🎓 DERIVATIVE RATE OF CHANGES QUIZ

Nigerian Curriculum SS2 Level

Instructions: Choose the best answer for each question

🟢 SECTION A: EASY QUESTIONS (5 marks each)

1. If s(t) = 3t represents distance in meters, the rate of change of distance with respect to time is:

a) 3t
b) 3
c) 6t
d) 0

Correct Answer: b) 3
The derivative of s(t) = 3t is ds/dt = 3. This represents constant velocity of 3 m/s.

2. A particle moves such that x(t) = 2t + 5. Its velocity at any time t is:

a) 2t
b) 2
c) 5
d) 2t + 5

Correct Answer: b) 2
Velocity = dx/dt = d/dt(2t + 5) = 2. The constant term disappears when differentiated.

3. If temperature T(t) = 20 + 3t, the rate of temperature change per hour is:

a) 20°C/hr
b) 3°C/hr
c) 23°C/hr
d) 3t°C/hr

Correct Answer: b) 3°C/hr
dT/dt = d/dt(20 + 3t) = 3. Temperature increases at constant rate of 3°C per hour.

4. If y = 4x², then dy/dx equals:

a) 4x
b) 8x
c) 4
d) x²

Correct Answer: b) 8x
Using power rule: d/dx(4x²) = 4 × 2x¹ = 8x. This represents the rate of change of y with respect to x.

5. A car's distance is d(t) = 10t. The car's speed is:

a) 10t
b) 10
c) 5t
d) 20t

Correct Answer: b) 10
Speed = dd/dt = d/dt(10t) = 10. The car moves at constant speed of 10 units per time.

🟡 SECTION B: MEDIUM QUESTIONS (6 marks each)

6. Ball height h(t) = 20t - 5t². The velocity at t = 2 seconds is:

a) 0 m/s
b) 10 m/s
c) 20 m/s
d) -10 m/s

Correct Answer: a) 0 m/s
v(t) = dh/dt = 20 - 10t. At t = 2: v(2) = 20 - 10(2) = 0 m/s. Ball momentarily stops at maximum height.

7. Population P(t) = 1000 + 50t + 2t². Growth rate after 3 years is:

a) 50 people/year
b) 62 people/year
c) 56 people/year
d) 1162 people/year

Correct Answer: b) 62 people/year
dP/dt = 50 + 4t. At t = 3: dP/dt = 50 + 4(3) = 50 + 12 = 62 people/year.

8. Balloon radius r(t) = 2√t cm. Rate of radius change at t = 4 seconds:

a) 0.5 cm/s
b) 1 cm/s
c) 2 cm/s
d) 4 cm/s

Correct Answer: a) 0.5 cm/s
dr/dt = d/dt(2√t) = 2 × (1/2)t^(-1/2) = 1/√t. At t = 4: dr/dt = 1/√4 = 1/2 = 0.5 cm/s.

9. Cost C(x) = 100 + 5x + 0.1x². Marginal cost at x = 20 items:

a) ₦5
b) ₦9
c) ₦4
d) ₦25

Correct Answer: b) ₦9
Marginal cost = dC/dx = 5 + 0.2x. At x = 20: dC/dx = 5 + 0.2(20) = 5 + 4 = ₦9.

10. Water volume V(t) = 3t² + 2t liters. Flow rate at t = 5 minutes:

a) 32 L/min
b) 30 L/min
c) 85 L/min
d) 6 L/min

Correct Answer: a) 32 L/min
Flow rate = dV/dt = 6t + 2. At t = 5: dV/dt = 6(5) + 2 = 30 + 2 = 32 L/min.

🔴 SECTION C: HARD QUESTIONS (9 marks each)

11. Particle position x(t) = t³ - 6t² + 9t. When is the particle at rest?

a) t = 1 and t = 3
b) t = 0 and t = 6
c) t = 2 and t = 4
d) Never at rest

Correct Answer: a) t = 1 and t = 3
v(t) = dx/dt = 3t² - 12t + 9. At rest when v(t) = 0: 3t² - 12t + 9 = 0 → t² - 4t + 3 = 0 → (t-1)(t-3) = 0.

12. Profit P(x) = -2x² + 80x - 300. Production level for maximum profit rate:

a) x = 20
b) x = 40
c) x = 80
d) x = 10

Correct Answer: a) x = 20
Marginal profit = dP/dx = -4x + 80. Maximum rate when d²P/dx² = 0, but for maximum profit: dP/dx = 0 → x = 20.

13. 10m ladder slides down wall. Bottom moves at 2 m/s. When bottom is 6m from wall, top slides down at:

a) 1.5 m/s
b) 2.4 m/s
c) 1.2 m/s
d) 3 m/s

Correct Answer: a) 1.5 m/s
Using x² + y² = 100, when x = 6, y = 8. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0. So dy/dt = -(x/y)(dx/dt) = -(6/8)(2) = -1.5 m/s.

14. Drug concentration C(t) = 5t/(t² + 1). Rate of change at t = 2 hours:

a) -0.6 mg/L/hr
b) 0.6 mg/L/hr
c) -1.2 mg/L/hr
d) 2 mg/L/hr

Correct Answer: c) -1.2 mg/L/hr
Using quotient rule: dC/dt = [5(t² + 1) - 5t(2t)]/(t² + 1)² = (5 - 5t²)/(t² + 1)². At t = 2: dC/dt = (5 - 20)/25 = -15/25 = -0.6 mg/L/hr.

15. Conical tank (vertex down) r = 3m, h = 6m. Water flows in at 2 m³/min. Water level rise rate when depth = 4m:

a) 1/2π m/min
b) 1/4π m/min
c) 1/8π m/min
d) 1/π m/min

Correct Answer: a) 1/2π m/min
V = (1/3)πr²h, r/h = 3/6 = 1/2, so r = h/2. V = (1/3)π(h/2)²h = πh³/12. dV/dt = πh²/4 × dh/dt. When h = 4: 2 = π(16)/4 × dh/dt → dh/dt = 2/(4π) = 1/2π m/min.