More Christmas Lights

You're probably thinking, what if Santa comes down the chimney, or through the plastic pipe leading to the furnace, survives and sees my Christmas tree, with the lights and garlands not exactly in the right place? Will he leave me a lump of coal, or just leave? But Christmas is for everyone, even the fastidious. What you need is a way to space the strings of lights evenly, vertically. This means that as you move from the bottom of the tree toward the top, the circumference of the tree gets smaller and so the slope of the string of lights must become steeper while the vertical distance between strings stays constant. The slope is the vertical distance divided by the circumference, which gets smaller, proportional to distance from the top. By having a uniform vertical distance between strings of lights, you have room to fill in the gaps evenly with garlands, ornaments, or simply to ensure that Santa is not offended.

The slope at any particular point along the path of the string of lights is $m=dy/dx$. The infinitesimal length of the path is $dl=\sqrt{d{x}^2+d{y}^2}$. Assuming the lights on a tree are on the lateral surface of a right circular cone, for a point along the path of a string of lights, the $y$ direction is along a line that includes the apex of the cone and the $x$ direction is perpendicular, along the circumference of a circle parallel to the base. Let $D$ represent the slant height of the cone. Let $y$ be a distance from the apex, in the range 0 to $D$. Let $M$ be the slope at the base of the cone and let $R$ be the radius of the base. In order to have a constant distance between turns of lights along the path, the slope, $m$, must vary with the circumference of the circle, $2\pi r$, where $r=Ry/D$, the radius of the circle and a linear function of the distance from the apex. The denominator of the slope represents the $x$ direction and must vary linearly with $y$, therefore $\frac{1}{m}=\frac{1}{M}\frac{y}{D}$ and so $dx=\frac{dy}{M}\frac{y}{D}$ and finally $dl=\frac{dy}{MD}\sqrt{y^2+M^2{D}^2}$. Integrating this to get the length, $l$, of the path of lights from the apex of the cone, occupying the lateral surface of a subcone with slant height, $d$: $$l=\frac{1}{MD}{\int }_0^d\sqrt{M^2{D}^2+y^2}dy$$
$$l=\frac{D}{2}[\frac{d}{D}\frac{\sqrt{M^2+{\left(\frac{d}{D}\right)}^2}}{M}+Mln(\frac{\frac{d}{D}+\sqrt{M^2+{\left(\frac{d}{D}\right)}^2}}{M}\left)\right]$$

1. Let $l=L$, the total length of the string of lights and let $d=D$, the total distance down the cone. Find $M$, the slope at the base of the cone.
2. Let $l$ be the length of the string of lights to a particular light. With $D$ given and having found $M$, find $d$, repeating to find $d$ for $l$ of each light.