Further complex numbers — Question 248

Original QCE Vault practice · 12 marks

Q248 · Practice questionTechnology-freeVery complex unfamiliar12 marks

QUESTION 248 (12 marks)

A scanner spot has complex position

z(t)=1+i+2\frac{1+it}{1-it}

, where

t\ge0

is time in seconds and the real and imaginary parts give coordinates in metres. A straight detection gate lies on

x+2y=7

A radius-two circle centred at (1,1), the starting point (3,1), and the gate x+2y=7; arrows show the direction of scanning. Crossing points are not labelled.

Write

z(t)=x(t)+iy(t)

in Cartesian form.

[3 marks]

Show that the spot remains on a circle and determine its speed as a function of time.

[3 marks]

Determine all times at which the spot crosses the gate.

[2 marks]

Find the exact distance travelled and the magnitude of the displacement between these crossings. Explain why they differ.

[4 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

12 marks

ANSWER

(a)

x=1+\frac{2(1-t^2)}{1+t^2}

y=1+\frac{4t}{1+t^2}

. (b)

(x-1)^2+(y-1)^2=4

; speed

4/(1+t^2)

m/s. (c)

t=1/3

and

t=1

s. (d) Distance

4\tan^{-1}(1/2)

m; displacement magnitude

4/\sqrt5

Worked solution

(a) Multiply the quotient by

(1+it)/(1+it)

. Then

(1+it)^2=1-t^2+2it

and

(1-it)(1+it)=1+t^2

. Separating parts gives the stated coordinates.

(b) The squared radius is

4[(1-t^2)^2+4t^2]/(1+t^2)^2=4

. Differentiation gives

x^{\prime}=-8t/(1+t^2)^2

and

y^{\prime}=4(1-t^2)/(1+t^2)^2

. Their squared sum is

16/(1+t^2)^2

, so the speed is

4/(1+t^2)

x+2y=7

gives

3+2(1-t^2+4t)/(1+t^2)=7

, hence

3t^2-4t+1=(3t-1)(t-1)=0

. Both roots satisfy

t\ge0

(d) The distance is

4\int_{1/3}^1dt/(1+t^2)=4[\tan^{-1}1-\tan^{-1}(1/3)]=4\tan^{-1}(1/2)

. The difference of angles lies in

(0,\pi/2)

, justifying this use of the tangent subtraction identity. The positions are

(13/5,11/5)

and

(1,3)

, so the displacement is

(-8/5,4/5)

with magnitude

4/\sqrt5

. The arc is longer than the chord joining its endpoints.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Use the conjugate to obtain denominator

1+t^2

[1 mark]

Part a: Obtain the real coordinate.

[1 mark]

Part a: Obtain the imaginary coordinate.

[1 mark]

Part b: Establish the circle equation.

[1 mark]

Part b: Differentiate both coordinates correctly.

[1 mark]

Part b: Combine components and simplify the speed.

[1 mark]

Part c: Obtain the quadratic equation for a crossing.

[1 mark]

Part c: Obtain both admissible times.

[1 mark]

Part d: Integrate speed over the correct interval.

[1 mark]

Part d: Obtain the exact inverse-tangent distance with the correct branch.

[1 mark]

Part d: Obtain the displacement magnitude from the two positions.

[1 mark]

Part d: Explain the difference using arc and chord lengths.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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Further complex numbers — Question 248

QUESTION 248 (12 marks)

Practice marking scheme

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