anonymous
  • anonymous
Spherical Geometry Let A = (\(\theta_{A}, \psi_{A}\)) be a point on the earth at latitude \(\psi_{A}\) and longitude \(\theta_{A}\). Let B = (\(\theta_{B}, \psi_{B}\)) be another point on the earth. Let R be the radius of the earth. Prove that the distance |AB| between A and B is given by \(\cos(\frac{|AB|}{R}) = \sin\psi_{A}\sin\psi_{B} + \cos\psi_{A}\cos\psi_{B}\cos(\theta_{B}-\theta_{A})\)
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Astrophysics
  • Astrophysics
@ganeshie8
Jhannybean
  • Jhannybean
@ikram002p @zzr0ck3r
ikram002p
  • ikram002p
this need a long type -,- (if u wanna prove an exist theorem) and i have to go now, unless u wanna it directly

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anonymous
  • anonymous
If you don't have the time it's okay. It seems like that the problem is asking to prove the law of cosines but with a twist to it. Unless it's simpler than that.
anonymous
  • anonymous
So, it looks really similar to: \(\cos(c) = \cos(a)\cos(b) + \sin(a)\sin(b)\cos(C)\) But of course the cosines and sines are flipflopped and the above formula assumed unit sphere. So knowing how to adapt the formula and then see how things got switched up like they did.
ganeshie8
  • ganeshie8
\[d^2 = (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\] change to spherical coordinates and simplify ?
anonymous
  • anonymous
So represent all the points like \((\rho, \theta, \phi\)) and see if I can get something?
ganeshie8
  • ganeshie8
Exactly! Also \(\rho \) = constant since we are "on" the sphere
ganeshie8
  • ganeshie8
we may first work the formula in regular spherical coordinates changing them to latitude/longitude in the end shouldn't be hard
ganeshie8
  • ganeshie8
No wait, that wont work. that formula still gives the direct distance between those two points
anonymous
  • anonymous
Oh. Lol, okay.
anonymous
  • anonymous
|dw:1438756659753:dw|
ganeshie8
  • ganeshie8
|dw:1438756776294:dw|
ganeshie8
  • ganeshie8
once we know the distance \(d\), its not hard to find the distance along that arc because we know that the shortest distance is always along the "circle whose center is same as the center of the sphere"
anonymous
  • anonymous
So we're looking at a triangle with a vertex at the origin, right?
ganeshie8
  • ganeshie8
Yep, isosceles triangle
ganeshie8
  • ganeshie8
the required distance on the sphere is simply \(Rt\) we can find the angle \(t\) using law of cosines
anonymous
  • anonymous
|dw:1438757025200:dw| Oh, so we do want to use \(s = r\theta\) But we want law of cosines to get the \(\theta\) first?
ganeshie8
  • ganeshie8
Yes maybe call that angle between radii with some other name because you're already using \(\theta\) for latitude
anonymous
  • anonymous
|dw:1438757317770:dw|
anonymous
  • anonymous
\(d^{2} = R^{2} + R^{2} - 2R^{2}\cos(t)\) \(\frac{2R^{2}-d^{2}}{2R^{2}} = \cos(t)\)
anonymous
  • anonymous
\(\vec{A}\cdot\vec{B} = ||\vec{A}||||\vec{B}|| \cos(t)\) Or maybe this idea. Not that it does anything better, lol.
ganeshie8
  • ganeshie8
|dw:1438757456725:dw|
anonymous
  • anonymous
But even then, I don't know how it works where you'd translate that to the actual angles on the triangle. I mean, that is just finding the angle between the vector representations of the points. Not that we need it for this problem, but I wouldn't know how to use that to get the angles of the triangle. Oh, okay, back to spherical coordinates.
anonymous
  • anonymous
|dw:1438757852912:dw| So if we need \(\phi\) we want \(\psi = \frac{\pi}{2} - \phi\) I guess.
ganeshie8
  • ganeshie8
Consider two points \(A, B\) in space whose coordinates in cartesian and spherical are given by : \(A = (x_1,y_1,z_1) = (R, \theta_1,\phi_1)\) \(B = (x_2,y_2,z_2) = (R, \theta_2,\phi_2)\)
anonymous
  • anonymous
Oh, wait! So that's why the cosines and sines are flipped from the spherical law of cosines! \(z = \rho \cos(\phi)\) \( = \rho \cos(\frac{\pi}{2} - \psi) = \rho \sin(\psi)\)
ganeshie8
  • ganeshie8
Lets compute the direct distance between them in terms of angles : \[\begin{align}d^2 &= (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\\~\\ &=(R\sin \phi_2\cos\theta_2 -R\sin \phi_1\cos\theta_1 )^2 \\&+ (R\sin \phi_2\sin\theta_2 -R\sin \phi_1\sin\theta_1 )^2\\&+( R\cos\phi_2-R\cos\phi_1)^2\\~\\ &=\cdots \end{align}\] some simplification is possible here..
ganeshie8
  • ganeshie8
that conversion from regular spherical to lat/long looks good to me
anonymous
  • anonymous
A few things simplify if you expand that out, but its still nasty as a direct distance.
ganeshie8
  • ganeshie8
\[\begin{align}d^2 &= (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\\~\\ &=(R\sin \phi_2\cos\theta_2 -R\sin \phi_1\cos\theta_1 )^2 \\&+ (R\sin \phi_2\sin\theta_2 -R\sin \phi_1\sin\theta_1 )^2\\&+( R\cos\phi_2-R\cos\phi_1)^2\\~\\ &=2R^2[1-\sin\phi_2\sin\phi_1\cos(\theta_2-\theta_1)-\cos\phi_2\cos\phi_1] \end{align}\]
anonymous
  • anonymous
Okay, Im with ya there. So that's definitely close to what we need.
anonymous
  • anonymous
\(2R^{2}[1 - \sin(\frac{\pi}{2} - \psi_{2})\sin(\frac{\pi}{2} - \psi_{1})\cos(\theta_{2} - \theta_{1}) - \cos(\frac{\pi}{2} - \psi_{2})\cos(\frac{\pi}{2} -\psi_{1}) ]\) = \(2R^{2}[1 - \cos\phi_{2}\cos\phi_{1}\cos(\theta_{2}-\theta_{1}) - \sin\psi_{2}\sin\psi_{1} ]\)
ganeshie8
  • ganeshie8
we may simply equate that to another \(d^2\) that we get using cosine law in the triangle
anonymous
  • anonymous
Ah. With the angle t still or do we need \(\theta\) and/or \(\psi\)?
ganeshie8
  • ganeshie8
|dw:1438759481827:dw|
ganeshie8
  • ganeshie8
our goal is to get \(Rt\) in terms of angles
anonymous
  • anonymous
\(2R^{2} - 2R^{2}\cos(t) = 2R^{2}[1-\cos\psi_{2}\cos\psi_{1}\cos(\theta_{2}-\theta_{1}) - \sin\psi_{2}\sin\psi_{1}]\) \(\implies \cos(t) = \cos\psi_{2}\cos\psi_{1}cos(\theta_{2} - \theta_{1}) + \sin\psi_{2}\sin\psi_{1}\)
anonymous
  • anonymous
Oh, so then you have \(s = Rt \implies t = \frac{s}{R}\) . And the arc length would be the measure of the side, so can say s = \(|AB|\). Well, at least I hope I can just go ahead and make that claim, but seems legit :D
ganeshie8
  • ganeshie8
looks good! \(s = \text{shortest distance between A,B over sphere}=|AB|\)
anonymous
  • anonymous
Okay, awesome :D Do you think thatll be the approach to a lot of these spherical problems? Like, even though things are in terms of vectors or long/lat, just turn it all into spherical coordinates? Or are there different ways to do the algebra without leaving vectors and long/lat?
ganeshie8
  • ganeshie8
Honestly, I don't know if there are other simpler ways... I used spherical coordinates because thats the only thing I can think of haha!
ganeshie8
  • ganeshie8
Alternatively, we may directly optimize the arc length over sphere... there is some method in calculus of variations... but its not on top of my head at the moment..
ganeshie8
  • ganeshie8
does this make sense ? http://mathworld.wolfram.com/SphericalDistance.html
ganeshie8
  • ganeshie8
This shouldn't matter much but I think the relation between regular spherical coordinates and lat/long is \(\psi = \color{red}{|}\frac{\pi}{2} - \phi\color{red}{|}\) .
anonymous
  • anonymous
Yeah, it makes sense in context with everything else. Getting used to the idea that the sides are basically angles now, lol. So yeah, thats angle between vectors formula, but the "angle" is the distance. So maybe it wouldve worked out that way as well. Now I'll have to look through notes and see if I can find out how you get the angle between the actual sides. The angles we're getting are actually sides in disguise. Now need ways to get the actual angles, lol
ganeshie8
  • ganeshie8
https://en.wikipedia.org/wiki/Spherical_law_of_cosines

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