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### Section 9.2 : Tangents with Parametric Equations

3. Find the equation of the tangent line(s) to the following set of parametric equations at the given point.

$x = 2\cos \left( {3t} \right) - 4\sin \left( {3t} \right)\hspace{0.25in}y = 3\tan \left( {6t} \right)\mbox{ at }t = \frac{\pi }{2}$

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We’ll need the first derivative for the set of parametric equations. We’ll need the following derivatives,

$\frac{{dx}}{{dt}} = - 6\sin \left( {3t} \right) - 12\cos \left( {3t} \right)\hspace{0.5in}\frac{{dy}}{{dt}} = 18{\sec ^2}\left( {6t} \right)$

The first derivative is then,

$\frac{{dy}}{{dx}} = \frac{{\displaystyle \,\,\frac{{dy}}{{dt}}\,\,}}{{\displaystyle \,\,\frac{{dx}}{{dt}}\,\,}} = \frac{{18{{\sec }^2}\left( {6t} \right)}}{{ - 6\sin \left( {3t} \right) - 12\cos \left( {3t} \right)}} = \frac{{3{{\sec }^2}\left( {6t} \right)}}{{ - \sin \left( {3t} \right) - 2\cos \left( {3t} \right)}}$ Show Step 2

The slope of the tangent line at $$t = \frac{\pi }{2}$$ is then,

$m = {\left. {\frac{{dy}}{{dx}}} \right|_{t = \frac{\pi }{2}}} = \frac{{3{{\left( { - 1} \right)}^2}}}{{ - \left( { - 1} \right) - 2\left( 0 \right)}} = 3$

At $$t = \frac{\pi }{2}$$ the parametric curve is at the point,

$\left. {{x_{t\, = \,\frac{\pi }{2}}}} \right| = 2\left( 0 \right) - 4\left( { - 1} \right) = 4\hspace{0.25in}\,\,\left. {{y_{t\, = \,\frac{\pi }{2}}}} \right| = 3\left( 0 \right) = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}\left( {4,0} \right)$ Show Step 3

The (only) tangent line for this problem is then,

$y = 0 + 3\left( {x - 4} \right)\hspace{0.25in} \to \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{y = 3x - 12}}$