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A perturbation in the temperature of a stream leaving a chemical reactor follows a decaying sinusoidal variation, according to the following mathematical equation,
T(t)=5e^(-at ).sin⁡(wt)
where a and w are constant.
Then show that the temperature T (t) is maximum at time “t”. Also verify that t= 1/w.〖tan〗^(-1) (w/a)
Determine the maximum temperature value at t= 1/w.〖tan〗^(-1) (w/a).

Respuesta :

Derivating the function of the temperature and equating to 0, we can find the critical points:

[tex]T'(t)=-a\cdot5e^{-at}\cdot\sin(wt)+w\cos(wt)\cdot5e^{-at}\\\\ 0=5e^{-at}(-a\sin(wt)+w\cos(wt))[/tex]

As [tex]5e^{-at}\neq0[/tex]:

[tex]0=-a\sin(wt)+w\cos(wt)\iff a\sin(wt)=w\cos(wt)\iff \\\\\sin(wt)=\dfrac{w}{a}\cos(wt)\iff \tan(wt)=\dfrac{w}{a}\iff wt=\tan^{-1}\left(\dfrac{w}{a}\right)\iff \\\\\boxed{t=\dfrac{1}{w}\tan^{-1}\left(\dfrac{w}{a}\right)}[/tex]

Replacing:

[tex]T(t)=5e^{-at}\cdot\sin(wt)=5e^{-\frac{a}{w}\tan^{-1}\left(\frac{w}{a}\right)}\cdot\sin(\tan^{-1}\left(\dfrac{w}{a}\right))[/tex]

We can reach: [tex]\sin(\tan^{-1}\left(\dfrac{w}{a}\right))=\dfrac{w}{\sqrt{w^2+a^2}}[/tex]

Hence:

[tex]T(t)=\dfrac{5w}{\sqrt{w^2+a^2}}e^{-\frac{a}{w}\tan^{-1}\left(\frac{w}{a}\right)}[/tex]
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