Question

An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum values of I

Answer 1

Take the derivative with respect to t
$- w \sin(wt) + \sqrt{8} w cos(wt)$
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
$0 = -w \sin(wt) + \sqrt{8} w cos(wt)$
divide by w
$0 =- \sin(wt) + \sqrt{8} cos(wt)$
we add sin(wt) to both sides

$\sin(wt)= \sqrt{8} cos(wt)$
divide both sides by cos(wt)
$\frac{sin(wt)}{cos(wt)}= \sqrt{8} \\ \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\ \\ wt=arctan(2 \sqrt{2)}$ OR $wt=arctan( { \frac{1}{ \sqrt{2} } )$
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

$I=cos(2n \pi -2arctan( \sqrt{2} ))$
since 2npi is just the period of cos
$cos(2arctan( \sqrt{2} ))= \frac{-1}{3}$
substituting our second soultion we get
$I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))$
since 2npi is the period
$I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}$
so the maximum value =$\frac{1}{3}$
minimum value =$- \frac{1}{3}$