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College
Calculus and Analysis
Integral Calculus
Parametric equations, polar coordinates, and vector-valued functions

coexpennan

Answered question

2021-08-11

Find an equation of the following curve in polar coordinates and describe the curve. $x = (1 + \cos t) \cos t, y = (1 + \cos t) \sin t . 0 \leq t \leq 2 π$

Answer & Explanation

davonliefI

Skilled2021-08-12Added 79 answers

Solution:
1.Consider the given curves,

x = (1 + \cos t) \cos t

y = (1 + \cos t) \sin t

Square and add the above equations,

x^{2} + y^{2} = {(1 + \cos t)}^{2}

Divide the above equations,

\frac{y}{x} = \frac{(1 + \cos t) \sin t}{(1 + \cos t) \cos t} = \tan t

x^{2} + y^{2} = {(1 + \cos t)}^{2}

{(1 + \cos t)}^{2} = {(1 + \frac{1}{\sec t})}^{2}

{(1 + \frac{1}{\sec t})}^{2} = (1 + \frac{1}{\sqrt{1 + \tan^{2} t}})

Therefore,

x^{2} + y^{2} = {(1 + \frac{1}{\sqrt{1 + \tan^{2} t}})}^{2}

{(1 + \frac{1}{\sqrt{1 + {(\frac{y}{x})}^{2}}})}^{2}

{(1 + \frac{1}{\sqrt{1 + \frac{1}{\sqrt{\frac{x^{2} + y^{2}}{x^{2}}}}}})}^{2}

{(1 + \frac{x}{\sqrt{x^{2} + y^{2}}})}^{2}

3. For polar form, substitute

x = r \cos θ, y = r \cos θ, r^{2} = x^{2} + y^{2}, θ = \tan^{- 1} (\frac{y}{x}) {(1 + \frac{1}{\sqrt{1 + \frac{1}{\sqrt{\frac{x^{2} + y^{2}}{x^{2}}}}}})}^{2}

r^{2} = {(1 + \frac{r \cos θ}{{\sqrt{r}}^{2}})}^{2}

r^{2} = (1 + \cos θ)

Thus, the required solution is

r = 1 + \cos θ

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