Proof that the shortest distance between two point is a straight line

 Proving that the shortest distance between two points is indeed a straight line: . Now to Prove: (well known equation for a straight line in Cartesian coordinates) 1) Distance between two points is "L", which in turn is the following integral: 2) Which does not help much, so lets expand this into an integral to get something more interesting. 3) We also use Pythagoras theorem with the above: 4) Then we put 2) into 3), it can be seen that by expanding the right side and squaring the left we get 3) back: 5) Now we want a y(x) that causes "L" to be a minimum: x = independent variable y = dependant variable y(x) is curve that causes J to be extremum, so look for conditions of f(x,y,y') to find y(x) 6) Now, with y and y' as functions of a and x, we will vary the path with some restrictions (7)  7) Here are the restrictions: 8) For all this to be useful, we have to take the derivative of "J" with respect to "a" and set it to "zero" to get the minimum, which in turn will be the equation of a (hopefully) straight line. 9) This is the equation which we will take the derivative of, it is the same as 5), but with "a's" instead of "x's", which represent the varied path: 10) This is what the derivative looks like: 11) Some parts can be simplified: 12) Now to solve the integral, we will use "u" substitution as follows: 13) Since there is an addition inside the integral, it can be split into two parts for ease of use: 14) Which in turn becomes this mess.  But, the first half (df/dy part) goes to zero because the beginning part is the same as the end part: 15) The result of all this is the following, everything together again: 16) As said earlier, we set it to zero, the n(x) can be taken out of the brackets.  And the definite integral becomes a general integral: 17) This is turn comes out to be the following for the extremum.  Which is Euler's equation: 18) Now, length is the following: 19) And when this length is put into equation 17) we get the following: 20) And with further simplification: 21) And more: 22) And more: 23) And finally we get this: 24) And from very trivial differential equations, the following comes out: 25) And substituting some letters for others it gets the familiar form!: 1 800 295 5693

Patrick Blochle - Canisius College, Buffalo, NY