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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.

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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,

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