Finding Distance Between Polar Coordinates

Key Questions

  • Hello,

    • In a orthonormal basis, the distance between #A(x,y)# and #A'(x',y')# is

    #d = sqrt((x-x')^2 + (y-y')^2)#.

    • With polar coordinates, #A[t, theta]# and #A'[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'))#.

  • Let say you have points #A(r_1,θ_1),Β(r_2,θ_2)# you must convert them to cartesian coordinates #A(x_1,y_1),Β(x_2,y_2)# and then use the distance formula #D=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)# and #P_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))#

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