Solveeit Logo
Login

Question

Question: How do you convert \(r=\dfrac{1}{1-\cos \theta }\) in rectangular form?...

How do you convert r=11−cos⁡θr=\dfrac{1}{1-\cos \theta } in rectangular form?

Explanation

Solution

In the above problem, we have given the polar form and asked us to convert into rectangular form. Polar form contains (r,θ)\left( r,\theta \right) form and the rectangular form contains (x,y)\left( x,y \right). We know the conversion of (r,θ)\left( r,\theta \right) into (x,y)\left( x,y \right) as: x=rcos⁡θ,y=rsin⁡θx=r\cos \theta ,y=r\sin \theta so using these conversions, we can convert the given polar form into rectangular form.

Complete step by step solution:
The polar form given in the above problem is as follows:
r=11−cos⁡θr=\dfrac{1}{1-\cos \theta }
Now, cross – multiplying in the above equation we get,
⇒r(1−cos⁡θ)=1\Rightarrow r\left( 1-\cos \theta \right)=1
Multiplying r inside the bracket we get,
⇒r−rcos⁡θ=1............(1)\Rightarrow r-r\cos \theta =1............(1)
We know the conversions from polar form to rectangular form as follows:
x=rcos⁡θ; y=rsin⁡θ \begin{aligned} & x=r\cos \theta ; \\\ & y=r\sin \theta \\\ \end{aligned}
Also, we know the relation between x, y and r as follows:
x2+y2=r2{{x}^{2}}+{{y}^{2}}={{r}^{2}}
Taking square root in the above equation we get,
⇒x2+y2=r\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=r
Substituting the value of r&rcos⁡θr\And r\cos \theta in eq. (1) we get,
⇒x2+y2−x=1\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}-x=1
Adding x on both the sides of the above equation we get,
⇒x2+y2=1+x\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=1+x
Squaring on both the sides of the above equation we get,
⇒(x2+y2)2=(1+x)2 ⇒x2+y2=1+x2+2x \begin{aligned} & \Rightarrow {{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}={{\left( 1+x \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}=1+{{x}^{2}}+2x \\\ \end{aligned}
In the above equation, x2{{x}^{2}} will be cancelled out on both the sides of the equation we get,
y2=1+2x{{y}^{2}}=1+2x
Hence, we have converted the given polar form into rectangular form as follows:
y2=1+2x{{y}^{2}}=1+2x

Note: In the above solution, we have written a relation between x, y and r as follows:
x2+y2=r2{{x}^{2}}+{{y}^{2}}={{r}^{2}}
The proof of the above equation is that in the above solution we have written the relation between x, r and cos⁡θ\cos \theta . Also, we have written the relation between y, r and sin⁡θ\sin \theta .
x=rcos⁡θ.....(2) y=rsin⁡θ.......(3) \begin{aligned} & x=r\cos \theta .....(2) \\\ & y=r\sin \theta .......(3) \\\ \end{aligned}
Squaring both the equations and then adding them we get,
x2=r2cos⁡2θ y2=r2sin⁡2θ ⇒x2+y2=r2(cos⁡2θ+sin⁡2θ).........(4) \begin{aligned} & {{x}^{2}}={{r}^{2}}{{\cos }^{2}}\theta \\\ & {{y}^{2}}={{r}^{2}}{{\sin }^{2}}\theta \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right).........(4) \\\ \end{aligned}
In the above equation, we can use the following trigonometric identity which is equal to
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
Using the above relation in eq. (4) we get,
⇒x2+y2=r2(1) ⇒x2+y2=r2 \begin{aligned} & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}\left( 1 \right) \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}} \\\ \end{aligned}