Paul's Online Notes
Paul's Online Notes
Home / Calculus I / Applications of Derivatives / Critical Points
Show Mobile Notice Show All Notes Hide All Notes
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 views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 4-2 : Critical Points

13. Determine the critical points of \(V\left( t \right) = {\sin ^2}\left( {3t} \right) + 1\).

Show All Steps Hide All Steps

Start Solution

We’ll need the first derivative to get the answer to this problem so let’s get that.

\[V'\left( t \right) = 6\sin \left( {3t} \right)\cos \left( {3t} \right)\] Show Step 2

Recall that critical points are simply where the derivative is zero and/or doesn’t exist.

This derivative exists everywhere and so we don’t need to worry about that. Therefore, all we need to do is determine where the derivative is zero. So, all we need to do is solve the equation,

\[6\sin \left( {3t} \right)\cos \left( {3t} \right) = 0\hspace{0.25in} \to \hspace{0.25in}\sin \left( {3t} \right) = 0\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\cos \left( {3t} \right) = 0\] Show Step 3

So, we now need to solve these two trig equations.

From a quick look at a unit circle we can see that sine is zero at 0 and \(\pi \) and so all solutions to \(\sin \left( {3t} \right) = 0\) are then,

\[\begin{array}{*{20}{l}}{3t = 0 + 2\pi n}\\{3t = \pi + 2\pi n}\end{array}\hspace{0.5in} \to \hspace{0.5in}\begin{array}{*{20}{l}}{t = {\displaystyle \frac{2}{3}}\pi n}\\{t = {\displaystyle \frac{1}{3}}\pi + {\displaystyle \frac{2}{3}}\pi n}\end{array}\hspace{0.25in}\,n = 0, \pm 1, \pm 2, \pm , \ldots \]

Another look at a unit circle and we can see that cosine is zero at \({\frac{\pi}{2}}\) and \({\textstyle{{3\pi } \over 2}}\) and so all solutions to \(\cos \left( {3t} \right) = 0\) are then,

\[\begin{array}{*{20}{l}}{3t = {\displaystyle \frac{\pi}{2}} + 2\pi n}\\{3t = {\displaystyle \frac{3\pi}{2}} + 2\pi n}\end{array}\hspace{0.5in} \to \hspace{0.5in}\begin{array}{*{20}{c}}{t = {\displaystyle \frac{\pi}{6}} + {\displaystyle \frac{2}{3}}\pi n}\\{t = {\displaystyle \frac{\pi}{2}} + {\displaystyle \frac{2}{3}}\pi n}\end{array}\hspace{0.25in}\,n = 0, \pm 1, \pm 2, \pm , \ldots \]

Therefore, critical points of the function are,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = {\frac{2}{3}}\pi n,\,\,\,t = {\frac{1}{3}}\pi + {\frac{2}{3}}\pi n,\,\,\,\,t = {\frac{\pi}{6}} + {\frac{2}{3}}\pi n,\,\,\,\,\,t = {\frac{\pi}{2}} + {\frac{2}{3}}\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \pm , \ldots }}\]

If you don’t remember how to solve trig equations you should go back and review those sections in the Review Chapter of the notes.