# Find the equation in x and y for the line tangent to the curve in polar coordinates r = 2 at the...

## Question:

Find the equation in {eq}\,x\, {/eq} and {eq}\,y\, {/eq} for the line tangent to the curve in polar coordinates {eq}\,r = 2\, {/eq} at the value {eq}\displaystyle\,\theta = \frac{\pi}{3} {/eq}.

## The Tangent Line:

To find out the slope of the tangent line to the given polar curve, we'll use the formula: {eq}\displaystyle \dfrac{\ dy}{\ dx}= \frac{\dfrac{\ dr }{\ d\theta} \sin \theta + r \cos \theta }{\dfrac{\ dr }{\ d\theta} \cos \theta - r \sin \theta} {/eq}

Next, we'll compute the tangent point {eq}(x,y) {/eq} and use the point slope formula {eq}y=mx+b {/eq} to get the desired solution.

## Answer and Explanation:

We are given the polar curve {eq}\displaystyle r= 2 {/eq}

First, we need to find out the slope:

{eq}\displaystyle \dfrac{\ dr }{\ d\theta}=0 {/eq}

Using the fact: {eq}\displaystyle \dfrac{\ dy}{\ dx}= \frac{\dfrac{\ dr }{\ d\theta} \sin \theta + r \cos \theta }{\dfrac{\ dr }{\ d\theta} \cos \theta - r \sin \theta} {/eq}

Plug in {eq}\displaystyle r =2\ , \ \dfrac{\ dr }{\ d\theta}=0 {/eq} :

{eq}\Rightarrow \displaystyle \dfrac{\ dy}{\ dx}=\dfrac{0+2\cos \theta }{0 -2 \sin \theta} {/eq}

{eq}\Rightarrow \displaystyle \dfrac{\ dy}{\ dx}=- \cot \theta {/eq}

Plug in {eq}\theta = \frac{\pi}{3} {/eq}:

{eq}\Rightarrow \displaystyle \dfrac{\ dy}{\ dx}=-\cot \frac{\pi}{3} {/eq}

{eq}\Rightarrow \displaystyle m = \dfrac{\ dy}{\ dx}=- \dfrac{1}{\sqrt 3} {/eq}

Hence, the slope of the tangent line to the curve is: {eq}m =- \dfrac{1}{\sqrt 3} . {/eq}

Now {eq}x= r \cos \theta = 2 \cdot \dfrac{1}{2}=1 {/eq}

and {eq}y= r \sin \theta = 2 \cdot \dfrac{\sqrt 3 }{2}=\sqrt 3 {/eq}

The point slope formula of tangent line to the curve is {eq}y=mx+b {/eq}

Plug in {eq}m = - \dfrac{1}{\sqrt 3} {/eq} and the point {eq}(1 , \sqrt 3 ) {/eq}:

{eq}\sqrt 3 =- \dfrac{1}{\sqrt 3} +b \Rightarrow b=\sqrt 3+\dfrac{1}{\sqrt 3} = \dfrac{4}{\sqrt 3} {/eq}

Therefore, the equation of the tangent line to the curve {eq}\displaystyle r= 2 {/eq} at {eq}\theta= \pi /3 {/eq} is: {eq}{\boxed{ y =-\dfrac{1}{\sqrt 3} x+ \dfrac{4}{\sqrt 3} }} {/eq}

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from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11