00:01
Hi there, so for this problem, we are asked to find the exit length of the curve.
00:08
And the curve that we are given is when x is equal to the parameter t times the sign of t.
00:19
This and y is defined similar, but with a cosine function.
00:25
And the parameter t is between 0 and 1.
00:35
So with that said, we know that the length, the length of the curve, let's call that quantity l is the integral, well, as you have it defined it in there, is the integral from, in this case, from zero to one, of the derivative of the x function with respect to the parameter t and all of this to the square.
01:09
And this plus the derivative of the y component with respect to the parameter t and that also to the square this integrated over the parameter t.
01:22
So, well, you already obtained that, well, the derivative of x with respect to time, so since it is given in, it is the product between the parameter t and the sign of t, that is the derivative of a product.
01:37
So the first, we can derive first the parameter t live in the sign without a divide.
01:44
So that will be just simply the sign of t.
01:48
Now for the second in here, we will have t without a divide, the derivative of the sign function, which is the cosine function of t.
01:57
Now, for the derivative of y, we will have something similar, but in this case is the cosine function that we have in here.
02:06
So we will have the cosine of t and this minus, because the derivative of the cosine function is the minus sign function, the parameter t times the sine of t.
02:17
Now, once we set this, we just need to simply substitute this into the equation for the length.
02:25
So we will have the integral from 0 to 1 of, well, this, all of this to the square.
02:34
So we start with the derivative of x squared respect to the parameter t.
02:37
So there will be the sign of t plus t times the cosine of t and that due to the square.
02:49
This plus the cosine of t minus t, sine of t, sine of t, of t and that to the square and this integrated over the parameter t now let's just do the let's span this quadratic expression in here so for the first one in here we will have the sign square of the parameter t this plus two times the sign of t times the cosine of t, which is the crass product in that quadratic expression.
03:34
And for the last one, we will have the parameter t squared times the cosine square of the parameter t.
03:42
And then this plus for the other quadratic expression here, we will have the cosine square of the parameter t.
03:53
This minus two times.
03:55
Oh, sorry, i forgot in here also.
03:57
The parameter t, two times the parameter t.
04:02
Okay, so in here it is minus two times the parameter t times the cosine of t times the sign of t.
04:10
And finally, we will have plus parameter t squared times the sign of t that to the square as well, and then this integrated over the parameter t.
04:20
Now, as you can see, we can simplify this because we can cancel this whole term with this term in here.
04:29
So finally, let me just write it in this way, so you can see what we are going to do.
04:35
So we will have the sign square of the parameter t, this plus the cosine square of the parameter t.
04:46
You must recall that this is a trigonometric identity, and that is just simply equals to one.
04:54
And for the other term, we can take out the parameter t squared times and then this times the cosine square of the time.
05:08
This plus the sign square of the parameter t.
05:14
So as you can see, again, in here we will have another one.
05:19
So this.
05:22
Oh, sorry, i forgot.
05:24
I forgot something in here.
05:27
Oh, yes, i forgot that all of this is inside a square root.
05:32
So let me put it in here.
05:33
All of this is elevated to 1 divided by 2.
05:36
It's the same as just put in the square root.
05:39
All of this elevated to 1 divided by 2 and all of this is 1 divided by 2.
05:44
So lastly, we are going to have that the integral reduces to just simply the square root of 1 plus the time square.
05:57
From here on is where this problem gets a little tricky.
06:07
This integral is not that easy to solve, but because we need first to do some changes, some trigonometric change.
06:23
So first we are going to set that the parameter t is equal to the tangine of t -tem.
06:30
So what we are going to do is a trigonometry change, and with that later on, we just interchange what we obtain with this definition.
06:46
Okay.
06:47
So the differential in the parameter t is going to be then the derivative of the tangent function, which we know is the second s -square.
06:59
Of teta, this times the differential in teta.
07:03
So let's substitute that in here.
07:07
I'm going to leave this without the limits for now, and later on we just again come back to the parameter t, and then we can just evaluate this integral in that interval.
07:22
So, okay, now we will have the integral of the square root of 1 plus the tangent of theta square, this times the integral, the differential in the parameter t, which we set is the second square of theta, times the differential in tita.
07:46
Now there is a trigonometry property where one plus the tangent square of teta is just simply the second square of teta...