00:01
Okay, so you have the general idea of this down.
00:07
So what's happening here is you have y equals x squared over two.
00:16
So, d y, d x is x squared over two.
00:28
You're taking the derivative of it.
00:30
So that is one half times 2x which is x.
00:38
So d .y over dx squared is x squared.
00:47
And then you need to find the length.
00:53
You're going from negative 1 to 1 of the square root of 1 plus d .y over d x squared.
01:04
So this is the integral from negative 1 to 1.
01:07
Of the square root of 1 plus x squared which it looks like you have and then yes you let x be tangents of theta so then dx is sicken squared of theta d theta there's a dx here and a dx there i just dropped it by accident so then you're finding the integral you're doing a substitution, so you have to be careful here if we are going to do that substitution, because it will change the limits of integration.
02:17
So we'll just need to come back and fix these limits later.
02:22
So this is the square root of 1 plus tangent squared of theta, which is secant to theta.
02:34
So you have the integral of sikin cubed of theta, d theta.
02:50
And then from here, you can split it up.
02:54
It is secant times secant squared.
03:05
So this is equal to secant.
03:10
Of theta, tangent of theta minus the integral of secant of theta.
03:26
Yes, so i'm looking at yours...