How many VCAA Specialist Mathematics questions cover Calculus?

AusGrader has 240 VCAA Specialist Mathematics questions on Calculus, all with instant AI grading and detailed marking feedback.

VCE Specialist Mathematics — Calculus Questions & Answers

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

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

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

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

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).

Q13

2020

QCAA

Paper 1

4 marks

Q13

4 marks

The expected value of an exponential random variable $X$ with parameter $\lambda > 0$ can be determined using the rule

$E(X) = \int_0^\infty x\lambda e^{-\lambda x} dx$

Use integration by parts to determine $E(X)$ .
Express your answer in simplest form.

Reveal Answer

$E(X) = \int_0^\infty x \lambda e^{-\lambda x} dx$
$\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx$

$u = x \quad \frac{du}{dx} = 1$
$\frac{dv}{dx} = \lambda e^{-\lambda x} \quad v = -e^{-\lambda x}$

$E(X) = -xe^{-\lambda x}|_0^\infty + \int_0^\infty e^{-\lambda x} dx$
$= 0 + \int_0^\infty e^{-\lambda x} dx$
$= \frac{e^{-\lambda x}}{-\lambda}|_0^\infty$
$= 0 - \frac{1}{-\lambda}$
$= \frac{1}{\lambda}$

Marking Criteria

Descriptor	Marks
correctly determines $\frac{du}{dx}$ and $v$	1
substitutes into the integration by parts rule	1
calculates $-xe^{-\lambda x}\|_0^\infty$ to equal 0	1
shows that $E(X) = \frac{1}{\lambda}$	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

Q12

2024

VCAA

Paper 2

1 mark

Q12

1 mark

The position, $x$ metres, of a particle moving in a straight line from a fixed origin $O$ at time, $t$ seconds, is given by $x = e^{(k - 1)t}$ , where $k > 1$ .

The acceleration of the particle, in $\text{m s}^{-2}$ , when $x = k + 1$ is

A

$k^2 - 1$

B

$(k^2 - 1)(k + 1)$

C

$(k^2 - 1)(k - 1)$

D

$(k - 1)^2$

Reveal Answer

A

$k^2 - 1$

Incorrect. This represents the velocity of the particle when $x = k + 1$ , found by taking the first derivative $v = (k - 1)x$ , rather than the acceleration.

B

$(k^2 - 1)(k + 1)$

Incorrect. This expression does not match the acceleration formula $a = (k - 1)^2 x$ . It incorrectly multiplies the velocity by $x$ instead of taking the second derivative.

C

$(k^2 - 1)(k - 1)$

Correct Answer

Correct. The acceleration is the second derivative of position, $a = (k - 1)^2 x$ . Substituting $x = k + 1$ gives $a = (k - 1)^2(k + 1) = (k - 1)(k^2 - 1).

D

$(k - 1)^2$

Incorrect. This is the coefficient of $x$ in the acceleration equation $a = (k - 1)^2 x$ , but it fails to substitute the given value $x = k + 1$ .

Q5

2024

VCAA

Paper 1

3 marks

Q5

3 marks

The curve with equation $y = \sqrt{k - \frac{1}{x^2}}$ , for $1 \le x \le \frac{k}{2}$ where $k > 2$ , is rotated about the $x$ -axis to form a solid of revolution that has volume $\frac{7\pi}{2}$ units $^3$ .

Show that $k$ satisfies the equation $k^3 - 2k^2 - 9k + 4 = 0$ .

Reveal Answer

The volume $V$ is given by $V = \pi \int_1^k \left(k - \frac{1}{x^2}\right) dx$ .

$V = \pi \int_1^k \left(k - \frac{1}{x^2}\right) dx= \pi \left[kx + \frac{1}{x}\right]_1^k= \pi \left(\frac{1}{2}k^2 + \frac{2}{k} - k - 1\right)$

$\pi \left(\frac{1}{2}k^2 + \frac{2}{k} - k - 1\right) = \frac{7\pi}{2}\implies k^3 - 2k^2 - 9k + 4 = 0$

Marking Criteria

Descriptor	Marks
Sets up the correct definite integral for the volume	1
Correctly integrates the expression and substitutes the limits	1
Equates the evaluated integral to $\frac{7\pi}{2}$ and correctly rearranges to show $k^3 - 2k^2 - 9k + 4 = 0$	1

Q17

2020

QCAA

Paper 2

7 marks

Q17

7 marks

An object is released from rest at a height of 100 m above the ground.
The motion of the vertical descent of the object is modelled by

$v\frac{dv}{dx} = 9.8 - 0.004v^2 \quad (v \ge 0)$

where $v$ is the velocity (m s $^{-1}$ ) and $x$ is the displacement from the ground (m).
Determine the velocity of the object when it strikes the ground.

Reveal Answer

$v\frac{dv}{dx} = 9.8 - 0.004v^2, v > 0$
$\int \frac{v}{9.8 - 0.004v^2} dv = \int dx$
$\frac{-1}{0.008} \int \frac{-0.008v}{9.8 - 0.004v^2} dv = \int dx$
$-125 \ln|9.8 - 0.004v^2| = x + c$

Given $v = 0$ when $x = -100$
$-125 \ln|9.8| = -100 + c$
$c \approx -185.298$

$-125 \ln|9.8 - 0.004v^2| = x - 185.298$
Determining $v$ when $x = 0$
$-125 \ln|9.8 - 0.004v^2| = -185.298$

Using graph facility of GDC
$v \approx -36.7 \text{ ms}^{-1}$ or $v \approx 36.7 \text{ ms}^{-1}$
As $v > 0$ , the negative solution is rejected
$\therefore v \approx 36.7 \text{ ms}^{-1}$

Marking Criteria

Descriptor	Marks
correctly uses separation of variables	1
correctly develops the general solution of the differential equation	1
correctly uses the given position of the origin	1
uses the given condition to determine value for c	1
substitutes the displacement at impact to form an equation in terms of v	1
determines one reasonable solution of v	1
shows logical organisation communicating key steps	1

Q1

2020

QCAA

Paper 2

1 mark

Q1

1 mark

The position $x$ (m) at time $t$ (s) of a 7 kg particle moving in a straight line is given by

$x = 3t^3 - 5t^2 + 2t - 4 \text{ for } 0 \le t \le 10$

Determine the time when the particle has a momentum of 620 kg m s $^{-1}$ .

A

1.73 s

B

2.60 s

C

3.66 s

D

3.71 s

Reveal Answer

A

1.73 s

This time value is incorrect. It may result from a calculation error or solving an incorrect kinematic equation.

B

2.60 s

This is an incorrect answer. Substituting $t = 2.60$ s into the derived momentum equation yields a value significantly lower than 620 kg m s $^{-1}$ .

C

3.66 s

This value is close but incorrect. Using $t = 3.66$ s results in a momentum of approximately 602 kg m s $^{-1}$ , which is less than the required value.

D

3.71 s

Correct Answer

First, find the velocity function by differentiating the position: $v(t) = \frac{dx}{dt} = 9t^2 - 10t + 2$ . Then, use the momentum formula $p = mv$ to set up $620 = 7(9t^2 - 10t + 2)$ . Solving this quadratic equation for $t$ yields $t \approx 3.71$ s.

Q17

2025

QCAA

Paper 1

7 marks

Q17

7 marks

The radius of a cylinder decreases at a constant rate of $0.5 \text{ m s}^{-1}$ , while maintaining a constant height of four metres.

Given that the cylinder has an initial volume of $100\pi \text{ m}^3$ , determine the rate of change of the volume ( $\text{m}^3 \text{ s}^{-1}$ ) of the cylinder after four seconds.

Reveal Answer

Given $V = \pi r^2 h \Rightarrow V = 4\pi r^2$

Given $\frac{dr}{dt} = -0.5 \text{ m s}^{-1}$

$V = 4\pi r^2 \Rightarrow \frac{dV}{dt} = 8\pi r \frac{dr}{dt}$

At $t = 0$ , $V = 100\pi$ (given)

100\pi = 4\pi r^2 \Rightarrow r = 5 \text{ m} (\text{as } r > 0)

At $t = 4$ ,

r = 5 - 4 \times 0.5 = 3 \text{ m}

When $r = 3$ :

\frac{dV}{dt} = 8\pi \times 3 \times (-0.5) = -12\pi \text{ m}^3 \text{ s}^{-1}

Marking Criteria

Descriptor	Marks
correctly determines the rule for the volume of cylinder in terms of $r$	1
correctly expresses the rate of change of the radius as a mathematical expression	1
determines a general expression for $\frac{dV}{dt}$	1
determines initial value of $r$	1
determines value of $r$ after 4 seconds	1
determines value of $\frac{dV}{dt}$ when $t = 4$	1
shows logical organisation, having fully attempted the question	1

Q18

2023

QCAA

Paper 1

6 marks

Q18

6 marks

A particular solution to the differential equation $\frac{dy}{dx} = \frac{x}{(x^2+1)\tan(y)}$ , where $x \ge 0$ and $-\frac{\pi}{2} < y \le 0$ , passes through the origin.

Determine this solution in the form $x = f(y)$ . Leave your answer in simplified form.

Reveal Answer

$\frac{dy}{dx} = \frac{x}{(x^2+1)\tan(y)}$

$\int \tan(y) dy = \int \frac{x}{x^2+1} dx$

$-\int \frac{-\sin(y)}{\cos(y)} dy = \frac{1}{2} \int \frac{2x}{x^2+1} dx$

$-\ln|\cos(y)| = \frac{1}{2}\ln|x^2+1| + c$

Given $y=0$ when $x=0$ ,
$-\ln|\cos(0)| = \frac{1}{2}\ln|1| + c \implies c=0$

$\therefore \frac{1}{2}\ln|x^2+1| = -\ln|\cos(y)|$

$\ln\sqrt{x^2+1} = \ln\left|\frac{1}{\cos(y)}\right|$

$\sqrt{x^2+1} = |\sec(y)|$
$x^2+1 = \sec^2(y)$
$x^2 = \tan^2(y)$
$x = \pm \tan(y)$

As $x \ge 0, -\frac{\pi}{2} < y \le 0$
$x = -\tan(y)$

Marking Criteria

Descriptor	Marks
Correctly separates the variables	1
Applies suitable integration methods	1
Determines a value for the constant of integration	1
Determines an expression for a solution that does not contain logarithms	1
Expresses $x$ in terms of $y$	1
Evaluates the reasonableness of the results and expresses the solution in the form of $x = f(y)$ in simplified form	1

Q17

2024

QCAA

Paper 1

6 marks

Q17

6 marks

The acceleration ( $\text{m s}^{-2}$ ) of an object that moves in a straight line in an easterly direction over time $t$ for $0 \le t \le \frac{\pi}{6}$ seconds is given by $a = 2(1+v^2)$ , where $v$ is its velocity ( $\text{m s}^{-1}$ ).

The object is initially at rest at a position that is $\ln(\sqrt{2})$ metres west of the origin.

A student uses this information to calculate that the object is positioned at the origin when $t = \frac{\pi}{6}$ seconds.

Evaluate the reasonableness of the student's calculation.

Reveal Answer

Method 1

$a = 2(1+v^2)$
$\frac{dv}{dt} = 2(1+v^2)$

$\int \frac{1}{1+v^2} dv = \int 2 dt$
$\tan^{-1}(v) = 2t + c$

At $t=0, v=0: \tan^{-1}(0) = 0 + c$
$\therefore c = 0$

$\therefore \tan^{-1}(v) = 2t$
$v = \tan(2t)$
$x = \int v dt$
$= \int \tan(2t) dt$
$= -\frac{1}{2} \ln(\cos(2t)) + c$
$= -\frac{1}{2} \ln(\cos(2t)) + c$

Given $x = -\ln(\sqrt{2})$ when $t=0$
$-\ln(\sqrt{2}) = -\frac{1}{2} \ln(\cos(0)) + c$
$c = -\ln(\sqrt{2})$

$x = -\frac{1}{2} \ln(\cos(2t)) - \ln(\sqrt{2})$

When $t = \frac{\pi}{6}$
$x = -\frac{1}{2} \ln(\cos(\frac{\pi}{3})) - \ln(\sqrt{2})$
$= -\frac{1}{2} \ln(\frac{1}{2}) - \ln(\sqrt{2})$

$x = \ln((\frac{1}{2})^{-\frac{1}{2}}) - \ln(\sqrt{2})$
$= \ln(\sqrt{2}) - \ln(\sqrt{2})$
$= 0$
So the calculation is reasonable.

Marking Criteria

Descriptor	Marks
correctly determines a general solution to the differential equation in terms of v and time	1
determines an appropriate constant of integration	1
determines a general solution for displacement in terms of time	1
determines an appropriate constant of integration	1
determines a result for displacement when $t = \frac{\pi}{6}$ without a trigonometric term	1
provides an appropriate statement of reasonableness based on mathematical reasoning	1

VCAA Specialist Mathematics Calculus

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