Converting Polar Coordinates to Cartesian

The polar coordinates are defined in terms of $r$and $\theta$, where $r$is the distance of the point from the origin and $\theta$is the angle made with the positive $x$-axis.

Clearly, using trigonometry, if the Cartesian coordinates are $(x,y),$then

$\begin{array}{c}&x = r \cos \theta, &y = r \sin \theta. \end{array}$

If $3$is the distance of a point from the origin and $\压裂{\π}{3}$is the angle made with the positive $x$-axis, then what will be the Cartesian coordinates of the point?

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The Cartesian coordinates can be obtained from the distance $3$of the point from the origin and the angle $\压裂{\π}{3}$made with the positive $x$-axis. Using trigonometry, we have

$\begin{aligned} x &= 3\cos\frac{\pi}{3} \\ &= \frac{3}{2} \\ y &= 3\sin\frac{\pi}{3} \\ &= \frac{3\sqrt{3}}{2}. \end{aligned}$

Thus, the Cartesian coordinates of the point are $\displaystyle \left ( \frac{3}{2}, \frac{3\sqrt{3}}{2}\right ).$ $_\square$

Write the Cartesian coordinates $(2, 2\sqrt{3})$as polar coordinates.

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Let $a$denote the distance between the origin and the point $(2, 2\sqrt{3} ),$then

$a = \sqrt{ 2^2 + (2\sqrt{3})^2} = 4. \qquad (1)$

Now, let $\theta$denote the angle made with the positive $x$-axis, then we have

$\begin{aligned} \sin \theta &= \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \\ \cos \theta &= \frac{2}{4} = \frac{1}{2}, \end{aligned}$

which both imply $\theta=\frac{\pi}{3} .\qquad (2)$

Thus from $(1)$and $(2),$the polar coordinates of the Cartesian coordinates $(2, 2\sqrt{3} )$are $(a, \theta) = \left(4, \frac{\pi}{3}\right). \ _\square$

What are the Cartesian coordinates of the point whose distance from origin is $4$and the angle made with the positive $x$-axis is $\frac{\pi}{4} ?$

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Since the distance of the point from the origin is $4$and the angle made with the positive $x$-axis is $\frac{\pi}{4} ,$the Cartesian coordinates are

$\begin{aligned} x &= 4\cos\frac{\pi}{4} = 2\sqrt{2} \\ y &= 4\sin\frac{\pi}{4} = 2\sqrt{2}. \end{aligned}$

Thus, the answer is $\left(2\sqrt{2}, 2\sqrt{2} \right).$ $_\square$

Write the Cartesian coordinates $\left(\sqrt{3}, 1\right)$as polar coordinates.

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Let $a$denote the distance between the origin and the point $\left(\sqrt{3}, 1 \right),$then

$a = \sqrt{ (\sqrt{3})^2 + 1^2} = 2. \qquad (1)$

Now, let $\theta$denote the angle made with the positive $x$-axis, then we have

$\begin{aligned} \sin \theta &= \frac{1}{2} \\ \cos \theta &= \frac{\sqrt{3}}{2}, \end{aligned}$

which both imply $\theta=\frac{\pi}{6} . \qquad (2)$

Thus, from $(1)$and $(2),$the polar coordinates of the point $\left(\sqrt{3}, 1\right)$笛卡儿平面的 $(a, \theta) = \left(2, \frac{\pi}{6} \right). \ _\square$

What are the Cartesian coordinates of the point whose distance from origin is $10$and the angle made with the positive $x$-axis is $\frac{\pi}{2} ?$

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Since the distance of point from the origin is $10$and the angle made with the positive $x$-axis is $\frac{\pi}{2} ,$the Cartesian coordinates are

$\begin{aligned} x &= 10\cos\frac{\pi}{2} = 0 , \\ y &= 10\sin\frac{\pi}{2} = 10. \end{aligned}$

Thus, the answer is $(0, 10).$ $_\square$

ParallelAngleBisector

The radius of the above circle is $10$and the Cartesian coordinates of $A$is $(5, 5\sqrt{3} ).$If $\angle AOB$is $60^\circ ,$what are the Cartesian coordinates of $B?$

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From the cartesian coordinates of $A$ $(5,5\sqrt{3}) ,$we can get the angle between $\overline {OA}$and the positive $x$-axis using the polar coordinates of $A.$

Let $r\cos \theta$and $r\sin \theta$be the polar coordinates of $A,$then since $r$is $10,$the angle $\theta$is established as follows:

$\begin{aligned} 10 \cos \theta &= 5 \\ \cos \theta &= \frac{1}{2}, \\ 10 \sin \theta &= 5\sqrt{3} \\ \sin \theta &= \frac{\sqrt{3}}{2}, \\ \Rightarrow \theta &= 60^\circ. \end{aligned}$

Since the angle between $\overline {OA}$and the positive $x$-axis is $60^\circ ,$the angle between $\overline {OB}$and the positive $x$-axis is $60^\circ + 60^\circ = 120^\circ.$

Hence, the Cartesian coordinates of $B$are
$\begin{aligned} x&= 10 \cos 120^\circ = 10\left ( -\frac{1}{2} \right ) = -5, \\ y&= 10 \sin 120^\circ = 10\left( \frac{\sqrt{3}}{2} \right ) = 5\sqrt{3}. \ _\square \end{aligned}$