00:01
In this question, we are asked to calculate the arc length of the given curve over the given interval.
00:06
And let the first coordinate be x, the expression in front of i be equal to x.
00:17
We'll call the expression in front of j by y and the expression in front of k by z.
00:24
Then also let a be the left end point and b be the right end point of the interval.
00:32
Arc length equals to the integral from a to b of the square root of x prime squared plus y prime squared plus z prime squared dt.
00:56
And in our case, this equals to the interval from 0 to 5 of the square root.
01:06
Now x prime equals to 7, y prime equals to 6.
01:18
And z prime equals to 2t this equals to the interval from 0 to 5 or the square root of 49 plus 36 plus 4 t squared now 49 plus 36 equals to 85 right so we're going to get the interval from 0 to 5 or the square root of 85 plus 85 plus 4 t squared now i want to use the formula on the previous page at the bottom we want to use this formula and to do that we need to want the coefficient in front of t squared to be equal to 1 so we're going to factor out 4 to get the interval from 0 to 5 of 2 multiplied by the square root we are going to get 2 because when you factor out 4 out of the square root the square root of 4 equals to 2 and under the square root, we are going to get 85 over 4 plus t squared.
02:57
Now this looks like the integral from the formula, where a squared equals to 85 over 4.
03:07
This is a squared.
03:11
Therefore, according to the formula, we are going to get 2 multiplied by 1 .5t times square root of t squared plus a squared.
03:30
T squared plus 85 over 4 plus 1 half a squared times ln so plus 1 half times a squared and a squared equals to 85 over 4 multiplied by ln of t plus square root of t squared plus a squared t squared plus a squared t squared plus 85 over 4 and all this in substitution from 0 to 5 first of all we can cancel 2.
04:26
We can distribute 2 over parenthesis to cancel 2...