Paul's Online Notes
Paul's Online Notes
Home / Calculus I / Derivatives / Derivatives of Trig Functions
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 3-5 : Derivatives of Trig Functions

12. The position of an object is given by \(s\left( t \right) = 2 + 7\cos \left( t \right)\) determine all the points where the object is not moving.

Show Solution

We know that the object will not be moving if its velocity, which is simply the derivative of the position function, is zero. So, all we need to do is take the derivative, set it equal to zero and solve.

\[s'\left( t \right) = - 7\sin \left( t \right)\hspace{0.5in} \Rightarrow \hspace{0.5in} - 7\sin \left( t \right) = 0\]

So, for this problem the object will not be moving anywhere that sine is zero. From our recollection of the unit circle we know that will be at,

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