Simple minimum/maximum problem

  • Thread starter Thread starter John O' Meara
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 4K views
John O' Meara
Messages
325
Reaction score
0
The voltage E applied to the sending end of a high-pressure transmitting line is connected to the voltage e at the receiving end by the equation [tex]E^2 = ( e cos \theta + a)^2 + ( e sin\theta +b)^2\\[/tex], where a and b are constants.
Expand the right-hand side of this equation and by expressing [tex]a cos\theta + b sin\theta \\[/tex] in the form [tex]R cos(\theta + \alpha) \\[/tex] show that the maximum and minimum values of R as [tex]\theta[/tex] varies , are e +/- [tex]\sqrt{a^2 +b^2} \\[/tex]. ( obtaining these results by differentation is a much more difficult method and we would remind our readers that the calculus is not always the best way of solving a problem.
Expanding this I get: [tex]E^2= e^2 cos^2\theta +a^2 +2ae cos\theta + e^2 sin^2\theta + 2eb sin\theta +b^2 \\[/tex], which gives [tex]E^2 = e^2 +2e(a cos\theta + b sin\theta ) +a^2 + b^2 \\[/tex], which can be expressed as
[tex]e^2 + 2e(a cos\theta + b cos(\frac{\pi}{2} - \theta) ) + a^2 + b^2[/tex]. The thing is I do not know how to express R in terms of the coefficients a and b? Thanks for the help.
 
Physics news on Phys.org
You didn't do what you were asked to do.

You rewrote [itex]\sin\theta[/itex] as [itex]\cos(\pi/2-\theta)[/itex]. That is not what the problem says to do. The problem specifically says to write [itex]a\cos\theta+b\sin\theta[/itex] in the form [itex]R\cos(\theta+\alpha)[/tex]. Simply expand this latter form, equate to [itex]a\cos\theta+b\sin\theta[/itex], and solve for [itex]R[/itex] and [itex]\alpha[/itex].[/itex]
 
Expanding R [tex]cos(\theta + \alpha) \\[/tex] we get[tex]R( \cos\theta \sin\alpha - \sin\theta \sin\alpha) \\[/tex]. Now [tex]R \cos\theta\cos\alpha - R \sin\theta\sin\alpha = a \cos\theta + b \sin\theta \\[/tex] When [tex]\theta = 0 R = \frac{a}{\cos\alpha} \\[/tex] When [tex]\theta = \frac{\pi}{2} R = \frac{-b}{\sin\alpha}\\[/tex] Therefore [tex]\tan\alpha = \frac{-b}{a}[/tex]. I would appreciate help in getting R.
 
Last edited: