Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt ÷ dx/dt) and finding the parameter value at a given point. The algebra is routine for C4 level, though the second part requires solving 1/(1+t²) = 1/2 to find t. Slightly above average due to the multi-step nature and algebraic manipulation, but uses standard techniques with no novel insight required.
A curve has parametric equations
$$x = at^3, \quad y = \frac{a}{1+t^2},$$
where \(a\) is a constant.
Show that \(\frac{dy}{dx} = \frac{-2}{3t(1+t^2)^2}\).
Hence find the gradient of the curve at the point \((a, \frac{1}{2}a)\). [7]
A curve has parametric equations
$$x = at^3, \quad y = \frac{a}{1+t^2},$$
where $a$ is a constant.
Show that $\frac{dy}{dx} = \frac{-2}{3t(1+t^2)^2}$.
Hence find the gradient of the curve at the point $(a, \frac{1}{2}a)$. [7]
\hfill \mbox{\textit{OCR MEI C4 2009 Q5 [7]}}