N2022 P1 Q3

N2022 P1 Q3

Junior College 2
6 marks
Free

The parametric equations of a curve are \(x=\frac12(\mathrm{e}^{3t}+2\mathrm{e}^{-3t})\) and \(y=\frac12(\mathrm{e}^{3t}-2\mathrm{e}^{-3t})\).

  1. Using calculus, find the gradient of the normal to the curve at the point where \(t=\frac13\ln 2\).[3]
  2. By considering \(x^2\) and \(y^2\) or otherwise, find the cartesian equation of the curve, stating any restriction on the values of \(x\).[3]

Default solution

  1. \(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac32(\mathrm{e}^{3t}-2\mathrm{e}^{-3t}),\quad \frac{\mathrm{d}y}{\mathrm{d}t}=\frac32(\mathrm{e}^{3t}+2\mathrm{e}^{-3t})\)
    \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{e}^{3t}+2\mathrm{e}^{-3t}}{\mathrm{e}^{3t}-2\mathrm{e}^{-3t}}\)
    \(t=\frac13\ln 2\Rightarrow \mathrm{e}^{3t}=2,\ \mathrm{e}^{-3t}=\frac12\)
    \(\frac{\mathrm{d}y}{\mathrm{d}x}=3\Rightarrow \text{normal gradient}=-\frac13\)
  2. \(x^2-y^2=\frac14[(\mathrm{e}^{3t}+2\mathrm{e}^{-3t})^2-(\mathrm{e}^{3t}-2\mathrm{e}^{-3t})^2]=2\)
    \(\mathrm{e}^{3t}+2\mathrm{e}^{-3t}\geq 2\sqrt{2}\Rightarrow x\geq\sqrt{2}\) (AM–GM inequality)
    \(x^2-y^2=2,\quad x\geq\sqrt{2}\)
Answer:(a) \(-\frac13\) (b) \(x^2-y^2=2,\ x\geq\sqrt{2}\)

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