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Simple minimum/maximum problem

  1. Jul 13, 2007 #1
    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.
  2. jcsd
  3. Jul 13, 2007 #2

    D H

    Staff: Mentor

    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].
  4. Jul 13, 2007 #3


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  5. Jul 17, 2007 #4
    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: Jul 17, 2007
  6. Jul 17, 2007 #5


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    Ok, so a=R*cos(alpha) and b=-R*sin(alpha). Square and add both of those equations and use everybodies favorite trig identity.
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