
How do you convert $(0,\;2)$ from Cartesian to polar coordinates?
Answer
557.1k+ views
Hint:Cartesian coordinates $(x,\;y)$ can be converted into polar coordinates $(r,\;\theta )$ by following method: First find the value of “r” which represents the actual distance of the point from the origin, And then find the value of “$\theta $”, which represents the angle of the point with respect to horizontal axis in the anti clockwise direction having the origin as the vertex.
Complete step by step solution:
In order to convert $(0,\;2)$ from rectangular coordinate system (also known as Cartesian coordinates) to polar coordinates $(r,\;\theta )$, we have to first determine the value of “r” which is the distance between the origin and the given point.
We know that the distance between two points having coordinates $(a,\;b)\;{\text{and}}\;(c,\;d)$ can be
given as
$r = \sqrt {({a^2} - {c^2})({b^2} - {d^2})} $
So finding the distance between origin $(0,\;0)$ and the given point $(0,\;2)$
$
r = \sqrt {({0^2} - {0^2}) + ({0^2} - {2^2})} \\
r = \sqrt {0 + ( - {2^2})} \\
r = \sqrt 4 \\
r = \pm 2 \\
$
Since distance is a positive quantity it never takes negative values
$\therefore r = 2$
Now we will find value of $\theta $ which gives information about the angle by which the point is raised from the horizontal axis in the anti clockwise direction
We can find the value of angle of a point $(a,\;b)$ with respect to horizontal axis in clockwise direction having the origin as the vertex as follows
$
\tan \theta = \dfrac{b}{a} \\
\therefore \theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right) \\
$
Now finding the value of $\theta $ for the point $(0,\;2)$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{2}{0}} \right)$
Since the value $\left( {\dfrac{2}{0}} \right)$ is defined, but tangent function is always not defined at $\dfrac{\pi }{2}$
$\therefore \theta = \dfrac{\pi }{2}$
Then the required polar coordinates will be $(r,\;\theta ) \equiv \left( {2,\;\dfrac{\pi }{2}} \right)$
Note: We have learnt how to convert Cartesian coordinates into polar coordinates in this problem. But for vice-versa, that is for converting polar coordinates $(r,\;\theta )$ into Cartesian coordinates $(x,\;y)$ there is a direct formula given following:
$x = r\cos \theta \;{\text{and}}\;y = \sin \theta $
Complete step by step solution:
In order to convert $(0,\;2)$ from rectangular coordinate system (also known as Cartesian coordinates) to polar coordinates $(r,\;\theta )$, we have to first determine the value of “r” which is the distance between the origin and the given point.
We know that the distance between two points having coordinates $(a,\;b)\;{\text{and}}\;(c,\;d)$ can be
given as
$r = \sqrt {({a^2} - {c^2})({b^2} - {d^2})} $
So finding the distance between origin $(0,\;0)$ and the given point $(0,\;2)$
$
r = \sqrt {({0^2} - {0^2}) + ({0^2} - {2^2})} \\
r = \sqrt {0 + ( - {2^2})} \\
r = \sqrt 4 \\
r = \pm 2 \\
$
Since distance is a positive quantity it never takes negative values
$\therefore r = 2$
Now we will find value of $\theta $ which gives information about the angle by which the point is raised from the horizontal axis in the anti clockwise direction
We can find the value of angle of a point $(a,\;b)$ with respect to horizontal axis in clockwise direction having the origin as the vertex as follows
$
\tan \theta = \dfrac{b}{a} \\
\therefore \theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right) \\
$
Now finding the value of $\theta $ for the point $(0,\;2)$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{2}{0}} \right)$
Since the value $\left( {\dfrac{2}{0}} \right)$ is defined, but tangent function is always not defined at $\dfrac{\pi }{2}$
$\therefore \theta = \dfrac{\pi }{2}$
Then the required polar coordinates will be $(r,\;\theta ) \equiv \left( {2,\;\dfrac{\pi }{2}} \right)$
Note: We have learnt how to convert Cartesian coordinates into polar coordinates in this problem. But for vice-versa, that is for converting polar coordinates $(r,\;\theta )$ into Cartesian coordinates $(x,\;y)$ there is a direct formula given following:
$x = r\cos \theta \;{\text{and}}\;y = \sin \theta $
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