Day 9

Rates Of Change And Particle Motion

Interpreting Rates of Change Particle Motion (Rectilinear Motion)

Interpreting Rates of Change

The derivative isn't just a slope on a graph — it's a rate of change, and rates of change show up everywhere: geometry, physics, economics. Reading a derivative correctly in context (with the right units and the right sentence) is its own AP-tested skill, separate from just computing it.

Core Theorem

For

y=f(x)

\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}

is the instantaneous rate of change of

y

with respect to

x

.
In economics, if

C(x)

is the cost of producing

x

units, the marginal cost is

C'(x)

— approximately the cost of producing one additional unit:

C'(n) \approx C(n+1) - C(n)

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

The area of a circle in terms of its diameter is

A = \frac{\pi}{4}D^2

. Find

\frac{dA}{dD}

and evaluate it at

D=10

How to think about it

A derivative is a rate:

\frac{dA}{dD}

is how fast area changes per unit change in diameter. Differentiate normally, then read the number back in context with units.

Key idea: Differentiate with respect to $D$ just like any other power rule problem, then substitute.

Full solution

1. Step 1: Differentiate

\frac{dA}{dD} = \frac{\pi}{4}(2D) = \frac{\pi D}{2}

2. Step 2: Evaluate at D = 10

\frac{dA}{dD}\Big|_{D=10} = \frac{10\pi}{2} = 5\pi \approx 15.7

— the area grows by about

15.7

square units for every additional unit of diameter, at the instant

D=10

Worked Example 02Medium

A wire's mass from its left end to a point

x

meters along it is

m(x) = \sqrt{x}

. Find the linear density of the wire at

x=1

How to think about it

Key idea: Density is the rate of change of mass with respect to position: $\rho(x) = \frac{dm}{dx}$ .

Full solution

1. Step 1: Differentiate m(x)

\rho(x) = \frac{dm}{dx} = \frac{1}{2\sqrt{x}}

2. Step 2: Evaluate at x = 1

\rho(1) = \frac{1}{2\sqrt{1}} = \frac{1}{2}

(mass per unit length at that point).

Step-by-Step SOP

1
Identify the Two Quantities
Determine which variable is changing with respect to which other variable — that tells you what to differentiate and with respect to what.
2
Differentiate and Substitute
Find the derivative symbolically, then plug in the specific value asked for.
3
Translate Into a Sentence
If asked to interpret, state the rate in words with correct units and context — don't just leave a bare number.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

A factory's cost to produce

x

units is

C(x) = 10000 + 5x + 0.001x^2

dollars. Find the marginal cost

C'(500)

and explain what it means.

Need a hint?

Marginal cost is just

C'(x)

— differentiate, substitute, then translate the number into a sentence about producing one more unit.

Show solution

1. Step 1: Differentiate C(x)

C'(x) = 5 + 0.002x

2. Step 2: Evaluate at x = 500

C'(500) = 5 + 0.002(500) = 5 + 1 = 6

3. Step 3: Interpret

At a production level of 500 units, producing one additional (the 501st) unit costs approximately

6

dollars more.

Practice 02Medium

For the same cost function

C(x) = 10000 + 5x + 0.001x^2

, compare the marginal cost

C'(500)

to the actual cost of the 501st unit,

C(501) - C(500)

Need a hint?

Compute

C(500)

and

C(501)

directly, subtract, then compare to the

C'(500)

you already found.

Show solution

1. Step 1: Compute C(500) and C(501)

C(500) = 10000+2500+250 = 12750

C(501) = 10000+2505+0.001(501)^2 = 10000+2505+251.001 = 12756.001

2. Step 2: Find the Actual Cost Difference

C(501)-C(500) = 12756.001-12750 = 6.001

3. Step 3: Compare

C'(500) = 6

from the previous example, almost identical to the true difference of

6.001

— this is exactly why the derivative is used as a stand-in for marginal cost: it's a very close, instantaneous approximation of the discrete change.

Practice 03Easy

A cost function

C(x)

satisfies

C'(1000) = 9

. Write a sentence explaining what this number means in context.

Need a hint?

State the units (dollars per unit) and describe it as the approximate cost of one more unit at that production level.

Show solution

1. Interpretation

At a production level of 1000 units, the cost of producing one additional unit is approximately

9

dollars.

Common Pitfalls

⚠
Always Attach UnitsAP FRQs award points specifically for correct units when interpreting a rate of change — 'the area is increasing' is incomplete without 'square units per unit of diameter,' etc.
⚠
Marginal ≠ Total $C'(500)$ is the approximate cost of ONE more unit at that production level, not the total cost of producing 500 units — don't confuse the derivative with the original function.

Particle Motion (Rectilinear Motion)

When position is a function of time, its first and second derivatives aren't abstract slopes anymore — they're velocity and acceleration. Reading the signs of $v(t)$ and $a(t)$ together tells you the entire story of how a particle is moving.

Core Theorem

s(t)

is position, then velocity

v(t) = s'(t)

, acceleration

a(t) = v'(t) = s''(t)

, and jerk

j(t) = a'(t)

.
Speed is

|v(t)|

— note that displacement

\neq

distance traveled, just as velocity

\neq

speed.
The particle moves in the positive direction when

v(t) > 0

, negative direction when

v(t) < 0

, and is momentarily at rest when

v(t) = 0

.
The particle is speeding up when

v(t)

and

a(t)

have the SAME sign, and slowing down when they have OPPOSITE signs.

Watch the TikTok ExplanationPosition, Velocity, and Acceleration→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

A particle moves along a line with position

s(t) = t^3 - 6t^2 + 9t

(

t \ge 0

). Find its velocity and acceleration functions.

How to think about it

Position, velocity, acceleration form a derivative chain:

v = s'

a = v' = s''

. So the whole problem is just differentiating

s(t)

once, then again.

Key idea: Just differentiate once for velocity, and again for acceleration.

Full solution

1. Step 1: Differentiate for Velocity

v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)

2. Step 2: Differentiate Again for Acceleration

a(t) = v'(t) = 6t - 12 = 6(t-2)

Worked Example 02Medium

For the same particle

s(t) = t^3-6t^2+9t

, find all times when the particle is momentarily at rest.

How to think about it

Key idea: 'At rest' means velocity equals zero — set $v(t)=0$ and solve.

Full solution

1. Step 1: Set v(t) = 0

v(t) = 3(t-1)(t-3) = 0 \implies t=1 \text{ or } t=3

2. Step 2: Conclude

The particle is momentarily at rest at

t=1

and

t=3

seconds.

Step-by-Step SOP

1
Find v(t) and a(t)
Differentiate the position function once for velocity, twice for acceleration (or differentiate a given velocity function once for acceleration).
2
Build a Sign Chart
Mark the zeros of $v(t)$ and $a(t)$ on the same number line, splitting the domain into intervals.
3
Compare Signs on Each Interval
Same sign for $v$ and $a$ means speeding up; opposite signs mean slowing down; $v=0$ means momentarily at rest.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

For the same particle

s(t) = t^3-6t^2+9t

, determine the time intervals (

t \ge 0

) when the particle is moving in the positive direction.

Need a hint?

The particle moves in the positive direction whenever

v(t) > 0

— use the factored form and a sign chart around

t=1

and

t=3

Show solution

1. Step 1: Set Up the Sign Chart

v(t) = 3(t-1)(t-3)

is an upward-opening parabola in

t

with roots at

t=1, 3

, so

v(t) > 0

outside the roots and

v(t) < 0

between them.

2. Step 2: State the Intervals

v(t) > 0

[0,1)

and

(3,\infty)

— these are the intervals where the particle moves in the positive direction.

Practice 02Hard

For the same particle

s(t) = t^3-6t^2+9t

, determine when the particle is speeding up and when it's slowing down.

Need a hint?

Compare the signs of

v(t) = 3(t-1)(t-3)

and

a(t) = 6(t-2)

on each interval formed by

t=1,2,3

Show solution

1. Step 1: Build a Combined Sign Chart

The critical points from

v

and

a

split

[0,\infty)

into four intervals:

(0,1)

(1,2)

(2,3)

(3,\infty)

2. Step 2: Check Each Interval

(0,1)

v>0, a<0

(opposite) — slowing down.

(1,2)

v<0, a<0

(same) — speeding up.

(2,3)

v<0, a>0

(opposite) — slowing down.

(3,\infty)

v>0, a>0

(same) — speeding up.

3. Step 3: Conclude

Speeding up on

(1,2) \cup (3,\infty)

; slowing down on

(0,1) \cup (2,3)

Practice 03Medium

A particle's velocity is given directly as

v(t) = t^2 - 6t + 5

for

t \ge 0

. Find when the particle changes direction.

Need a hint?

A direction change happens where

v(t)

actually changes sign (crosses zero), not just touches it — factor

v(t)

and check the sign on both sides of each root.

Show solution

1. Step 1: Factor and Find Roots

v(t) = (t-1)(t-5) = 0 \implies t=1, 5

2. Step 2: Check the Sign Change

Since

v(t)

is an upward parabola with two distinct roots, it is positive for

t<1

, negative for

1<t<5

, and positive again for

t>5

— the sign genuinely flips at both roots.

3. Step 3: Conclude

The particle changes direction at

t=1

(moving forward to backward) and at

t=5

(backward to forward).

Practice 04Medium

A particle has position

s(t) = t^4 - 4t^3

. Determine whether the particle is speeding up or slowing down at

t=1

Need a hint?

Find

v(1)

and

a(1)

and compare their signs.

Show solution

1. Step 1: Find v(t) and a(t)

v(t) = 4t^3-12t^2

a(t) = 12t^2-24t

2. Step 2: Evaluate at t = 1

v(1) = 4-12 = -8

a(1) = 12-24 = -12

3. Step 3: Compare Signs and Conclude

Both

v(1)

and

a(1)

are negative — same sign — so the particle is speeding up at

t=1

Common Pitfalls

⚠
Velocity ≠ Speed, Displacement ≠ DistanceVelocity and displacement carry direction (sign); speed and distance traveled are always non-negative. Mixing these up is one of the most common AP point losses in this topic.
⚠
A Zero Velocity Isn't Always a Direction ChangeIf $v(t)$ touches zero but doesn't change sign (a repeated root), the particle momentarily stops but keeps moving the same direction — always double check the sign on both sides.
⚠
Speeding Up/Slowing Down Needs BOTH v and aYou cannot determine whether a particle is speeding up from acceleration alone — you must compare the sign of $a(t)$ to the sign of $v(t)$ at that same instant.

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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Worked Examples

1. Step 1: Differentiate

2. Step 2: Evaluate at D = 10

1. Step 1: Differentiate m(x)

2. Step 2: Evaluate at x = 1

Identify the Two Quantities

Differentiate and Substitute

Translate Into a Sentence

Practice — Your Turn

1. Step 1: Differentiate C(x)

2. Step 2: Evaluate at x = 500

3. Step 3: Interpret

1. Step 1: Compute C(500) and C(501)

2. Step 2: Find the Actual Cost Difference

3. Step 3: Compare

1. Interpretation

Worked Examples

1. Step 1: Differentiate for Velocity

2. Step 2: Differentiate Again for Acceleration

1. Step 1: Set v(t) = 0

2. Step 2: Conclude

Find v(t) and a(t)

Build a Sign Chart

Compare Signs on Each Interval

Practice — Your Turn

1. Step 1: Set Up the Sign Chart

2. Step 2: State the Intervals

1. Step 1: Build a Combined Sign Chart

2. Step 2: Check Each Interval

3. Step 3: Conclude

1. Step 1: Factor and Find Roots

2. Step 2: Check the Sign Change

3. Step 3: Conclude

1. Step 1: Find v(t) and a(t)

2. Step 2: Evaluate at t = 1

3. Step 3: Compare Signs and Conclude

Gary Chang