Finding Distance Between Polar Coordinates
Key Questions
-
Hello,
- In a orthonormal basis, the distance between
A(x,y)A(x,y) andA'(x',y') is
d = sqrt((x-x')^2 + (y-y')^2) .- With polar coordinates,
A[t, theta] andA'[r',theta'] , you have to write the relations :
x = r cos theta, y = r sin theta
x' = r' cos theta', y' = r' sin theta' ,So,
d = sqrt((r cos theta - r' cos theta')^2 + (r sin theta - r' sin theta')^2 ) Develop, and use the formula
cos^2 x + sin^2 x = 1 . So you get :d = sqrt(r^2 - 2 rr' (cos theta cos theta' + sin theta sin theta')+ r'^2) Finally, you know that
cos theta cos theta' + sin theta sin theta' = cos(theta - theta') , therefore,d = sqrt(r^2 + r'^2 - 2rr' cos(theta - theta')) . - In a orthonormal basis, the distance between
-
Let say you have points
A(r_1,θ_1),Β(r_2,θ_2) you must convert them to cartesian coordinatesA(x_1,y_1),Β(x_2,y_2) and then use the distance formulaD=sqrt((x_2-x_1)^2+(y_2-y_1)^2) -
Answer:
See below.
Explanation:
Given in cartesian coordinates.
P_1=(x_1,y_1) andP_2= (x_2,y_2) the transition formulas
{(x=r cos theta),(y=r sin theta):} then
(x_1,y_1) rArr (r_1 cos theta_1, r_1 sin theta_1)
(x_2,y_2) rArr (r_2 cos theta_2, r_2 sin theta_2) so
d = sqrt((x_1-x_2)^2+(y_1-y_2)^2) rArr sqrt((r_1 costheta_1-r_2 cos theta_2)^2+(r_1 sin theta_1-r_2 sin theta_2)^2) then
d = sqrt(r_1^2+r_2^2-2r_1r_2(cos theta_1 cos theta_2+sin theta_1 sin theta_2)) = sqrt(r_1^2+r_2^2-2r_1r_2cos (theta_1 -theta_2))