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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

Respuesta :

chi3fcomm5nd
take the derivative with respect to t
[tex] - w \sin(wt) + \sqrt{8} w cos(wt)[/tex]
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
[tex]0 = -w \sin(wt) + \sqrt{8} w cos(wt)[/tex]
divide by w
[tex]0 =- \sin(wt) + \sqrt{8} cos(wt)[/tex]
we add sin(wt) to both sides

[tex]\sin(wt)= \sqrt{8} cos(wt)[/tex]
divide both sides by cos(wt)
[tex] \frac{sin(wt)}{cos(wt)}= \sqrt{8} \\ \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\ \\ wt=arctan(2 \sqrt{2)} [/tex] OR[tex] \\ [/tex] [tex]wt=arctan( { \frac{1}{ \sqrt{2} } ) [/tex]
(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

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


ernikhil

The maximum value of the electric current is [tex]I_{max}=\dfrac{1}{3}[/tex]

the minimum value of the electric current is [tex]I_{min}=-\dfrac{1}{3}[/tex]

The given expression for electric current is [tex]I=\cos(\omega t)+\sqrt{8}\sin(\omega t)[/tex] where [tex]\omega \neq 0[/tex].

Differentiate the given function with respect to [tex]t[/tex]-

[tex]\dfrac{dI}{dt}=\dfrac{d(\cos(\omega t)+\sqrt 8 \sin(\omega t))}{dt}\\\dfrac{dI}{dt}=-\omega \sin(\omega t)+\sqrt 8 \omega \cos(\omega t)[/tex]

Equate the derivative to 0-

[tex]\dfrac{dI}{dt}=0\\-\omega \sin(\omega t)+\sqrt 8 \omega \cos(\omega t)=0\\\sin(\omega t)=\sqrt 8 \cos (\omega t)\\\tan (\omega t)=\sqrt 8\\\omega t=\tan^{-1}\sqrt 8\\\omega t=2n\pi-2\tan^{-1}\sqrt 2}\\or,\\\omega t=2n\pi+2\tan^{-1}\sqrt {1/2}}[/tex]

[tex]2n\pi[/tex] is the period of sine and the cosine trigonometric function.

So, the maximum value of the electric current is

[tex]I_{max}=\cos(2\tan^{-1}\sqrt {1/2})\\I_{max}=\dfrac{1}{3}[/tex]

Also, the minimum value of the electric current is

[tex]I_{min}=\cos(2\tan^{-1}\sqrt {2})\\I_{min}=-\dfrac{1}{3}[/tex]

Learn more about maximum and minimum values of the functions here:

https://brainly.com/question/13581879?referrer=searchResults

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