Maclaurin Series: Maclaurin + sketch + approximation — TMJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
An architect designs a flower bed for a garden. The flower bed is enclosed by the curve \(C\) with equation \[ \left(\dfrac{x+y}{2}\right)^2 + \left(\dfrac{y-x}{4}\right)^2 = 1. \] The region in the first quadrant bounded by \(C\), the \(x\)-axis and the lines \(x = 0\) and \(x = 0.5\) is to be planted with tulips.
- By finding the Maclaurin series expansion of \(y\) up to and including the term in \(x^2\), where \(y > 0\), find an approximation for the area to be planted with tulips.
The curve \(C\) is obtained by rotating an ellipse \(S\) about the origin in the \(x\)-\(y\) plane. It is known that the curve \(C\) can be expressed by the following parametric equations \[ x = \cos t - 2\sin t, \quad y = \cos t + 2\sin t, \] where \(-\pi < t \leq \pi\).
- Let \(L\) be the distance between the origin and a point on \(C\). By using differentiation, determine the exact maximum and minimum values of \(L\). You do not need to show that these values are maximum and minimum.
- Write down a possible cartesian equation of \(S\). Hence, determine the total area of the flower bed.
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(a) At \(x = 0\): \[\begin{aligned} \left(\frac{y}{2}\right)^2 + \left(\frac{y}{4}\right)^2 &= 1\\ \frac{5y^2}{16} &= 1\\ y &= \pm\frac{4}{\sqrt{5}} \end{aligned}\] Since the region lies in the first quadrant, \(y > 0\), take \(y = \dfrac{4}{\sqrt{5}}\).
Expanding the equation: \[\begin{aligned} 4(x+y)^2 + (y-x)^2 &= 16 \\ 4x^2 + 8xy + 4y^2 + y^2 - 2xy + x^2 &= 16 \\ 5x^2 + 6xy + 5y^2 &= 16. \end{aligned}\]
Differentiating w.r.t. \(x\): \[ 10x + 6y + 6x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 10y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0. \quad\cdots(\ast) \]
Differentiating again: \[ 10 + 12\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6x\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 10\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 10y\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 0. \]
At \(x = 0\), \(y = \dfrac{4}{\sqrt{5}}\): From \((\ast)\): \(6\left(\dfrac{4}{\sqrt{5}}\right) + 10\left(\dfrac{4}{\sqrt{5}}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{5}\).
From the second derivative equation: \[ 10 + 12\left(-\tfrac{3}{5}\right) + 10\left(-\tfrac{3}{5}\right)^2 + 10\left(\tfrac{4}{\sqrt{5}}\right)\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 0. \] \[ 10 - \dfrac{36}{5} + \dfrac{90}{25} + \dfrac{40}{\sqrt{5}}\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 0 \Rightarrow \dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -\dfrac{4\sqrt{5}}{25}. \]
Thus \[ y \approx \dfrac{4}{\sqrt{5}} - \dfrac{3}{5}x - \dfrac{2}{5\sqrt{5}}x^2 + \cdots \]
Approximate area: \[ \int_0^{0.5} y\,\mathrm{d}x \approx \int_0^{0.5} \dfrac{4}{\sqrt{5}} - \dfrac{3}{5}x - \dfrac{2}{5\sqrt{5}}x^2\,\mathrm{d}x \approx 0.812~\mathrm{m^2}. \]
(b) \[\begin{aligned} L &= \sqrt{(\cos t - 2\sin t)^2 + (\cos t + 2\sin t)^2} \\ &= \sqrt{\cos^2 t - 4\sin t\cos t + 4\sin^2 t + \cos^2 t + 4\sin t\cos t + 4\sin^2 t} \\ &= \sqrt{2\cos^2 t + 8\sin^2 t} \\ &= \sqrt{2 + 6\sin^2 t}. \end{aligned}\] \[ \dfrac{\mathrm{d}L}{\mathrm{d}t} = \dfrac{6\sin t\cos t}{\sqrt{2 + 6\sin^2 t}}. \] Setting \(\dfrac{\mathrm{d}L}{\mathrm{d}t} = 0\): \(\sin t = 0\) or \(\cos t = 0\), giving \(t \in \left\{-\tfrac{\pi}{2}, 0, \tfrac{\pi}{2}, \pi\right\}\).
- When \(t = \pm\tfrac{\pi}{2}\): \(L = \sqrt{2 + 6} = 2\sqrt{2}\).
- When \(t = 0\) or \(\pi\): \(L = \sqrt{2}\).
Maximum \(L = 2\sqrt{2}\). Minimum \(L = \sqrt{2}\).
(c) A possible cartesian equation of \(S\): \[ \dfrac{x^2}{8} + \dfrac{y^2}{2} = 1 \quad \text{or} \quad \dfrac{x^2}{2} + \dfrac{y^2}{8} = 1. \]
Using \(\dfrac{x^2}{2} + \dfrac{y^2}{8} = 1\), total area of the flower bed: \[ \text{Area} = 4\int_0^{\sqrt{2}} \sqrt{8 - 4x^2}\,\mathrm{d}x \approx 12.6~\mathrm{m^2}. \]
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