Secondary 3 Elementary Mathematics Calculus Quiz

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Secondary 3 Elementary Mathematics Quiz - Calculus

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

Duration: 45 Minutes
Total Marks: 40

Instructions:

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

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Section A: Estimation of Gradient (10 Marks)

1. The diagram below shows part of the curve $y = x^2$. Point $P$ has coordinates $(2, 4)$ and point $Q$ has coordinates $(2.1, 4.41)$.

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

<br> <br> <br>

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

<br> <br>

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]

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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]

<br> <br> <br>

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Section B: Differentiation Rules (18 Marks)

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

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

<br>

(b) $y = 7$
[1]

<br>

(c) $y = 4x^{-2}$
[2]

<br> <br>

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]

<br>

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]

<br> <br> <br> <br> <br> <br>

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]

<br> <br> <br> <br>

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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]

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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]

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

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(c) Calculate the acceleration of the object when $t = 4$.
[2]

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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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Secondary 3 Elementary Mathematics Quiz - Calculus (Answer Key)

Total Marks: 40

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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)

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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)

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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]

(c) Acceleration $a = \frac{dv}{dt} = 6t - 12$ When $t = 4$, $a = 6(4) - 12 = 24 - 12 = 12$ m/s$^2$
[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]