Secondary 2 Mathematics Calculus Quiz

Secondary 2 Mathematics Quiz - Calculus (Rates of Change & Kinematics) - Answer Key

Note to Students: This topic introduces the concept of differentiation, which finds the rate of change.

• Displacement ($s$) $\xrightarrow{\text{differentiate}}$ Velocity ($v = \frac{ds}{dt}$) $\xrightarrow{\text{differentiate}}$ Acceleration ($a = \frac{dv}{dt}$).
• Power Rule: If $y = ax^n$, then $\frac{dy}{dx} = anx^{n-1}$.
• Constant Rule: The derivative of a constant is 0.

Section A: Short Questions

1. Answer: 19 m/s

• Working: $s = 3t^2 - 5t$ Velocity $v = \frac{ds}{dt} = 2(3)t^{2-1} - 1(5)t^{1-1} = 6t - 5$. When $t = 4$, $v = 6(4) - 5 = 24 - 5 = 19$.
• Concept: Velocity is the first derivative of displacement with respect to time.

2. Answer: $12x^2 - 2$

• Working: $y = 4x^3 - 2x + 7$ $\frac{dy}{dx} = 3(4)x^{3-1} - 1(2)x^{1-1} + 0$ $\frac{dy}{dx} = 12x^2 - 2$.
• Concept: Apply power rule to each term. The derivative of the constant 7 is 0.

3. Answer: $3x^2$

• Working: Volume of cube $V = x^3$. $\frac{dV}{dx} = 3x^{3-1} = 3x^2$.
• Concept: Differentiating volume with respect to side length gives the rate of change of volume per unit change in side.

4. Answer: 2 m/s$^2$

• Working: $s = t^2 + 10t$ Velocity $v = \frac{ds}{dt} = 2t + 10$. Acceleration $a = \frac{dv}{dt} = 2$.
• Concept: Acceleration is the derivative of velocity (second derivative of displacement). Since $v$ is linear, $a$ is constant.

5. Answer: 2

• Working: $y = x^2 - 4x + 3$ Gradient function $\frac{dy}{dx} = 2x - 4$. At $x = 3$, Gradient $= 2(3) - 4 = 6 - 4 = 2$.
• Concept: The derivative $\frac{dy}{dx}$ represents the gradient of the tangent to the curve at a specific $x$ value.

6. Answer: $-\frac{1}{4}$ or $-0.25$

• Working: $y = x^{-2}$ $\frac{dy}{dx} = -2x^{-3} = -\frac{2}{x^3}$. When $x = 2$, $\frac{dy}{dx} = -\frac{2}{2^3} = -\frac{2}{8} = -\frac{1}{4}$.
• Concept: Rewrite $\frac{1}{x^2}$ as $x^{-2}$ to apply the power rule.

7. Answer: $10\pi$ cm

• Working: $A = \pi r^2$ $\frac{dA}{dr} = 2\pi r$. When $r = 5$, $\frac{dA}{dr} = 2\pi(5) = 10\pi$.
• Concept: Rate of change of area with respect to radius.

8. Answer: $t = \frac{1}{\sqrt{6}}$ (or approx $0.408$)

• Working: $s = 2t^3 - t$ $v = \frac{ds}{dt} = 6t^2 - 1$. At rest, $v = 0 \Rightarrow 6t^2 - 1 = 0 \Rightarrow t^2 = \frac{1}{6}$. Since $t > 0$, $t = \sqrt{\frac{1}{6}} = \frac{1}{\sqrt{6}}$.
• Concept: "At rest" means velocity is zero. Time must be positive.

9. Answer: $20x^3 - 6x + 2$

• Working: $y = 5x^4 - 3x^2 + 2x - 1$ $\frac{dy}{dx} = 4(5)x^3 - 2(3)x + 1(2) - 0$ $\frac{dy}{dx} = 20x^3 - 6x + 2$.
• Concept: Standard polynomial differentiation.

10. Answer: 20 m/s

• Working: $h = 20t - 5t^2$ Velocity $v = \frac{dh}{dt} = 20 - 10t$. Initial velocity is at $t = 0$. $v(0) = 20 - 10(0) = 20$.
• Concept: Initial value implies substituting $t=0$ into the velocity equation.

Section B: Structured Questions

11. Kinematics of a Particle

(a) Expression for velocity [2 marks]

• Answer: $v = 3t^2 - 12t + 9$
• Working: $s = t^3 - 6t^2 + 9t$ $v = \frac{ds}{dt} = 3t^2 - 12t + 9$.
• Marking: 1 mark for correct power rule application, 1 mark for final expression.

(b) Times when particle is at rest [2 marks]

• Answer: $t = 1$ s and $t = 3$ s
• Working: At rest, $v = 0$. $3t^2 - 12t + 9 = 0$ Divide by 3: $t^2 - 4t + 3 = 0$ Factorise: $(t - 3)(t - 1) = 0$ $t = 1$ or $t = 3$.
• Marking: 1 mark for setting $v=0$ and solving quadratic, 1 mark for both correct values.

(c) Acceleration when $t = 1$ [2 marks]

• Answer: $-6$ m/s$^2$
• Working: $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 9) = 6t - 12$. When $t = 1$, $a = 6(1) - 12 = -6$.
• Marking: 1 mark for finding $a(t)$, 1 mark for substitution and final answer.

12. Marginal Cost

(a) Marginal Cost [2 marks]

• Answer: $\frac{dC}{dx} = 0.02x + 5$
• Working: $C = 0.01x^2 + 5x + 100$ $\frac{dC}{dx} = 2(0.01)x + 5 = 0.02x + 5$.
• Marking: 1 mark for differentiation, 1 mark for correct coefficients.

(b) Estimate increase in cost [2 marks]

• Answer: $7$ dollars
• Working: Marginal cost at $x = 100$ is $\frac{dC}{dx}\big|_{x=100} = 0.02(100) + 5 = 2 + 5 = 7$. This represents the approximate cost of producing the next item (101st item). Alternatively, $\Delta C \approx \frac{dC}{dx} \times \Delta x = 7 \times 1 = 7$.
• Marking: 1 mark for evaluating derivative at $x=100$, 1 mark for interpreting as the increase for 1 unit.

13. Leaking Tank

(a) Rate of change at $t=5$ [2 marks]

• Answer: $-40$ cm$^3$/min
• Working: $V = 1000 - 50t + t^2$ $\frac{dV}{dt} = -50 + 2t$. When $t = 5$, $\frac{dV}{dt} = -50 + 2(5) = -50 + 10 = -40$.
• Marking: 1 mark for derivative, 1 mark for correct substitution.

(b) Increasing or Decreasing? [1 mark]

• Answer: Decreasing.
• Reasoning: The rate of change $\frac{dV}{dt}$ is negative ($-40$), which indicates the volume is reducing.
• Marking: 1 mark for correct conclusion with reference to the negative sign.

(c) Time when rate is zero [2 marks]

• Answer: $t = 25$ minutes
• Working: Set $\frac{dV}{dt} = 0$. $-50 + 2t = 0$ $2t = 50$ $t = 25$.
• Marking: 1 mark for setting equation to 0, 1 mark for correct solution.

14. Stationary Points

(a) Coordinates of stationary points [3 marks]

• Answer: $(1, -2)$ and $(-1, 2)$
• Working: $y = x^3 - 3x$ $\frac{dy}{dx} = 3x^2 - 3$. At stationary points, $\frac{dy}{dx} = 0$. $3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1$ or $x = -1$. When $x = 1$, $y = 1^3 - 3(1) = -2$. Point: $(1, -2)$. When $x = -1$, $y = (-1)^3 - 3(-1) = -1 + 3 = 2$. Point: $(-1, 2)$.
• Marking: 1 mark for finding $x$ values, 1 mark for finding corresponding $y$ values, 1 mark for correct coordinate pairs.

(b) Nature of stationary points [2 marks]

• Answer: $(1, -2)$ is a minimum point; $(-1, 2)$ is a maximum point.
• Working: Second derivative $\frac{d^2y}{dx^2} = 6x$. At $x = 1$, $\frac{d^2y}{dx^2} = 6(1) = 6 > 0$ (Positive $\rightarrow$ Minimum). At $x = -1$, $\frac{d^2y}{dx^2} = 6(-1) = -6 < 0$ (Negative $\rightarrow$ Maximum).
• Marking: 1 mark for correct test/application, 1 mark for correct classification of both points.

15. Related Rates (Rectangle)

(a) Formula for Area [1 mark]

• Answer: $A = l \times w$ or $A = lw$
• Marking: 1 mark for correct formula.

(b) Rate of increase of Area [3 marks]

• Answer: 20 cm$^2$/s
• Working: We need $\frac{dA}{dt}$. Using the product rule concept (or chain rule expansion for Sec 2 extension): $\frac{dA}{dt} = l \frac{dw}{dt} + w \frac{dl}{dt}$. Given: $\frac{dl}{dt} = 2$, $\frac{dw}{dt} = 1$, $l = 10$, $w = 5$. $\frac{dA}{dt} = 10(1) + 5(2)$ $\frac{dA}{dt} = 10 + 10 = 20$.
• Marking: 1 mark for identifying correct rates/variables, 1 mark for substitution into formula, 1 mark for final answer.

16. Particle Motion (Velocity Given)

(a) Acceleration when $t=2$ [2 marks]

• Answer: $0$ m/s$^2$
• Working: $v = 3t^2 - 12t + 9$ $a = \frac{dv}{dt} = 6t - 12$. When $t = 2$, $a = 6(2) - 12 = 12 - 12 = 0$.
• Marking: 1 mark for differentiation, 1 mark for correct substitution.

(b) Displacement from $t=0$ to $t=2$ [2 marks]

• Answer: $2$ m
• Working: Note: In Secondary 2, if integration is not covered, this question might rely on provided antiderivative rules or specific context. However, assuming standard calculus progression: $s = \int v \, dt = t^3 - 6t^2 + 9t + C$. Given $s=0$ when $t=0$, $C=0$. So $s(t) = t^3 - 6t^2 + 9t$. When $t=2$, $s(2) = 2^3 - 6(2)^2 + 9(2) = 8 - 24 + 18 = 2$. Alternative if integration not taught: This question tests the reverse concept. If strictly differentiation only, this question would be adjusted. Assuming basic integration knowledge or provided formula: Displacement change = $s(2) - s(0)$.
• Marking: 1 mark for correct antiderivative/expression, 1 mark for final value.

17. Sphere Surface Area

(a) Find $\frac{dS}{dr}$ [1 mark]

• Answer: $8\pi r$
• Working: $S = 4\pi r^2$ $\frac{dS}{dr} = 2 \times 4\pi r = 8\pi r$.
• Marking: 1 mark for correct derivative.

(b) Rate of increase of Surface Area [3 marks]

• Answer: $24\pi$ cm$^2$/s (or approx $75.4$ cm$^2$/s)
• Working: $\frac{dS}{dt} = \frac{dS}{dr} \times \frac{dr}{dt}$. $\frac{dS}{dr} = 8\pi r$. Given $\frac{dr}{dt} = 0.5$ and $r = 3$. $\frac{dS}{dt} = (8\pi \times 3) \times 0.5$ $\frac{dS}{dt} = 24\pi \times 0.5 = 12\pi$. Correction in working: $8\pi(3) = 24\pi$. $24\pi \times 0.5 = 12\pi$. Let's re-calculate: $\frac{dS}{dr} = 8\pi r$. At $r=3$, $\frac{dS}{dr} = 24\pi$. $\frac{dS}{dt} = 24\pi \times 0.5 = 12\pi$. Answer: $12\pi$ cm$^2$/s.
• Marking: 1 mark for chain rule setup, 1 mark for substitution, 1 mark for final answer.

18. Curve Stationary Points

(a) Find $\frac{dy}{dx}$ [1 mark]

• Answer: $6x^2 - 18x + 12$
• Working: $y = 2x^3 - 9x^2 + 12x$ $\frac{dy}{dx} = 6x^2 - 18x + 12$.
• Marking: 1 mark for correct differentiation.

(b) Solve $\frac{dy}{dx} = 0$ [2 marks]

• Answer: $x = 1$ and $x = 2$
• Working: $6x^2 - 18x + 12 = 0$ Divide by 6: $x^2 - 3x + 2 = 0$ Factorise: $(x - 2)(x - 1) = 0$ $x = 1$ or $x = 2$.
• Marking: 1 mark for solving quadratic, 1 mark for both values.

(c) $y$-coordinate when $x=2$ [1 mark]

• Answer: $4$
• Working: Substitute $x=2$ into original equation: $y = 2(2)^3 - 9(2)^2 + 12(2)$ $y = 2(8) - 9(4) + 24$ $y = 16 - 36 + 24 = 4$.
• Marking: 1 mark for correct substitution and answer.

19. Bacteria Growth

(a) Rate of growth when $t=3$ [2 marks]

• Answer: $30$ bacteria/hour
• Working: $N = 100 + 5t^2$ Rate $\frac{dN}{dt} = 10t$. When $t = 3$, $\frac{dN}{dt} = 10(3) = 30$.
• Marking: 1 mark for derivative, 1 mark for substitution.

(b) Time when rate is 40 [2 marks]

• Answer: $t = 4$ hours
• Working: Set $\frac{dN}{dt} = 40$. $10t = 40$ $t = 4$.
• Marking: 1 mark for setting up equation, 1 mark for solution.

20. Circular Ripples

(a) Formula for Area [1 mark]

• Answer: $A = \pi r^2$
• Marking: 1 mark for correct formula.

(b) Rate of increase of Area [3 marks]

• Answer: $40\pi$ cm$^2$/s (or approx $125.7$ cm$^2$/s)
• Working: $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$. $\frac{dA}{dr} = 2\pi r$. Given $\frac{dr}{dt} = 2$ and $r = 10$. $\frac{dA}{dt} = (2\pi \times 10) \times 2$ $\frac{dA}{dt} = 20\pi \times 2 = 40\pi$.
• Marking: 1 mark for chain rule/derivative of area, 1 mark for substitution, 1 mark for final answer.