Post deleted by user.

Printable View

- Aug 13th 2007, 10:34 PMmathsBMaths Problems
Post deleted by user.

- Aug 13th 2007, 10:55 PMCaptainBlack
See attachment.

Now the square of the path length is:

$\displaystyle

l^2=(1+x^2) +((1-x)^2+1)

$

Now the minimum path length corresponds to a minimum of the square path length, so we differentiate $\displaystyle l^2$ with respect to $\displaystyle x$ and set this to zero.

$\displaystyle

\frac{d}{dx}l^2=2x-2(1-x)=4x-2 =0

$

so the required $\displaystyle x=1/2$ and the length is:

$\displaystyle

l=\sqrt{5}

$

RonL - Aug 13th 2007, 11:26 PMmathsB
Post deleted by user.

- Aug 14th 2007, 03:08 AMmathsB
Post deleted by user.

- Aug 14th 2007, 04:01 AMCaptainBlack
- Aug 20th 2007, 05:15 AMCaptainBlack