Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 4

The double-angle identity eliminates the parameter: $y=1+\cos 2\theta =2{\cos }^2\theta =2x^2$. As $\theta$ increases from zero to $\pi$, $x$ decreases from one to minus one. The positive area below this parabola is thereforeEquivalently, the parameter integral is $-{\int }_0^{\pi }(1+\cos 2\theta )(-\sin \theta )d\theta$. The leading minus sign corrects for traversal from right to left; the signed integral $\int ydx$ along that traversal would be negative.

For $0<\theta <\pi$, the chain rule for a parametric curve givesAt $\theta =0$ the quotient of parameter derivatives is formally $0/0$, but the simplified expression has limit $16$. Thus the limiting tangent at $(x,y)=(1,1)$ has gradient 16. This is an endpoint limit of the given open parameter interval, or the ordinary tangent of the smooth polynomial extension $y=8x^4-8x^2+1$.

The fourth Chebyshev polynomial eliminates the parameter:Its derivatives are $y^{\prime }=16x(2x^2-1)$ and $y^{\prime \prime }=16(6x^2-1)$. The interior stationary points are a local maximum $(0,1)$ at $\theta =\pi /2$, and two minima $(\pm 1/\sqrt{2},-1)$ at $\theta =\pi /4,3\pi /4$. At the maximum $y^{\prime \prime }=-16<0$, while at either minimum $y^{\prime \prime }=32>0$.

The two inflection points areThe curvature changes sign there. The endpoint limits are $(\pm 1,1)$, at $\theta =0,\pi$; they are excluded from the open curve. The central maximum and the two minima are also the attained global extremal values on that open curve.