QCE Specialist Maths: Mathematical induction and trigonometric proofs Q253

Q253 · Practice questionTechnology-activeVery complex unfamiliar13 marks

QUESTION 253 (13 marks)

A game has centre

\mathbf c=(1,-1)

and checkpoints

\mathbf p_0=\mathbf c+(8,0)

\mathbf p_{n+1}=\mathbf c+A(\mathbf p_n-\mathbf c)

, where

A=\frac14\begin{pmatrix}1&-\sqrt3\\\sqrt3&1\end{pmatrix}

. A player visits the checkpoints in order along straight segments. Coordinates are in metres. You may use

\sum_{n=0}^\infty r^n=1/(1-r)

for

|r|<1

Checkpoint spiral about C=(1,-1), showing the first straight segments as solid lines, a dashed smooth spiral, and labelled early points P0, P1 and P2. Later point labels are omitted to avoid crowding.

Prove by induction that

A^n=2^{-n}\begin{pmatrix}\cos(n\pi/3)&-\sin(n\pi/3)\\\sin(n\pi/3)&\cos(n\pi/3)\end{pmatrix}

for every

n\ge0

[4 marks]

Determine the exact total length of the infinitely many straight segments.

[3 marks]

A smooth alternative path is

z(t)=1-i+8e^{-(\ln2)t}\operatorname{cis}(\pi t/3)

t\ge0

. Verify that it passes through every checkpoint at the corresponding integer time and determine its speed.

[3 marks]

Determine the exact length of the smooth path and compare the two lengths numerically to three decimal places.

[3 marks]

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WORKED SOLUTION

Practice marking scheme

13 marks

ANSWER

(a) The stated matrix-power formula. (b)

L_{\mathrm{straight}}=8\sqrt3

m. (c)

z(n)

corresponds to

\mathbf p_n

; speed

8e^{-(\ln2)t}\sqrt{(\ln2)^2+\pi^2/9}

. (d)

L_{\mathrm{smooth}}=\frac{8\sqrt{(\ln2)^2+\pi^2/9}}{\ln2}

m; it is longer.

Worked solution

(a) For

n=0

, both sides are the identity. Assume

A^k=2^{-k}R(k\pi/3)

, where

R(\theta)

is the displayed rotation matrix. Since

A=\tfrac12R(\pi/3)

, multiplication and the angle-addition identities give

A^{k+1}=2^{-(k+1)}R((k+1)\pi/3)

. Thus the formula holds for all

n\ge0

(b) At checkpoint

n

, the distance from the centre is

8\,2^{-n}

. The next radius has half that length and angle

\pi/3

to it. The cosine rule gives segment length

4\sqrt3\,2^{-n}

. Summing gives

4\sqrt3/(1-1/2)=8\sqrt3

n

e^{-(\ln2)n}=2^{-n}

, so the position matches the matrix-power formula. Writing

a=\ln2

and

b=\pi/3

, differentiation of

(8e^{-at}\cos bt,8e^{-at}\sin bt)

gives components whose squared sum is

64e^{-2at}(a^2+b^2)

. Taking the positive square root yields the speed.

(d) Integrating the speed over

t\ge0

gives

L_{\mathrm{smooth}}=8\sqrt{a^2+b^2}/a

. Numerically the straight path has length

13.856

m and the smooth path has length

14.494

m. The smooth path is therefore longer, although both visit the same checkpoints.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Check the base case and state the hypothesis.

[1 mark]

Part a: Identify the scale factor and rotation in

A

[1 mark]

Part a: Multiply using the angle-addition identities to prove the next case.

[1 mark]

Part a: Give a valid induction conclusion.

[1 mark]

Part b: Obtain the radius at checkpoint

n

[1 mark]

Part b: Obtain the segment length using the angle or the matrix difference.

[1 mark]

Part b: Sum the convergent series exactly.

[1 mark]

Part c: Verify all integer-time checkpoint positions.

[1 mark]

Part c: Differentiate the two Cartesian components.

[1 mark]

Part c: Combine them to obtain the speed.

[1 mark]

Part d: Form the improper speed integral.

[1 mark]

Part d: Evaluate its exact convergent value.

[1 mark]

Part d: Compare both lengths at the requested accuracy.

[1 mark]

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Mathematical induction and trigonometric proofs — Question 253

QUESTION 253 (13 marks)

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