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Question: How do you convert \[r = \sin (\theta ) + 1\] to rectangular form?...

How do you convert r=sin⁡(θ)+1r = \sin (\theta ) + 1 to rectangular form?

Explanation

Solution

We multiply the given equation by ‘r’ which converts one side of the equation as r2{r^2}. Use the conversion formula of polar coordinates into rectangular coordinates and write the equation.

  • A complex number can be represented using rectangular coordinates (x,y)(x,y) and polar coordinates (r,θ)(r,\theta ).
  • Rectangular coordinates (x,y)(x,y) can be converted into polar coordinates (r,θ)(r,\theta ) using the conversion formulas:
    x=rcos⁡θ;y=rsin⁡θx = r\cos \theta ;y = r\sin \theta
    And r2=x2+y2{r^2} = {x^2} + {y^2}
    i.e. r=x2+y2r = \sqrt {{x^2} + {y^2}}

Complete step by step solution:
We are given the equation r=sin⁡(θ)+1r = \sin (\theta ) + 1
Multiply the both sides of equation by ‘r’
⇒r×r=r(sin⁡(θ)+1)\Rightarrow r \times r = r\left( {\sin (\theta ) + 1} \right)
Multiply terms outside the bracket with terms inside the bracket on right hand side of the equation
⇒r2=rsin⁡(θ)+r\Rightarrow {r^2} = r\sin (\theta ) + r
Now substitute r2=x2+y2{r^2} = {x^2} + {y^2}on left hand side of the equation, r=x2+y2r = \sqrt {{x^2} + {y^2}} and rsin⁡θ=yr\sin \theta = yon right hand side of the equation.
⇒x2+y2=y+x2+y2\Rightarrow {x^2} + {y^2} = y + \sqrt {{x^2} + {y^2}}
∴\therefore Rectangular form of r=sin⁡(θ)+1r = \sin (\theta ) + 1 is x2+y2=y+x2+y2{x^2} + {y^2} = y + \sqrt {{x^2} + {y^2}}

Note: Many students make the mistake of writing the given equation by separating the variables to different sides. Then they try to square and convert into rectangular form. Keep in mind we know the direct values of conversions, so we directly try to form those terms in the equation.
Alternate method:
We have r=sin⁡(θ)+1r = \sin (\theta ) + 1
Square both sides of the equation
⇒r2=(sin⁡(θ)+1)2\Rightarrow {r^2} = {\left( {\sin (\theta ) + 1} \right)^2}
Use the identity (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
⇒r2=sin⁡2θ+1+2sin⁡θ\Rightarrow {r^2} = {\sin ^2}\theta + 1 + 2\sin \theta
Now we know that x=rcos⁡θ;y=rsin⁡θx = r\cos \theta ;y = r\sin \theta
And r2=x2+y2{r^2} = {x^2} + {y^2}
i.e. r=x2+y2r = \sqrt {{x^2} + {y^2}}
So, we can convert the equation r2=sin⁡2θ+1+2sin⁡θ{r^2} = {\sin ^2}\theta + 1 + 2\sin \theta into rectangular form
⇒x2+y2=(yr)2+1+2(yr)\Rightarrow {x^2} + {y^2} = {\left( {\dfrac{y}{r}} \right)^2} + 1 + 2\left( {\dfrac{y}{r}} \right)
⇒x2+y2=y2r2+1+2yr\Rightarrow {x^2} + {y^2} = \dfrac{{{y^2}}}{{{r^2}}} + 1 + \dfrac{{2y}}{r}
⇒x2+y2=y2+r2+2yrr2\Rightarrow {x^2} + {y^2} = \dfrac{{{y^2} + {r^2} + 2yr}}{{{r^2}}}
Substitute the value of r2=x2+y2{r^2} = {x^2} + {y^2} and r=x2+y2r = \sqrt {{x^2} + {y^2}}
⇒x2+y2=y2+x2+y2+2yx2+y2x2+y2\Rightarrow {x^2} + {y^2} = \dfrac{{{y^2} + {x^2} + {y^2} + 2y\sqrt {{x^2} + {y^2}} }}{{{x^2} + {y^2}}}
Cross multiply the denominator from right hand side of the equation to numerator of left hand side of the equation
⇒(x2+y2)(x2+y2)=x2+2y2+2yx2+y2\Rightarrow \left( {{x^2} + {y^2}} \right)\left( {{x^2} + {y^2}} \right) = {x^2} + 2{y^2} + 2y\sqrt {{x^2} + {y^2}}
We can add the powers in left hand side as the values in the product are same
⇒(x2+y2)2=x2+y2+y2+2yx2+y2\Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {x^2} + {y^2} + {y^2} + 2y\sqrt {{x^2} + {y^2}}
Now we observe on the right hand side of the equation that
⇒(x2+y2)2=(x2+y2)2+y2+2yx2+y2\Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {\left( {\sqrt {{x^2} + {y^2}} } \right)^2} + {y^2} + 2y\sqrt {{x^2} + {y^2}}
Which on applying property (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
⇒(x2+y2)2=(x2+y2+y)2\Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {\left( {\sqrt {{x^2} + {y^2}} + y} \right)^2}
Take square root on both sides and cancel square root by square power
⇒x2+y2=x2+y2+y\Rightarrow {x^2} + {y^2} = \sqrt {{x^2} + {y^2}} + y
∴\therefore Rectangular form of r=sin⁡(θ)+1r = \sin (\theta ) + 1is x2+y2=y+x2+y2{x^2} + {y^2} = y + \sqrt {{x^2} + {y^2}}