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### Section 9.2 : Tangents with Parametric Equations

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

$x = {t^2} - 2t - 11\hspace{0.25in}y = t{\left( {t - 4} \right)^3} - 3{t^2}{\left( {t - 4} \right)^2} + 7\mbox{ at } \left( { - 3,7} \right)$

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Start Solution

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

\begin{align*}\frac{{dx}}{{dt}} & = 2t - 2\\ & \\ \frac{{dy}}{{dt}} & = {\left( {t - 4} \right)^3} + 3t{\left( {t - 4} \right)^2} - 6t{\left( {t - 4} \right)^2} - 6{t^2}\left( {t - 4} \right) = {\left( {t - 4} \right)^3} - 3t{\left( {t - 4} \right)^2} - 6{t^2}\left( {t - 4} \right)\end{align*}

The first derivative is then,

$\frac{{dy}}{{dx}} = \frac{{\,\,\frac{{dy}}{{dt}}\,\,}}{{\,\,\frac{{dx}}{{dt}}\,\,}} = \frac{{{{\left( {t - 4} \right)}^3} - 3t{{\left( {t - 4} \right)}^2} - 6{t^2}\left( {t - 4} \right)}}{{2t - 2}}$
Hint : Don’t forget that because the derivative we found above is in terms of $$t$$ we need to determine the value(s) of $$t$$ that put the parametric curve at the given point.
Show Step 2

Okay, the derivative we found above is in terms of $$t$$ and we we’ll need to next determine the value(s) of $$t$$ that put the parametric curve at $$\left( { - 3,7} \right)$$.

This is easy enough to do by setting the $$x$$ and $$y$$ coordinates equal to the known parametric equations and determining the value(s) of $$t$$ that satisfy both equations.

Doing that gives,

\begin{align*} - 3 & = {t^2} - 2t - 11\\ 0 & = {t^2} - 2t - 8\\ 0 & = \left( {t - 4} \right)\left( {t + 2} \right)\hspace{0.5in}\hspace{0.25in}\,\,\, \to \hspace{0.5in}\hspace{0.5in}t = - 2,t = 4\end{align*} \begin{align*}7 & = t{\left( {t - 4} \right)^3} - 3{t^2}{\left( {t - 4} \right)^2} + 7\\ 0 & = {\left( {t - 4} \right)^2}\left[ {t\left( {t - 4} \right) - 3{t^2}} \right]\\ 0 & = {\left( {t - 4} \right)^2}\left[ { - 4t - 2{t^2}} \right]\\ 0 & = - 2t{\left( {t - 4} \right)^2}\left[ {2 + t} \right]\hspace{0.5in}\hspace{0.25in}\,\,\,\,\,\, \to \hspace{0.5in}\hspace{0.25in}\,\,\,t = - 2,t = 0,t = 4\end{align*}

We can see from this list that the parametric curve will be at $$\left( { - 3,7} \right)$$ for $$t = - 2$$ and $$t = 4$$.

Show Step 3

From the previous step we can see that we will in fact have two tangent lines at the point. Here are the slopes for each tangent line.

The slope of the tangent line at $$t = - 2$$ is,

$m = {\left. {\frac{{dy}}{{dx}}} \right|_{t = - 2}} = - 24$

and the slope of the tangent line at $$t = 4$$ is,

$m = {\left. {\frac{{dy}}{{dx}}} \right|_{t = 4}} = 0$ Show Step 4

The tangent line for $$t = - 2$$ is then,

$y = 7 - 24\left( {x + 3} \right)\hspace{0.25in} \to \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{y = - 24x - 65}}$

The tangent line for $$t = 4$$ is then,

$y = 7 - \left( 0 \right)\left( {x + 3} \right)\hspace{0.25in} \to \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{y = 7}}$

Do not get excited about the second tangent line! It is just saying that the second tangent line is a horizontal line.