Calculus II - Tangents with Parametric Equations

Show Mobile Notice

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 9.2 : Tangents with Parametric Equations

Back to Problem List

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

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

Show All Steps Hide All Steps

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,

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