Secondary 3 Elementary Mathematics Calculus Quiz

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Secondary 3 Elementary Mathematics Quiz - Calculus

Name: ______________________________ Class: ______________ Date: ______________ Score: ______ / 40

Duration: 50 minutes

Total Marks: 40

Instructions:

• Answer ALL questions.
• Show all working clearly. Marks will be awarded for correct working even if the final answer is wrong.
• Write your answers in the spaces provided.
• The use of calculators is allowed unless otherwise stated.
• Give non-exact answers correct to 3 significant figures unless otherwise stated.
• This quiz tests your understanding of differentiation, gradients of curves, and rates of change.

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Section A: Short Answer Questions (10 marks)

Questions 1–5. Each question carries 2 marks.

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1. Differentiate the following with respect to $x$:

(a) $y = 5x^3$ \hfill [1]

(b) $y = 4x^2 - 7x + 3$ \hfill [1]

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2. Find the gradient of the curve $y = 2x^2 - 3x + 1$ at the point where $x = 2$. \hfill [2]

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3. A curve has equation $y = x^3 - 6x^2 + 9x$. Find $\dfrac{dy}{dx}$. \hfill [2]

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4. The equation of a curve is $y = \dfrac{4}{x^2}$. Express $y$ in index form and hence find $\dfrac{dy}{dx}$. \hfill [2]

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5. Find the gradient of the tangent to the curve $y = 3x^2 + 2x - 5$ at the point $(-1, -4)$. \hfill [2]

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Section B: Structured Questions (20 marks)

Questions 6–15. Each question carries 2 marks.

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6. Given $y = 6x^4 - 3x^2 + x - 8$, find:

(a) $\dfrac{dy}{dx}$ \hfill [1]

(b) the value of $\dfrac{dy}{dx}$ when $x = 1$ \hfill [1]

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7. The displacement $s$ metres of a particle at time $t$ seconds is given by $s = 3t^2 - 12t + 5$.

(a) Find an expression for the velocity $v$ of the particle. \hfill [1]

(b) Find the velocity when $t = 3$ seconds. \hfill [1]

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8. Find the coordinates of the point on the curve $y = x^2 - 4x + 7$ where the gradient is zero. \hfill [2]

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9. The equation of a curve is $y = 2x^3 - 9x^2 + 12x - 4$.

(a) Find $\dfrac{dy}{dx}$. \hfill [1]

(b) Find the gradient of the curve at the point $(2, 0)$. \hfill [1]

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10. A curve is given by $y = x^3 - 3x$. Find the values of $x$ at the points where the gradient of the curve is 0. \hfill [2]

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11. The cost $C$ dollars of producing $x$ items is given by $C = 0.01x^3 - 0.6x^2 + 15x + 200$. Find the rate of change of cost when $x = 10$. \hfill [2]

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12. Given that $f(x) = 5x^3 - 2x^2 + 7$, find $f'(x)$ and hence evaluate $f'(-1)$. \hfill [2]

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13. The area $A$ cm$^2$ of a circle is increasing at a rate of $12\pi$ cm$^2$/s. Given $A = \pi r^2$, find the rate at which the radius is increasing when $r = 3$ cm. \hfill [2]

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14. Find the equation of the tangent to the curve $y = x^2 - 2x + 3$ at the point where $x = 1$. \hfill [2]

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15. The volume $V$ cm$^3$ of a sphere is given by $V = \dfrac{4}{3}\pi r^3$. Find the rate of change of volume with respect to the radius when $r = 5$ cm. \hfill [2]

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Section C: Application and Problem Solving (10 marks)

Questions 16–20. Each question carries 2 marks.

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16. A rectangular enclosure is to be fenced on three sides, with a wall forming the fourth side. If the total length of fencing available is 40 m, and the side perpendicular to the wall has length $x$ metres:

(a) Show that the area $A$ of the enclosure is given by $A = 40x - 2x^2$. \hfill [1]

(b) Find the value of $x$ that gives the maximum area. \hfill [1]

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17. The height $h$ metres of a ball thrown vertically upwards at time $t$ seconds is given by $h = 20t - 5t^2$.

(a) Find an expression for the velocity of the ball. \hfill [1]

(b) Find the maximum height reached by the ball. \hfill [1]

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18. The equation of a curve is $y = x^3 - 6x^2 + 9x + 2$.

(a) Find $\dfrac{dy}{dx}$. \hfill [1]

(b) Determine the nature of the stationary points of the curve. \hfill [1]

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19. A cylindrical tank of radius 4 cm is being filled with water at a rate of 50 cm$^3$/s. Given that the volume of a cylinder is $V = \pi r^2 h$, find the rate at which the height of water is increasing. \hfill [2]

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20. The surface area $S$ cm$^2$ of a cube with side length $x$ cm is given by $S = 6x^2$. The volume of the cube is $V = x^3$.

(a) Find $\dfrac{dV}{dx}$ and $\dfrac{dS}{dx}$. \hfill [1]

(b) Find the rate of change of volume with respect to surface area when $x = 2$ cm. That is, find $\dfrac{dV}{dS}$ when $x = 2$. \hfill [1]

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Secondary 3 Elementary Mathematics Quiz - Calculus

Answer Key

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Section A: Short Answer Questions

1.

(a) $\dfrac{dy}{dx} = 15x^2$ \hfill [1]

(b) $\dfrac{dy}{dx} = 8x - 7$ \hfill [1]

Working: Apply the power rule $\dfrac{d}{dx}(x^n) = nx^{n-1}$ to each term.

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2. Gradient = 5 \hfill [2]

Working: $\dfrac{dy}{dx} = 4x - 3$ At $x = 2$: $\dfrac{dy}{dx} = 4(2) - 3 = 8 - 3 = 5$

Marking: [1] for correct derivative, [1] for correct substitution and answer.

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3. $\dfrac{dy}{dx} = 3x^2 - 12x + 9$ \hfill [2]

Working: $\dfrac{d}{dx}(x^3) = 3x^2$, $\dfrac{d}{dx}(-6x^2) = -12x$, $\dfrac{d}{dx}(9x) = 9$

Marking: [1] for each pair of correct terms (or equivalent).

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4. $y = 4x^{-2}$, $\dfrac{dy}{dx} = -8x^{-3}$ or $\dfrac{dy}{dx} = -\dfrac{8}{x^3}$ \hfill [2]

Working: Rewrite: $y = 4x^{-2}$ $\dfrac{dy}{dx} = 4 \times (-2)x^{-3} = -8x^{-3} = -\dfrac{8}{x^3}$

Marking: [1] for correct index form, [1] for correct derivative.

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5. Gradient = −4 \hfill [2]

Working: $\dfrac{dy}{dx} = 6x + 2$ At $x = -1$: $\dfrac{dy}{dx} = 6(-1) + 2 = -6 + 2 = -4$

Marking: [1] for correct derivative, [1] for correct substitution and answer.

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Section B: Structured Questions

6.

(a) $\dfrac{dy}{dx} = 24x^3 - 6x + 1$ \hfill [1]

(b) $\dfrac{dy}{dx}\big|_{x=1} = 24(1)^3 - 6(1) + 1 = 24 - 6 + 1 = \mathbf{19}$ \hfill [1]

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7.

(a) $v = \dfrac{ds}{dt} = 6t - 12$ \hfill [1]

(b) At $t = 3$: $v = 6(3) - 12 = 18 - 12 = \mathbf{6}$ m/s \hfill [1]

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8. Coordinates: (2, 3) \hfill [2]

Working: $\dfrac{dy}{dx} = 2x - 4$ Set $\dfrac{dy}{dx} = 0$: $2x - 4 = 0 \Rightarrow x = 2$ $y = (2)^2 - 4(2) + 7 = 4 - 8 + 7 = 3$

Marking: [1] for setting derivative = 0 and finding $x = 2$, [1] for finding $y = 3$ and writing coordinates.

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9.

(a) $\dfrac{dy}{dx} = 6x^2 - 18x + 12$ \hfill [1]

(b) At $x = 2$: $\dfrac{dy}{dx} = 6(4) - 18(2) + 12 = 24 - 36 + 12 = \mathbf{0}$ \hfill [1]

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10. $x = \mathbf{1}$ and $x = \mathbf{-1}$ \hfill [2]

Working: $\dfrac{dy}{dx} = 3x^2 - 3$ Set $\dfrac{dy}{dx} = 0$: $3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$

Marking: [1] for correct derivative and setting = 0, [1] for both correct values.

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11. Rate of change = 6 dollars per item \hfill [2]

Working: $\dfrac{dC}{dx} = 0.03x^2 - 1.2x + 15$ At $x = 10$: $\dfrac{dC}{dx} = 0.03(100) - 1.2(10) + 15 = 3 - 12 + 15 = 6$

Marking: [1] for correct derivative, [1] for correct substitution and answer.

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12. $f'(x) = 15x^2 - 4x$, $f'(-1) = \mathbf{19}$ \hfill [2]

Working: $f'(x) = 15x^2 - 4x$ $f'(-1) = 15(1) - 4(-1) = 15 + 4 = 19$

Marking: [1] for correct derivative, [1] for correct evaluation.

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13. Rate = 2 cm/s \hfill [2]

Working: $\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$ $12\pi = 2\pi(3)\dfrac{dr}{dt}$ $12\pi = 6\pi \dfrac{dr}{dt}$ $\dfrac{dr}{dt} = 2$ cm/s

Marking: [1] for correct differentiation of $A = \pi r^2$ with respect to $t$, [1] for correct answer.

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14. Equation of tangent: $\mathbf{y = 2}$ \hfill [2]

Working: At $x = 1$: $y = 1 - 2 + 3 = 2$, so point is $(1, 2)$ $\dfrac{dy}{dx} = 2x - 2$ At $x = 1$: gradient $= 2(1) - 2 = 0$ Tangent is horizontal through $(1, 2)$: $y = 2$

Marking: [1] for finding the point and gradient = 0, [1] for correct equation.

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15. Rate of change = $100\pi$ cm$^3$/cm (or cm$^2$) \hfill [2]

Working: $\dfrac{dV}{dr} = 4\pi r^2$ At $r = 5$: $\dfrac{dV}{dr} = 4\pi(25) = 100\pi$

Marking: [1] for correct derivative, [1] for correct evaluation.

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Section C: Application and Problem Solving

16.

(a) Side parallel to wall $= 40 - 2x$; Area $= x(40 - 2x) = 40x - 2x^2$ ✓ \hfill [1]

(b) $x = \mathbf{10}$ \hfill [1]

Working: $\dfrac{dA}{dx} = 40 - 4x$ Set $\dfrac{dA}{dx} = 0$: $40 - 4x = 0 \Rightarrow x = 10$ $\dfrac{d^2A}{dx^2} = -4 < 0$, so maximum confirmed.

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17.

(a) Velocity $= \dfrac{dh}{dt} = 20 - 10t$ m/s \hfill [1]

(b) Maximum height = 20 m \hfill [1]

Working: At max height, $v = 0$: $20 - 10t = 0 \Rightarrow t = 2$ $h = 20(2) - 5(4) = 40 - 20 = 20$ m

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18.

(a) $\dfrac{dy}{dx} = 3x^2 - 12x + 9$ \hfill [1]

(b) Stationary points: $3x^2 - 12x + 9 = 0 \Rightarrow x^2 - 4x + 3 = 0 \Rightarrow (x-1)(x-3) = 0$ $x = 1$: $y = 1 - 6 + 9 + 2 = 6$, point $(1, 6)$ $x = 3$: $y = 27 - 54 + 27 + 2 = 2$, point $(3, 2)$

Second derivative: $\dfrac{d^2y}{dx^2} = 6x - 12$ At $x = 1$: $\dfrac{d^2y}{dx^2} = 6 - 12 = -6 < 0$ → maximum at $(1, 6)$ At $x = 3$: $\dfrac{d^2y}{dx^2} = 18 - 12 = 6 > 0$ → minimum at $(3, 2)$ \hfill [1]

Marking for (b): [1] for finding both stationary points and correctly determining their nature.

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19. Rate = $\dfrac{25}{8\pi}$ cm/s (or approximately 0.995 cm/s) \hfill [2]

Working: $V = \pi r^2 h = \pi(16)h = 16\pi h$ $\dfrac{dV}{dt} = 16\pi \dfrac{dh}{dt}$ $50 = 16\pi \dfrac{dh}{dt}$ $\dfrac{dh}{dt} = \dfrac{50}{16\pi} = \dfrac{25}{8\pi} \approx 0.995$ cm/s

Marking: [1] for correct differentiation/substitution, [1] for correct answer.

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20.

(a) $\dfrac{dV}{dx} = 3x^2$, $\dfrac{dS}{dx} = 12x$ \hfill [1]

(b) $\dfrac{dV}{dS} = \dfrac{dV/dx}{dS/dx} = \dfrac{3x^2}{12x} = \dfrac{x}{4}$ At $x = 2$: $\dfrac{dV}{dS} = \dfrac{2}{4} = \mathbf{0.5}$ \hfill [1]

Marking for (a): [1] for both derivatives correct. Marking for (b): [1] for correct application of chain rule and answer.

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