Answered: "Parametric to polar equations Find an equation of..."

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

Mylo O'Moore

Answered question

2021-01-08

Parametric to polar equations Find an equation of the following curve in polar coordinates and describe the curve.

x = (1 + c o s t) c o s t, y = (1 + c o s t) s i n t, 0 \leq t \leq 2 π

Answer & Explanation

tabuordg

Skilled2021-01-09Added 99 answers

Step 1

Consider the given curves, $x = (1 + c o s t) c o s t$
$y = (1 + c o s t) s i n t$ Square and add the above equations, $x^{2} + y^{2} = (1 + c o s t)^{2}$ Divide the above equations, $\frac{x}{y} = \frac{(1 + c o s t) s i n t}{(1 + c o s t) c o s t} = t a n t$

Step 2

$x^{2} + y^{2} = (1 + c o s t)^{2}$
$(1 + c o s t)^{2} = {\frac{1 + (1}{s e c t)}}^{2}$
$(1 + \frac{1}{s e c t})^{2} = (1 + \frac{1}{\sqrt{(} 1 + t a n_{t}^{2})})^{2}$ Therefore, $x^{2} + y^{2} = (1 + \frac{1}{\sqrt{(} 1 + t a n_{t}^{2})})^{2}$
$= (1 + \frac{1}{(\sqrt{1} + \frac{1}{\sqrt{1} + (\frac{x}{y})^{2})})^{2}})$
$= (1 + \frac{1}{\sqrt{(} (\frac{x^{2} + y^{2}}{x^{2}})}))^{2}$
$= (1 + \frac{x}{\sqrt{(x^{2} + y^{2})}})^{2}$

Step 3

For polar form, substitute $x = r c o s θ, y = r s i n θ, r^{2} = x^{2} + y^{2}, θ = t a n^{(} - 1) (\frac{y}{x})$
$x^{2} + y^{2} (1 + \frac{x}{\sqrt{(} (x^{2} + y^{2}))})^{2}$
$r^{2} = \frac{(}{+ (r c o s θ} s q r t (r^{2}))^{2}$
$r^{2} = (1 + c o s θ)^{2}$ Thus, the required solution is $r = 1 + c o s θ$

Jeffrey Jordon

Expert2021-10-14Added 2605 answers

Answer is given below (on video)

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