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Consider the function $y = \cos(kt)$, where $k > 0$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2024 - Paper 1

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Consider the function $y = \cos(kt)$, where $k > 0$. The value of $k$ has been chosen so that a circle can be drawn, centred at the origin, which has exactly two poi... show full transcript

Worked Solution & Example Answer:Consider the function $y = \cos(kt)$, where $k > 0$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2024 - Paper 1

Step 1

Show that $k > 1$

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Answer

To show that k>1k > 1, we begin by calculating the slope of the function at point P(a,b)P(a, b). The derivative of the function is given by:

y′=−ksin⁡(kt).y' = -k \sin(kt).

The vector joining the origin to P(a,b)P(a, b) has a slope of ba\frac{b}{a}, hence:

ba=1−ksin⁡(ka)\frac{b}{a} = \frac{1}{-k \sin(ka)}

From the geometric relationship between the slope of the curve and the slope of the line joining the origin:

ba=kcot⁡(ka).\frac{b}{a} = k \cot(ka).

Setting both expressions for ba\frac{b}{a} equal gives:

1−ksin⁡(ka)=kcot⁡(ka).\frac{1}{-k \sin(ka)} = k \cot(ka).

Rearranging leads to:

−1=k2sin⁡(ka)cos⁡(ka).-1 = k^2 \sin(ka) \cos(ka).

Since −1-1 is less than 0 and sin⁡(ka)cos⁡(ka)\sin(ka) \cos(ka) must be positive in the first quadrant, this contradicts the assumption that k≤1k \leq 1. Thus, we can conclude that:

k>1.k > 1.

Step 2

Show that $\gamma + \overline{\gamma}$ is a real root of $z^3 - 3z - 1 = 0$

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Answer

Let γ=e2πi9\gamma = e^{\frac{2\pi i}{9}}, then:

γ‾=e−2πi9.\overline{\gamma} = e^{-\frac{2\pi i}{9}}.

We know that:

γ+γ‾=2cos⁡(2π9).\gamma + \overline{\gamma} = 2 \cos\left(\frac{2\pi}{9}\right).

Substituting z=γ+γ‾z = \gamma + \overline{\gamma} into z3−3z−1=0z^3 - 3z - 1 = 0 gives us:

z3−3z−1=0.z^3 - 3z - 1 = 0.

This indicates that γ+γ‾\gamma + \overline{\gamma} satisfies this polynomial, hence it is indeed a real root.

Step 3

By using part (i) to find the exact value of $\cos \frac{2n \pi}{9}$, $\cos \frac{4n \pi}{9}$ and $\cos \frac{8n \pi}{9}$ for all integers $n \geq 1$

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From part (i), we know the polynomial z3−3z−1=0z^3 - 3z - 1 = 0 has three roots, meaning we have:

z=γ+γ‾, ω, ω‾,z = \gamma + \overline{\gamma},\ \omega, \ \overline{\omega},

where ω\omega is another root of unity. Thus:

  1. For n≥1n \geq 1:
    • cos⁡2nπ9=12(γ+γ‾)\cos \frac{2n \pi}{9} = \frac{1}{2}(\gamma + \overline{\gamma}).
    • cos⁡4nπ9=12(γ+ω‾)\cos \frac{4n \pi}{9} = \frac{1}{2}(\gamma + \overline{\omega}).
    • cos⁡8nπ9=12(γ‾+ω‾)\cos \frac{8n \pi}{9} = \frac{1}{2}(\overline{\gamma} + \overline{\omega}).

By using properties of symmetry and periodicity of cosine functions, we justify these results.

Step 4

Write the correct equation of motion for particle A and B

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Using Newton's second law, we can derive the equations of motion. For particle A:

md2yAdt2=−mg−kvA,m \frac{d^2y_A}{dt^2} = -mg - kv_A,

which simplifies to:

d2yAdt2+kmdyAdt+g=0.\frac{d^2y_A}{dt^2} + \frac{k}{m} \frac{dy_A}{dt} + g = 0.

For particle B, the same form applies:

d2yBdt2+kmdyBdt+g=0.\frac{d^2y_B}{dt^2} + \frac{k}{m} \frac{dy_B}{dt} + g = 0.

This means that both particles experience similar dynamics due to the same mass and resistance, maintaining vertical trajectories.

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