A curve $C$ has parametric equations $x=2{{e}^{t}}$, $y={{t}^{3}}-t$, where $-1\le t\le \frac{3}{2}$

Question

A curve $C$ has parametric equations $x=2{{e}^{t}}$, $y={{t}^{3}}-t$, where $-1\le t\le \frac{3}{2}$. Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $t$. Hence find the exact equations of the normals to the curve which are parallel to the $y$-axis. [3] Sketch the curve $C$. [1] Find the equation of the tangent to $C$ at the point$\left( 2{{e}^{a}},{{a}^{3}}-a \right)$. Find the possible exact values of $a$ such that the tangent cuts the $y$-axis at $y=1$. [4]

Accepted Answer

(i) $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3t^2-1}{2\mathrm e^t}$, normals $x=2\mathrm e^{1/\sqrt3}$ and $x=2\mathrm e^{-1/\sqrt3}$ (iii) $y=\dfrac{3a^2-1}{2\mathrm e^a}x+a^3-3a^2-a+1$, $a=0$ or $a=\dfrac{3-\sqrt{13}}2$

2018 TJC P2 Q1

2018 TJC P2 Q1

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