# Find the equation in x and y for the line tangent to the curve in polar coordinates r = 3 - 6...

## Question:

Find the equation in {eq}\,x\, {/eq} and {eq}\,y\, {/eq} for the line tangent to the curve in polar coordinates {eq}\,r = 3 - 6\sin\theta\, {/eq} at the value {eq}\,\theta = 0 {/eq}.

## The Tangent Line:

The point slope formula for the tangent line to the curve is {eq}y=mx+b {/eq}. Tangent line touches the curve at a specific point, which is called tangent point.

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 use the trigonometric values {eq}\sin 0=0 , \cos 0=1 {/eq}.

We are given the polar curve {eq}\displaystyle r = 3 - 6\sin\theta\ {/eq}

First, we need to find out {eq}\displaystyle \dfrac{\ dy}{\ dx} {/eq}

Take the derivative:

{eq}\displaystyle \dfrac{\ dr }{\ d\theta}= -6\cos \theta {/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 =3 - 6\sin\theta \ , \ \dfrac{\ dr }{\ d\theta}=- 6\cos \theta {/eq} :

{eq}\Rightarrow \displaystyle \dfrac{\ dy}{\ dx}=\dfrac{- 6\cos \theta\sin \theta +( 3 - 6\sin\theta) \cos \theta }{ - 6\cos \theta \cos \theta-( 3 - 6\sin\theta) \sin \theta} {/eq}

{eq}\Rightarrow \displaystyle \dfrac{\ dy}{\ dx}=\dfrac{-6\cos \theta \sin \theta +3 \cos \theta - 6\sin\theta \cos \theta}{ -6\cos^2 \theta - 3 \sin \theta+6\sin^2 \theta } {/eq}

{eq}\Rightarrow \displaystyle \dfrac{\ dy}{\ dx}=\dfrac{-12 \cos \theta \sin \theta +3 \cos \theta }{ -6\cos 2 \theta - 3 \sin \theta} {/eq}

Plug in {eq}\theta = 0 {/eq}:

{eq}\Rightarrow \displaystyle m=\dfrac{0+3 } { -6 -0} {/eq}

{eq}\Rightarrow \displaystyle m =-\dfrac{1}{2} {/eq}

Now {eq}x= r \cos \theta = (3-\sin \theta) \cos \theta {/eq}, at {eq}\theta=0 {/eq} {eq}x=3 {/eq}

and {eq}y= r \sin \theta =(3-\sin \theta) \sin \theta {/eq}, at {eq}\theta=0 {/eq} {eq}y=0 {/eq}

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

Plug in {eq}m = - \dfrac{1}{2} {/eq} and the point {eq}(3,0) {/eq}:

{eq}0=- \dfrac{3}{2} +b \Rightarrow b= \dfrac{3}{2} {/eq}

Therefore, the equation of the tangent line to the curve {eq}\displaystyle r = 3 - 6\sin\theta {/eq} at {eq}\theta= 0 {/eq} is: {eq}{\boxed{ y =-\dfrac{1}{2} x+ \dfrac{3}{2} }} {/eq} 