# Simple minimum/maximum problem

1. Jul 13, 2007

### John O' Meara

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 $$E^2 = ( e cos \theta + a)^2 + ( e sin\theta +b)^2\\$$, where a and b are constants.
Expand the right-hand side of this equation and by expressing $$a cos\theta + b sin\theta \\$$ in the form $$R cos(\theta + \alpha) \\$$ show that the maximum and minimum values of R as $$\theta$$ varies , are e +/- $$\sqrt{a^2 +b^2} \\$$. ( 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: $$E^2= e^2 cos^2\theta +a^2 +2ae cos\theta + e^2 sin^2\theta + 2eb sin\theta +b^2 \\$$, which gives $$E^2 = e^2 +2e(a cos\theta + b sin\theta ) +a^2 + b^2 \\$$, which can be expressed as
$$e^2 + 2e(a cos\theta + b cos(\frac{\pi}{2} - \theta) ) + a^2 + b^2$$. The thing is I do not know how to express R in terms of the coefficients a and b? Thanks for the help.

2. Jul 13, 2007

### D H

Staff Emeritus
You didn't do what you were asked to do.

You rewrote $\sin\theta$ as $\cos(\pi/2-\theta)$. That is not what the problem says to do. The problem specifically says to write $a\cos\theta+b\sin\theta$ in the form $R\cos(\theta+\alpha)[/tex]. Simply expand this latter form, equate to [itex]a\cos\theta+b\sin\theta$, and solve for $R$ and $\alpha$.

3. Jul 13, 2007

### Dick

4. Jul 17, 2007

### John O' Meara

Expanding R $$cos(\theta + \alpha) \\$$ we get$$R( \cos\theta \sin\alpha - \sin\theta \sin\alpha) \\$$. Now $$R \cos\theta\cos\alpha - R \sin\theta\sin\alpha = a \cos\theta + b \sin\theta \\$$ When $$\theta = 0 R = \frac{a}{\cos\alpha} \\$$ When $$\theta = \frac{\pi}{2} R = \frac{-b}{\sin\alpha}\\$$ Therefore $$\tan\alpha = \frac{-b}{a}$$. I would appreciate help in getting R.

Last edited: Jul 17, 2007
5. Jul 17, 2007

### Dick

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