QCAA Specialist Mathematics Alternative Sequence Vector calculus

4 sample questions with marking guides and sample answers

Q3
2021
QCAA
Paper 1
1 mark
Q3
1 mark

An object has a velocity v(t)=e−2t i^+(1t)k^v(t)=e^{-2t}\,\hat{\mathbf{i}}+\left(\dfrac{1}{t}\right)\hat{\mathbf{k}}, where tt represents time (t>0)(t>0).

The displacement r(t)r(t) of the object could be

A

−2e−2t i^+ln⁡(t) k^-2e^{-2t}\,\hat{\mathbf{i}}+\ln(t)\,\hat{\mathbf{k}}

B

−2e−2t i^−1t2 k^-2e^{-2t}\,\hat{\mathbf{i}}-\dfrac{1}{t^2}\,\hat{\mathbf{k}}

C

−12e−2t i^+ln⁡(t) k^-\dfrac{1}{2}e^{-2t}\,\hat{\mathbf{i}}+\ln(t)\,\hat{\mathbf{k}}

D

−12e−2t i^−1t2 k^-\dfrac{1}{2}e^{-2t}\,\hat{\mathbf{i}}-\dfrac{1}{t^2}\,\hat{\mathbf{k}}

Reveal Answer
A

−2e−2t i^+ln⁡(t) k^-2e^{-2t}\,\hat{\mathbf{i}}+\ln(t)\,\hat{\mathbf{k}}

This option incorrectly differentiates the i^\hat{\mathbf{i}} component instead of integrating it. The integral of e−2te^{-2t} is −12e−2t-\frac{1}{2}e^{-2t}, not −2e−2t-2e^{-2t}.

B

−2e−2t i^−1t2 k^-2e^{-2t}\,\hat{\mathbf{i}}-\dfrac{1}{t^2}\,\hat{\mathbf{k}}

This option represents the acceleration of the object, which is found by taking the derivative of the velocity vector, rather than the integral.

C

−12e−2t i^+ln⁡(t) k^-\dfrac{1}{2}e^{-2t}\,\hat{\mathbf{i}}+\ln(t)\,\hat{\mathbf{k}}

Correct Answer

Displacement is the integral of velocity. Integrating each component of v(t)v(t) yields ∫e−2tdt=−12e−2t\int e^{-2t} dt = -\frac{1}{2}e^{-2t} and ∫1tdt=ln⁡(t)\int \frac{1}{t} dt = \ln(t) for t>0t>0.

D

−12e−2t i^−1t2 k^-\dfrac{1}{2}e^{-2t}\,\hat{\mathbf{i}}-\dfrac{1}{t^2}\,\hat{\mathbf{k}}

While the i^\hat{\mathbf{i}} component is correctly integrated, the k^\hat{\mathbf{k}} component is incorrectly differentiated. The integral of 1t\frac{1}{t} is ln⁡(t)\ln(t), not −1t2-\frac{1}{t^2}.

Q19
2021
QCAA
Paper 1
7 marks
Q19
7 marks

The velocity vectors of two objects A and B (in m s−1\mathrm{m\ s^{-1}}) at time tt (in s) are given respectively by

vA=6sin⁡(3t) i^+6cos⁡(3t) j^v_A=6\sin(3t)\,\hat{\mathbf{i}}+6\cos(3t)\,\hat{\mathbf{j}}

vB=cos⁡(t) i^−sin⁡(t) j^v_B=\cos(t)\,\hat{\mathbf{i}}-\sin(t)\,\hat{\mathbf{j}}

Objects A and B are initially at (−2,0,2)(-2,0,2) and (0,1,−1)(0,1,-1) respectively. Determine the position of Object A when it is 4 metres away from Object B for the first time.

Reveal Answer

vA=6sin⁡(3t)ı^+6cos⁡(3t)ȷ^\boldsymbol{v}_A = 6\sin(3t)\hat{\boldsymbol{\imath}} + 6\cos(3t)\hat{\boldsymbol{\jmath}}
vB=cos⁡(t)ı^−sin⁡(t)ȷ^\boldsymbol{v}_B = \cos(t)\hat{\boldsymbol{\imath}} - \sin(t)\hat{\boldsymbol{\jmath}}

rA=∫vA dt=−2cos⁡(3t)ı^+2sin⁡(3t)ȷ^+cA\boldsymbol{r}_A = \int \boldsymbol{v}_A \, dt = -2\cos(3t)\hat{\boldsymbol{\imath}} + 2\sin(3t)\hat{\boldsymbol{\jmath}} + \boldsymbol{c}_A
When t=0t = 0
−2ı^+2k^=−2cos⁡(0)ı^+2sin⁡(0)ȷ^+cA⇒cA=2k^-2\hat{\boldsymbol{\imath}} + 2\hat{\boldsymbol{k}} = -2\cos(0)\hat{\boldsymbol{\imath}} + 2\sin(0)\hat{\boldsymbol{\jmath}} + \boldsymbol{c}_A \Rightarrow \boldsymbol{c}_A = 2\hat{\boldsymbol{k}}
∴rA=−2cos⁡(3t)ı^+2sin⁡(3t)ȷ^+2k^\therefore \boldsymbol{r}_A = -2\cos(3t)\hat{\boldsymbol{\imath}} + 2\sin(3t)\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}}

rB=∫vB dt=sin⁡(t)ı^+cos⁡(t)ȷ^+cB\boldsymbol{r}_B = \int \boldsymbol{v}_B \, dt = \sin(t)\hat{\boldsymbol{\imath}} + \cos(t)\hat{\boldsymbol{\jmath}} + \boldsymbol{c}_B
When t=0t = 0
ȷ^−k^=sin⁡(0)ı^+cos⁡(0)ȷ^+cB⇒cB=−k^\hat{\boldsymbol{\jmath}} - \hat{\boldsymbol{k}} = \sin(0)\hat{\boldsymbol{\imath}} + \cos(0)\hat{\boldsymbol{\jmath}} + \boldsymbol{c}_B \Rightarrow \boldsymbol{c}_B = -\hat{\boldsymbol{k}}
∴rB=sin⁡(t)ı^+cos⁡(t)ȷ^−k^\therefore \boldsymbol{r}_B = \sin(t)\hat{\boldsymbol{\imath}} + \cos(t)\hat{\boldsymbol{\jmath}} - \hat{\boldsymbol{k}}

rB−rA\boldsymbol{r}_B - \boldsymbol{r}_A
=(sin⁡(t)ı^+cos⁡(t)ȷ^−k^)…= (\sin(t)\hat{\boldsymbol{\imath}} + \cos(t)\hat{\boldsymbol{\jmath}} - \hat{\boldsymbol{k}}) \dots
⋯−(−2cos⁡(3t)ı^+2sin⁡(3t)ȷ^+2k^)\dots - (-2\cos(3t)\hat{\boldsymbol{\imath}} + 2\sin(3t)\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}})
=(sin⁡(t)+2cos⁡(3t))ı^+(cos⁡(t)−2sin⁡(3t))ȷ^−3k^= (\sin(t) + 2\cos(3t))\hat{\boldsymbol{\imath}} + (\cos(t) - 2\sin(3t))\hat{\boldsymbol{\jmath}} - 3\hat{\boldsymbol{k}}

∣rB−rA∣=sin⁡2(t)+4sin⁡(t)cos⁡(3t)+4cos⁡2(3t)+…|\boldsymbol{r}_B - \boldsymbol{r}_A| = \sqrt{\sin^2(t) + 4\sin(t)\cos(3t) + 4\cos^2(3t) + \dots}
…cos⁡2(t)−4cos⁡(t)sin⁡(3t)+4sin⁡2(3t)+9‾\overline{\dots \cos^2(t) - 4\cos(t)\sin(3t) + 4\sin^2(3t) + 9}
=14−4(sin⁡(3t)cos⁡(t)−cos⁡(3t)sin⁡(t))= \sqrt{14 - 4(\sin(3t)\cos(t) - \cos(3t)\sin(t))}
=14−4(sin⁡(3t−t))= \sqrt{14 - 4(\sin(3t - t))}
=14−4sin⁡(2t)= \sqrt{14 - 4\sin(2t)}

Given ∣rB−rA∣=4|\boldsymbol{r}_B - \boldsymbol{r}_A| = 4
14−4sin⁡(2t)=4\sqrt{14 - 4\sin(2t)} = 4
sin⁡(2t)=−12\sin(2t) = -\frac{1}{2}
2t=7π62t = \frac{7\pi}{6}
t=7π12t = \frac{7\pi}{12} s (first positive solution)

Position of A
rA=−2cos⁡(3t)ı^+2sin⁡(3t)ȷ^+2k^\boldsymbol{r}_A = -2\cos(3t)\hat{\boldsymbol{\imath}} + 2\sin(3t)\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}}
=−2cos⁡(7π4)ı^+2sin⁡(7π4)ȷ^+2k^= -2\cos\left(\frac{7\pi}{4}\right)\hat{\boldsymbol{\imath}} + 2\sin\left(\frac{7\pi}{4}\right)\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}}
=−2ı^−2ȷ^+2k^= -\sqrt{2}\hat{\boldsymbol{\imath}} - \sqrt{2}\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}} (m)

Marking Criteria
DescriptorMarks

correctly determines the expression for the position of Object A

1

correctly determines the expression for the position of Object B

1

determines an expression to represent the relative position of Objects A and B

1

determines an expression to represent the distance (or square of the distance) between the objects

1

uses a trigonometric identity to determine an expression in terms of a single trigonometric function that represents the distance (or square of the distance) between the objects

1

determines the first time that Object A is 4 metres away from Object B

1

determines position of Object A

1
Q10
2021
QCAA
Paper 2
1 mark
Q10
1 mark

The respective positions r\boldsymbol{r} over time t(t≥0)t (t \geq 0) of two objects are given by

r1=4e2−ti^+(t−2)j^\boldsymbol{r}_1 = 4e^{2-t}\hat{i} + (t-2)\hat{j}
r2=2ti^+(4−t2)j^\boldsymbol{r}_2 = 2t\hat{i} + (4-t^2)\hat{j}

Determine the coordinates of the point where the objects collide.

A

(0,4)(0, 4)

B

(2,0)(2, 0)

C

(2,4)(2, 4)

D

(4,0)(4, 0)

Reveal Answer
A

(0,4)(0, 4)

This point corresponds to the position of the second object at t=0t=0, but the first object is at (4e2,−2)(4e^2, -2) at this time, so no collision occurs here.

B

(2,0)(2, 0)

This incorrectly uses the time of collision (t=2t=2) as the x-coordinate. The actual x-coordinate at t=2t=2 is 44.

C

(2,4)(2, 4)

This incorrectly uses the time of collision (t=2t=2) as the x-coordinate and the initial y-coordinate of the second object as the y-coordinate.

D

(4,0)(4, 0)

Correct Answer

Setting the y-components equal gives t−2=4−t2t-2 = 4-t^2, which simplifies to t2+t−6=0t^2+t-6=0. The only valid solution for t≥0t \geq 0 is t=2t=2. Substituting t=2t=2 into either position vector yields the collision point (4,0)(4, 0).

Q17
2022
QCAA
Paper 1
7 marks
Q17

The Cartesian equation of an ellipse is x29+y216=1\frac{x^2}{9} + \frac{y^2}{16} = 1.

Q17a
3 marks

Determine the gradient of the ellipse at the point in quadrant 2 where y=22y = 2\sqrt{2}.

Reveal Answer

At y=22y = 2\sqrt{2}
x29+(22)216=1\frac{x^2}{9} + \frac{(2\sqrt{2})^2}{16} = 1
x2=92⇒x=±32x^2 = \frac{9}{2} \Rightarrow x = \pm \frac{3}{\sqrt{2}}

Given the point on the ellipse is in quadrant 2,
x=−322x = -\frac{3\sqrt{2}}{2}

Determining gradient of the ellipse
x29+y216=1\frac{x^2}{9} + \frac{y^2}{16} = 1
2x9+2ydydx16=0\frac{2x}{9} + \frac{2y \frac{dy}{dx}}{16} = 0
dydx=−16x9y\frac{dy}{dx} = -\frac{16x}{9y}

Determining gradient at the required point
dydx=−16(−322)9(22)=43\frac{dy}{dx} = -\frac{16\left(-\frac{3\sqrt{2}}{2}\right)}{9(2\sqrt{2})} = \frac{4}{3}

Marking Criteria
DescriptorMarks

correctly determines the xx-coordinate of the required point in quadrant 2

1

determines a general expression for the gradient of the ellipse in terms of xx and yy

1

determines gradient of the ellipse at required point

1
Q17b
4 marks

The ellipse can be shown to be represented by the parametric equations x=3cos⁡(θ)x = 3\cos(\theta) and y=4sin⁡(θ)y = 4\sin(\theta), where 0≤θ≤2π0 \leq \theta \leq 2\pi.

Use the parametric equations to verify your result from Question 17a). Do not show the conversion from the parametric equations into the Cartesian equation.

Reveal Answer

Determining gradient of ellipse
x=3cos⁡(θ)⇒dxdθ=−3sin⁡(θ)x = 3\cos(\theta) \Rightarrow \frac{dx}{d\theta} = -3\sin(\theta)
y=4sin⁡(θ)⇒dydθ=4cos⁡(θ)y = 4\sin(\theta) \Rightarrow \frac{dy}{d\theta} = 4\cos(\theta)

Determining θ\theta for the point on the ellipse
x=3cos⁡(θ)x = 3\cos(\theta): At x=−322⇒cos⁡(θ)=−22x = -\frac{3\sqrt{2}}{2} \Rightarrow \cos(\theta) = -\frac{\sqrt{2}}{2}
y=4sin⁡(θ)y = 4\sin(\theta): At y=22⇒sin⁡(θ)=22y = 2\sqrt{2} \Rightarrow \sin(\theta) = \frac{\sqrt{2}}{2}

dydx=dydθdxdθ=4cos⁡(θ)−3sin⁡(θ)\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{4\cos(\theta)}{-3\sin(\theta)}

At the point on the ellipse
dydx=4(−22)−3(22)=43\frac{dy}{dx} = \frac{4\left(-\frac{\sqrt{2}}{2}\right)}{-3\left(\frac{\sqrt{2}}{2}\right)} = \frac{4}{3}

The result from 17a) is verified.

Marking Criteria
DescriptorMarks

correctly determines expressions for dxdθ\frac{dx}{d\theta} and dydθ\frac{dy}{d\theta}

1

determines expressions for cos⁡(θ)\cos(\theta) and sin⁡(θ)\sin(\theta)

1

determines a general expression for the gradient of the ellipse in terms of θ\theta

1

verifies gradient of the ellipse result from 17a)

1

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