what would be a distance formula between two points given in polar form?

Question

I need a formula using polar form i.e. (r1, theta 1) and (r2, theta 2). The more simplified the formula is the better :)
oh and heres a trig identity that may be useful:
cos(theta 1)cos(theta 2) + sin(theta 1)sin(theta 2)= cos(theta1-theta2)

Answer 1

Algebraically you just plug it into the distance formula:

http://cheeser1.slyip.com/interspace/distance.gif

This also has geometric meaning, since what you're doing amounts to the law of cosines. Connect each point to the origin, and connect the two of them. That's a triangle with two known sides (r1 and r2) and a little computation tells us that the angle between them is t1 - t2.

http://cheeser1.slyip.com/interspace/cosines.gif

That means that, from this geometric standpoint, all you have to do is wave your hands at the law of cosines and you're done:

d^2 = r1^2 + r2^2 - 2r1 r2 cos(t1-t2)

Answer 2

The distance will be root of
r1^2 + r2^2 - 2*r1*r2*(cos(theta1-theta2)).
You know the formula of the cos thing already.

Answer 3

u must be knowing the distance formula right???


d= sqroot [ (x2-x1) ^2 + (y2-y1) ^2 ]
= sqroot [ (r2cos th1 - r2 costh2)^2 + ( r2sin th1 - r2 sinth2)^2 ]


Expand the terms inside sqroot to get
Bring r1 ^ 2 terms together
Bring r2 ^ 2 terms together
and 2 r1 r2 terms together

and simplify their coefficients using identities
sin sq th + cos sq th =1
and the one u mentioned is used to simplify the third term to give cos(th1-th2)

finally

d= sqroot (r1 ^2 + r2 ^ 2 + 2 *r1 *r2 *cos(th1 - th2))

Answer 4

Monatomic Molecule could be particularly complicated and can appear as if a misnomer (think of of hydrogen bonding - they don't seem to be bonds, they're intramolecular forces consequently it particularly is a misnomer), even though it isn't any longer there to misinform. you're easily incorrect in calling HCl a diatomic molecule. inspite of the reality that the be conscious di does mean 2 and atomic does seek advice from atoms, that's no longer what diatomic potential interior the international of chemistry. Diatomic Molecules seek advice from the seven components that, while via themselves (no longer covalently or ionically bonded to different components), proceed to be in a molecule and not as an atom. those 7 molecules are Hydrogen, Nitrogen, Oxygen, Chlorine, Fluorine, Bromine, and Iodine. None of those molecules are ever got here across as a unmarried atom in nature. an incredible occasion is the synthesis of water 2H2 + O2 --> 2H2O As you will locate Hydrogen and Oxygen might desire to be H2 and O2 interior the equation using fact they do no longer exist as purely H and O - they're diatomic. The term Monatomic refers to those gases that don't seem in diatomic sort (to illustrate Argon (Ar). purely deliver me a message in case you go with greater explanation : )

Answer 5

x1 = r1 * cos(theta1)
y1 = r1 * sin(theta1)
x2 = r2 * cos(theta2)
y2 = r2 * sin(theta2)

D = sqrt((x1-x2)^2+(y1-y2)^2)
=sqrt((r1* cos(theta1) - r2 * cos(theta2))^2 + (r1 * sin(theta1) - r2 * sin(theta2))^2)

This is getting messy, let's do it in parts
There is an (r1 cos(theta1))^2 from the first part, and and (r1 sin(theta1))^2 in the second part, this becomes just r1^2.
There is an (r2 cos(theta2))^2 from the first part, and and (r2 sin(theta2))^2 in the second part, this becomes just r2^2.
The only stuff left is -2*r1*r2*cos(theta1)cos(theta2) and -2*r1*r2*sin(theta1)sin(theta2).

So we have:
D = sqrt(r1^2 + r2^2 -2*r1*r2(cos(theta1)cos(theta2) + sin(theta1)sin(theta2)))
= sqrt(r1^2 + r2^2 -2*r1*r2*cos(theta1-theta2))

Answer 6

=sqrt(r1^2+r2^2 -2r1r2cos(t1-t2)