2025 TMJC P1 Q11

2025 TMJC P1 Q11

15 marks

An architect designs a flower bed for a garden. The flower bed is enclosed by the curve \(C\) with equation
\(\left( \frac{x+y}{2} \right)^2 + \left( \frac{y-x}{4} \right)^2 = 1\).

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.

  1. 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.[7]

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

\[x = \cos t - 2\sin t, y = \cos t + 2\sin t,\]

where \(-\pi < t \le \pi\).

  1. 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.[5]
  2. Write down a possible cartesian equation of \(S\). Hence, determine the total area of the flower bed.[3]

Solution:

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Answer:(a) \(y=\frac4{\sqrt5}-\frac35x-\frac{2}{5\sqrt5}x^2+\cdots\), area \(0.812\) square units. (b) Maximum \(2\sqrt2\), minimum \(\sqrt2\). (c) One possible ellipse: \(\frac{x^2}{2}+\frac{y^2}{8}=1\); area \(4\pi\approx12.6\) square units.

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