Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. Arc length. If F prime is continuous on the closed and bounded interval AB, then the length of the curve Y equal F of X from the point A to the point B is just L equal the integral A to B Square root 1 + F prime of X ^2 DX. Our different notation is just A to B sqrt 1 + d YD X ^2 times DX. Also, if G prime's continuous on CD, then the length of the curve X = g of Y from A to B is L equaling C to D of the square root 1 + g prime y ^2 d Y or different notation C to D of square root 1 + d XD y ^2 d Y. So basically it's just saying that we can look at this length in terms of DX's or dy's, and sometimes one's undefined and one is defined. So we need to be careful and look to see which one makes the most sense. What we're really doing is we're looking at some smooth continuous curve and we're going to subdivide them into line segments. And what we're going to do is we're going to use triangles to find the length of each of these line segments, and then we're going to add up all of those line segment lengths. So if we look at one single arc, one single line segment on an arc, we can see our Pythagorean theorem here that says the delta Y sub K ^2 plus delta X sub K ^2 has got to equal L sub K ^2, or we could solve for L sub K. So now what we're going to do is we're going to add up all of those hypotenuse, all of those lengths of those triangles. So K equal 1 to north of L sub K is going to equal K equal 1 to north of all of those square roots of delta X sub K ^2 plus delta Y sub K ^2. Now remember that a derivative F prime of CK is just really the change of Y over the change of X. So we could think of that change of Y sub K being F prime C of K times the delta X sub K. So if we do straight substitution, now we're going to take out this delta Y sub K and we're going to put in what it equals right here. And then we're going to realize that there's a delta X sub K in both the terms that's squared. So we're going to factor it out. So if I take out this delta X sub K ^2, now the square root of squared just cancels. So we're going to put the delta X sub K down at the end. So now this is going to simplify to K equal 1, the summation K equal 1 to north square root 1 + F prime of C sub K ^2 delta X sub K by the Riemann sum, I'm going to take the limit as N goes to Infinity. So I'm going to take all of these summations of a bunch of NS. And if I do the limit as N goes to Infinity one side, we're going to do it on the other by the reem and sum. When we look at this right hand side, we can rewrite this as an integral. So we're going to go from A to B square root 1 + F prime of X ^2 DX. And we know that F prime of X we could think of as DYDX. And that's nice because sometimes that notation will help us keep in mind what we're doing. There's a DX in the denominator here, and I had DX on the outside here. There was a DY in the denominator here, and there's a DY on the outside here. So let's look at a different way to write this. A more generic formula is to think of the integral of DS where S is the length of the arc. So we could think of the integral of DS equaling the integral of the square root DX squared plus D y ^2. Basically just our Pythagorean theorem, right? So here if we thought about taking each of these terms inside the square root and dividing by DX squared. But we can't just divide something by DX squared without keeping it equivalent. So we would multiply by DX squared. So now we'd have DX over DX which is 1 + d Y over DX squared. And then the square root of a square would bring this DX on the outside. So if we started with the same thing, we could come up with the generic for the DY, the integral DS equaling the integral of the square root DX squared plus D y ^2. If we took this inside piece and divided everything through by D y ^2. Now, but if we divide by D y ^2, we've got to multiply by D y ^2. So DX over D y ^2, D y / d Y turns into 1 ^2, which is still one. And then the square root of D y ^2 is just DY, and that comes out. So these are two formulas that are in terms of DX and DY. But if we just remember this one generic one, we can come up with those formulas pretty easily. Let's look at some examples. If we let Y equal X to the three halves 0 less than or equal to X less than or equal to 4:00, we'd find that DYDX i.e. bring down the exponent to the one less power is 3 halves X 1/2. So we're going to use the formula that says the integral of sqrt 1 + d YD X ^2 times DX because we could find our DYDX easily. So we'd have zero to four. The square root 1 + 3 halves X to the half squared DX. If we simplify that up, we get sqrt 1 + 9 fourths X DX. We don't know how to find the integral of that, but we could do AU substitution. So if we did AU substitution, we would let U equal 1 + 9 fourths X. So DU is 9 fourths DX. If we go ahead and change our bounds, U of 0 is one and U of four is 10. Remember, we're just sticking in these bounds of 0 and four into this U equation. So 1 + 9 fourths times 0 is one, 1 + 9 fourths times four is 10. If we figure out that DU is 9 fourths DX, then we know that DX is 4 ninths. DU should have maybe left a little more room in there. So now we're just going to do a substitution. We're going to have instead of zero to four, it's one to 10. Instead of the DX, we're going to put in four ninths DU. And instead of sqrt 1 + 9, four six, it's just going to turn into square root of U. So now we're going to add 1 to the exponent, divide by the new exponent and put in our bounds. So we can see that it turns into 820 sevenths times the quantity 10 root 10 -, 1. If we look at another One X equal Y to the three halves over 3 -, y to the one half one less than or equal to Y less than or equal to 9. So DXDY bring down the exponent and then to the one less power. So we get 1/2 Y to the three halves minus Y 1/2 Y to the negative 1/2. So this time because we have a DXDY, we're going to use our generic equation that has the integral of square root 1 + d XD y ^2 times the DY. So from here 1 to 9 square root 1 plus this whole quantity squared DY, we actually have to foil this out. So first, outer, inner last. And what happens is the negative 1/2 to the Y 0, negative 1/4, Y to the 0 -, 1/4 Y to the zero is going to give us a negative 1/2 in there. But then if we have a +1 and a negative 1/2, that's going to give us a positive half. So we foiled this out. And then we're going to combine like terms Y to the zeros, just one. So 1 -, 1/4 -, 1/4 is a positive half. Now we're going to realize that these three terms are really basically these terms here. But the middle term instead of a -, 1/2 is now a + 1/2. So it's going to re factor. And instead of a minus here, it's going to turn into a plus. So now we have a quantity that's square, rooted and squared. Square root and square are going to cancel. So we're going to get the one to nine of 1/2 Y to the half plus 1/2 Y to the negative half DY finding the antiderivative, adding one to the exponent divided by the new exponent, adding one to the exponent divided by the new exponent. Sticking in our bounds 9 to one, we get 32 thirds units. It's units because it's a length, we're talking an arc length. So back here, this should have said units. So let's look at one more. X equal 1/3 Y to the three halves minus Y to the 1/2. That's the one we just did. Why did the three halves? OK, so let's look at X = 0 to the Y square root secant squared t -, 1 DT negative π thirds less than or equal to Y less than or equal to π fours. So I need our DXDY, and that's going to equal putting in the upper bound times the derivative of the upper bound minus the lower bound times the derivative lower bound. Remember the fundamental theorem of calculus here? So the square root of secant squared y -, 1, secant squared y -, 1 is one of our Pythagorean identities. It's equal to tangent squared Y. And the square root of tangent squared Y is really just going to be tangent Y. But we have to pay attention that on our bounds of negative π thirds to π force tangent Y wasn't always positive. So we're going to use absolute value so that then we can force that into being. The square root of a square is always going to be positive, so the length is going to equal negative π thirds to π fours of sqrt 1 plus my DX dy squared times dy. So sqrt 1 plus tangent squared Y. We know that one plus tangent squared Y is really secant squared Y, and the square root of a square is going to leave a secant Y. At this point, we actually don't know how to evaluate this, but it's a technology question. So we're just going to grab our calculators and plug it in and we're going to get approximately 2.2 units as its length. Thank you and have a wonderful day.