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Question

Question: How do you convert \[{{r}^{2}}=\sin \left( 2\theta \right)\]into rectangular form?...

How do you convert r2=sin(2θ){{r}^{2}}=\sin \left( 2\theta \right)into rectangular form?

Explanation

Solution

In order to find the solution of the above question that is to convert the equationr2=sin(2θ){{r}^{2}}=\sin \left( 2\theta \right) into the rectangular form, apply the following formulas that is x2+y2=r2{{x}^{2}}+{{y}^{2}}={{r}^{2}},x=rcos(θ)x=r\cos \left( \theta \right) and y=rsin(θ)y=r\sin \left( \theta \right) to substitute rr and θ\theta in the equation r2=sin(2θ){{r}^{2}}=\sin \left( 2\theta \right) into the variables xx and yy. Simplify the expression and get the answer.

Complete step by step solution:
Given equation in the question in as follows:
r2=sin(2θ){{r}^{2}}=\sin \left( 2\theta \right)
Clearly the above equation is in the polar form and according to the question, we have to convert it into the rectangular form.
Before this simplify the equation by applying the formula that is sin(2θ)=2sin(θ)cos(θ)\sin \left( 2\theta \right)=2\sin \left( \theta \right)\cos \left( \theta \right), we get:
r2=2sin(θ)cos(θ)\Rightarrow {{r}^{2}}=2\sin \left( \theta \right)\cos \left( \theta \right)
Now apply the following formulas to convert the above equation into the rectangular form that is x2+y2=r2{{x}^{2}}+{{y}^{2}}={{r}^{2}},x=rcos(θ)x=r\cos \left( \theta \right) and y=rsin(θ)y=r\sin \left( \theta \right) then substitute rr and θ\theta in the equation r2=sin(2θ){{r}^{2}}=\sin \left( 2\theta \right) into the variables xx and yy, we get:
x2+y2=2(xr)(yr)\Rightarrow {{x}^{2}}+{{y}^{2}}=2\left( \dfrac{x}{r} \right)\left( \dfrac{y}{r} \right)
After this simplify the brackets and multiply the terms in numerator with each other and in denominator with each other
x2+y2=2xyr2\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{2xy}{{{r}^{2}}}
Now apply the formula in the above equation that is x2+y2=r2{{x}^{2}}+{{y}^{2}}={{r}^{2}} to get:
x2+y2=2xyx2+y2\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{2xy}{{{x}^{2}}+{{y}^{2}}}
Simplify the above equation further by cross multiplying the term in the denominator to the term in the left-hand-side of the equation, we get:
(x2+y2)(x2+y2)=2xy\Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)=2xy
As you can see both the terms in brackets are similar therefore, we can add their powers that is we can write (x2+y2)(x2+y2)=(x2+y2)1+1=(x2+y2)2\left( {{x}^{2}}+{{y}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( {{x}^{2}}+{{y}^{2}} \right)}^{1+1}}={{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}. Now applying this in the above equation we get:
(x2+y2)2=2xy\Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=2xy
After this expand the left-hand-side with help of this identity that is (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab where a=x2a={{x}^{2}} and b=y2b={{y}^{2}}, we get:
x4+y4+2x2y2=2xy\Rightarrow {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}=2xy
Simplify it further we get:
x4+y4+2x2y22xy=0\Rightarrow {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}-2xy=0
Therefore, rectangular form of the given equation r2=sin(2θ){{r}^{2}}=\sin \left( 2\theta \right) is x4+y4+2x2y22xy=0{{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}-2xy=0.

Note: Students make mistakes while writing the right equation while converting it into the rectangular form and by not applying the formulas correctly. One should be aware about the rectangular coordinates, the polar coordinates and coordinate conversion equations that is x2+y2=r2{{x}^{2}}+{{y}^{2}}={{r}^{2}},x=rcos(θ)x=r\cos \left( \theta \right) and y=rsin(θ)y=r\sin \left( \theta \right).