Lesson 25: Unit 4–5 Review — Applications of Derivatives

(a) Opening Question

A particle moves along the x-axis with velocity v(t) = t² - 6t + 8 for t ≥ 0, where t is in seconds and v is in meters per second.

Before reading on, try all four:

1. At what times is the particle at rest?
2. On what interval(s) is the particle moving left?
3. Is the particle speeding up or slowing down at t = 1? Justify.
4. The acceleration is a(t) = 2t - 6. What is the particle's minimum velocity on [0, 5], and how do you know it's a minimum?

These four questions touch four different Unit 4–5 ideas — solving for rest (motion), sign analysis (intervals), the speeding-up/slowing-down test (do velocity and acceleration agree in sign?), and a closed-interval minimum (a candidate-test/EVT idea). One velocity function, four tools. That is exactly what the Unit 4–5 portion of the exam does: it hands you one object and asks you to read it from several angles. Hold your answers — we confirm them in Core Concepts.

(b) Core Concepts

This is a synthesis lesson, not a re-teach. The single most valuable skill across Units 4–5 is diagnosis: reading a problem and naming the tool before touching algebra. Get the tool right and the execution is the easy part.

The decision guide: "What is the question asking?"

The exam will not label which tool to use. Your first move on every problem is to fill in the left column.

The two "trap doors" that cost the most points

Related rates — substitute LAST. Differentiate the geometric relationship with respect to t while every quantity is still a variable, then plug in the instantaneous values. If you substitute a value that is changing before differentiating, you freeze it and the derivative is wrong.

Motion — speeding up ⇔ v and a have the SAME sign. Not "a > 0." A particle with v < 0 and a < 0 is speeding up (moving left, getting faster). Speed = |v| is increasing exactly when v · a > 0.

Justification templates (memorize the structure, not the words)

These are the sentences that earn the justification points. Each names the trigger (what you observed) and the conclusion.

Relative maximum (First Derivative Test):

f has a relative maximum at x = c because f' changes from positive to negative at x = c.

Relative minimum:

f has a relative minimum at x = c because f' changes from negative to positive at x = c.

Inflection point:

f has a point of inflection at x = c because f'' changes sign at x = c (equivalently, f' changes from increasing to decreasing or vice versa). Note: f''(c) = 0 alone is not enough — the sign must change.

Absolute extremum on a closed interval (Candidates Test):

On [a, b], the absolute maximum of f is the largest of the values of f at the critical points and at the endpoints. List them, compare, name the winner.

Mean Value Theorem:

Since f is continuous on [a, b] and differentiable on (a, b), by the Mean Value Theorem there exists at least one c in (a, b) such that f'(c) = [f(b) − f(a)] / (b − a).

Second Derivative Test (for a max/min at a critical point c where f'(c) = 0):

Since f'(c) = 0 and f''(c) < 0, f has a relative maximum at x = c. (If f''(c) > 0, relative minimum. If f''(c) = 0, the test is inconclusive — fall back to the First Derivative Test.)

The f ↔ f' ↔ f'' relationship (the spine of Unit 5)

The most common exam error lives in this table: when the problem hands you the graph of f', an x-intercept of f' where the sign changes is a relative extremum of f — not of f'. And where f' itself has a peak or valley, that is an inflection point of f (because f' is changing direction, so f'' changes sign). Read the axis label first: are you looking at f, or at f'?

Now resolve the Opening Question

v(t) = t² − 6t + 8 = (t − 2)(t − 4).

1. At rest: v = 0 at t = 2 and t = 4 s.
2. Moving left (v < 0): the upward parabola is negative between its roots, so on (2, 4).
3. At t = 1: v(1) = 1 − 6 + 8 = 3 > 0; a(1) = 2(1) − 6 = −4 < 0. Opposite signs ⇒ slowing down.
4. Minimum velocity on [0, 5]: a(t) = 2t − 6 = 0 at t = 3. Compare v at the critical point and endpoints: v(0) = 8, v(3) = 9 − 18 + 8 = −1, v(5) = 25 − 30 + 8 = 3. The minimum velocity is −1 m/s at t = 3 — it is the minimum because it is the least of the values at the one interior critical point and the two endpoints (Candidates Test).

(c) Worked Examples

Example 1 — Related Rates (Unit 4) — [NO CALC]

Problem. Air is pumped into a spherical balloon so that its volume increases at dV/dt = 100 cm³/s. How fast is the radius increasing when the radius is 5 cm? (V = (4/3)πr³)

Strategy. Two quantities (V, r) related by a formula; one rate given, one wanted. Related rates. Differentiate w.r.t. t first; substitute r = 5 last.

Solution.

V = (4/3)πr³
dV/dt = 4πr² · dr/dt        (chain rule — note the dr/dt)
100 = 4π(5)² · dr/dt
100 = 100π · dr/dt
dr/dt = 1/π ≈ 0.318 cm/s

Justification / units. When r = 5 cm, the radius increases at 1/π ≈ 0.318 cm/s. The chain-rule factor dr/dt is what carries the time dependence — dropping it is the classic error.

Example 2 — Optimization with justification (Unit 5) — [NO CALC]

Problem. A rectangular pen is built against a straight river (no fence needed on the river side) using 200 m of fencing for the other three sides. Find the dimensions giving maximum area, and justify that it is a maximum.

Strategy. Optimization: write area in one variable using the constraint, find the critical point, justify with the Second Derivative Test.

Solution. Let x = the two sides perpendicular to the river, y = the side parallel to it. Constraint: 2x + y = 200, so y = 200 − 2x.

A(x) = x·y = x(200 − 2x) = 200x − 2x²,   0 < x < 100
A'(x) = 200 − 4x = 0  ⟹  x = 50
A''(x) = −4 < 0

Justification. Since A'(50) = 0 and A''(x) = −4 < 0, by the Second Derivative Test A has a relative maximum at x = 50; as the only critical point on the open interval, it is the absolute maximum. Then y = 200 − 100 = 100. Dimensions: 50 m × 100 m, area 5000 m².

Example 3 — Analysis from the graph of f' (Unit 5) — [NO CALC]

Problem. The graph of f', the derivative of f, is shown below. f is continuous on [−4, 4].

(i) Where does f have relative extrema?

f' changes − → + at x = −2 ⇒ relative minimum at x = −2.

f' changes + → − at x = 2 ⇒ relative maximum at x = 2.

(At x = 0, f' does not change sign — it stays positive — so f has no extremum there.)

(ii) Where is f concave up? Inflection points?

f'' is the slope of f'. f' is increasing on (−4, 0) and decreasing on (0, 4), so f'' > 0 on (−4, 0) (concave up) and f'' < 0 on (0, 4) (concave down). f has an inflection point at x = 0, because f'' changes sign there (the graph of f' changes from increasing to decreasing).

Justification (model). "f has a relative maximum at x = 2 because f' changes from positive to negative there. f has a point of inflection at x = 0 because f' changes from increasing to decreasing, so f'' changes sign."

Example 4 — Motion interpretation (Unit 4) — [CALC]

Problem. A particle moves so that its velocity is v(t) = t·cos(t) for 0 ≤ t ≤ 4 (t in seconds, v in m/s). At t = 2, is the particle speeding up or slowing down? Justify with units.

Strategy. Speeding up ⇔ v and a share a sign. Need v(2) and a(2) = v'(2).

Solution.

v(2) = 2·cos(2) ≈ 2(−0.4161) ≈ −0.832 m/s
v'(t) = cos(t) − t·sin(t)         (product rule)
a(2) = cos(2) − 2·sin(2) ≈ −0.4161 − 2(0.9093) ≈ −2.235 m/s²

TI-84 check: nDeriv(X cos(X), X, 2) ≈ −2.235.

Justification. At t = 2, v(2) ≈ −0.832 m/s < 0 and a(2) ≈ −2.235 m/s² < 0. Since velocity and acceleration have the same sign, the particle is speeding up at t = 2. (It is moving in the negative direction and its speed is increasing.)

(d) Common Mistakes

Writing "it's a max" with no justification. Saying a point is a maximum earns the answer, not the justification point. Always name the trigger: "f' changes from positive to negative" or "f''(c) < 0." On the FRQ, the justification is usually worth as much as the answer.

Confusing f with f' (and f' with f''). When the graph shown is f', an x-intercept (with sign change) is an extremum of f, and a peak/valley of the graph is an inflection point of f. Read the axis label before you analyze.

Substituting too early in related rates. Plugging in the instantaneous value before differentiating turns a changing quantity into a constant, killing its rate. Differentiate the general relationship first; substitute only at the end.

"Speeding up = a > 0." Speed is |v|. The particle speeds up when v and a have the same sign, slows down when they have opposite signs. Check both signs, never just acceleration.

Calling every f''(c) = 0 an inflection point. f''(c) = 0 is only a candidate. You must verify f'' actually changes sign at c. (E.g., f(x) = x⁴ has f''(0) = 0 but no inflection point.)

(e) Practice Problems

(Justification) A particle moves with velocity v(t) = (t − 1)(t − 5) for t ≥ 0. Determine whether the particle is speeding up or slowing down at t = 2, and justify your answer.

(Justification) Let f be continuous on [0, 4] with f(0) = 1 and f(4) = 9. State the Mean Value Theorem conclusion for f on [0, 4], including the conditions required, and give the guaranteed value of f'(c).

(f) AP Exam Focus — Free Response

[NO CALC] — 9 points total

A particle moves along the x-axis. The graph of its velocity v(t), for 0 ≤ t ≤ 8 seconds, consists of line segments and is shown below. The particle is at position x(0) = 2 meters at time t = 0.

(a) (2 pts) At what time(s) t in (0, 8) does the particle change direction? Justify your answer.

(b) (2 pts) Is the speed of the particle increasing or decreasing at t = 5? Justify your answer.

(c) (3 pts) Find the position x(8) of the particle at time t = 8. Show the work that leads to your answer.

(d) (2 pts) The acceleration a(t) = v'(t). Find a(3), and using correct units, interpret its meaning in the context of the particle's motion.

Model Solution

(a) The particle changes direction where v changes sign.

v > 0 on (0, 4) and v < 0 on (4, 8), so v changes from positive to negative at t = 4. (At t = 2, v = 2 ≠ 0 — a corner of the graph, not a sign change.)

The particle changes direction at t = 4 only, because v(t) changes from positive to negative there.

(b) Speed is increasing when v and a have the same sign. At t = 5: v(5) = −1 < 0 (on the segment from (4,0) to (6,−2)), and a(5) = v'(5) = slope of that segment = (−2 − 0)/(6 − 4) = −1 < 0.

The speed is increasing at t = 5, because v(5) < 0 and a(5) < 0 have the same sign.

(c) Displacement is ∫₀⁸ v(t) dt, the signed area between v and the t-axis.

∫₀⁴ v dt = area of triangle above axis = ½(4)(2) = 4
∫₄⁸ v dt = −(area of triangle below axis) = −½(4)(2) = −4
∫₀⁸ v dt = 4 + (−4) = 0
x(8) = x(0) + ∫₀⁸ v dt = 2 + 0 = 2 meters

x(8) = 2 meters.

(d) a(3) = v'(3) = slope of the segment from (2, 2) to (4, 0) = (0 − 2)/(4 − 2) = −1 m/s².

a(3) = −1 m/s². At t = 3 seconds, the particle's velocity is decreasing at a rate of 1 m/s per second. (Since v(3) = 1 > 0 and a(3) < 0, the particle is moving in the positive direction but slowing down.)

Scoring Commentary

Where students lose points:

• (a): Listing t = 2 because the graph "peaks" there. A peak in v is not a direction change — v must cross zero. This is the f vs f' confusion applied to motion.
• (b): Concluding from a(5) < 0 alone that the particle slows down. With v < 0, same signs mean speeding up. Must compare both signs.
• (c): Forgetting the initial position (answering 0 instead of 2), or treating the area below the axis as positive (answering 8).
• (d): Giving the number with no units or no contextual interpretation. Both are required for the second point.