Question

How to convert $$r^2=\sin \theta$$ from polar to rectangular form?

Answer 1

Hint: Here, we will use the rectangular coordinates and simplify it to find the value of $$r^2$$. Then we will rewrite the given equation and substitute the obtained value and rectangular coordinate to simplify it further. Then by using the trigonometric identity and exponent rules we will convert the equation into the rectangular form.

Formula Used:
We will use the following formula:
1. The square of the sum of the numbers is given by $${\left(a+b\right)}^2=a^2+b^2+2ab$$.
2. Product rule of exponents:$$a^m\times a^n=a^{m+n}$$
3. Power rule for Exponents: $${\left(a^m\right)}^n=a^{mn}$$

Complete Step by Step Solution:
We are given the equation in a Polar form $$r^2=\sin \theta$$.
Now, multiplying both the sides of the equation with $$r$$, we get
$$\Rightarrow r^2\cdot r=r\cdot \sin \theta$$ ……………………………………………….$$\left(1\right)$$
We know that in Rectangular form $$x=r\cos \theta$$ and $$y=r\sin \theta$$.
Now squaring and adding the coordinates of the Rectangular form, we get
$$x^2+y^2=r^2{\cos }^2\theta +r^2{\sin }^2\theta$$
By taking out the common factor, we get
 $$\Rightarrow x^2+y^2=r^2\left({\cos }^2\theta +{\sin }^2\theta \right)$$
Now, by using the Trigonometric Identity $${\cos }^2\theta +{\sin }^2\theta =1$$, we get
$$\Rightarrow x^2+y^2=r^2\left(1\right)$$
$$\Rightarrow x^2+y^2=r^2$$ ……………………………………………$$\left(2\right)$$
By substituting above equation in equation $$\left(1\right)$$, we get
$$\left(x^2+y^2\right)\cdot r=y$$
Dividing both sides by $$r$$, we get
$$\Rightarrow \left(x^2+y^2\right)={\displaystyle \frac{y}{r}}$$…………………………………….$$\left(3\right)$$
Taking square root on both the sides of the equation $$\left(2\right)$$, we get
$$r=\pm \sqrt{x^2+y^2}$$
Substituting the value of $$r$$ in the equation $$\left(3\right)$$, we get
$$\left(x^2+y^2\right)={\displaystyle \frac{y}{\pm \sqrt{x^2+y^2}}}$$
Now, by squaring on both the sides, we get
$$\Rightarrow {\left(x^2+y^2\right)}^2={\displaystyle \frac{y^2}{x^2+y^2}}$$
Now, by using the algebraic identity $${\left(a+b\right)}^2=a^2+b^2+2ab$$, we get
$$\Rightarrow {\left(x^2\right)}^2+{\left(y^2\right)}^2+2\left(x^2\right)\left(y^2\right)={\displaystyle \frac{y^2}{x^2+y^2}}$$
Using the Power rule for exponents $${\left(a^m\right)}^n=a^{mn}$$, we get
$$\Rightarrow \left(x^4\right)+\left(y^4\right)+2\left(x^2\right)\left(y^2\right)={\displaystyle \frac{y^2}{x^2+y^2}}$$
Multiplying $$x^2+y^2$$ on both the sides, we get
$$\Rightarrow \left[\left(x^4\right)+\left(y^4\right)+2\left(x^2\right)\left(y^2\right)\right]\left(x^2+y^2\right)=y^2$$
Now, by using the FOIL method, we get
$$\Rightarrow x^2\left(\left(x^4\right)+\left(y^4\right)+2\left(x^2\right)\left(y^2\right)\right)+y^2\left(\left(x^4\right)+\left(y^4\right)+2\left(x^2\right)\left(y^2\right)\right)=y^2$$
Simplifying the equation, we get
$$\Rightarrow \left(\left(x^2\cdot x^4\right)+\left(x^2\cdot y^4\right)+2\left(x^2\cdot x^2\right)\left(y^2\right)\right)+\left(\left(y^2\cdot x^4\right)+\left(y^2\cdot y^4\right)+2\left(x^2\right)\left(y^2\cdot y^2\right)\right)=y^2$$
Now, by using product rule of exponents $$a^m\times a^n=a^{m+n}$$, we get
$$\Rightarrow \left(\left(x^6\right)+\left(x^2{y}^4\right)+2\left(x^4\right)\left(y^2\right)\right)+\left(\left(y^2{x}^4\right)+\left(y^6\right)+2\left(x^2\right)\left(y^4\right)\right)=y^2$$
Rewriting the equation, we get
$$\Rightarrow y^2=x^6+x^2{y}^4+2x^4{y}^2+y^2{x}^4+y^6+2x^2{y}^4$$
Adding the like terms, we get
$$\Rightarrow y^2=x^6+3x^2{y}^4+3x^4{y}^2+y^6$$

Therefore, the rectangular form of $$r^2=\sin \theta$$ is $$y^2=x^6+3x^2{y}^4+3x^4{y}^2+y^6$$.

Note:
We know that Polar coordinates are used to specify only two dimensions whereas Rectangular Coordinates which is also called as Cartesian Coordinates is used to specify three dimensions in a plane. FOIL method is a method of multiplying the binomials by multiplying the first terms, then the outer terms, then the inner terms and at last the last terms. Using this we can easily combine like terms.