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Which is the polar form of the parametric equations x = 2t and y = t^2? a) r = 4 \tan^2 \theta \\...

Question:

Which is the polar form of the parametric equations x = 2t and {eq}y = t^2 {/eq}?

{eq}a) r = 4 \tan^2 \theta \\ b) r = 4 \sec^2 \theta \\ c) r = 4 \sec \theta \\ d) r = 4 \tan \theta \sec \theta {/eq}

Parametric Equations :

We are given the parametric equations and we need to find out the polar equation. In order to do that we'll first compute the Cartesian equation and then convert it into a polar equation.

To solve, we'll apply the formulas {eq}x = r \cos(\theta) \ , \ y= r \sin(\theta) {/eq} and use the trigonometric ratios {eq}\dfrac{ \sin(\theta) }{\cos (\theta)}= \tan(\theta) , \dfrac{1}{\cos (\theta)}= \sec(\theta) {/eq} to get the desired solution.

Answer and Explanation:

We are given the parametric equations {eq}x=2t ,\ y=t^2 {/eq}, which gives {eq}t= \dfrac{x}{2} , t = \sqrt y {/eq}


Eliminate the parameter t, we'll get {eq}\dfrac{x}{2}= \sqrt y {/eq}


Simplify and we'll get {eq}x^2=4y {/eq}


Now we need to convert the equation {eq}x^2=4y {/eq} to an equation in polar coordinates.


Apply {eq}x = r \cos(\theta) \ , \ y= r \sin(\theta) {/eq}

{eq}\Rightarrow (r \cos (\theta) )^2= 4 r \sin(\theta) {/eq}

{eq}\Rightarrow r^2 \cos^2(\theta) = 4 r \sin (\theta) {/eq}

{eq}\Rightarrow r \cos^2(\theta) = 4 \sin (\theta) {/eq}

Isolate r:

{eq}\Rightarrow r = \dfrac{ 4\sin(\theta) }{\cos^2 (\theta)} {/eq}

{eq}\Rightarrow r = 4 \dfrac{ \sin(\theta) }{\cos (\theta)} \cdot \dfrac{1}{\cos (\theta)} {/eq}


Apply the trigonometric ratios {eq}\dfrac{ \sin(\theta) }{\cos (\theta)}= \tan(\theta) , \dfrac{1}{\cos (\theta)}= \sec(\theta) {/eq}

{eq}\Rightarrow r = 4 \tan(\theta) \cdot \sec(\theta) {/eq}


Therefore the solution is: {eq}{\boxed{ r=4 \tan(\theta) \cdot \sec(\theta) .}} {/eq}


ANSWER- d)


Learn more about this topic:

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Evaluating Parametric Equations: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 3
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