Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 11

Differentiate the numerator and denominator using the product rule and quotient rule:The numerator simplifies to $-11x^3-6x^2$, giving

For the real inverse hyperbolic tangent, take $a\ne 0$ and $|x/a|<1$. Its logarithmic expression givesThe same derivative holds on a differentiable complex branch away from its singularities.

First take $a>0$ and $|x|<a$, and put $u=\arcsin (x/a)$, $q=\sqrt{a^2-x^2}$. Then $u^{\prime }=1/q$. The chain rule givesConsequentlywhich proves the identity on the open domain of differentiability. The squared inverse sine is unchanged when $a$ is replaced by $-a$, so the identity also holds for negative $a$ with $|x|<|a|$. The value $a=0$ is excluded.

An axial section gives similar right triangles. At height $x$ above the base, the cone's remaining height is $h-x$, so the cylinder's radius is $r=(h-x)\tan \alpha$. Therefore its volume isAssume $h>0$ and $0<\alpha <\pi /2$ for a nondegenerate cone. The geometry and the dimensionless optimization curves are illustrated below.

At fixed $h,\alpha$, the derivative is $V^{\prime }=\pi {\tan }^2\alpha (h-x)(h-3x)$. It is positive for $0<x<h/3$ and negative for $h/3<x<h$. The endpoint volumes are zero. Thus the unique positive-volume global maximum is atIts value is $4\pi h^3{\tan }^2\alpha /27$, whereas the cone volume is $\pi h^3{\tan }^2\alpha /3$. HenceThis is the volume case of the optimal cylinder inscribed in a cone.

The curved surface area excludes the two end disks, so $A=2\pi rx=2\pi \tan \alpha x(h-x)$. Completing the square givesThusIt differs from the volume optimum because the objective contains one power of the radius rather than two.