Derivatives and shortest length

[mscbuck]

Homework Statement

A straight line is drawn from the point (0,a) to horizontal axis, and then back to (1,b). Prove that the total length is shortest when the angles $$\alpha$$ and $$\beta$$ are the same.

2. Homework Equations /graphs

[PLAIN]http://dl.dropbox.com/u/23215/Graph.jpg

The Attempt at a Solution

Hey all, just having some trouble with this question. I thought about attacking it first by finding a maximum/minimum area of the triangle, since we probably can't assume at this point (other than the 90 degree angle with the y-axis) that $$\alpha$$ or $$\beta$$ are 30/60/90.) Is that a correct step to take, or should I start concerning myself with the angles right away?

Any help is appreciated! Thanks again

 

[micromass]

I wouldn't immediately bother about the angles.

Lets first try to see what the total length could be, given a point (x,0)

The length from (0,a) to (x,0) is $$\sqrt{a^2+x^2}$$.
The length from (x,0) to (1,b) is $$\sqrt{(x-1)^2+b^2}$$

So the total length is $$\sqrt{a^2+x^2}+\sqrt{(x-1)^2+b^2}$$.
This gives a function
$$f(x)=\sqrt{a^2+x^2}+\sqrt{(x-1)^2+b^2}$$.

The goal is now to find the point x in which f reaches a minimum.

 

[mscbuck]

Hi micromass,

I actually had gotten up to that point, I probably should've written some of it down, but I ignored it because I kept getting led nowhere. But perhaps it's because I was using the information wrong.

I would assume since we are finding a minimum that I'd like to find any critical points of that function where f'(x) = 0, but I keep getting thrown because I keep finding my only critical point is at x = 0 and I'm not sure what I"m supposed to do with it (or if that's even right)

Thank you for your help!

 

[micromass]

Can you show me your work in arriving to x=0? Maybe I can find where it went wrong...

 

[mscbuck]

Here is what I have:

$$f(x)=\sqrt{a^2+x^2}+\sqrt{(x-1)^2+b^2}$$

$$f'(x) = x/(\sqrt{a^2+x^2}) + (x-1)/(\sqrt{(x-1)^2+b^2} = 0$$

Then I squared both sides to get rid of any square roots if need be, but from there I'm kind of stuck deciding what algebra to use. I found my mistake with x=0, so I'm still looking for others

 

[micromass]

That derivative seems good. So we need to have f'(x)=0.
This is equivalent to

$$x/\sqrt{a^2+x^2} = (1-x)/\sqrt{(x-1)^2+b^2}$$

Which is equivalent to

$$x\sqrt{(x-1)^2+b^2} = (1-x)\sqrt{a^2+x^2}$$

Now you simply need to square and solve for x.

 

[mscbuck]

It appears from that that I have received x = (a/b) + 1 as my final answer? Does this appear to be correct?