Finding Tangent Lines

Process

Example: Tangent to f(x)=x2f(x)=x^2f(x)=x2 at x=3x=3x=3

Example: Tangent to f(x)=xf(x)=\sqrt{x}f(x)=x​ at x=4x=4x=4

Linear Approximation Preview

Using the Tangent Line to Estimate

Example

Secant Line Approximation (Less Accurate)

Physical Interpretation

Computing IROC

Method 1: Difference Quotient Limit

Method 2: Shrinking Intervals

Interpreting IROC

Sign of IROC

Magnitude of IROC

Example: Population Growth

Positive vs Negative Velocity

Motion Example

Ball Thrown Upward

Average vs Instantaneous Velocity

Displacement vs Total Distance

Displacement

Total Distance Traveled

Example

Key Insight

Marginal Analysis (Economics)

Worked Examples

Example 1: Population Growth

Example 2: Cooling

Example 3: Profit

Interpreting Rates in Context

Units Matter!

Answering Rate Questions

Related Rates Preview

Example 1: $f(x) = x^2$

Example 2: $f(x) = 3x + 5$

Example 3: $f(x) = 1/x$

Example: Tangent to $f(x)=x^2$ at $x=3$

Example: Tangent to $f(x)=\sqrt{x}$ at $x=4$