Length of Path Along a Cone Surface

Observe that the height of the path (h) purely depends on its arc length (s) with the relationship $\frac{dh}{ds} = tan(10º)$ , so WLOG we may treat the path as the hypotenuse of right triangle with incline 10º and height of length 50. Hence the answer is $\frac{50}{sin50º}\approx\boxed{287.9385}$ .

From the symmetry of the cone, it is convenient to use cylindrical coordinates to describe the surface of the cone. The equation of the surface is given by

$\mathbf{p}(r, t, z) = ( r \cos t, r \sin t, z )$

Where I have replaced the polar angle $\phi$ with the letter $t$ .

Now, for a cone with height H and radius of base R, and using similar triangles, we deduce that

$(H - z) / r = H / R$

Hence $r = R/H (H - z) = R (1 - z/H)$

Therefore, the equation of the surface of the cone is

$\mathbf{p} = ( R(1 - z/H) \cos t, R(1 - z/H) \sin t, z )$

Now we'll create a path on the cone surface, so $z$ is a function of the angle $t$ .

The only constraint we have is that the angle between the tangent vector to the track and the z-axis is equal to $a_0$ , where $a_0 = 90^{\circ} - 10^{\circ} = 80^{\circ}$ . Differentiating the position vector $\mathbf{p}$ with respect to the polar angle $t$ , we get,

$\dfrac{d \mathbf{p}}{dt} = \dot{\mathbf{p}} = ( -R/H \dot{z} \cos t - R(1 - z/H) \sin t, -R/H \dot{z} \sin t + R(1 - z/H) \cos t, \dot{z} )$

where $\dot{z} = \dfrac{dz}{dt}$ ; then, using the dot product with the $z$ unit vector, the angle that the tangent vector (which is $\dot{\mathbf{p}}$ ) makes with the $z$ -axis is $\psi$ , where

$\cos \psi = \cos a_0 = \dfrac{ \dot{z} }{ \sqrt{ {\dot{z}}^2 ( (R/H)^2 + 1 ) + R^2 (1 - z/H)^2 } }$

this determines $\dot{z}$ in terms of $z, t$ . Namely, by squaring and cross multiplying,

$\cos^2 a_0 . { {\dot{z}}^2 (1 + (R/H)^2 ) + R^2 (1 - z/H)^2 } = {\dot{z}}^2$

Solving for $\dot{z}$ , one obtains,

${\dot{z}}^2 ( 1 - \cos^2 a_0 (1 + (R/H)^2 ) ) = \cos^2 a_0 R^2 (1 - z/H)^2$

so that,

$\dot{z} = k_0 R (1 - z/H)$

where $k_0 = \cos a_0 / (1 - \cos^2 a_0 (1 + (R/H)^2) )^{(1/2)}$

We can now solve for $z(t)$ , by separating the variables $z$ and $t$ . We have,

$\dfrac{dz}{(1 - z/H)} = k_0 R dt$

Integrating from $z = 0$ to $z$ , and $t = 0$ to $t$

$- H \ln(1 - z/H) = k_0 R t$

so that,

$1 - z/H = e^{- k_0 (R/H) t }$

and finally,

$z = H (1 - e^{- k_0 (R/H) t} )$

Now, the length of the path is given by,

$L = \displaystyle \int_{t = 0}^{\infty} | \dot{\mathbf{p}} | dt$

$= \displaystyle \int_{t = 0}^{\infty} \sqrt{ (1 + (R/H)^2 ) . ( k_0 R e^{-k_0 (R/H) t} )^2 + R^2 ( e^{-k_0 (R/h) t} )^2 }$

$= \displaystyle \int_{t = 0}^{\infty} \sqrt{ (1 + (R/H)^2 ) . {k_0}^2 R^2 + R^2 } e^{-k_0 R/H t }$

$= \dfrac{1}{k_0 (R/H) } \sqrt{ (1 + (R/H)^2 ) {k_0}^2 R^2 + R^2 }$

$= \dfrac{H}{ k_0} \sqrt{ (1 + (R/H)^2 ) {k_0}^2 + 1 }$

This, when simplified, turns out to be,

$L = \dfrac{ H }{ \cos a_0 } = \dfrac{50}{ \cos 80^{\circ} } = \dfrac{50}{\sin 10^{\circ} } = 287.9385$