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:

<br /> f(x)=\sqrt{a^2+x^2}+\sqrt{(x-1)^2+b^2}<br />

<br /> f&#039;(x) = x/(\sqrt{a^2+x^2}) + (x-1)/(\sqrt{(x-1)^2+b^2} = 0<br />

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?