- #1
John O' Meara
- 330
- 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.
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.