Lesson 17: Rates of Change in Context

(a) Opening Question

A particle moves along a horizontal line so that its position (in meters) at time t (in seconds, t ≥ 0) is

s(t) = t² - 4t + 3

Without calculus vocabulary yet, answer these:

1. Where does the particle start (at t = 0)?
2. The velocity is the rate of change of position. Compute s'(t). At what time is the velocity zero?
3. At t = 0, is the particle moving left (toward smaller s) or right (toward larger s)?
4. Here is a tempting trap. At t = 0 the velocity is -4 m/s. A student writes, "The speed is -4 m/s." Is that statement correct? If not, fix it.

Spend two minutes before reading on. The heart of this lesson is the difference between velocity (which carries a sign and a direction) and speed (which never does), plus a single sign rule that tells you whether a moving object is gaining or losing speed. Master that rule and the units habit below, and you will collect points that many students leave on the table.

(Answers: starts at s(0) = 3 m; s'(t) = 2t - 4, zero at t = 2; at t=0, v = -4 < 0, so moving left; speed is |-4| = 4 m/s — speed is never negative.)

(b) Core Concepts

The derivative is a rate of change

Every derivative you have computed is secretly a rate of change. If y = f(x), then f'(a) is the instantaneous rate of change of y with respect to x at x = a. In an applied problem this is the single most tested idea in Unit 4, and the AP readers score it strictly. Two things must be right:

• The units. The units of f'(a) are always (units of output) per (unit of input). If W(t) measures water in liters and t is in minutes, then W'(5) is in liters per minute.
• The interpretation sentence. You must say what the number means, not just restate the symbol. A full-credit interpretation names the rate, the value, the units, and the moment.

Interpret f'(a) with units — worked template.

Suppose C(q) gives the cost, in dollars, of producing q widgets, and C'(200) = 1.75.

Reading: C'(200) is the instantaneous rate of change of cost with respect to quantity when 200 widgets have been produced.

Units: dollars per widget.

Sentence: "When 200 widgets have been produced, the cost is increasing at a rate of 1.75 dollars per widget." Equivalently, producing approximately one more widget raises the cost by about \$1.75. This C'(q) is exactly what economists call marginal cost.

Notice what a full answer avoids: it never says "the cost is 1.75" (that confuses the rate with the amount) and it never omits the units.

Rectilinear motion: the position–velocity–acceleration family

A particle moving along a straight line (a number line) has a position function s(t). From it we build a family of functions by differentiating:

position      s(t)
velocity      v(t) = s'(t)
acceleration  a(t) = v'(t) = s''(t)
speed         |v(t)|

• Velocity v(t) = s'(t) is a signed rate. Its sign is the direction of motion:

- v(t) > 0 → moving in the positive direction (right / up).

- v(t) < 0 → moving in the negative direction (left / down).

- v(t) = 0 → the particle is at rest (instantaneously stopped). These times are where the particle can change direction.

• Acceleration a(t) = v'(t) is the rate of change of velocity. It tells you how the velocity is changing, not how fast the particle is going.
• Speed is |v(t)|, the magnitude of velocity. Speed is never negative. A particle with velocity -7 m/s has speed 7 m/s.

A particle is at rest when v(t) = 0. It changes direction at a time where v(t) = 0 and v changes sign there (a velocity of zero alone is not enough — v could touch zero and keep its sign).

Speeding up vs. slowing down — the sign rule

This is the trap question every year. Whether a particle is speeding up or slowing down depends on how velocity and acceleration relate, not on the sign of either one alone.

The sign rule. A particle is

speeding up when v(t) and a(t) have the SAME sign (both positive or both negative), and

slowing down when v(t) and a(t) have OPPOSITE signs.

Why is this true? Speed is |v|. The speed is increasing exactly when the velocity is moving away from zero. If v > 0 and a > 0, velocity is positive and growing more positive — moving away from zero — so speed increases. If v < 0 and a < 0, velocity is negative and becoming more negative — also moving away from zero — so speed increases again. When the signs disagree, acceleration pulls the velocity back toward zero, so speed drops. (Equivalently: the product v(t)·a(t) > 0 means speeding up; v(t)·a(t) < 0 means slowing down.)

Example. Let v(t) = t² - 4 and a(t) = 2t for t ≥ 0. At t = 1: v(1) = -3 < 0 and a(1) = 2 > 0. Opposite signs → slowing down. At t = 3: v(3) = 5 > 0 and a(3) = 6 > 0. Same sign → speeding up. Notice that at t = 1 the acceleration is positive even though the particle is slowing down — proof that "positive acceleration" does not mean "speeding up."

Calculator skills for motion (Part B / Section II Part A)

When a calculator is allowed you do not have to differentiate by hand to evaluate velocity at a point or to find when the particle is at rest.

TI-84 — velocity at a point (numerical derivative):
  MATH → 8:nDeriv(  →  nDeriv(s(T), T, 5)
  gives v(5) = s'(5).

TI-84 — when is the particle at rest (solve v(t)=0):
  Y1 = derivative or known v(t); use 2nd→CALC→2:zero, or
  MATH → Solver/  PlySmlt2, to solve v(t) = 0 on the interval.

Acceleration at a point is the numerical derivative of velocity: nDeriv(v(T), T, 5) gives a(5). Report calculator answers to three decimal places unless told otherwise, and still write a justification sentence — the reader scores the reasoning, not the keystrokes.

Other applied rates

The same interpretation discipline applies everywhere derivatives appear:

• Marginal cost / marginal revenue. If R(q) is revenue, R'(q) is marginal revenue in dollars per unit — the approximate revenue from the next unit.
• Science rates. If T(t) is temperature in °C and t is in minutes, T'(t) is in °C per minute; if P(h) is pressure in kilopascals at altitude h km, P'(h) is in kPa per km.
• Flow / accumulation rates. If V(t) is volume in gallons and t in hours, V'(t) is in gallons per hour. A positive value means the volume is increasing; a negative value means it is decreasing (draining).

In every case: output-units per input-unit, plus a sentence about the meaning.

(c) Worked Examples

Example 1 — Analyzing particle motion (foundation) · [NO CALC]

Problem. A particle moves along the x-axis with position s(t) = t³ - 6t² + 9t for t ≥ 0, where s is in meters and t in seconds.

(i) Find v(t) and a(t). (ii) When is the particle at rest? (iii) On what intervals is the particle moving left? (iv) When is the particle speeding up?

Strategy. Differentiate twice, factor for sign analysis, then apply the sign rule.

Solution.

v(t) = s'(t) = 3t² - 12t + 9 = 3(t - 1)(t - 3)
a(t) = v'(t) = 6t - 12 = 6(t - 2)

(ii) v(t) = 0 at t = 1 and t = 3. The particle is at rest at t = 1 s and t = 3 s.

(iii) Moving left means v(t) < 0. The factors (t-1)(t-3) are negative between the roots, so v < 0 on 1 < t < 3. The particle moves left on the interval (1, 3).

(iv) Sign chart:

The particle is speeding up on (1, 2) and on (3, ∞).

Justification (model language). "On (1, 2), v(t) < 0 and a(t) < 0; because velocity and acceleration have the same sign, the speed is increasing, so the particle is speeding up."

Example 2 — Interpret a derivative with units (procedural) · [CALC]

Problem. A tank is being filled with water. The volume of water (in liters) after t minutes is V(t) = 50 + 12t - 0.2t² for 0 ≤ t ≤ 30.

(i) Find V'(t) and V'(10). (ii) Using correct units, interpret V'(10) in the context of the problem. (iii) Is the tank filling or draining at t = 10?

Strategy. Differentiate, evaluate, then write a complete interpretation sentence with units.

Solution.

V'(t) = 12 - 0.4t
V'(10) = 12 - 0.4(10) = 12 - 4 = 8

(Calculator check: nDeriv(50+12T-0.2T², T, 10) returns 8.)

(ii) Interpretation. V'(10) = 8 liters per minute. "At time t = 10 minutes, the volume of water in the tank is increasing at a rate of 8 liters per minute." (Units: liters per minute = output unit per input unit.)

(iii) Since V'(10) = 8 > 0, the volume is increasing, so the tank is filling at t = 10.

Example 3 — Marginal analysis (procedural) · [NO CALC]

Problem. A company's revenue from selling q units is R(q) = 80q - 0.05q² dollars.

(i) Find the marginal revenue function R'(q). (ii) Compute and interpret R'(300). (iii) Use the marginal revenue to estimate the revenue from selling the 301st unit.

Solution.

R'(q) = 80 - 0.1q
R'(300) = 80 - 0.1(300) = 80 - 30 = 50

(ii) R'(300) = 50 dollars per unit. "When 300 units have been sold, revenue is increasing at a rate of 50 dollars per unit." This is the marginal revenue at q = 300.

(iii) Marginal revenue approximates the revenue from the next unit: selling the 301st unit adds approximately \$50 in revenue. (Exact check: R(301) - R(300) = (80·301 - 0.05·301²) - (80·300 - 0.05·300²) = 19549.95 - 19500 = 49.95, confirming the ≈ $50 estimate.)

Example 4 — Motion from a velocity expression (AP level) · [CALC]

Problem. A particle moves along a line with velocity v(t) = t·cos(t) - 1 for 0 ≤ t ≤ 4 (t in seconds, v in m/s).

(i) Find the time(s) in (0, 4) at which the particle is at rest. (ii) Find the acceleration at t = 3 and state its units. (iii) At t = 3, is the particle speeding up or slowing down? Justify.

Solution.

(i) Solve v(t) = 0 with the calculator: graph Y1 = X·cos(X) - 1 and use 2nd → CALC → 2:zero on (0, 4). The solutions are t ≈ 1.380 and t ≈ 3.323 seconds. The particle is at rest at t ≈ 1.380 s and t ≈ 3.323 s.

(ii) a(3) = v'(3). Use nDeriv(X·cos(X) - 1, X, 3) ≈ -2.566. So a(3) ≈ -2.566 m/s².

(iii) Compute v(3) = 3·cos(3) - 1 ≈ 3(-0.98999) - 1 ≈ -3.970. So v(3) ≈ -3.970 < 0 and a(3) ≈ -2.566 < 0.

Justification. "At t = 3, v(3) < 0 and a(3) < 0; velocity and acceleration have the same sign, so the speed is increasing and the particle is speeding up."

(d) Common Mistakes

Confusing velocity with speed.

What students do: Report a negative speed, e.g., "speed = -5 m/s," or treat v(t) and speed as the same thing.

Why it's wrong: Speed is the magnitude |v(t)| and can never be negative; velocity carries the sign that encodes direction.

Fix: Speed = |v(t)|. If v = -5, then speed = 5. Use "velocity" when direction matters, "speed" only for magnitude.

Thinking "positive acceleration = speeding up."

What students do: Conclude the particle speeds up whenever a(t) > 0 (or slows down whenever a(t) < 0).

Why it's wrong: Speeding up depends on the relationship between v and a, not on a's sign alone. A particle with v < 0 and a > 0 is slowing down even though acceleration is positive.

Fix: Apply the sign rule — same sign ⇒ speeding up; opposite signs ⇒ slowing down (equivalently, v·a > 0 ⇒ speeding up).

Omitting units in an interpretation.

What students do: Write "V'(10) = 8, the volume is increasing" with no units.

Why it's wrong: The AP rubric awards the interpretation point only when units are present and correct. A rate with no units is incomplete.

Fix: Always attach (output unit) per (input unit) — here, "8 liters per minute."

Saying "speed" when you mean "velocity" (and vice versa) in justifications.

What students do: Write "the speed is negative so it moves left," or "the velocity is increasing" when they mean the speed is increasing.

Why it's wrong: The grader reads literally. "Speed increasing" and "velocity increasing" are different claims; only the velocity sign indicates direction.

Fix: Direction ↔ sign of velocity. Speeding up/slowing down ↔ whether speed (|v|) increases or decreases.

Treating every v(t) = 0 as a direction change.

What students do: Assume the particle reverses direction at every time velocity is zero.

Why it's wrong: The velocity must actually change sign. If v touches zero without changing sign, the particle keeps going the same way.

Fix: Check the sign of v on both sides before claiming a direction change.

(e) Practice Problems

For multiple choice, choose the single best answer. Calculator labels match AP rules.

(Short answer)The temperature of a cup of coffee is H(t) °C after t minutes, and H'(8) = -2.5. Interpret H'(8) in context, with units, in a complete sentence.

(Short answer)A drone rises so that its height is h(t) = 30t - 5t² + 0.1t³ meters, t in seconds, 0 ≤ t ≤ 10. (a) Find h'(4) and h''(4) (use nDeriv if you wish). (b) Is the drone speeding up or slowing down at t = 4? Justify using velocity and acceleration.

(Short answer)A particle moves with position s(t) = t³ - 3t² for t ≥ 0. Find all times at which the particle changes direction, and justify that a change of direction occurs at each such time.

(Justification)A particle has v(2) = 5 m/s and a(2) = -3 m/s². A student claims, "Since acceleration is negative, the particle is slowing down." State whether the conclusion is correct and give the correct justification.

(Justification)Profit from selling q units is P(q) = -0.02q² + 15q - 500 dollars. (a) Find and interpret P'(250) with units. (b) Explain what it means, in business terms, that P'(q) = 0 at q = 375.

(f) AP Exam Focus

FRQ — Particle Motion (Section II style). Calculator NOT permitted. Total: 9 points.

A particle moves along a horizontal line. For 0 ≤ t ≤ 6, the particle's position is given by

s(t) = t³ - 9t² + 24t

where s is measured in meters and t is measured in seconds.

(a) Find the velocity v(t) and the acceleration a(t) of the particle. (2 points)

(b) Find all times t in the open interval (0, 6) at which the particle is at rest, and find all intervals on which the particle is moving to the left. Justify your answer. (3 points)

(c) Determine whether the particle is speeding up or slowing down at t = 3.5. Justify your answer. (2 points)

(d) Find v(5) and interpret its meaning in the context of the problem, using correct units. (2 points)

Model Solution

(a) Differentiate the position function.

v(t) = s'(t) = 3t² - 18t + 24 = 3(t - 2)(t - 4)
a(t) = v'(t) = 6t - 18 = 6(t - 3)

(b) The particle is at rest where v(t) = 0:

3(t - 2)(t - 4) = 0  ⇒  t = 2  and  t = 4

The particle is at rest at t = 2 s and t = 4 s.

The particle moves to the left where v(t) < 0. Since v(t) = 3(t-2)(t-4), the product (t-2)(t-4) is negative between its roots:

v(t) < 0  for  2 < t < 4

The particle moves to the left on the interval (2, 4), because v(t) < 0 there.

(c) Evaluate velocity and acceleration at t = 3.5:

v(3.5) = 3(3.5 - 2)(3.5 - 4) = 3(1.5)(-0.5) = -2.25 < 0
a(3.5) = 6(3.5 - 3) = 6(0.5) = 3 > 0

At t = 3.5, v(3.5) < 0 and a(3.5) > 0. Because the velocity and acceleration have opposite signs, the speed is decreasing, so the particle is slowing down at t = 3.5.

(d) Evaluate velocity at t = 5:

v(5) = 3(5 - 2)(5 - 4) = 3(3)(1) = 9

v(5) = 9 meters per second. Interpretation: "At time t = 5 seconds, the particle is moving to the right (positive direction) at a velocity of 9 meters per second." (Units: meters per second.)

Scoring Commentary — where students lose points

• Part (a) [2 pts]: 1 pt for a correct v(t), 1 pt for a correct a(t). A common loss: differentiating s only once and calling the result acceleration. Factoring v(t) is optional but pays off in parts (b) and (c).
• Part (b) [3 pts]: 1 pt for the rest times t = 2, 4; 1 pt for the interval (2, 4); 1 pt for the justification v(t) < 0. Students who give the interval but never reference the sign of velocity lose the justification point. Stating v < 0 on (2,4) is the load-bearing phrase.
• Part (c) [2 pts]: 1 pt for both signs (v < 0, a > 0); 1 pt for the conclusion with reasoning ("opposite signs ⇒ slowing down"). A bare "slowing down" with no sign comparison earns 0 of the 2. Saying "acceleration is positive so it speeds up" is the classic wrong answer and earns nothing.
• Part (d) [2 pts]: 1 pt for v(5) = 9; 1 pt for an interpretation that includes units (meters per second) and the meaning (moving right / in the positive direction). Dropping the units or writing "the position is 9" forfeits the interpretation point.

Justification language that earns full credit: "Because v and a have opposite signs at t = 3.5, the particle is slowing down." Vague phrasing like "it's slowing because of the numbers" earns no justification credit.