f'(a) as a rate of change in context, with correct units.s(t), velocity v(t) = s'(t), acceleration a(t) = s''(t), and speed |v(t)|, and determine direction and when a particle is at rest.If y = f(x), then f'(a) is the instantaneous rate of change of y with respect to x at x = a. Two things must be right in an interpretation:
V(t) is water in liters and t is in minutes, V'(10) is in liters per minute.C'(200) = 1.75 (with C in dollars, q in widgets) is: "When 200 widgets have been produced, cost is increasing at about 1.75 dollars per widget." This is marginal cost. Never say "the cost is 1.75" — that confuses the rate with the amount.A particle on a line has position s(t), and by differentiating:
velocity v(t) = s'(t) (signed: direction of motion)
acceleration a(t) = v'(t) = s''(t)
speed |v(t)| (magnitude, never negative)
v(t) > 0 → moving in the positive direction; v(t) < 0 → negative direction.v(t) = 0 → the particle is at rest, a place where it can change direction.|v(t)|. A particle with velocity −7 m/s has speed 7 m/s.A particle changes direction at a time where v(t) = 0 and v actually changes sign there. A velocity that touches zero without changing sign is not a direction change.
This is the classic trap. Whether a particle speeds up depends on how velocity and acceleration relate, not on either sign alone.
A particle is speeding up when
v(t)anda(t)have the same sign, and slowing down when they have opposite signs.
Speed is |v|, which grows when velocity moves away from zero. If v > 0 and a > 0, velocity grows more positive — speed rises. If v < 0 and a < 0, velocity grows more negative — speed rises again. When the signs disagree, acceleration pulls velocity back toward zero and speed drops. Equivalently: v·a > 0 means speeding up. Note that positive acceleration alone does not mean speeding up — a particle with v < 0 and a > 0 is slowing down.
A related-rates problem gives the rate of one quantity and asks for the rate of a related one. The two are linked by an equation, so their rates are linked through the chain rule. If A depends on r and r depends on t:
dA/dt = (dA/dr)·(dr/dt)
The one essential habit: differentiate first, substitute second. If you plug a changing value in before differentiating, you freeze that quantity into a constant and its rate vanishes.
t.t, treating each variable as a function of time (chain rule → each gets a d(...)/dt factor).A 10-ft ladder leans against a wall; its base is pulled away at 2 ft/s. How fast is the top sliding down when the base is 6 ft out?
[GRAPH: Right triangle for the sliding ladder
- Vertical wall along the y-axis from (0,0) to (0,y)
- Horizontal ground along the x-axis from (0,0) to (x,0)
- Hypotenuse (the ladder, length 10) from (x,0) to (0,y); right angle at the origin
- Arrow on the foot in the +x direction labeled dx/dt = 2 ft/s
- Arrow on the top pointing downward labeled dy/dt = ?]
The ladder length is constant, so by the Pythagorean theorem x² + y² = 100. Differentiate with respect to t:
2x·(dx/dt) + 2y·(dy/dt) = 0
At x = 6: 36 + y² = 100 → y = 8. Substitute x = 6, y = 8, dx/dt = 2:
2(6)(2) + 2(8)·(dy/dt) = 0 ⟹ 24 + 16·(dy/dt) = 0 ⟹ dy/dt = −3/2
The top slides down at 1.5 ft/s. The negative sign matches the physical picture — a good final check.
The volume V = (1/3)πr²h has two changing variables. Keeping both forces the product rule and an unknown dr/dt. Instead, use the tank's fixed proportions (similar triangles) to write r in terms of h first. For a cone of height 12 and top radius 6, r/h = 6/12 = 1/2, so r = h/2:
V = (1/3)π(h/2)²h = πh³/12 ⟹ dV/dt = (πh²/4)·(dh/dt)
Substituting an exact geometric identity like r = h/2 before differentiating is legitimate — it holds for all time. Freezing a value (like h = 4) before differentiating is not.
f'(a) is a rate: report (output unit) per (input unit) and state its meaning, not just its value.|v| and is never negative; acceleration is s''.v and a have the same sign; slowing down ⟺ opposite signs. Positive acceleration alone proves nothing.t, giving each variable a d(...)/dt factor.r = h/2) before differentiating.s(t) = 2t³ − 9t² + 12t. Its velocity is1. B) 6t² − 18t + 12. v(t) = s'(t): differentiate term by term. Distractors: A) mis-differentiated −9t² as −9t instead of −18t. C) gives a(t) = s''(t), the acceleration. D) antidifferentiated instead of differentiating. E) dropped the constant term +12 (from 12t). Fix: velocity is the first derivative of position; apply the power rule to every term.
2. D) t = 1 and t = 2. v = 6t² − 18t + 12 = 6(t − 1)(t − 2) = 0 → t = 1, 2. Distractors: A) dropped a root. B) and E) set the wrong expression to zero (e.g., acceleration 12t − 18 = 0 gives t = 1.5). C) added a spurious t = 0. Fix: "at rest" means v(t) = 0; factor the velocity and take all roots.
v = −8 m/s, its speed is3. A) 8 m/s. Speed = |v| = |−8| = 8. Distractors: B) reported velocity as speed, but speed is never negative. C) confused "at rest" with a nonzero velocity. D) squared the velocity. E) speed is fully determined by |v|. Fix: speed is the magnitude of velocity — drop the sign.
v(t) < 0 and a(t) > 0. The particle is4. C) slowing down and moving left. v < 0 means moving in the negative direction (left); v < 0 and a > 0 are opposite signs, so slowing down. Distractors: A) misread the velocity sign as rightward and the pair as speeding up. B) correct motion state but wrong direction. D) opposite signs mean slowing, not speeding. E) v < 0 is nonzero, so not at rest. Fix: direction comes from the sign of v; speeding up/slowing down comes from whether v and a share a sign.
C(q) is the cost in dollars to produce q items and C'(150) = 4, the best interpretation is5. D) When 150 items are produced, cost is increasing at about \$4 per additional item. C'(150) is marginal cost — a rate in dollars per item. Distractors: A) and B) confuse the rate with a total cost. C) multiplies the rate by 150, which a derivative does not report. E) reports an average, not the instantaneous rate. Fix: a derivative is (output unit) per (input unit) at one instant — here dollars per additional item.
r = 10 cm?6. C) 60π cm²/s. A = πr² → dA/dt = 2πr(dr/dt) = 2π(10)(3) = 60π. Distractors: A) used πr(dr/dt), dropping the factor 2. B) an arithmetic slip. D) computed πr² forgetting dr/dt. E) used 2π(dr/dt) without the r. Fix: differentiate A = πr² to 2πr·(dr/dt); keep the chain-rule factor.
r = 3 cm? (V = (4/3)πr³)7. E) 72π cm³/s. V = (4/3)πr³ → dV/dt = 4πr²(dr/dt) = 4π(9)(2) = 72π. Distractors: A) dropped dr/dt = 2 (computed 4π·9 = 36π). B) used (4/3)πr²(dr/dt), forgetting the 3 from the power rule. C) halved the coefficient. D) used r instead of r². Fix: d/dt[(4/3)πr³] = 4πr²(dr/dt) — the 3 cancels the 1/3, and the dr/dt stays.
8. B) −5/6 ft/s. x² + y² = 169; at x = 5, y = 12. From 2x(dx/dt) + 2y(dy/dt) = 0: dy/dt = −x(dx/dt)/y = −(5)(2)/12 = −5/6. Distractors: A) used 2y without cancelling the 2 correctly. C) copied dx/dt. D) inverted the ratio x/y → y/x. E) used the hypotenuse 13 as the denominator. Fix: find the missing side from the Pythagorean relation, then solve dy/dt = −x(dx/dt)/y.
9. A) 10/3 ft/s. Similar triangles give 15/(x + s) = 6/s, so s = (2/3)x and ds/dt = (2/3)(5) = 10/3. Distractors: B) 25/3 is the speed of the shadow tip (dx/dt + ds/dt), not the shadow length. C) reports the person's speed. D) forgot to multiply by dx/dt. E) an unrelated value. Fix: shadow length rate is ds/dt; the tip's rate is d(x + s)/dt — different quantities.
10. D) 100 mph. With z² = x² + y²: z(dz/dt) = x(dx/dt) + y(dy/dt). At x = 40, y = 30, z = 50: 50(dz/dt) = 40(80) + 30(60) = 5000, so dz/dt = 100. Distractors: A) added the distances 30 + 40. B) added the speeds 60 + 80. C) reported z itself. E) an arithmetic slip. Fix: differentiate z² = x² + y² to z(dz/dt) = x(dx/dt) + y(dy/dt) and solve for dz/dt.
r = 8 cm? (A = 4πr²)11. E) −12.8π cm²/s. A = 4πr² → dA/dt = 8πr(dr/dt) = 8π(8)(−0.2) = −12.8π. Distractors: A) dropped the negative sign — the radius is decreasing. B) used 4πr(dr/dt) (wrong coefficient). C) used 2πr(dr/dt), the circle formula. D) both wrong coefficient and wrong sign. Fix: surface area A = 4πr² differentiates to 8πr(dr/dt); keep the sign of dr/dt.
r = 4 into A = πr² and then differentiates with respect to t. Which statement best describes the result?12. E) The π(4)² becomes a constant with derivative 0, so the term carrying dr/dt disappears and dA/dt = 0. Substituting r = 4 before differentiating freezes the radius, and a constant has derivative 0. Distractors: A) the method is invalid, not valid. B) the failure is not about units. C) the collapse happens regardless of dr/dt. D) the term does not merely become "too large" — it vanishes entirely. Fix: differentiate first, substitute second; never plug a changing value in before applying the chain rule.
1. B) 6t² − 18t + 12. v(t) = s'(t): differentiate term by term. Distractors: A) mis-differentiated −9t² as −9t instead of −18t. C) gives a(t) = s''(t), the acceleration. D) antidifferentiated instead of differentiating. E) dropped the constant term +12 (from 12t). Fix: velocity is the first derivative of position; apply the power rule to every term.
2. D) t = 1 and t = 2. v = 6t² − 18t + 12 = 6(t − 1)(t − 2) = 0 → t = 1, 2. Distractors: A) dropped a root. B) and E) set the wrong expression to zero (e.g., acceleration 12t − 18 = 0 gives t = 1.5). C) added a spurious t = 0. Fix: "at rest" means v(t) = 0; factor the velocity and take all roots.
3. A) 8 m/s. Speed = |v| = |−8| = 8. Distractors: B) reported velocity as speed, but speed is never negative. C) confused "at rest" with a nonzero velocity. D) squared the velocity. E) speed is fully determined by |v|. Fix: speed is the magnitude of velocity — drop the sign.
4. C) slowing down and moving left. v < 0 means moving in the negative direction (left); v < 0 and a > 0 are opposite signs, so slowing down. Distractors: A) misread the velocity sign as rightward and the pair as speeding up. B) correct motion state but wrong direction. D) opposite signs mean slowing, not speeding. E) v < 0 is nonzero, so not at rest. Fix: direction comes from the sign of v; speeding up/slowing down comes from whether v and a share a sign.
5. D) When 150 items are produced, cost is increasing at about \$4 per additional item. C'(150) is marginal cost — a rate in dollars per item. Distractors: A) and B) confuse the rate with a total cost. C) multiplies the rate by 150, which a derivative does not report. E) reports an average, not the instantaneous rate. Fix: a derivative is (output unit) per (input unit) at one instant — here dollars per additional item.
6. C) 60π cm²/s. A = πr² → dA/dt = 2πr(dr/dt) = 2π(10)(3) = 60π. Distractors: A) used πr(dr/dt), dropping the factor 2. B) an arithmetic slip. D) computed πr² forgetting dr/dt. E) used 2π(dr/dt) without the r. Fix: differentiate A = πr² to 2πr·(dr/dt); keep the chain-rule factor.
7. E) 72π cm³/s. V = (4/3)πr³ → dV/dt = 4πr²(dr/dt) = 4π(9)(2) = 72π. Distractors: A) dropped dr/dt = 2 (computed 4π·9 = 36π). B) used (4/3)πr²(dr/dt), forgetting the 3 from the power rule. C) halved the coefficient. D) used r instead of r². Fix: d/dt[(4/3)πr³] = 4πr²(dr/dt) — the 3 cancels the 1/3, and the dr/dt stays.
8. B) −5/6 ft/s. x² + y² = 169; at x = 5, y = 12. From 2x(dx/dt) + 2y(dy/dt) = 0: dy/dt = −x(dx/dt)/y = −(5)(2)/12 = −5/6. Distractors: A) used 2y without cancelling the 2 correctly. C) copied dx/dt. D) inverted the ratio x/y → y/x. E) used the hypotenuse 13 as the denominator. Fix: find the missing side from the Pythagorean relation, then solve dy/dt = −x(dx/dt)/y.
9. A) 10/3 ft/s. Similar triangles give 15/(x + s) = 6/s, so s = (2/3)x and ds/dt = (2/3)(5) = 10/3. Distractors: B) 25/3 is the speed of the shadow tip (dx/dt + ds/dt), not the shadow length. C) reports the person's speed. D) forgot to multiply by dx/dt. E) an unrelated value. Fix: shadow length rate is ds/dt; the tip's rate is d(x + s)/dt — different quantities.
10. D) 100 mph. With z² = x² + y²: z(dz/dt) = x(dx/dt) + y(dy/dt). At x = 40, y = 30, z = 50: 50(dz/dt) = 40(80) + 30(60) = 5000, so dz/dt = 100. Distractors: A) added the distances 30 + 40. B) added the speeds 60 + 80. C) reported z itself. E) an arithmetic slip. Fix: differentiate z² = x² + y² to z(dz/dt) = x(dx/dt) + y(dy/dt) and solve for dz/dt.
11. E) −12.8π cm²/s. A = 4πr² → dA/dt = 8πr(dr/dt) = 8π(8)(−0.2) = −12.8π. Distractors: A) dropped the negative sign — the radius is decreasing. B) used 4πr(dr/dt) (wrong coefficient). C) used 2πr(dr/dt), the circle formula. D) both wrong coefficient and wrong sign. Fix: surface area A = 4πr² differentiates to 8πr(dr/dt); keep the sign of dr/dt.
12. E) The π(4)² becomes a constant with derivative 0, so the term carrying dr/dt disappears and dA/dt = 0. Substituting r = 4 before differentiating freezes the radius, and a constant has derivative 0. Distractors: A) the method is invalid, not valid. B) the failure is not about units. C) the collapse happens regardless of dr/dt. D) the term does not merely become "too large" — it vanishes entirely. Fix: differentiate first, substitute second; never plug a changing value in before applying the chain rule.