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Calculate the distance between these two points (sphr. and cyl. coordinates)

by Tanegashima
Tags: coordinates, distance, points, sphr
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Tanegashima
#1
Nov20-11, 12:05 AM
P: 3
1. The problem statement, all variables and given/known data

Calculate the distance between these two points:

(3;π/2;−1) and (5;3π/2;5) (cylindrical coordinates)
(10;π/4;3π/4) and (5;π/6;7π/4) (spherical coordinates)


Do I need to put them in cartesian coordinates and continue the calc. or can I do with integrals?


2. Relevant equations

->dl = dr r + r d∅ ∅ + dz ^k

->dl = dr r + r dθ θ + r sin (∅) ∅


3. The attempt at a solution
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SammyS
#2
Nov20-11, 01:11 AM
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Quote Quote by Tanegashima View Post
1. The problem statement, all variables and given/known data

Calculate the distance between these two points:

(3;π/2;−1) and (5;3π/2;5) (cylindrical coordinates)
(10;π/4;3π/4) and (5;π/6;7π/4) (spherical coordinates)


Do I need to put them in Cartesian coordinates and continue the calc. or can I do with integrals?


2. Relevant equations

->dl = dr r + r d∅ ∅ + dz ^k

->dl = dr r + r dθ θ + r sin (∅) ∅


3. The attempt at a solution
Hi Tanegashima. Welcome to PF.

It's likely easier in Cartesian coordinates . But for the first one, what is the distance from (3, π/2) to (5, 3π/2) in polar coordinates?
Tanegashima
#3
Nov20-11, 10:31 AM
P: 3
Thanks, but can anyone provide me with a sample in sph.c. or cyl.c.?


Re: (2, π)

SammyS
#4
Nov20-11, 03:39 PM
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Calculate the distance between these two points (sphr. and cyl. coordinates)

No. The distance from (3, π/2) to (5, 3π/2) is 8.

Therefore, the distance (horizontal) from (3, π/2, -1) to (5, 3π/2, -1) is 8 units.

Of course the distance from (5, 3π/2, -1) to (5, 3π/2, 5) is 6 units.

These two distances are perpendicular to each other.


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