While looking for examples non-linear functions, I thought: If a "normal", cylinder shaped candle burns down at a constant pace (which means its height can be modelled by a linear function), then what function describes the height of a burning sphere-shaped candle with radius r that burns down completely in t minutes? Since its narrower at the top and bottom, the candle burns down faster in the beginning and in the end and slower in the middle. A graph of the function can be seen in this pdf:
http://groolfs.de/klasse9pdf/FunktionenFortsetzung.pdf
Any advice would be appreciated! And in the same vein, what is the function if the candle is cone-shaped with radius r and height h and burns down completely in t minutes?
The best model is to assume that equal volumes of wax melt in equal times. Let the rate at which the wax melts be $v$. Consider a candle of height $H$ with a cross sectional area of $A(h)$ for $h\in[0,H]$. The volume of wax $\mathrm{d}V$ in a strip $[h,h+\mathrm{d}h]$ is then simply $$\mathrm{d}V=A(h)\mathrm{d}h$$ Since we stated before that the rate at which wax melts is $v$, this means that $$-v=\frac{\mathrm{d}V}{\mathrm{d}t}$$ Hence, $$\frac{\mathrm{d}V}{\mathrm{d}t}=-v=A(h)\frac{\mathrm{d}h}{\mathrm{d}t}$$ So what we have is a first order autonomous ODE (IVP): $$\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{-v}{A(h)}~~|~~h(0)=H$$ So we can express the solution implicitly as $$-vt+c=\int A(h)\mathrm{d}h$$ In general you'll need to invert the expression obtained to get a solution for $h(t)$ - typically this will be quite hard, even in simple cases like a spherical or conical candle.
EDIT: An illustration of the spherical candle.
Imagine we have a spherical candle of radius $1$. Let's say for arguments sake $v=1$. Let $h(t)$ be the position of the flame, with $0$ at the ground and $2$ at the top of the candle. The area function is $$A(h)=\pi \cdot r(h)^2$$ And of course geometry tells us that $$r(h)=\sqrt{1-(h-1)^2}=\sqrt{2h-h^2}$$ Verify here. So our area function is $$A(h)=\pi h\left(2-h\right)$$ Now our IVP is $$\dot{h}=\frac{-1}{\pi h\left(2-h\right)}~~|~~h(0)=1.999$$ Our initial condition is theoretically $h(0)=2$, but because of the singularity in the RHS at $h=2$, we have to pick some value just under $2$. Not bothered to work out any integrals, here is a plot of the numerical solution:
Numerical techniques show that $h(4.189)\approx 0$.
Here is also the conical case. Let the cone have height $1$ and radius $1$ and again let $v=1$. In this example $r(h)=1-h$ So $$A(h)=\pi (1-h)^2$$ So our IVP is $$\dot{h}=\frac{-1}{\pi (1-h)^2}~~|~~h(0)=0.999$$
Once again numerical techniques show that $h(1.0472)\approx 0$.