How do you convert ${r^2} = \sin 2\left( \theta \right)$ to rectangular form?
Answer
631.8k+ views
Hint: This problem deals with converting the given polar coordinates into rectangular coordinates. Given the polar coordinate $\left( {r,\theta } \right)$, write $x = r\cos \theta $ and $y = r\sin \theta $. Evaluate $\cos \theta $ and $\sin \theta $. Multiply $\cos \theta $ by $r$ to find the x-coordinate of the rectangular form. Multiply $\sin \theta $ by $r$ to find the y-coordinate of the rectangular form.
Complete step-by-step solution:
The given polar coordinates form of the equation is ${r^2} = \sin 2\theta $.
Let $x = r\cos \theta $ and $y = r\sin \theta $
From here $\cos \theta = \dfrac{x}{r}$ and $\sin \theta = \dfrac{y}{r}$
Then as we know according to the basic trigonometric identity that, as given below:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
Now multiply the above equation with ${r^2}$, on both sides, as shown below:
$ \Rightarrow {r^2}{\sin ^2}\theta + {r^2}{\cos ^2}\theta = {r^2}$
We know that $x = r\cos \theta $, now squaring this equation on both sides gives:
$ \Rightarrow {x^2} = {r^2}{\cos ^2}\theta $
Now we know that $y = r\sin \theta $, now squaring this equation on both sides gives:
$ \Rightarrow {y^2} = {r^2}{\sin ^2}\theta $
Hence ${x^2} + {y^2} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta $
$\therefore {x^2} + {y^2} = {r^2}$
From the trigonometric identity we know that $\sin 2\theta $ is expressed as:
$ \Rightarrow \sin 2\theta = 2\sin \theta \cos \theta $
Given that ${r^2} = \sin 2\theta $, now substituting everything we obtained in this expression, as given below:
$ \Rightarrow {r^2} = \sin 2\theta $
$ \Rightarrow {r^2} = 2\sin \theta \cos \theta $
We also obtained that$\cos \theta = \dfrac{x}{r}$ and $\sin \theta = \dfrac{y}{r}$, also ${r^2} = {x^2} + {y^2}$, substituting them in the above equation as shown below:
$ \Rightarrow {x^2} + {y^2} = 2\left( {\dfrac{x}{r}} \right)\left( {\dfrac{y}{r}} \right)$
$ \Rightarrow {x^2} + {y^2} = \dfrac{{2xy}}{{{r^2}}}$
We know that ${r^2} = {x^2} + {y^2}$, hence substituting it in the denominator of the right hand side of the above equation:
$ \Rightarrow {x^2} + {y^2} = \dfrac{{2xy}}{{{x^2} + {y^2}}}$
\[ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 2xy\]
Simplifying the left hand side of the above equation, gives:
\[ \Rightarrow {x^4} + {y^4} + 2{x^2}{y^2} = 2xy\]
\[ \Rightarrow {x^4} + {y^4} + 2{x^2}{y^2} - 2xy = 0\]
Converting ${r^2} = \sin 2\left( \theta \right)$ rectangular form is \[{x^4} + {y^4} + 2{x^2}{y^2} - 2xy = 0\].
Note: Please note that rectangular coordinates or Cartesian coordinates come in the form of (x, y). Polar coordinates come in the form of $\left( {r,\theta } \right)$. Instead of moving out from the origin using horizontal and vertical lines, we instead pick the angle $\theta $, which is in the direction, and then move out from the origin a certain distance $r$.
Complete step-by-step solution:
The given polar coordinates form of the equation is ${r^2} = \sin 2\theta $.
Let $x = r\cos \theta $ and $y = r\sin \theta $
From here $\cos \theta = \dfrac{x}{r}$ and $\sin \theta = \dfrac{y}{r}$
Then as we know according to the basic trigonometric identity that, as given below:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
Now multiply the above equation with ${r^2}$, on both sides, as shown below:
$ \Rightarrow {r^2}{\sin ^2}\theta + {r^2}{\cos ^2}\theta = {r^2}$
We know that $x = r\cos \theta $, now squaring this equation on both sides gives:
$ \Rightarrow {x^2} = {r^2}{\cos ^2}\theta $
Now we know that $y = r\sin \theta $, now squaring this equation on both sides gives:
$ \Rightarrow {y^2} = {r^2}{\sin ^2}\theta $
Hence ${x^2} + {y^2} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta $
$\therefore {x^2} + {y^2} = {r^2}$
From the trigonometric identity we know that $\sin 2\theta $ is expressed as:
$ \Rightarrow \sin 2\theta = 2\sin \theta \cos \theta $
Given that ${r^2} = \sin 2\theta $, now substituting everything we obtained in this expression, as given below:
$ \Rightarrow {r^2} = \sin 2\theta $
$ \Rightarrow {r^2} = 2\sin \theta \cos \theta $
We also obtained that$\cos \theta = \dfrac{x}{r}$ and $\sin \theta = \dfrac{y}{r}$, also ${r^2} = {x^2} + {y^2}$, substituting them in the above equation as shown below:
$ \Rightarrow {x^2} + {y^2} = 2\left( {\dfrac{x}{r}} \right)\left( {\dfrac{y}{r}} \right)$
$ \Rightarrow {x^2} + {y^2} = \dfrac{{2xy}}{{{r^2}}}$
We know that ${r^2} = {x^2} + {y^2}$, hence substituting it in the denominator of the right hand side of the above equation:
$ \Rightarrow {x^2} + {y^2} = \dfrac{{2xy}}{{{x^2} + {y^2}}}$
\[ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 2xy\]
Simplifying the left hand side of the above equation, gives:
\[ \Rightarrow {x^4} + {y^4} + 2{x^2}{y^2} = 2xy\]
\[ \Rightarrow {x^4} + {y^4} + 2{x^2}{y^2} - 2xy = 0\]
Converting ${r^2} = \sin 2\left( \theta \right)$ rectangular form is \[{x^4} + {y^4} + 2{x^2}{y^2} - 2xy = 0\].
Note: Please note that rectangular coordinates or Cartesian coordinates come in the form of (x, y). Polar coordinates come in the form of $\left( {r,\theta } \right)$. Instead of moving out from the origin using horizontal and vertical lines, we instead pick the angle $\theta $, which is in the direction, and then move out from the origin a certain distance $r$.
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