00:01
We are given a curve with a specified sector as its domain, and we are asked to find the length of this curve.
00:12
The curve is r equals theta squared, and the sector is theta's from 0 to 2 pi.
00:22
If we treat r as a function of theta, say f of theta, then we have that f prime of theta, which is drd theta, is equal to 2 theta.
00:40
And theta is continuous, so it follows that f prime is continuous.
00:47
Therefore, the length of this curve is well defined, and is given by the formula, l equals the integral from 0 to 2 pi of square root of r squared, plus d -theta squared, which is equal to, well, d theta, the integral from 0 to 2 pi of square root of theta to the fourth plus 4 theta squared d -theta, which is equal to factoring out of theta squared, we get integral from 0 to 2 pi of theta -squared of theta -squared, square -rooted is the absolute value of theta, times the square root of theta squared plus 4 d -theta.
02:34
And we know that since theta lies between 0 and 2 pi, it's positive.
02:40
So the absolute value of theta is simply theta.
02:51
In order to find an anti -derivative for this, we'll use the substitution rule...