2023 NJC P1 Q11

2023 NJC P1 Q11

12 marks
Prelims

Baby Peter is pulling a toy wagon on a flat ground which is modelled by the \(xy\)-coordinate plane. Peter is initially at the origin \(O\) and the toy wagon is located on the \(y\)-axis at \({{W}_{0}}\left( 0,6 \right)\). The toy wagon is tethered to him with a rope of length \(6\) units. As he moves in the direction of the positive \(x\)-axis (given by point \(P\)), he pulls the toy wagon (given by the point \({{W}_{1}}\)) using the rope \({{W}_{1}}P\) which is kept taut throughout the motion. The path of the toy wagon is given by curve \(C\). It is known that the rope is always tangent to \(C\).

  1. Show that a differential equation for \(C\) can be expressed as
    \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\alpha y}{\sqrt{36-{{y}^{2}}}}\),
    where \(\alpha \) is a constant to be determined.[2]
  2. Using the substitution \({{w}^{2}}=36-{{y}^{2}}\) where \(w>0\), show that the differential equation in (a) can be reduced to
    \(\frac{\mathrm{d}w}{\mathrm{d}x}=\frac{36-{{w}^{2}}}{{{w}^{2}}}\).[2]
  3. Find \(x\) in terms of \(y\).[6]
  4. Determine the behaviour of \(x\) as \(y\) approaches \(0\).[2]

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Answer:(c) \(x=-\sqrt{36-{{y}^{2}}}+3\ln \frac{6+\sqrt{36-{{y}^{2}}}}{6-\sqrt{36-{{y}^{2}}}}\) \text{ or } \(x=-\sqrt{36-{{y}^{2}}}+3\ln \left( 6+\sqrt{36-{{y}^{2}}} \right)-3\ln \left( 6-\sqrt{36-{{y}^{2}}} \right)\)

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