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Page Layout: The image appears to be a page from a textbook or notebook.
Title: "6.17-1" is at the top.
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Diagram: A cone is drawn to illustrate the problem.
Calculations: Detailed calculations involving calculus and differential equations are shown.
Textual Content:
6.17-1
Find the geodesics on the cone whose equation in cylindrical polar coordinates is z = λρ. [Let the required curve have the form φ = φ(ρ).] Check your result for the case that λ → 0.
Geodesics on a cone?
Let curve have form φ = φ(ρ)
[z = \lambda\rho]
In cylindrical coords,
a small move in z-direction adds dist. Δz
a small move in ρ-direction adds dist. Δρ
a small move in φ-direction adds dist. ρΔφ
So:
[\Delta s = \sqrt{(\Delta z)^2 + (\Delta \rho)^2 + (\rho \Delta \phi)^2}]
In cylindrical coords,
[z = \lambda\rho \Rightarrow \Delta z = \lambda \Delta \rho]
So we have:
[\Delta s = \sqrt{(\lambda \Delta \rho)^2 + (\Delta \rho)^2 + (\rho \Delta \phi)^2}]
[= \sqrt{(\lambda^2 + 1)(\Delta \rho)^2 + (\rho \Delta \phi)^2}]
[= \Delta \rho \sqrt{(\lambda^2 + 1) + (\rho \frac{\Delta \phi}{\Delta \rho})^2}]
[\Delta s = \Delta \rho \sqrt{(\lambda^2 + 1) + (\rho \phi'(ρ))^2}]
A geodesic should extremize distance, i.e.,
[L = \int_{\rho_1}^{\rho_2} ds = \int_{\rho_1}^{\rho_2} d\rho \sqrt{(\lambda^2 + 1) + (\rho \phi'(ρ))^2}]
should be stationary for geodesic.
We can apply the Euler-Lagrange equations:
[\frac{\partial f}{\partial \phi} - \frac{d}{d\rho} \left(\frac{\partial f}{\partial \phi'}\right) = 0]
where (f(\phi, \phi'; \rho) = \sqrt{(\lambda^2 + 1) + (\rho \phi'(ρ))^2})
Observe:
[\frac{\partial f}{\partial \phi} = 0]
Hence, some constant
[\frac{d}{d\rho} \left[\frac{\partial f}{\partial \phi'}\right] = 0 \Rightarrow \frac{\partial f}{\partial \phi'} = C]
6.17-2
[c = \frac{\partial f}{\partial \phi'} = \frac{\partial}{\partial \phi'} \left[(\lambda^2 + 1) + \rho^2(\phi')^2\right]^{1/2}]
[= \frac{\rho^2 \phi'}{[(\lambda^2 + 1) + \rho^2(\phi')^2]^{1/2}}]
[c^2[(\lambda^2 + 1) + \rho^2(\phi')^2] = \rho^4(\phi')^2]
[(\phi')^2[\rho^4 - c^2 \rho^2] = c^2(\lambda^2 + 1)]
[(\phi')^2 \rho^2 [\rho^2 - c^2] = c^2(\lambda^2 + 1)]
[(\phi')^2 = \frac{c^2(\lambda^2 + 1)}{\rho^2 [\rho^2 - c^2]}]
[\phi' = \frac{c\sqrt{\lambda^2 + 1}}{\rho\sqrt{\rho^2 - c^2}}]
[\phi(\rho) = \phi_0 + \sqrt{\lambda^2 + 1} \int d\rho \frac{c}{\rho\sqrt{\rho^2 - c^2}}]
Interpretation
This is on the inverse sine book: (cos^{-1}(x))
(\phi(\rho) = \phi_0 + \sqrt{\lambda^2 + 1} cos^{-1}(\frac{c}{\rho}))
For λ → 0, i.e., the cone approaches the x-y plane (z = 0):
[\phi(\rho) = \phi_0 + cos^{-1}(\frac{c}{\rho})]
[cos(\phi(\rho) - \phi_0) = \frac{c}{\rho}]
[\rho = \frac{c}{cos(\phi(\rho) - \phi_0)}]
What does this look like?
If (\phi_0 = 0), i.e., (x = c cos(\phi)), we'd recognize it as a circle.