Standard +0.3 This is a standard parametric second derivative question requiring the chain rule formula d²y/dx² = (d/dt(dy/dx))/(dx/dt) and routine trigonometric differentiation. While it involves multiple steps and trigonometric simplification, it follows a well-practiced procedure with no novel insight required, making it slightly easier than average for Further Maths students.
The curve \(C\) is defined parametrically by
$$x = 2\cos^3 t \quad \text{and} \quad y = 2\sin^3 t, \quad \text{for } 0 < t < \frac{1}{2}\pi.$$
Show that, at the point with parameter \(t\),
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{1}{6}\sec^4 t \cosec t.$$ [4]
The curve $C$ is defined parametrically by
$$x = 2\cos^3 t \quad \text{and} \quad y = 2\sin^3 t, \quad \text{for } 0 < t < \frac{1}{2}\pi.$$
Show that, at the point with parameter $t$,
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{1}{6}\sec^4 t \cosec t.$$ [4]
\hfill \mbox{\textit{CAIE FP1 2015 Q1 [4]}}