00:02
All right.
00:03
Let's take a look at this.
00:04
Find the arc length for y equals a square root of quantity 2 minus x squared for x being greater than or equal to zero and less than or equal to the square root of 2.
00:16
So if we just look at the function, y is the square root of 2 minus x squared.
00:23
If we were to square both sides and bringing the variables over to the other side, we'd have x squared plus y squared is 2.
00:32
So that's just a circle of radius square root of two.
00:37
But since we're using the square root function and it's a positive square root function, that's the top half of the circle, radius square root of two.
00:48
And since they're specifically interested in being greater than or equal to zero, less than or equal to square root of two, they're looking for this quarter circle.
01:00
So just this piece of the circle right here, which is a quarter of a circle.
01:05
Of radius squared of 2.
01:07
And let's just do this part ahead of time.
01:10
So we know whether we get the right answer in the end.
01:13
Oops, that's a part of a circle.
01:16
I didn't realize it would erase that.
01:20
So we have one quarter of a circle.
01:25
And the circumference of a circle is 2 pi r.
01:29
But here our radius is square root of 2.
01:33
So this quarter of the full circle circumference looks like it would be, we simplify that a little bit, square root of two over two, pot.
01:45
Now, see what we get through integration.
01:48
All right, the integration is going to be a bit messier.
01:51
Remember, substantially messier.
01:54
The arc length formula, a to b, square root.
02:02
So, there's your a and b.
02:04
Square root of one plus the derivative squared, dx.
02:10
So let's find our derivative.
02:13
Let's use a different color.
02:14
Y prime.
02:16
It's a square root.
02:17
That's a power function, one -half power.
02:20
So we'll get one -half.
02:22
It's a chain -rule problem also.
02:24
So 2 minus x squared, drop to the negative 1 -half by the chain -rule times the derivative of the inside function.
02:31
That's just a quadratic polynomial.
02:34
Derivative will be negative 2x.
02:37
And we can simplify that.
02:39
So remember negative power.
02:41
Means it's in the denominator.
02:43
I can divide off the twos.
02:45
So we get x over, well, technically negative x over, the square root of 2 minus x squared.
02:54
All right, well, we're going to square that and add one, and that will be our integran.
02:58
So the integral we're trying to solve is from zero to the square root of 2.
03:05
The square root of 1 plus this thing squared.
03:11
So 1 plus x squared, and we square a fraction of the top and square of the bottom.
03:17
2 minus x squared, all the x.
03:20
And let's clean up that integran a little bit more.
03:25
0 to a square root of 2.
03:29
1 we can write as when we common denominate 2 minus x squared over 2 minus x squared.
03:34
So we'll get a square root of 2 minus x squared plus x squared from the other one over the common denominator.
03:42
Denominator of 2 minus x squared.
03:46
And then obviously we can cancel out these x squares and split the square root over the numerator and denominator.
03:54
So these are good notation, dx.
03:57
So 0 to square root of two...