In cylindrical polar coordinates, the curve (ρ(θ), θ, αρ(θ)) lies on the surface of the cone z = αρ. Show that geodesics (curves of minimum length joining two points) on the cone satisfy ρ^4 = c²[β²ρ′² + ρ²], where c is an arbitrary constant, but β has to have a particular value. Determine the form

Question

In cylindrical polar coordinates, the curve [latex](\rho(θ), θ, \alpha \rho(θ))[/latex] lies on the surface of the cone [latex]z = \alpha \rho[/latex]. Show that geodesics (curves of minimum length joining two points) on the cone satisfy

[latex]{\rho}^{4} = {c}^{2}[{β}^{2}{\rho}^{\prime 2}+{\rho}^{2}][/latex],

where c is an arbitrary constant, but β has to have a particular value. Determine the form of [latex]\rho(θ)[/latex] and hence find the equation of the shortest path on the cone between the points [latex](R,−{θ}_{0}, \alpha R)[/latex] and [latex](R, {θ}_{0}, \alpha R)[/latex].

[ You will find it useful to determine the form of the derivative of [latex]\cos^{−1} {({u}^{−1})}[/latex].]

Accepted Answer

In cylindrical polar coordinates the element of length is given by

[latex](d s)^2=(d ho)^2+( ho d heta)^2+(d z)^2 ,[/latex]

and the total length of a curve between two points parameterised by [latex] heta_0 ext { and } heta_1[/latex] is

[latex]\begin{aligned}s & =\int_{ heta_0}^{ heta_1} \sqrt{\left(\frac{d ho}{d heta} ight)^2+ ho^2+\left(\frac{d z}{d heta} ight)^2} d heta \& =\int_{ heta_0}^{ heta_1} \sqrt{ ho^2+\left(1+\alpha^2 ight)\left(\frac{d ho}{d heta} ight)^2} d heta, ext { since } z=\alpha ho .\end{aligned}[/latex]

Since the independent variable θ does not occur explicitly in the integrand, a first integral of the E–L equation is

[latex]\sqrt{ ho^2+\left(1+\alpha^2 ight) ho^{\prime 2}}- ho^{\prime} \frac{\left(1+\alpha^2 ight) ho^{\prime}}{\sqrt{ ho^2+\left(1+\alpha^2 ight) ho^{\prime 2}}}=c .[/latex]

After being multiplied through by the square root, this can be arranged as follows:

[latex]\begin{aligned} ho^2+\left(1+\alpha^2 ight) ho^{\prime 2}-\left(1+\alpha^2 ight) ho^{\prime 2} & =c \sqrt{ ho^2+\left(1+\alpha^2 ight) ho^{\prime 2}}, \ ho^4 & =c^2\left[ ho^2+\left(1+\alpha^2 ight) ho^{\prime 2} ight] .\end{aligned}[/latex]

This is the given equation of the geodesic, in which c is arbitrary but β² must have the value 1 + α².

Guided by the hint, we first determine the derivative of [latex]y(u)=\cos ^{-1}\left(u^{-1} ight)[/latex]:

[latex]\frac{d y}{d u}=\frac{-1}{\sqrt{1-u^{-2}}} \frac{-1}{u^2}=\frac{1}{u \sqrt{u^2-1}}.[/latex]

Now, returning to the geodesic, rewrite it as

[latex]\begin{aligned} ho^4-c^2 ho^2 & =c^2 \beta^2 ho^{\prime 2}, \ ho\left( ho^2-c^2 ight)^{1 / 2} & =c \beta \frac{d ho}{d heta}.\end{aligned}[/latex]

Setting ρ = cu,

[latex]\begin{aligned}u c^2\left(u^2-1 ight)^{1 / 2} & =c^2 \beta \frac{d u}{d heta} ,\d heta & =\frac{\beta d u}{u\left(u^2-1 ight)^{1 / 2}},\end{aligned}[/latex]

which integrates to

[latex] heta=\beta \cos ^{-1}\left(\frac{1}{u} ight)+k ,[/latex]

using the result from the hint.

Since the geodesic must pass through both [latex]\left(R,- heta_0, \alpha R ight) ext { and }\left(R, heta_0, \alpha R ight)[/latex], we must have k = 0 and

[latex]\cos \frac{ heta_0}{\beta}=\frac{c}{R} .[/latex]

Further, at a general point on the geodesic,

[latex]\cos \frac{ heta}{\beta}=\frac{c}{ ho}.[/latex]

Eliminating c then shows that the geodesic on the cone that joins the two given points is

[latex] ho( heta)=\frac{R \cos \left( heta_0 / \beta ight)}{\cos ( heta / \beta)}.[/latex]

Question 22.7: In cylindrical polar coordinates, the curve (ρ(θ), θ, αρ(θ))...

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