00:01
Okay, what we want to do is we want to be able to find a curve that goes through 0 -1 and whose length is equal to the integral from 1 to 2 of the square root of 1 plus 1 over y to the 4th, d .y.
00:27
Okay, so a couple things that we need to think about is that when we've, when we set up the integral length that this 1 plus y to the fourth is actually equal to f prime of y squared.
00:48
And so that is equal to 1 over y to the fourth.
00:52
And so f prime of y is equal to 1 over y squared or plus and minus that.
01:01
But since i am looking at positive y values here and positive here, i'm just going to use the positive.
01:12
And so now i know that my function is the integral of its derivative.
01:24
So this is going to be the integral of y to the negative 2, d .y, which is actually negative to y to the negative 1 .5 .1.
01:34
First, nope, i didn't do that right.
01:43
Let's see, negative 2, i add that would be negative 1, and then that's going to be a negative.
01:51
That's just going to be a negative 1 because negative 2 plus 1.
01:55
So that would be negative plus some c value, which is negative 1 over y plus c.
02:06
So there is my function, but we do know that it goes to the point 0 .1...