Using differential equations to determine the number of rolls on a roll of toilet paper

Question

Puzzle: A role of toilet paper has $180$ sheets on it. The outside is covered with exactly two sheets. The inside around the cardboard cylinder is covered by exactly one. Question of the puzzle: how many layers of toilet paper are on the roll of toilet paper?

The "Given" Solution: One way to solve this is by saying that the average round is covered by $1.5$ sheets, so therefore the answer is $120$ (I have no source for whether this is actually correct)

I Tried: I tried to solve it with a differential equation, but ultimately failed: Let $S$ be the number of toilet paper sheets on the roll, and $n$ the number of rotations.

I think the number of sheets per rotation depends linearly on the number of rotations at a given point, because with every rotation the toilet role becomes more thick, so: $$\frac{\mathrm{d}S}{\mathrm{d}n}=kn.$$

This equation is separable, so $$dS=(kn)\,\mathrm{d}n.$$

Integrate to get $$S=\frac{1}{2}kn^2+C.$$

Now we need to find the values of constants $k$ and $C$: we know that when $S=1$ then $n=1$ and also when $S=180$ then $\frac{dS}{dn}=2$.

But now I am stuck.

My Question: What would be the correct way of solving this problem using differential equations?

This question is also linked to this question: Toilet paper puzzle (question 2)

Answer 1

Use the "sheets" as units of length, so the total length is $L=180$, the "inner" radius is $r_i = 1/(2\pi)$, the "external" one is $r_e = 2/(2\pi)$. Assume that the paper draws a spiral and makes $n$ loops (not necessarily an integer). Then, we parametrize this spiral in terms of the angle $t \in [0, 2 n \pi]$ (I use rad for angles, so that $n$ is the number of layers). The radius is slowly increasing from $r_i$ to $r_e$, say linearly: $$ r(t) = r_i +(r_e-r_i) \frac{t}{2 n \pi} $$ We have $$ L = \int_0^{2 \pi n } dt \sqrt{ r(t)^2 + r'(t)^2} \approx \int_0^{2 \pi n } dt \, r(t) = n \pi (r_e + r_i) $$ This gives that the number of loops (or layers) is $$ n \approx \frac{L}{\pi (r_e + r_i)} = 2L/3 = 120 $$ Note: the approximation is valid if the spiral is "spiralling out" slowly. Otherwise the line of reasoning is the same, but you have to deal with more complex calculations (i.e. taking the full square root term in the integral).

PS: it is not explicitly with differential equations, but there is an integral. Solving the integral above is equivalent to "integrate" the ODE $$ \frac{dL(n)}{dn} \approx 2 \pi \, \frac{r_e + r_i}{2} $$ You have to integrate this simple ODE with the initial condition $L(0) = 0$ till you touch the value $L(n) =180$.

Answer 2

Your equation $\frac {dS}{dR}=kR$ is not correct because at $R=0$ you would have $\frac {dS}{dR}=0$. Either you need to measure $R$ from the center and start the wrap at an $r \gt 0$ (which I would recommend) or you need to measure $R$ from the start of the wrap and then $\frac {dS}{dR}=k(R+r)$ where $r$ is the starting radius.

Let us use your solution $S=\frac 12kR^2+C$. When $R=r$ we have $S=0$, so $S=\frac 12k(R^2-r^2)$. Then we have $S=180$ when $R=2r$, so $180=\frac 12k\cdot 3r^2, 2=k\cdot 2r.$ This gives $k=\frac 1r, 180=\frac 32r, r=120$ and the outer radius is $2r=240$ so there are $120$ turns.

Answer 3

The following is simple lateral area calculation:

Let outer,inner radii be (R,r) respectively. Since circumferences $2 \pi r , 2 \pi R $ proportional to $(r,R)$are given double,

$$ \dfrac{R}{r}=2 \tag1$$ It is given that total toilet paper length when unrolled $$ L= 180\, (2 \pi r)\tag2$$

Let effective toilet paper thickness be $t$. The lateral visible layered packed area of paper roll cylinder between radii $(R,r)$ is $$ L\cdot t= \pi (R^2-r^2) \tag3$$ Eliminate L between (2) and (3) and simplify and factor it $$ 360 r= \dfrac {(R-r)(R+r)}{t} \tag4$$ Plug in from(1) and simplify to eliminate R $$ 120 =\dfrac{r}{t} \tag5$$ Number of layers by tight roll winding the layers $$n = \dfrac{R-r}{t} = \dfrac{r}{t} \tag6 $$ and lastly from (5) and (6) $$ n= 120 \tag7$$

And yes, it comes to the same result by integrating linear variation to second degree with smaller radius of roll boundary condition: $$S=\frac{1}{2}kR^2-\frac{1}{2}kr^2,\,k= 2 \pi, $$ in other words effectively same as (3) with (2).

Answer 4

I now understand that one cannot solve this puzzle using continuous functions. We can make an approximation however. This is how that would work:

$S$=total amount of sheets

$n$=number of completed rounds

We know that the growth of the amount of Sheets (S) per Round (n) is constant. Why? Because it is depending om the circumference which grows linearly with the radius with $2\pi r$.

Therefore the second derivative of $S(n)$ is constant: $S''(n)=C_1$

The first derivative is therefore: $S'(n)=C_1n+C_2$

In round $0$ the amount of sheets per round is $1$, so we can approximate: $S'(0)=C_2\approx 1$

In the last round (when we used up all $180$ sheets of toilet paper) the amount of sheets per round is $2$, so we can approximate: $S'(n_{180})=C_1 n_{180}+1\approx 2$

And we can conclude that $C_1=\frac{1}{n_{180}}$

So:

$$\bbox[5px,border:2px solid black]{S'(n)=\frac{1}{n_{180}}n+1}$$

We can integrate to find: $S(n)=\int S'(n)dn=\frac{n^2}{2n_{180}}+n+C_3$

$S(0)=C_3=1$

So:

$$\bbox[5px,border:2px solid black]{S(n)=\frac{n^2}{2n_{180}}+n+1}$$

Now we can take $S(n_{180})=\frac{n_{180}^2}{2n_{180}}+n_{180}+1=180$

To find that after $180$ sheets, the approximate amount of completed rounds is $n_{180}=\frac{358}{3}=119\frac{1}{3}$

Now we can also take the inverse to find that:

$$\bbox[5px,border:2px solid black]{n(S)=\sqrt{\frac{716}{3}S+14001\frac{7}{9}}-\frac{358}{3}}$$

With a derivative of:

$$\bbox[5px,border:2px solid black]{n'(S)=\frac{358}{3\sqrt{\frac{716}{3}S+14001\frac{7}{9}}}}$$

Now I see why $S'(n)$ is lineair, but $n'(S)$ is not lineair! Which was what I was trying to get my head around when asking this question.