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

by Tanegashima
Tags: coordinates, distance, points, sphr
 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 ûr + r d∅ û∅ + dz ^k ->dl = dr ûr + r dθ ûθ + r sin (∅) û∅ 3. The attempt at a solution
Emeritus
HW Helper
PF Gold
P: 7,819
 Quote by Tanegashima 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 ûr + r d∅ û∅ + dz ^k ->dl = dr ûr + r dθ ûθ + r sin (∅) û∅ 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?
 P: 3 Thanks, but can anyone provide me with a sample in sph.c. or cyl.c.? Re: (2, π)
 Emeritus Sci Advisor HW Helper PF Gold P: 7,819 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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