# A calculus of the absurd

### Chapter C Assorted problems

Solutions to problems which have yet to be categorised.

##### C.0.1 Cones problem

Problem: Two similar cones, $$X$$ and $$Y$$, have surface areas $$270\text {cm}^2$$ and $$120\text {cm}^2$$ respectively.

The volume of cone $$X$$ is $$1215\text {cm}^3$$.

Show that the volume of cone Y is $$360 \text {cm}^3$$.

Solution:

We have two cones, let’s let $$X$$ be this one, with radius $$r$$ and height $$h$$ Then we can draw $$Y$$ (which is smaller than $$X$$, but not drawn to scale here). As they are similar cones, the radius of $$Y$$ is equal to $$ar$$ (where $$a$$ is the scale factor between the two cones) and the height of $$Y$$ is equal to $$ah$$ First, the volume of a cone is $$\frac {\pi r^2 h}{3}$$ and the surface area of a cone is $$\pi r (r + \sqrt {h^2 + r^2})$$. These can be derived using integration (but not in Single Maths A Level at least).

Using this, we can write this equation relating the radius and height of $$X$$ to the value given in the question

\begin{equation} \pi r(r + \sqrt {h^2 + r^2}) = 270 \end{equation}

We can also do the same for $$Y$$

\begin{equation} \pi a r(ar + \sqrt []{(ah)^2+(ar)^2}) = 120 \end{equation}

This can be simplified a bit by factoring the $$a^2$$ from the square rooted part of the equation, giving

\begin{equation} \pi a r(ar + \sqrt {a^2}\sqrt []{(h)^2+(r)^2}) = 120 \end{equation}

and thus that

\begin{equation} \pi a^2 r(r + \sqrt {(h)^2+(r)^2}) = 120 \end{equation}

We can solve this for $$a$$ by dividing the two equations, giving that

\begin{equation} \frac { \pi a^2 r(r + \sqrt {(h)^2+(r)^2}) }{\pi r(r + \sqrt {h^2 + r^2})} = \frac {120}{270} \end{equation}

and thus, after cancelling, that

\begin{equation} a^2 = \frac {9}{4} \end{equation}

which means that

\begin{equation} a = \frac {3}{2} \end{equation}

Given this, we can then turn to the volume formula.

We know that the volume of $$X$$ is

\begin{equation} \label {volume of x} \pi r^2 h = 1215 \end{equation}

and the formula for the volume of $$Y$$ is equal to

\begin{equation} \pi (ar)^2 (ah) = a^3 \pi r^2 h \end{equation}

Thus, by multiplying Equation C.8 through by $$a^3$$, we have

\begin{equation} a^3 \pi r^2 h = a^3 1215 \end{equation}

The left-hand side is just the volume of $$Y$$, and the right-hand side is just

\begin{align} a^3 1215 &= \left (\frac {2}{3}\right )^3 \cdot 1215 \\ &= 360 \end{align}

Which is what we needed to show.