How is this equation related?

[KFS]

In Example 2, everything was fine, what I was doing matched perfectly with the book until the part "Setting this equal to zero gives *equation*". How is this? The expression on the left has no p, how can it have a p on the right? If the derivative dS/dx is set to 0 how is an x on the right, at the other side of =? I understand the rest. It's just that I don't see how one expression results in the other. That's all. Thank you.

P.S: Sorry if images are annoying, english is not my first language, so this way it's easier for me to explain and for you to understand.

 

[Dr.Peterson]

The expression you have at that point is [MATH]\frac{x}{\sqrt{1 + x^2}} + \frac{x - p}{\sqrt{(x-p)^2 + q^2}}[/MATH].

Setting that equal to zero gives the equation [MATH]\frac{x}{\sqrt{1 + x^2}} + \frac{x - p}{\sqrt{(x-p)^2 + q^2}} = 0[/MATH].

That equation quickly becomes [MATH]\frac{x}{\sqrt{1 + x^2}} = -\frac{x - p}{\sqrt{(x-p)^2 + q^2}}[/MATH].

In this case, an image is the best thing you can do.

 

[KFS]

The expression you have at that point is [MATH]\frac{x}{\sqrt{1 + x^2}} + \frac{x - p}{\sqrt{(x-p)^2 + q^2}}[/MATH].

Setting that equal to zero gives the equation [MATH]\frac{x}{\sqrt{1 + x^2}} + \frac{x - p}{\sqrt{(x-p)^2 + q^2}} = 0[/MATH].

That equation quickly becomes [MATH]\frac{x}{\sqrt{1 + x^2}} = -\frac{x - p}{\sqrt{(x-p)^2 + q^2}}[/MATH].

In this case, an image is the best thing you can do.

Thank you for the reply, but how did you do that? Solving for x? Multyplying both sides by the denominator of one of the expressions on the left?

 

[willyengland]

AFAI understand the point is that they DON'T solve for x, but use some trigonometry trick, to get the solution.
But you did not copy the next page of the book, which seems to explain it.
Probably something to the effect that both lines are perpendicular to each other ...

 

[KFS]

AFAI understand the point is that they DON'T solve for x, but use some trigonometry trick, to get the solution.
But you did not copy the next page of the book, which seems to explain it.
Probably something to the effect that both lines are perpendicular to each other ...

Here's the next page if that's of some help. But what's the process to equal both expressions?

 

[HallsofIvy]

Thank you for the reply, but how did you do that? Solving for x? Multyplying both sides by the denominator of one of the expressions on the left?

What, exactly, are you trying to do? Your original question was just "how to go from \(\displaystyle \frac{x}{1+ x^2}+ \frac{x- p}{\sqrt{(x- p)^2+ q^2}}\) to \(\displaystyle \frac{x}{1+ x^2}= -\frac{x- p}{\sqrt{(x- p)^2+ q^2}}\) and Dr. Peterson explained that: set the expression equal to 0 and subtract \(\displaystyle \frac{x- p}{\sqrt{(x- p)^2+ q^2}}\) from both sides. Now you appear to be asking how to solve for x. There is no easy way to do that and the paper you link tells you not to do that!

 

[willyengland]

But what's the process to equal both expressions?

It is used so that you see that each side represents a Sinus.
And this leads you to the law of reflection.

But i don't see how this helps to find x.

 

[Dr.Peterson]

Although they don't carry it out, you could finish solving for x by writing the equation of the line from (0,-1) to (p,q), and determining its x-intercept. Apparently they are satisfied just describing the location geometrically. (That would not be sufficient if you were trying to actually solve the problem as stated; they are just using it as something to discuss, I guess.)

 

[KFS]

Thanks everyone for the answers.

 

[willyengland]

you could finish solving for x by writing the equation of the line from (0,-1) to (p,q), and determining its x-intercept.

Ha, ha, that's cool!