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Let \(A = (3, \theta_1)\) and \(B = (9, \theta_2)\) in polar coordinates. If \(\theta_1 - \theta_2 = \frac{\pi}{2},\) then \(AB\) find the distance 

 Apr 20, 2020
 #1
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By polar coordinates, AB = sqrt(3^2 + 3*9 + 9^2) = 3*sqrt(13).

 Apr 21, 2020
 #2
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Let \(A = (3, \theta_1)\) and \(B = (9, \theta_2)\) in polar coordinates. If  \(\theta_1 - \theta_2 = \dfrac{\pi}{2}\), then  find the distance \(AB\)

 

Formula:
We have two points \((r_1,\ \theta_1)\) and \(( r_2,\ \theta_2)\).
The distance between two polar coordinates \(\mathbf{d =\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1})}\)

 

\(\begin{array}{|rcll|} \hline d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1) } \\ d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos\Big(-(\theta_1 - \theta_2)\Big) } \\ d &=& \sqrt{r_1^2+r_2^2+2r_1r_2\cos(\theta_1 - \theta_2) } \\ d &=& \sqrt{3^2+9^2+2*3*9\cos\left( \dfrac{\pi}{2} \right) } \\ d &=& \sqrt{9+81+54\cos\left( \dfrac{\pi}{2} \right) } \quad | \quad \cos\left( \dfrac{\pi}{2} \right) = 0 \\ d &=& \sqrt{9+81+0 } \\ d &=& \sqrt{90} \\ d &=& \sqrt{9*10} \\ d &=& \sqrt{3^2*10} \\ \mathbf{d} &=& \mathbf{3\sqrt{10}} \\ \hline \end{array} \)

 

laugh

 Apr 21, 2020
edited by heureka  Apr 21, 2020
edited by heureka  Apr 29, 2020
edited by heureka  Apr 29, 2020

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