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### Section 9.2 : Tangents with Parametric Equations

5. Find the values of $$t$$ that will have horizontal or vertical tangent lines for the following set of parametric equations.

$x = {t^5} - 7{t^4} - 3{t^3}\hspace{0.25in}y = 2\cos \left( {3t} \right) + 4t$

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Start Solution

We’ll need the following derivatives for this problem.

$\frac{{dx}}{{dt}} = 5{t^4} - 28{t^3} - 9{t^2}\hspace{0.5in}\hspace{0.25in}\frac{{dy}}{{dt}} = - 6\sin \left( {3t} \right) + 4$ Show Step 2

We know that horizontal tangent lines will occur where $$\frac{{dy}}{{dt}} = 0$$, provided $$\frac{{dx}}{{dt}} \ne 0$$ at the same value of $$t$$.

So, to find the horizontal tangent lines we’ll need to solve,

$- 6\sin \left( {3t} \right) + 4 = 0\hspace{0.5in}\,\,\, \to \hspace{0.5in}\,\,\,\,\sin \left( {3t} \right) = \frac{2}{3}\hspace{0.25in}\,\,\,\,\,\, \to \hspace{0.25in}\,\,\,\,\,\,3t = {\sin ^{ - 1}}\left( {\frac{2}{3}} \right) = 0.7297$

Also, a quick glance at a unit circle we can see that a second angle is,

$3t = \pi - 0.7297 = 2.4119$

All possible values of $$t$$ that will give horizontal tangent lines are then,

$\begin{array}{*{20}{c}}{3t = 0.7297 + 2\pi n}\\{3t = 2.4119 + 2\pi n}\end{array}\hspace{0.25in}\, \to \hspace{0.25in}\,\,\,\,\,\begin{array}{*{20}{c}}{t = 0.2432 + \frac{2}{3}\pi n}\\{t = 0.8040 + \frac{2}{3}\pi n}\end{array}\,,\,\,\,\,\,n = 0, \pm 1, \pm 2, \pm 3, \ldots$

Note that we don’t officially know these do in fact give horizontal tangent lines until we also determine that $$\frac{{dx}}{{dt}} \ne 0$$ at these points. We’ll be able to determine that after the next step.

Show Step 3

We know that vertical tangent lines will occur where $$\frac{{dx}}{{dt}} = 0$$, provided $$\frac{{dy}}{{dt}} \ne 0$$ at the same value of $$t$$.

So, to find the vertical tangent lines we’ll need to solve,

\begin{align*}5{t^4} - 28{t^3} - 9{t^2} & = 0\\ {t^2}\left( {5{t^2} - 28t - 9} \right) & = 0\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\,t = 0,\,\,\,t = \frac{{28 \pm \sqrt {964} }}{{10}}\,\,\,\,\, \to \,\,\,\,\,\,t = 0,\,\,\,t = - 0.3048,\,\,\,t = 5.9048\end{align*} Show Step 4

From a quick inspection of the two lists of $$t$$ values from Step 2 and Step 3 we can see there are no values in common between the two lists. Therefore, any values of $$t$$ that gives $$\frac{{dy}}{{dt}} = 0$$ will not give $$\frac{{dx}}{{dt}} = 0$$ and visa-versa.

Therefore the values of $$t$$ that gives horizontal tangent lines are,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\,\begin{array}{*{20}{c}}{t = 0.2432 + \frac{2}{3}\pi n}\\{t = 0.8040 + \frac{2}{3}\pi n}\end{array}\,,\,\,\,\,\,n = 0, \pm 1, \pm 2, \pm 3, \ldots }}$

The values of $$t$$ that gives vertical tangent lines are,

$\require{bbox} \bbox[2pt,border:1px solid black]{{t = 0,\,\,\,t = - 0.3048,\,\,\,t = 5.9048}}$