Question

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

Answer 1

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

$f''(t) = -ω^2IR\sin ωt$ and $f''(\frac{\pi }{2ω}+\frac{\pi n}{\omega}) = -ω^2IR\sin ω(\frac{\pi }{2ω}+\frac{\pi n}{\omega}) = -ω^2IR\sin (\frac{\pi }{2}+\pi n),$

we conclude that $f''(\frac{\pi }{2ω}+\frac{2\pi k}{\omega}) = = -ω^2IR\sin (\frac{\pi }{2}+2\pi k) = -ω^2IR\sin (\frac{\pi }{2})= -ω^2IR<0$

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