# Find an equation for the line tangent to the curve at the point defined by the given value of t ...

## Question:

Find an equation for the line tangent to the curve at the point defined by the given value of {eq}t{/eq}

- {eq}x=\csc t, y=6 \cot t, t =\frac{\pi}{3}{/eq}

A) {eq}y=-12x+6\sqrt3{/eq} {eq}\enspace \enspace \enspace {/eq}B) {eq}y=2\sqrt3x-12{/eq} {eq}\enspace \enspace \enspace {/eq}C) {eq}y=12x-6\sqrt3{/eq} {eq}\enspace \enspace \enspace {/eq}D) {eq}y=12x+2\sqrt3{/eq}

(Choose the correct one)

## Parametrics and Cartesian Equations:

To move from an equation in parametrics to an equation in Cartesian we have to eliminate the parameter and then rewrite it in terms of the Cartesian variables.

## Answer and Explanation:

First, with the value of the function and the first derivative, we obtain the tangent line in parametrics:

{eq}x = \csc t,y = 6\cot t,t = \frac{\pi }{3}\\ t = \frac{\pi }{3} \to x = \csc \frac{\pi }{3} = \frac{2}{{\sqrt 3 }},y = 6\cot \frac{\pi }{3} = 6 \cdot \frac{1}{{\sqrt 3 }} = \frac{6}{{\sqrt 3 }}\\ \\ x' = - \frac{1}{{{{\sin }^2}t}}\cos t,y' = 6\left( { - \frac{1}{{{{\sin }^2}t}}} \right)\\ t = \frac{\pi }{3} \to x' = - \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}\left( {\frac{1}{2}} \right) = - \frac{2}{3},y' = 6\left( { - \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} \right) = - 8\\ \\ \left\{ \begin{array}{l} x = \frac{2}{{\sqrt 3 }} - \frac{2}{3}t\quad \left( 1 \right)\\ y = \frac{6}{{\sqrt 3 }} - 8t\quad \left( 2 \right) \end{array} \right. {/eq}

Once the parameter is eliminated, we obtain the equation of the tangent line:

{eq}\frac{1}{{12}}\left( 2 \right) - \left( 1 \right)\\ \frac{1}{{12}}y - x = \frac{1}{{12}}\left( {\frac{6}{{\sqrt 3 }} - 8t} \right) - \left( {\frac{2}{{\sqrt 3 }} - \frac{2}{3}t} \right)\\ \frac{1}{{12}}y - x = \frac{{ - 3}}{{2\sqrt 3 }}\\ \left[ {\frac{1}{{12}}y - x = \frac{{ - 3}}{{2\sqrt 3 }}} \right]12\\ y - 12x = - \frac{{18}}{{\sqrt 3 }}\\ y = 12x - \frac{{18}}{{\sqrt 3 }}\\ y = 12x - 6\sqrt 3 {/eq}

So, the correct option is C)

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#### Learn more about this topic:

from Precalculus: High School

Chapter 24 / Lesson 3