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Secondary 3 Elementary Mathematics Calculus Quiz

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Questions

Secondary 3 Elementary Mathematics Quiz - Calculus

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 40

Duration: 45 Minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all necessary working clearly. Marks may be awarded for method even if the final answer is incorrect.
Non-exact numerical answers should be given to 3 significant figures unless otherwise specified.
The use of an approved scientific calculator is expected.

Section A: Estimation of Gradient (10 Marks)

1. The diagram below shows part of the curve $y = x^2$ .

Image pending generation for this question.

Point $P$ has coordinates $(2, 4)$ and point $Q$ has coordinates $(2.1, 4.41)$ .

(a) Calculate the gradient of the chord $PQ$ .
[2]

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(b) Explain what happens to the gradient of the chord $PQ$ as point $Q$ moves closer to point $P$ along the curve.
[1]

Answer space

2. The graph of $y = 6x - x^2$ is shown below. Estimate the gradient of the curve at the point where $x = 1$ by drawing a suitable tangent.

(a) Draw the tangent to the curve at $x = 1$ on the grid provided (assume standard grid).
[1]

(b) Using two points on your tangent, calculate the estimated gradient.
[2]

Answer space

3. A curve passes through the points $A(1, 3)$ and $B(1.01, 3.0401)$ . Estimate the gradient of the curve at point $A$ .
[2]

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4. The position $s$ (in metres) of a particle at time $t$ (in seconds) is given by $s = t^2 + 2t$ . Estimate the instantaneous velocity of the particle at $t = 3$ by calculating the average velocity between $t = 3$ and $t = 3.001$ .
[2]

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5. A curve passes through points $C(3, 9)$ and $D(3.001, 9.006001)$ . Calculate the gradient of the chord $CD$ and use it to estimate the gradient of the curve at $x=3$ .
[3]

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Section B: Differentiation Rules (18 Marks)

6. Differentiate the following expressions with respect to $x$ :

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

(b) $y = 7$
[1]

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7. Given that $y = 3x^2 - 5x + 2$ , find:

(a) $\frac{dy}{dx}$
[2]

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(b) The value of $\frac{dy}{dx}$ when $x = 4$ .
[1]

8. Find the derivative of $f(x) = \frac{x^3 + 2x}{x}$ , where $x \neq 0$ .
[2]

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9. The equation of a curve is $y = 2x^3 - 9x^2 + 12x$ . Find the coordinates of the stationary points on the curve.
[4]

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10. Determine whether the stationary point at $x = 2$ for the curve $y = x^3 - 3x^2 + 4$ is a maximum or a minimum point. Show your working.
[3]

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11. A function is defined by $g(x) = \sqrt{x}$ . Find the value of $g'(4)$ .
[2]

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

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Section C: Applications of Calculus (12 Marks)

13. The gradient of a curve is given by $\frac{dy}{dx} = 6x - 4$ . The curve passes through the point $(1, 5)$ . Find the equation of the curve.
[3]

Answer space

14. The displacement $s$ metres of a moving object from a fixed point $O$ at time $t$ seconds is given by $s = t^3 - 6t^2 + 9t$ .

(a) Find an expression for the velocity $v$ m/s of the object at time $t$ .
[1]

(b) Find the time(s) when the object is at rest.
[2]

Answer space

15. A rectangular enclosure is to be built using 20 metres of fencing for three sides, with the fourth side being an existing wall. Let $x$ be the width of the enclosure perpendicular to the wall.

(a) Show that the area $A$ of the enclosure is given by $A = 20x - 2x^2$ .
[2]

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(b) Find the value of $x$ that maximizes the area.
[2]

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16. The volume $V$ cm $^3$ of water in a tank at time $t$ minutes is given by $V = 100 - 2t^2$ . Find the rate at which the volume is decreasing when $t = 3$ .
[2]

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17. The cost $C$ of producing $x$ items is given by $C = 100 + 5x + 0.1x^2$ . Find the marginal cost when $x = 10$ .
[1]

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18. A particle moves such that its velocity $v$ m/s at time $t$ seconds is $v = 4t - t^2$ . Find the acceleration of the particle when $t = 2$ .
[2]

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19. The height $h$ metres of a ball thrown upwards is given by $h = 20t - 5t^2$ . Find the maximum height reached by the ball.
[3]

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20. The population $P$ of a bacteria culture at time $t$ hours is modelled by $P = 100e^{0.5t}$ . Find the rate of growth of the population when $t = 2$ . Give your answer to 3 significant figures.
[2]

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Answers

Secondary 3 Elementary Mathematics Quiz - Calculus (Answer Key)

Total Marks: 40

Section A: Estimation of Gradient

1. (a) Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4.41 - 4}{2.1 - 2} = \frac{0.41}{0.1} = 4.1$
[2] (1 for substitution, 1 for answer)

(b) The gradient of the chord approaches the gradient of the tangent at $P$ (or the instantaneous rate of change at $P$ ).
[1]

2. (a) Tangent drawn at $x=1$ . It should touch the curve only at that point and follow the slope.
[1]

(b) Answers vary depending on drawing accuracy. Expected gradient: $\frac{dy}{dx} = 6 - 2x$ . At $x=1$ , gradient $= 4$ . Accept range $3.5$ to $4.5$ if working shows correct calculation from tangent points.
[2] (1 for valid points from tangent, 1 for calculation)

3. Gradient $\approx \frac{3.0401 - 3}{1.01 - 1} = \frac{0.0401}{0.01} = 4.01$
[2]

4. $s(3) = 3^2 + 2(3) = 9 + 6 = 15$ $s(3.001) = (3.001)^2 + 2(3.001) = 9.006001 + 6.002 = 15.008001$ Average velocity $= \frac{15.008001 - 15}{3.001 - 3} = \frac{0.008001}{0.001} = 8.001$ m/s
[2] (1 for substitution, 1 for answer)

5. Gradient of chord $CD = \frac{9.006001 - 9}{3.001 - 3} = \frac{0.006001}{0.001} = 6.001$ Estimated gradient at $x=3$ is $6$ .
[3] (1 for substitution, 1 for chord gradient, 1 for estimate)

Section B: Differentiation Rules

6. (a) $\frac{dy}{dx} = 15x^2$
[1]

(b) $\frac{dy}{dx} = 0$
[1]

(c) $\frac{dy}{dx} = 4(-2)x^{-3} = -8x^{-3}$ or $-\frac{8}{x^3}$
[2] (1 for power rule application, 1 for simplification)

7. (a) $\frac{dy}{dx} = 6x - 5$
[2]

(b) When $x = 4$ , $\frac{dy}{dx} = 6(4) - 5 = 24 - 5 = 19$
[1]

8. Simplify first: $f(x) = \frac{x(x^2 + 2)}{x} = x^2 + 2$ (for $x \neq 0$ ) $f'(x) = 2x$
[2] (1 for simplification, 1 for differentiation)

9. Stationary points occur when $\frac{dy}{dx} = 0$ . $\frac{dy}{dx} = 6x^2 - 18x + 12$ $6x^2 - 18x + 12 = 0$ Divide by 6: $x^2 - 3x + 2 = 0$ $(x - 1)(x - 2) = 0$ $x = 1$ or $x = 2$

When $x = 1$ , $y = 2(1)^3 - 9(1)^2 + 12(1) = 2 - 9 + 12 = 5$ . Point: $(1, 5)$ When $x = 2$ , $y = 2(2)^3 - 9(2)^2 + 12(2) = 16 - 36 + 24 = 4$ . Point: $(2, 4)$

Coordinates: $(1, 5)$ and $(2, 4)$
[4] (1 for derivative, 1 for solving x, 1 for each correct coordinate pair)

10. $\frac{dy}{dx} = 3x^2 - 6x$ $\frac{d^2y}{dx^2} = 6x - 6$ At $x = 2$ , $\frac{d^2y}{dx^2} = 6(2) - 6 = 6$ . Since $\frac{d^2y}{dx^2} > 0$ , the point is a Minimum.
[3] (1 for 2nd derivative, 1 for substitution, 1 for conclusion)

11. $g(x) = x^{1/2}$ $g'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ $g'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{2(2)} = \frac{1}{4}$ or $0.25$
[2]

12. $y = x^2 + 2x \Rightarrow \frac{dy}{dx} = 2x + 2$ At $x = 1$ , gradient $m = 2(1) + 2 = 4$ . Point: $x=1, y=1^2+2(1)=3$ . Point is $(1,3)$ . Equation: $y - 3 = 4(x - 1)$ $y = 4x - 4 + 3$ $y = 4x - 1$
[3] (1 for gradient, 1 for point, 1 for equation)

Section C: Applications of Calculus

13. $y = \int (6x - 4) dx = 3x^2 - 4x + c$ Substitute $(1, 5)$ : $5 = 3(1)^2 - 4(1) + c$ $5 = 3 - 4 + c$ $5 = -1 + c \Rightarrow c = 6$ Equation: $y = 3x^2 - 4x + 6$
[3] (1 for integration, 1 for finding c, 1 for final equation)

14. (a) $v = \frac{ds}{dt} = 3t^2 - 12t + 9$
[1]

(b) At rest, $v = 0$ . $3t^2 - 12t + 9 = 0$ $t^2 - 4t + 3 = 0$ $(t - 3)(t - 1) = 0$ $t = 1$ s or $t = 3$ s
[2]

15. (a) Perimeter of 3 sides = $2x + l = 20 \Rightarrow l = 20 - 2x$ Area $A = x \cdot l = x(20 - 2x) = 20x - 2x^2$
[2]

(b) For maximum area, $\frac{dA}{dx} = 0$ . $\frac{dA}{dx} = 20 - 4x$ $20 - 4x = 0 \Rightarrow 4x = 20 \Rightarrow x = 5$
[2]

16. Rate of change $\frac{dV}{dt} = -4t$ When $t = 3$ , $\frac{dV}{dt} = -4(3) = -12$ The volume is decreasing at a rate of $12$ cm $^3$ /min.
[2] (1 for derivative, 1 for correct rate and unit/direction)

17. Marginal Cost $= \frac{dC}{dx} = 5 + 0.2x$ When $x = 10$ , $MC = 5 + 0.2(10) = 5 + 2 = 7$ .
[1]

18. Acceleration $a = \frac{dv}{dt} = 4 - 2t$ When $t = 2$ , $a = 4 - 2(2) = 0$ m/s $^2$ .
[2]

19. $v = \frac{dh}{dt} = 20 - 10t$ At max height, $v = 0 \Rightarrow 20 - 10t = 0 \Rightarrow t = 2$ . Max height $h = 20(2) - 5(2)^2 = 40 - 20 = 20$ m.
[3] (1 for derivative, 1 for time, 1 for height)

20. $\frac{dP}{dt} = 100(0.5)e^{0.5t} = 50e^{0.5t}$ When $t = 2$ , $\frac{dP}{dt} = 50e^{1} = 50e \approx 135.914$ Rate of growth $\approx 136$ per hour.
[2]