Rates 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

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 $[1, 4]$:

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

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!

Secant Lines

Drawing a Secant Line

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

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

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

For $f(x) = x^2$:

• 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.

Average Rate of Change Quiz 🎯

Compute the AROC:

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

2) $f(x) = x^2 - 1$ on $[0, 3]$:

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

Rates Concepts 🔽

Exit Quiz ✅

🔬 The Difference Quotient

Part 2 of 7

From AROC to Instant Rate

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

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

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.

The Big Idea

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

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 $-1/x^2$.

Simplification Strategies

Step-by-Step Process

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

Common Expansion Patterns

• $(x+h)^2 = x^2 + 2xh + h^2$
• $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$
• $\sqrt{x+h}$: rationalize with conjugate

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 🔽

Exit Quiz ✅

📐 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:

1. Secant through $(a, f(a))$ and $(b, f(b))$ — wide interval
2. Move $b$ closer to $a$ — secant rotates
3. In the limit as $b \to a$ — secant BECOMES the tangent

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

Why This Matters

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

1. Compute $f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$ (the slope)
2. 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 $x=4$: slope $= \frac{1}{4}$.

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 $x=4$:

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

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 🔽

Exit Quiz ✅

⚡ Instantaneous Rate of Change

Part 4 of 7

AROC vs IROC

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 $f(x)=x^2$:

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
• $|f'(a)| = 0$: 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.

IROC Quiz 🎯

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 🔽

Exit Quiz ✅

🚗 Motion & Velocity Applications

Part 5 of 7

Position, Velocity, Acceleration

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

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? $s(t)=0$: $-16t^2+64t+80=0$ $t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1)=0 \Rightarrow t=5$ 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
• Instantaneous velocity at $t=1$: $v(1) = -32+64 = 32$ ft/s (upward)

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 🔽

Exit Quiz ✅

📊 Real-World Rate Applications

Part 6 of 7

Rates of Change in Context

Rate of change applies to any quantity that varies:

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

Answering Rate Questions

Always include:

1. Value: the numerical rate
2. Units: output/input
3. 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 $dr/dt$ — this is related rates 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 🔽

Exit Quiz ✅

🏆 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

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

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

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

Known DQ Results

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

1. ✅ Functions & graphs (completed)
2. ✅ Limits (computed and understood)
3. ✅ Rates of change (AROC → IROC)
4. ➡️ Next: Derivatives (formalized IROC)
5. ➡️ Then: Integrals (reverse of derivatives)
6. ➡️ 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 $x=3$: $y = 6x - ?$

Synthesis 🔽

Exit Quiz ✅