How many VCAA Specialist Mathematics questions cover Calculus?

AusGrader has 438 VCAA Specialist Mathematics questions on Calculus, all with instant AI grading and detailed marking feedback.

VCE Specialist Mathematics — Calculus Questions & Answers

Q11

2024

VCAA

Paper 2

1 mark

Q11

1 mark

If $I_n = \int_1^e x^2 \left(\log_e x\right)^n dx$ where $n \in N$ , then for $n \ge 2$ , $I_n$ is equal to

A

$-\frac{n}{3} I_{n-1}$

B

$\frac{x^3 \left(\log_e x\right)^n}{3} - \frac{n}{3} I_{n-1}$

C

$\frac{e^3}{3} - \frac{1}{3} I_{n-1}$

D

$\frac{e^3}{3} - \frac{n}{3} I_{n-1}$

E

$\frac{e^3}{3} - \frac{1}{3} - \frac{n}{3} I_{n-1}$

Reveal Answer

A

$-\frac{n}{3} I_{n-1}$

This option is incorrect because it misses the evaluated boundary term $\frac{e^3}{3}$ that comes from applying the limits of integration to the $uv$ term in integration by parts.

B

$\frac{x^3 \left(\log_e x\right)^n}{3} - \frac{n}{3} I_{n-1}$

This represents the indefinite integral, but the question asks for a definite integral evaluated from $1$ to $e$ .

C

$\frac{e^3}{3} - \frac{1}{3} I_{n-1}$

This option is incorrect because it misses the factor of $n$ that comes from applying the chain rule when differentiating $u = (\log_e x)^n$ .

D

$\frac{e^3}{3} - \frac{n}{3} I_{n-1}$

Correct Answer

Using integration by parts with $u = (\log_e x)^n$ and $dv = x^2 dx$ , we get $\left[ \frac{x^3}{3} (\log_e x)^n \right]_1^e - \frac{n}{3} \int_1^e x^2 (\log_e x)^{n-1} dx$ , which evaluates to $\frac{e^3}{3} - \frac{n}{3} I_{n-1}$ .

E

$\frac{e^3}{3} - \frac{1}{3} - \frac{n}{3} I_{n-1}$

This option incorrectly evaluates the lower limit of integration; since $\log_e 1 = 0$ , the lower limit term evaluates to $0$ , not $\frac{1}{3}$ .

Q3

2025

VCAA

Paper 2

10 marks

Q3

A tank initially contains $5 \text{ kg}$ of salt dissolved in $3000 \text{ litres}$ of water. Salty water that contains $0.1 \text{ kg}$ of salt per litre of water enters the tank at a rate of $20 \text{ litres per minute}$ . The solution is kept thoroughly mixed and drains from the tank via a tap at the same rate of $20 \text{ litres per minute}$ .

Q3a

1 mark

By considering concentration, explain whether the quantity of salt in the tank increases with time.

Reveal Answer

The initial concentration $= \frac{5}{3000} = \frac{1}{600}$ which is smaller than the incoming concentration
$= 0.1 \text{ kg/litre}$ ; hence, the quantity of salt increases.

Marking Criteria

Descriptor	Marks
Explains that the quantity of salt increases by correctly comparing the initial concentration ( $\frac{1}{600} \text{ kg/L}$ ) to the incoming concentration ( $0.1 \text{ kg/L}$ )	1

Q3b

1 mark

Let $Q$ denote the quantity of salt, in kilograms, in the tank at time $t \text{ minutes}$ .

Show that $Q$ satisfies the differential equation $\frac{dQ}{dt} = \frac{300 - Q}{150}$ .

Reveal Answer

$\frac{dQ}{dt} = 0.1 \times 20 - \frac{Q}{3000} \times 20 = \frac{300}{150} - \frac{Q}{150}$

Marking Criteria

Descriptor	Marks
Shows the correct development of the differential equation by subtracting the rate out ( $\frac{Q}{3000} \times 20$ ) from the rate in ( $0.1 \times 20$ )	1

Q3c

2 marks

Using Euler's method with a step size of $15 \text{ minutes}$ , find $Q(30)$ , the approximate quantity of salt in the tank after $30 \text{ minutes}$ .

Give your answer in kilograms, correct to two decimal places.

Reveal Answer

61.05

$Q_{n+1} = Q_n + 15 \times \frac{300 - Q_n}{150}$
$Q(15) = \frac{69}{2}$
$Q(30) = 61.05$

Marking Criteria

Descriptor	Marks
Calculates the correct value for $Q(15)$ ( $\frac{69}{2}$ ) or demonstrates correct application of Euler's method	1
Calculates the correct final answer of $61.05$	1

Q3d

3 marks

Use calculus to solve the differential equation $\frac{dQ}{dt} = \frac{300 - Q}{150}$ , expressing $Q$ in terms of $t$ .

Reveal Answer

\begin{align*} \frac{dQ}{dt} &= \frac{300 - Q}{150}\\ t &= \int \frac{150}{300 - Q} dQ\\ t &= -150 \log_e(|300 - Q|) + c\\ c &= 150 \log_e(295)\\ t &= 150 \log_e\left(\frac{295}{300 - Q}\right)\\ Q &= 300 - 295 e^{-\frac{t}{150}} \end{align*}

Marking Criteria

Descriptor	Marks
Correctly integrates the differential equation to find an expression involving a constant of integration (e.g., $t = -150 \log_e(\|300 - Q\|) + c$ )	1
Correctly uses the initial condition to evaluate the constant of integration	1
Correctly expresses $Q$ in terms of $t$ as $Q = 300 - 295 e^{-\frac{t}{150}}$	1

Q3e

1 mark

What value does the quantity of salt in the tank approach as time approaches infinity?

Give your answer in kilograms.

Reveal Answer

300 kg

Marking Criteria

Descriptor	Marks
States the correct limiting value of $300$	1

Q3f

1 mark

Find the time taken for the quantity of salt in the tank to reach $100 \text{ kg}$ .

Reveal Answer

$t = 150 \log_e\left(\frac{59}{40}\right)$

Marking Criteria

Descriptor	Marks
Calculates the correct exact time of $150 \log_e\left(\frac{59}{40}\right)$	1

Q3g

1 mark

When the quantity of salt in the tank reaches $100 \text{ kg}$ , the tap draining the tank is turned off. Assume that the tank does not overflow and there is no change to the inflow rate.

After the tap is turned off, how many minutes does it take for the concentration of salt in the tank to reach $\frac{1}{20} \text{ kg L}^{-1}$ ?

Reveal Answer

50

This can be found by equating the concentration to the given value $\frac{2t + 100}{3000 + 20t} = \frac{1}{20}$ .

Marking Criteria

Descriptor	Marks
Calculates the correct time of $50$ minutes	1

Q13

2024

VCAA

Paper 2

1 mark

Q13

1 mark

A racing car is travelling on a straight horizontal track at a velocity of $80 \text{ m s}^{-1}$ when the brakes are applied at time $t = 0$ seconds. From time $t = 0$ to the moment the car stops, the acceleration of the car is given by $a(t) = -(6t^2 + t) \text{ m s}^{-2}$ .

During this time period, the distance the car travels, correct to the nearest metre, is

A

188

B

199

C

213

D

260

E

267

Reveal Answer

A

188

This is incorrect. It likely results from an error in integrating the acceleration function or evaluating the definite integral at the wrong stopping time.

B

199

Correct Answer

Integrating acceleration gives velocity $v(t) = -2t^3 - 0.5t^2 + 80$ . Setting $v(t) = 0$ gives a stopping time of $t \approx 3.342$ s. Integrating $v(t)$ from $t = 0$ to $3.342$ yields a distance of approximately $199$ m.

C

213

This is incorrect. It represents a miscalculation of the definite integral of the velocity function over the time interval.

D

260

This is incorrect. It may result from an algebraic mistake when finding the constant of integration or solving for the stopping time.

E

267

This is incorrect. This value is approximately $80 \times 3.342$ , which would be the distance traveled if the car maintained its initial velocity of $80 \text{ m s}^{-1}$ without decelerating.

Q6

2025

VCAA

Paper 1

4 marks

Q6

4 marks

Find the volume of the solid of revolution formed when the area between the curve

y = \sqrt{\frac{\arctan(x)}{1+x^2}}

and the $x$ -axis from $x = 1$ to $x = \sqrt{3}$ is rotated about the $x$ -axis.

Give your answer in the form $\frac{a\pi^b}{c}$ , where $a, b, c \in Z^+$ .

Reveal Answer

$V = \pi \int_1^{\sqrt{3}} \frac{\arctan(x)}{1+x^2} dx$

Method 1 (substitution)
$u = \arctan(x), \ du = \frac{dx}{1+x^2}$ .

\begin{align*} V &= \pi \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} u \ du\\ &= \pi \left[ \frac{u^2}{2} \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\\ &= \frac{\pi^3}{2} \left( \frac{1}{9} - \frac{1}{16} \right)\\ &= \frac{7\pi^3}{288} \end{align*}

Marking Criteria

Descriptor	Marks
Sets up the correct definite integral for the volume of revolution, including the factor of $\pi$	1
Identifies the correct substitution $u = \arctan(x)$ and corresponding $du = \frac{dx}{1+x^2}$	1
Correctly changes the limits of integration to $\frac{\pi}{4}$ and $\frac{\pi}{3}$ and finds the antiderivative	1
Evaluates the integral to obtain the correct final answer $\frac{7\pi^3}{288}$	1

Q12

2020

VCAA

Paper 2

1 mark

Q12

1 mark

If $\frac{dy}{dx}=e^{\cos(x)}$ and $y_0=e$ when $x_0=0$ , then, using Euler's formula with step size $0.1$ , $y_3$ is equal to

A

$e+0.1\left(1+e^{\cos(0.1)}\right)$

B

$e+0.1\left(1+e^{\cos(0.1)}+e^{\cos(0.2)}\right)$

C

$e+0.1\left(e+e^{\cos(0.1)}+e^{\cos(0.2)}\right)$

D

$e+0.1\left(e^{\cos(0.1)}+e^{\cos(0.2)}+e^{\cos(0.3)}\right)$

E

$e+0.1\left(1+e^{\cos(0.1)}+e^{\cos(0.2)}+e^{\cos(0.3)}\right)$

Reveal Answer

A

$e+0.1\left(1+e^{\cos(0.1)}\right)$

This option calculates $y_2$ instead of $y_3$ and incorrectly evaluates the first derivative term $e^{\cos(0)}$ as $1$ instead of $e$ .

B

$e+0.1\left(1+e^{\cos(0.1)}+e^{\cos(0.2)}\right)$

This option incorrectly evaluates the first derivative term $e^{\cos(0)}$ as $1$ instead of $e$ .

C

$e+0.1\left(e+e^{\cos(0.1)}+e^{\cos(0.2)}\right)$

Correct Answer

Using Euler's method, $y_3 = y_0 + h(f(x_0) + f(x_1) + f(x_2))$ . Substituting the given values yields $e + 0.1(e^{\cos(0)} + e^{\cos(0.1)} + e^{\cos(0.2)})$ , and since $e^{\cos(0)} = e$ , this is the correct expression.

D

$e+0.1\left(e^{\cos(0.1)}+e^{\cos(0.2)}+e^{\cos(0.3)}\right)$

This option incorrectly evaluates the derivative at $x_1, x_2,$ and $x_3$ instead of starting at $x_0$ .

E

$e+0.1\left(1+e^{\cos(0.1)}+e^{\cos(0.2)}+e^{\cos(0.3)}\right)$

This option calculates $y_4$ instead of $y_3$ and incorrectly evaluates the first derivative term $e^{\cos(0)}$ as $1$ .

Q16

2023

VCAA

Paper 2

1 mark

Q16

1 mark

A mass of $5\text{ kg}$ sitting at rest on a smooth horizontal surface is acted on by a horizontal force of $10\text{ N}$ .
How far, in metres, will the mass move from its original position in $3$ seconds?

A

$2.25$

B

$3$

C

$4.5$

D

$9$

E

$45$

Reveal Answer

A

$2.25$

Incorrect. This result comes from incorrectly calculating acceleration as $m/F = 5/10 = 0.5\text{ m/s}^2$ instead of $F/m$ .

B

$3$

Incorrect. This error occurs if you forget to square the time in the kinematic equation, calculating $s = \frac{1}{2}at$ instead of $s = \frac{1}{2}at^2$ .

C

$4.5$

Incorrect. This answer might result from incorrectly calculating the acceleration as $1\text{ m/s}^2$ or mistakenly dividing the final correct distance by $2$ .

D

$9$

Correct Answer

Correct. First, find acceleration using $a = F/m = 10/5 = 2\text{ m/s}^2$ . Then, use the kinematic equation $s = ut + \frac{1}{2}at^2$ with $u=0$ to find $s = 0 + \frac{1}{2}(2)(3)^2 = 9\text{ m}$ .

E

$45$

Incorrect. This mistake happens if you use the force ( $10\text{ N}$ ) instead of the calculated acceleration ( $2\text{ m/s}^2$ ) in the kinematic equation.

Q9

2021

VCAA

Paper 2

1 mark

Q9

1 mark

Which one of the following derivatives corresponds to a graph of $f$ that has no points of inflection?

A

$f'(x) = 2(x - 3)^2 + 5$

B

$f'(x) = 2(x - 3)^3 + 5$

C

$f'(x) = \frac{5}{2}(x - 3)^2$

D

$f'(x) = \frac{1}{2}(x - 3)^2 - 5$

E

$f'(x) = (x - 3)^3 - 12x$

Reveal Answer

A

$f'(x) = 2(x - 3)^2 + 5$

The second derivative is $f''(x) = 4(x - 3)$ , which changes sign at $x = 3$ . This indicates that the graph of $f$ has a point of inflection at $x = 3$ .

B

$f'(x) = 2(x - 3)^3 + 5$

Correct Answer

The second derivative is $f''(x) = 6(x - 3)^2$ . Because this expression is squared, it is always non-negative and never changes sign, meaning the graph of $f$ has no points of inflection.

C

$f'(x) = \frac{5}{2}(x - 3)^2$

The second derivative is $f''(x) = 5(x - 3)$ , which changes sign at $x = 3$ . This means the graph of $f$ has a point of inflection at $x = 3$ .

D

$f'(x) = \frac{1}{2}(x - 3)^2 - 5$

The second derivative is $f''(x) = x - 3$ , which changes sign at $x = 3$ . This indicates a point of inflection on the graph of $f$ at $x = 3$ .

E

$f'(x) = (x - 3)^3 - 12x$

The second derivative is $f''(x) = 3(x - 3)^2 - 12$ . This expression changes sign at $x = 1$ and $x = 5$ , indicating that the graph of $f$ has two points of inflection.

Q12

2025

VCAA

Paper 2

1 mark

Q12

1 mark

An object starts with an initial velocity of $4 \text{ m s}^{-1}$ at time $t = 0 \text{ seconds}$ and moves in a straight line from a fixed point $O$ .
At time $t \text{ seconds}$ its displacement is $x \text{ metres}$ from $O$ and its velocity is $v \text{ m s}^{-1}$ where $\frac{dv}{dx} = \frac{4 + v^2}{2v^2}$ .

The object's velocity, in metres per second, at time $t = \log_e(10)$ is

A

$\sqrt{6}$

B

$6$

C

$14$

D

$196$

Reveal Answer

A

$\sqrt{6}$

This is incorrect and likely results from incorrectly setting up the differential equation or making an arithmetic error when solving for $v$ .

B

$6$

This is incorrect and might come from an arithmetic error when solving $4 + v^2 = 200$ or evaluating the constant of integration.

C

$14$

Correct Answer

Correct. By applying the chain rule $\frac{dv}{dt} = v \frac{dv}{dx}$ , we get $\frac{dv}{dt} = \frac{4+v^2}{2v}$ . Integrating yields $\ln(4+v^2) = t + \ln(20)$ , which gives $v=14$ when $t=\ln(10)$ .

D

$196$

This is the value of $v^2$ , not $v$ . You must take the square root of $196$ to find the final velocity.

Q7

2021

VCAA

Paper 1

3 marks

Q7

3 marks

A particle moves in a straight line so that its acceleration is $(5 + 6x) \text{ ms}^{-2}$ , where its displacement from a fixed origin is $x$ metres.

If its velocity is $4 \text{ ms}^{-1}$ when $x = 0$ , find its velocity, where $v > 0$ , in metres per second, when $x = 2$ .

Reveal Answer

Use $v\frac{dv}{dx} = 5 + 6x$ to obtain

$\frac{1}{2}v^2 = 5x + 3x^2 + c$

When $x = 0$ , $v = 4$ . Therefore $c = 8$ and $v^2 = 10x + 6x^2 + 16$

When $x = 2$ , $v^2 = 20 + 24 + 16 = 60$ and since $v > 0$ , $v = \sqrt{60} = 2\sqrt{15}$

Marking Criteria

Descriptor	Marks
Integrates $v\frac{dv}{dx} = 5 + 6x$ to obtain $\frac{1}{2}v^2 = 5x + 3x^2 + c$	1
Evaluates the constant of integration using the initial conditions	1
Calculates the correct velocity when $x = 2$	1

Q6

2025

QCAA

Paper 2

1 mark

Q6

1 mark

Determine the gradient of the tangent to $y^2 = 4x$ when $y = 1$ .

A

4

B

2

C

0.5

D

0.25

Reveal Answer

A

4

Incorrect. This is the derivative of the right side ( $4x$ ) with respect to $x$ , but it fails to apply implicit differentiation to the $y^2$ term.

B

2

Correct Answer

Correct. Using implicit differentiation, $2y \frac{dy}{dx} = 4$ , which simplifies to $\frac{dy}{dx} = \frac{2}{y}$ . Substituting $y = 1$ gives a gradient of 2.

C

0.5

Incorrect. This calculates $\frac{dx}{dy} = \frac{y}{2} = 0.5$ , which is the reciprocal of the gradient. The gradient of the tangent must be $\frac{dy}{dx}$ .

D

0.25

Incorrect. This is the $x$ -coordinate of the point on the curve when $y = 1$ (since $1^2 = 4x \implies x = 0.25$ ), not the gradient of the tangent line.

Q3

2024

VCAA

Paper 2

10 marks

Q3

A pollutant, at time $t = 0$ days, begins to enter a pond of still, unpolluted water at a rate of $\frac{dV}{dt} = \frac{8t}{240 + 5t^4}$ , where $V$ is the volume of pollutant, in cubic metres, in the pond after $t$ days.

The pollutant does not dissolve or mix, and spreads across the pond, maintaining the shape of a thin circular disc of radius $r(t)$ metres and constant depth of $1$ millimetre.

Q3a

1 mark

What is the maximum rate, in cubic metres per day, at which the pollutant will enter the pond, and for what value of $t$ will this rate occur?

Reveal Answer

The maximum rate occurs when $\frac{d^2V}{dt^2} = \frac{1920 - 120t^4}{(240+5t^4)^2} = 0$ , so $1920 - 120t^4 = 120(16-t^4) = 0$ , so as $t > 0$ then $t = 2$ , at which time $\frac{dV}{dt} = \frac{16}{240+5 \cdot 2^4} = \frac{16}{320} = \frac{1}{20}$ cubic metres per day.

$0.05 = \frac{1}{20}$ (cubic metres per day) when $t = 2$

Marking Criteria

Descriptor	Marks
Calculates the correct maximum rate of $\frac{1}{20}$ (or $0.05$ ) cubic metres per day and the correct time $t = 2$	1

Q3b

3 marks

At what rate is the radius of the disc increasing after $t = 4$ days, where it may be assumed that the radius of the disc is $6.54$ m?

Give your answer in metres per day correct to two decimal places.

Reveal Answer

Let $r$ be the radius in metres, so $V = \frac{\pi r^2}{1000}$ and hence $\frac{dV}{dt} = \frac{\pi}{1000} 2r \frac{dr}{dt} = \frac{\pi}{500} r \frac{dr}{dt}$ .

When $t = 4$ we have $\frac{dV}{dt} = \frac{4 \cdot 8}{240 + 5 \cdot 4^4} = \frac{32}{1520} = \frac{2}{95}$ .

Also substituting $r = 6.54$ we obtain $\frac{2}{95} = \frac{\pi}{500} 6.54 \frac{dr}{dt}$ which yields $\frac{dr}{dt} = \frac{1000}{95\pi \times 6.54} \approx 0.51$
$\frac{dr}{dt} = 0.51$

Marking Criteria

Descriptor	Marks
Establishes a correct relationship between the rates of change, such as $\frac{dV}{dt} = \frac{\pi}{500} r \frac{dr}{dt}$	1
Calculates the correct rate of change of volume at $t = 4$ , $\frac{dV}{dt} = \frac{2}{95}$	1
Calculates the correct rate of change of the radius, $0.51$	1

Q3c (i)

1 mark

Use the substitution $u = \sqrt{5} t^2$ to express $\int \frac{8t}{240 + 5t^4} \, dt$ as an integral involving only the variable $u$ .

Reveal Answer

We have $\frac{du}{dt} = 2\sqrt{5} t$ and $u^2 = 5t^4$ , so $t \ dt = \frac{du}{2\sqrt{5}}$ and $\int \frac{8t}{240+5t^4} dt = \int \frac{8}{240+u^2} \frac{du}{2\sqrt{5}} = \frac{4}{\sqrt{5}} \int \frac{du}{240+u^2}$

Marking Criteria

Descriptor	Marks
Correctly expresses the integral in terms of $u$ , $\frac{4}{\sqrt{5}} \int \frac{1}{240+u^2} du$	1

Q3c (ii)

1 mark

Hence, or otherwise, find, in terms of $t$ , the total volume $V \text{ m}^3$ of pollutant that has entered the pond after $t$ days.

Give your answer in the form $\frac{1}{a\sqrt{b}} \arctan\left(\frac{t^c}{d\sqrt{b}}\right)$ , where $a, b, c, d \in Z^+$ .

Reveal Answer

\begin{align*} V &= \int_0^t \frac{8t'}{240+5t'^4} dt'\\ &= \frac{4}{\sqrt{5}} \int_0^{\sqrt{5}t^2} \frac{1}{240+u^2} du\\ &= \frac{4}{\sqrt{5 \cdot 240}} \int_0^{\sqrt{5}t^2} \frac{\sqrt{240}}{240+u^2} du \\ &= \frac{1}{\sqrt{75}} \left[ \arctan\left(\frac{u}{\sqrt{240}}\right) \right]_0^{\sqrt{5}t^2}\\ &= \frac{1}{5\sqrt{3}} \arctan\left(\frac{\sqrt{5}t^2}{\sqrt{240}}\right) \end{align*}

Marking Criteria

Descriptor	Marks
Correctly evaluates the integral and expresses the volume in the required form, $\frac{1}{5\sqrt{3}} \arctan\left(\frac{t^2}{4\sqrt{3}}\right)$	1

Q3d

2 marks

What surface area of the pond would the coverage of the pollutant approach?

Give your answer in square metres correct to two decimal places.

Reveal Answer

Since volume = surface area $\times \frac{1}{1000}$ , surface area is given by $1000 V = \frac{200}{\sqrt{3}} \arctan\left(\frac{t^2}{4\sqrt{3}}\right)$ .

As $t \to \infty$ the surface area approaches $\frac{200}{\sqrt{3}} \cdot \frac{\pi}{2} = \frac{100\pi}{\sqrt{3}} = 181.38$

Marking Criteria

Descriptor	Marks
Correctly identifies the limit of the arctan function as $t \to \infty$ is $\frac{\pi}{2}$ to find the limiting surface area	1
Calculates the correct limiting surface area, $181.38$	1

Q3e

2 marks

The clean-up of the pond begins after five days, where the pollutant is removed at a constant rate of $0.05$ cubic metres per day until the pond is free of pollutant. However, efforts to stem the flow are unsuccessful and the pollutant continues to enter the pond at a rate of $\frac{8t}{240 + 5t^4}$ cubic metres per day.

After how many days, from the start of the clean-up, will the pond be free of pollutant?
Give your answer in days correct to one decimal place.

Reveal Answer

The volume of the pollutant after 5 days is $V(5)$

To find the number of days needed for the pond to be free of pollutant solve

$0 = \int_5^{t+5} \left(\frac{8x}{240+5x^4} - 0.05\right) dx + V(5) \Rightarrow t = 3.4$ days

Marking Criteria

Descriptor	Marks
Sets up a correct equation to solve for the time, accounting for the 5-day delay and the constant removal rate, such as $0 = \int_5^{t+5} \left(\frac{8x}{240+5x^4} - 0.05\right) dx + V(5)$	1
Calculates the correct number of days, $3.4$	1

Q12

2021

SCSA

Paper 2

6 marks

Q12

The horizontal displacement of a Ferris wheel cabin exhibits simple harmonic motion. The maximum horizontal speed is $\frac{\pi}{2}$ metres per second and its period of motion is exactly 60 seconds.

Let $x(t) = A\cos(nt)$ be the horizontal displacement after $t$ seconds.

Q12a

3 marks

Determine the values of $A$ and $n$ .

Reveal Answer

Period $T = 60 = \frac{2\pi}{n} \implies \therefore n = \frac{\pi}{30} \quad (0.1047....)$

$v(t) = -A\left(\frac{\pi}{30}\right)\sin\left(\frac{\pi t}{30}\right)$

$\therefore \text{Max } v = A\left(\frac{\pi}{30}\right) = \frac{\pi}{2} \implies \therefore A = 15$

Hence $x(t) = 15\cos\left(\frac{\pi t}{30}\right)$

Marking Criteria

Descriptor	Marks
determines $n$ correctly	1
differentiates and forms the correct expression for the maximum speed	1
determines $A$ correctly	1

Q12b

3 marks

Determine the horizontal acceleration, correct to the nearest $0.001\text{ m/s}^2$ , when the horizontal displacement is 10 metres.

Reveal Answer

Condition for S.H.M. is $a = -\left(\frac{\pi}{30}\right)^2 x$

$\therefore$ When $x = 10 \quad a = -\left(\frac{\pi^2}{900}\right)(10) = -0.10966 ... \text{ metres/sec}^2$

i.e. acceleration is $-0.110 \text{ m/sec}^2$ (3 d.p.)

Marking Criteria

Descriptor	Marks
applies the condition for S.H.M. correctly	1
substitutes $x = 10$ correctly	1
determines the acceleration correct to $0.001 \text{ m/sec}^2$	1

Q2

2021

VCAA

Paper 1

3 marks

Q2

3 marks

Evaluate $\int_0^1 \frac{2x+1}{x^2+1}\,dx$ .

Reveal Answer

Write the integrand as the sum of two rational functions:
$\int_0^1 \frac{2x}{x^2+1} dx + \int_0^1 \frac{1}{x^2+1} dx = \left[\log_e(x^2+1)\right]_0^1 + \left[\arctan(x)\right]_0^1$

Answer is:
$\log_e(2) + \frac{\pi}{4}$

Marking Criteria

Descriptor	Marks
Splits the integrand into $\frac{2x}{x^2+1}$ and $\frac{1}{x^2+1}$ or correctly finds one antiderivative	1
Correctly finds both antiderivatives, obtaining $\log_e(x^2+1) + \arctan(x)$	1
Correctly evaluates the definite integral to obtain the final answer $\log_e(2) + \frac{\pi}{4}$	1

Q18

2021

QCAA

Paper 1

6 marks

Q18

6 marks

This differential equation can be used to determine the current $I$ (amperes) at time $t$ (seconds) with voltage $V$ (volts) in an electric circuit containing a resistance $R$ (ohms):

$k \frac{dI}{dt} + RI = V$

where $k$ , $R$ and $V$ are positive constants and $t \geq 0$ .

Assuming that there is no current in the electric circuit initially, show that the size of the current can never be greater than $\frac{V}{R}$ .

Reveal Answer

$k \frac{dI}{dt} + RI = V \Rightarrow k \frac{dI}{dt} = V - RI$
$\int \frac{k}{V - RI} dI = \int 1 dt$
$-\frac{k}{R} \ln|V - RI| = t + c$
Given $I = 0$ when $t = 0$
$c = -\frac{k}{R} \ln(V)$ (as $V > 0$ )
$-\frac{k}{R} \ln|V - RI| = t - \frac{k}{R} \ln(V)$
$\ln|V - RI| = -\frac{R}{k}t + \ln(V)$
$V - RI = e^{-\frac{R}{k}t + \ln(V)}$
$V - RI = V e^{-\frac{R}{k}t}$
For all $t, e^{-\frac{R}{k}t} > 0 \Rightarrow V - RI > 0 \Rightarrow V > RI \Rightarrow I < \frac{V}{R}$
So, the size of the current can never be greater than $\frac{V}{R}$ .

Marking Criteria

Descriptor	Marks
correctly uses the separation of variables method to set up indefinite integrals	1
develops a general solution of the differential equation	1
uses the given condition to determine expression for the constant of integration	1
rearranges relationship to express $\ln\|V - RI\|$ as the subject of the equation	1
expresses relationship as an exponential function	1
considers value of $I$ over time to determine the required limit	1

Q13

2025

VCAA

Paper 2

1 mark

Q13

1 mark

From an open window, a person projects a ball vertically up using an outstretched arm so the ball does not strike any part of the building. The point of projection of the ball is $50 \text{ m}$ above the ground and its velocity of projection is $20 \text{ m s}^{-1}$ .

The time, in seconds, it takes for the ball to reach the tray of a truck that is $1 \text{ m}$ above the ground directly below the point of projection is closest to

A

$1.72$

B

$5.80$

C

$5.83$

D

$1.75$

Reveal Answer

A

$1.72$

This is the magnitude of the negative root of the kinematic equation, which represents the time if the trajectory was extended backwards before projection.

B

$5.80$

Correct Answer

Using the kinematic equation $\Delta y = ut + \frac{1}{2}at^2$ with a displacement of $\Delta y = -49 \text{ m}$ (since the tray is $1 \text{ m}$ above the ground), $u = 20 \text{ m s}^{-1}$ , and $a = -9.8 \text{ m s}^{-2}$ yields $t \approx 5.80 \text{ s}$ .

C

$5.83$

This is the time it would take for the ball to reach the ground ( $\Delta y = -50 \text{ m}$ ), failing to account for the truck tray being $1 \text{ m}$ above the ground.

D

$1.75$

This is the magnitude of the negative root if the displacement was incorrectly set to $-50 \text{ m}$ (reaching the ground instead of the truck tray).

VCAA Specialist Mathematics Calculus

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