Find out the answer to "Find an equation of the following curve in..."

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

Braxton Pugh

Answered question

2021-08-30

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

Ayesha Gomez

Skilled2021-08-31Added 104 answers

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, $\frac{y}{x} = \frac{(1 + \cos t) \sin t}{(1 + \cos t) \cos t} = \tan t$
$x^{2} + y^{2} = {(1 + \cos t)}^{2}$
${(1 + \cos t)}^{2} = {(1 + \frac{1}{\sec t})}^{2}$
${(1 + \frac{1}{\sec t})}^{2} = {(1 + \frac{1}{\sqrt{1 + \tan^{2} t}})}^{2}$
Therefore, $x^{2} + y^{2} = {(1 + \frac{1}{\sqrt{1 + \tan^{2} t}})}^{2}$
$= (1 + {\frac{1}{\sqrt{1 + {(\frac{y}{x})}^{2}}}}^{2}) .$
$(1 + {\frac{1}{\sqrt{\frac{x^{2} + y^{2}}{x^{2}}}}}^{2}) .$
$(1 + \frac{x}{\sqrt{x^{2} + y^{2}}}) .$
For polar form, substitute $x = r \cos θ, y = r \sin θ, r^{2} = x^{2} + y^{2}, θ = \tan^{- 1} (\frac{y}{x})$
$x^{2} + y^{2} = {(1 + \frac{x}{\sqrt{x^{2} + y^{2}}})}^{2}$
$r^{2} = {(1 + \frac{r \cos θ}{\sqrt{r^{2}}})}^{2}$
$r^{2} = {(1 + \cos θ)}^{2}$ Thus, required solution is $r = 1 + \cos θ$

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