Question #292344

What is the value of the input Variable at which the maxima of function V = f(t) = IRsinωt occurs


1
Expert's answer
2022-02-02T11:45:04-0500

Let us find the value of the input variable at which the maximum of function V=f(t)=IRsinωtV = f(t) = IR\sinωt occurs. It follows that f(t)=ωIRcosωt=0f'(t) = ωIR\cosωt=0 implies ωt=π2+πnωt=\frac{\pi }2+\pi n, and thus t=π2ω+πnω.t=\frac{\pi }{2ω}+\frac{\pi n}{\omega}. Since

f(t)=ω2IRsinωtf''(t) = -ω^2IR\sin ωt and f(π2ω+πnω)=ω2IRsinω(π2ω+πnω)=ω2IRsin(π2+πn),f''(\frac{\pi }{2ω}+\frac{\pi n}{\omega}) = -ω^2IR\sin ω(\frac{\pi }{2ω}+\frac{\pi n}{\omega}) = -ω^2IR\sin (\frac{\pi }{2}+\pi n),

we conclude that f(π2ω+2πkω)==ω2IRsin(π2+2πk)=ω2IRsin(π2)=ω2IR<0f''(\frac{\pi }{2ω}+\frac{2\pi k}{\omega}) = = -ω^2IR\sin (\frac{\pi }{2}+2\pi k) = -ω^2IR\sin (\frac{\pi }{2})= -ω^2IR<0

and f(π2ω+π(2k+1)ω)==ω2IRsin(π2+π(2k+1))=ω2IRsin(π2)=ω2IR>0.f''(\frac{\pi }{2ω}+\frac{\pi (2k+1)}{\omega}) = = -ω^2IR\sin (\frac{\pi }{2}+\pi (2k+1)) = ω^2IR\sin (\frac{\pi }{2})= ω^2IR>0. Therefore, the points tk=π2ω+2πkω,t_k=\frac{\pi }{2ω}+\frac{2\pi k}{\omega}, where kN,k\in \N, is the values of the input variable at which the maximum of function V=f(t)=IRsinωtV = f(t) = IR\sinωt occurs.


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