(a)[4]
Show that $\frac{dy}{dx} = -\frac{1}{2}\cos^3 2t$.
(b)[4]
Hence find the equation of the normal to the curve at the point where $t=\tfrac{1}{8}\pi$. Give your answer in the form $y=mx+c$.
Mathematics 9709 · AS & A Level · Differentiation
Differentiation — practice question
The curve is given in parametric form by $x = \tan^2 2t$ and $y = \cos 2t$, where $0 < t \leq \frac{1}{4}\pi$.
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