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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.
\[x = 2\cos \left( {3t} \right) - 4\sin \left( {3t} \right)\hspace{0.25in}y = 3\tan \left( {6t} \right)\mbox{ at }t = \frac{\pi }{2}\]Show All Steps Hide All Steps
Start SolutionWe’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 2The 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 3The (only) tangent line for this problem is then,
\[y = 0 + 3\left( {x - 4} \right)\hspace{0.25in} \to \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{y = 3x - 12}}\]