How do you convert $\left( -4,3 \right)$ into polar coordinates?
Answer
581.4k+ views
Hint: In this question we have to convert the given Cartesian coordinates to the polar coordinates. We will convert the Cartesian coordinates $\left( x,y \right)$ to polar coordinates $\left( r,\theta \right)$ of a point using the following relation
$r=\sqrt{{{x}^{2}}+{{y}^{2}}}$
$\theta =atan2\left( y,x \right)$
Complete step-by-step answer:
We have been given a Cartesian coordinates $\left( -4,3 \right)$.
We have to convert the given Cartesian coordinates to polar coordinates.
In order to convert first of all we will find $r$ as the distance between origin and the Cartesian point as $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ where $\left( x,y \right)$ are Cartesian coordinates and $\theta $ is the angle the ray joining the origin and the point makes with positive x-axis.
$\theta =atan2\left( y,x \right)$
Here the function $atan2\left( y,x \right)$ called 2-argument inverse tangent and is defined as
$\theta =\left\{ \begin{align}
& {{\tan }^{-1}}\left( \dfrac{y}{x} \right)\text{ if }x>0 \\
& {{\tan }^{-1}}\left( \dfrac{y}{x} \right)\text{+}\pi \text{ if }x<0\text{ and y}\ge 0 \\
& {{\tan }^{-1}}\left( \dfrac{y}{x} \right)-\pi \text{ if }x<0\text{ and y}<0 \\
& \dfrac{\pi }{2}\text{ if }x=0\text{ and y}>0 \\
& -\dfrac{\pi }{2}\text{ if }x=0\text{ and y}>0 \\
& undefined\text{ if }x=0\text{ and y=}0 \\
\end{align} \right\}$
We have given the Cartesian coordinates $\left( x,y \right)=\left( -4,3 \right)$.
Now, we can find the value of r by using the relation $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Now, substituting the values we will get
$\Rightarrow r=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$
Simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow r=\sqrt{16+9} \\
& \Rightarrow r=\sqrt{25} \\
& \Rightarrow r=5 \\
\end{align}$
Now, we can find the value of $\theta $ as $\theta ={{\tan }^{-1}}\left( -\dfrac{3}{4} \right)$
Simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow \theta =-{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \\
& \Rightarrow \theta =-36.86{}^\circ \\
\end{align}$
Now, adding $180{}^\circ $ to the obtained value we will get
$\begin{align}
& \Rightarrow \theta =-36.86+180{}^\circ \\
& \Rightarrow \theta =143.13{}^\circ \\
\end{align}$
So the polar coordinates are$\left( 5,143.13{}^\circ \right)$.
Hence we get the required polar coordinates as $\left( 5,143.13{}^\circ \right)$.
Note: Here in this question the angle we get is negative and the point lies in the second quadrant so we need to add $180{}^\circ $ to the obtained value of the angle. Also remember that the angle is always measured in degree.
$r=\sqrt{{{x}^{2}}+{{y}^{2}}}$
$\theta =atan2\left( y,x \right)$
Complete step-by-step answer:
We have been given a Cartesian coordinates $\left( -4,3 \right)$.
We have to convert the given Cartesian coordinates to polar coordinates.
In order to convert first of all we will find $r$ as the distance between origin and the Cartesian point as $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ where $\left( x,y \right)$ are Cartesian coordinates and $\theta $ is the angle the ray joining the origin and the point makes with positive x-axis.
$\theta =atan2\left( y,x \right)$
Here the function $atan2\left( y,x \right)$ called 2-argument inverse tangent and is defined as
$\theta =\left\{ \begin{align}
& {{\tan }^{-1}}\left( \dfrac{y}{x} \right)\text{ if }x>0 \\
& {{\tan }^{-1}}\left( \dfrac{y}{x} \right)\text{+}\pi \text{ if }x<0\text{ and y}\ge 0 \\
& {{\tan }^{-1}}\left( \dfrac{y}{x} \right)-\pi \text{ if }x<0\text{ and y}<0 \\
& \dfrac{\pi }{2}\text{ if }x=0\text{ and y}>0 \\
& -\dfrac{\pi }{2}\text{ if }x=0\text{ and y}>0 \\
& undefined\text{ if }x=0\text{ and y=}0 \\
\end{align} \right\}$
We have given the Cartesian coordinates $\left( x,y \right)=\left( -4,3 \right)$.
Now, we can find the value of r by using the relation $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Now, substituting the values we will get
$\Rightarrow r=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$
Simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow r=\sqrt{16+9} \\
& \Rightarrow r=\sqrt{25} \\
& \Rightarrow r=5 \\
\end{align}$
Now, we can find the value of $\theta $ as $\theta ={{\tan }^{-1}}\left( -\dfrac{3}{4} \right)$
Simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow \theta =-{{\tan }^{-1}}\left( \dfrac{3}{4} \right) \\
& \Rightarrow \theta =-36.86{}^\circ \\
\end{align}$
Now, adding $180{}^\circ $ to the obtained value we will get
$\begin{align}
& \Rightarrow \theta =-36.86+180{}^\circ \\
& \Rightarrow \theta =143.13{}^\circ \\
\end{align}$
So the polar coordinates are$\left( 5,143.13{}^\circ \right)$.
Hence we get the required polar coordinates as $\left( 5,143.13{}^\circ \right)$.
Note: Here in this question the angle we get is negative and the point lies in the second quadrant so we need to add $180{}^\circ $ to the obtained value of the angle. Also remember that the angle is always measured in degree.
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