A basic math question about the current in a capacitor

[Adir_Sh]

The current in a capacitor is calculated from

[itex]i_{C}(t)=C \dfrac{dv_{C}(t)}{dt}[/itex]​

when [itex]i_{C}(t)[/itex] is the current in the capacitor, [itex]t[/itex] is the time variable, [itex]C[/itex] is the capacitance and [itex]v_{C}(t)[/itex] is the capacitor's voltage.
This equation can also be expressed, using [itex]\omega t[/itex] as the variable, as

[itex]i_{C}(\omega t)=\omega C \dfrac{dv_{C}(\omega t)}{d(\omega t)}[/itex]​

Now my basic question is why there's a multiplication in [itex]\omega[/itex]? That is, why is it correct to multiply when changing in independent variable and multiplying in by frequency [itex]\omega[/itex]? Seemingly, the new variable, [itex]\omega t[/itex], is derived in the same manner as [itex]t[/itex] is derived when the function uses only [itex]t[/itex] as the independent variable.
I don't quite understand this little issue. Hope someone could explain that in a sentence.

Thanks in advance!

 

[tiny-tim]

Hi Adir_Sh!

Now my basic question is why there's a multiplication in [itex]\omega[/itex]?

chain rule …

dv_C/dt

= dv_C/d(ωt)*d(ωt)/dt

= dv_C/d(ωt)*ω ​

 

[rude man]

Realize that d(wt) = w dt and you have your answer.