Finding and minimizing the length of a string wrapped around a cylinder.

Question

A string of length $l$ is wrapped around a cylinder of diameter $d$ and height $h$. The string does $n$ turns and starts at one end of the cylinder, ending at the top. The pitch of the resulting helix is constant.

1. Find $l$ in terms of $h$, $d$ and $n$.

2. Supposing the volume of the cylinder and the number $n$ of turns remain constant, show that there exist $h$ and $d$ that minimize the length of the string, and find their values.

First I find a parametrization for the helix:

$$\gamma(t) = \bigg(\frac{d}{2} \cos(t), \frac{d}{2} \sin(t), \frac{1}{2\pi}\frac{h}{n}t\bigg)$$

Then I find the length:

$$ L(\gamma) = \int_{0}^{2\pi n} \| \gamma'(t) \|dt$$

And obtain:

$$ l(h, d, n) = 2\pi n \bigg[ \frac{d^2}{4} + \bigg(\frac{1}{2\pi}\frac{h}{n}\bigg)^2\bigg]^{1/2} $$

If $V = \pi (\frac{d^2}{4})h$ remains constant then $\frac{d^2}{4} = \frac{V}{\pi h}$. Also since $n$ remains constant, we can treat $l$ as a function of $h$.

We have: $$ l(h) = 2\pi n \bigg[ \frac{V}{\pi h} + \bigg(\frac{1}{2\pi}\frac{h}{n}\bigg)^2 \bigg]^{1/2} $$

Differentiating and setting to $0$:

$$ l'(h) = \frac{\pi n}{\bigg[ \frac{V}{\pi h} + \bigg(\frac{1}{2\pi}\frac{h}{n}\bigg)^2 \bigg]^{1/2} } \bigg( \frac{1}{\pi}\frac{h}{n} - \frac{V}{\pi h^2} \bigg) = 0$$

Then:

$$ \frac{1}{\pi}\frac{h}{n} = \frac{V}{\pi h^2} $$

We get:

$$ h = (nV)^\frac{1}{3} $$

And:

$$ d = \frac{2}{\sqrt{\pi}} \frac{1}{(nV)^\frac{1}{6}}$$

Are these correct? I'm not sure if I'm supposed to include $V$ as a constant in the answers, or if there's some way to leave it out. I'm also not sure if there' any way to show that the values I found are minimums other than taking the (very complicated) second derivative.

Thanks.

Answer 1

Assuming that your calculations are correct, finding the extremum of$$L = 2\pi n \bigg[ \frac{V}{\pi h} + \bigg(\frac{1}{2\pi}\frac{h}{n}\bigg)^2 \bigg]^{1/2}$$ is the same as finding the extremum of$$Z=L^2 = (2\pi n)^2 \bigg[ \frac{V}{\pi h} + \bigg(\frac{1}{2\pi}\frac{h}{n}\bigg)^2 \bigg]$$ So, considering the term in brackets anddiffernetiating with respect to $h$ $$\bigg[ \frac{V}{\pi h} + \bigg(\frac{1}{2\pi}\frac{h}{n}\bigg)^2 \bigg]'=\frac{h}{2 \pi ^2 n^2}-\frac{V}{\pi h^2}$$ It seems that we do not agree for the derivative.

If I am not mistaken, solving for $h$ leads to $$h=\sqrt[3]{2 \pi n^2 V}$$

Now, $$\bigg[ \frac{V}{\pi h} + \bigg(\frac{1}{2\pi}\frac{h}{n}\bigg)^2 \bigg]''=\frac{2 V}{\pi h^3}+\frac{1}{2 \pi ^2 n^2}$$ which is positive; so we have the minimum of $Z$ and $L$.