Christmas Tree Problem: String around Cone This problem came up when I was trying to figure out how long of a string of lights I need for my Christmas tree. It was easy to estimate and obviously this isn't for use in the real world, but I found it a lot harder than I thought it would be at first.
A cone has a base radius $3$ ft and height of $8$ ft. A line ("string") is wrapped around the cone so that if you go straight up from any point on the string along the surface of the cone, the vertical distance (not distance along the cone's surface) to the next intersection of the string is $1$ ft.
The string is wrapped counterclockwise, looking from the top. As a more stringent restriction, a section of the string that encompasses some angle $\theta$ around the surface of the cone, the change in height from the bottom point to the top point is given by:
$$h = \frac{\theta}{2\pi}$$
What is the length of the string?
My first try at this problem assumed that the angle of the string relative to the horizontal would always be a constant. I quickly realized the string would have to get steeper as it goes up the cone ("tree"), in order to keep up with the shrinking circumferennce. I pretty much banged my head into a wall from there because I have no clue how to factor that in.
 A: To go from the bottom of the tree to the top $h$ increases from $0$ to $8$ feet, so  $\theta$ increases from $0$ to $16\pi$.  We can define $r(\theta)$ as the radius at a given $\theta$.  It decreases linearly from $3$ to $0$, so $r(\theta)=3-\frac 3{16\pi}\theta$.  The derivative of arc length with respect to $\theta$ can be found by Pythagoras.  In $\Delta \theta$ we go up $\frac {\Delta \theta}{2 \pi}$ and we go around $r \Delta \theta=\left(3-\frac 3{16\pi}\theta\right)\Delta \theta$  Now we can integrate
$$s=\int_0^{16\pi}\sqrt{\frac 1{4\pi^2}+\left(3-\frac 3{16\pi}\theta\right)^2}d\theta$$
Alpha tells me the integral is about $76.2746$.  I hit the time limit before it would give me the symbolic answer.  This seems reasonable as $8$ turns at the average radius of $1.5$ feet would give $75.4$ feet and the rise will add a bit.
A: Please see this link to my calculation of Christmas tree light string length.
The length of Christmas Tree Lights wrapping a conical tree from bottom to top with uniform vertical spacing.
(Reproduced below.)
Our $20$-year old artificial Christmas tree has built-in lights, but the top and bottom thirds are burnt out. So I decided to buy several (much cheaper) new "tree light" strings (during a sale after the end of last Christmas) the majority of which would continue to work even if a light were to burn out (wired in parallel, I assume).
The bottom tree boughs are $4$-feet in diameter, $r_0 = 2$-feet, the top tree boughs are $1$-foot in diameter, $r_h = 0.5$-foot, and the vertical height distance between the top and bottom boughs is $h = 5.5$-feet. The top center spike rises, but will not include lights.
If I want to wrap the tree from bottom to top, with $1$-foot vertical spacing between wraps, what length of light string is required?
This is how I attacked the problem.
I related all variables in terms of the number of turns wrapping the tree from bottom to top.
The height (of light string wrapping) is $5.5$-feet with $1$-foot vertical spacing, thus there are $5.5$-turns = $11 \cdot \pi$ radians, of rotation as you progress from bottom to top.
I let $\theta$ (the angle) be the variable of integration from $0$ to $11 \cdot \pi$.
I, imaginatively, affixed a sector of cylindrical shell element at the end of a bough at the middle height of the tree. The light strand would wrap diagonally through the middle of this element. The edges of the element represent: the differential arc length component, $ds$, the differential height component, $dh$, and the differential radius component, $dr$.
The differential diagonal length, $dl$, in terms of $d\theta$ would be $$\frac{dl}{d\theta} = \sqrt{(ds_\theta)^2 + (dh_\theta)^2 + (dr_\theta)^2}$$
$r$ as a function of theta: $$r_\theta = r_0 - \frac{( r_0 - r_h ) \cdot \theta}{11 \cdot \pi}$$
$\frac{dr_\theta}{d\theta}:$ $$dr_\theta = - \frac{ r_0 - r_h }{11 \cdot \pi} \cdot d\theta$$
$h$ as a function of theta:$$h_\theta = \frac{5.5 \cdot \theta}{11 \cdot \pi}$$
$\frac{dh_\theta}{d\theta}:$ $$dh_\theta = \frac{5.5}{11 \cdot \pi} \cdot d\theta$$
$s$, arc length, as a function of theta:$$s = r \cdot \theta$$
$\frac{ds_\theta}{d\theta}:$ $$ds_\theta = r_\theta \cdot d\theta$$
$$ds_\theta = (r_0 - \frac{( r_0 - r_h ) \cdot \theta}{11 \cdot \pi}) \cdot d\theta$$
Plugging the above into an HP-$42$S simulator on my phone and integrating $dl$ as $\theta$ revolves from $0$ to $11 \cdot \pi$: $$L=\int_0^{11 \cdot \pi} \frac{dl}{d\theta} \cdot d\theta$$yields, $L\approx43.6$-feet. By varying the vertical spacing this allows one to evenly accommodate a modularly assembled light string length. (Note this also alters the total number of turns.)
Before I had a chance to test the modification, my wife said, "No.", and we returned the new "tree light" sets and got a new Christmas Tree (last year's model) on sale after last Christmas instead.$$\prod_0^{n=2}(Ho_n) = Ho \cdot Ho \cdot Ho$$
