Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Parametric equations 2. Functions that have continuous derivatives on a closed and bounded interval AB that are not simultaneous zero are said to be continuously differentiable and the curve defined by them as smooth. So if X = F of T&Y equal G of T on a less than or equal to T less than or equal to B, we had the chain rule that says dy DT would just equal dy DX times DX DT. Or we could think of dy DX as being dy DT divided by DX DT. If we wanted to take the 2nd derivative, we take the derivative of the derivative. So notation wise it would be d ^2 y over the quantity of DX squared. But what we're really doing is we're taking the derivative of the first derivative, so DDX of Y prime. Now remember a minute ago we had DDX of Y equaling DY over DT divided by DX over DT. So if we replace this Y here with Y prime, we can see that we would then get DY prime DT over DX DT. So the second derivative is just the first derivative divided by that DX DT tangent to a curve. An example tangent to a curve. If we have X equals sine T&Y equal cosine T at some location, let's call it T equal π thirds DX DT is cosine T and dy DT is negative sine T. So at T equal π thirds we can see dy DX is just negative sine T divided by cosine T, which we know is negative tangent T. So the slope at π thirds is dy DX equaling negative tangent of π thirds or negative root 3. The point would be cosine or sine of π thirds which is root 3 / 2, cosine of π thirds which is 1/2. So the tangent to the curve at time equal π thirds is y -, 1/2 equaling negative root 3. The quantity X minus root 3 / 2. If we ask for the 2nd derivative, we would come back here and we would take the derivative of the derivative. Negative tangent T would be negative secant squared T, but then that has to get divided by the DX DT again. So divided by cosine T or negative secant cubed T If we wanted to find the length of a curve. What we're going to do is we're going to think about adding up a bunch of line segments. So if we look at this, we can see that L of X by the Pythagorean theorem is really just the square root of delta X sub K ^2 plus delta Y sub K ^2. Somewhere in here there's a point FTK 1 GTK 1 up to FTKGTK. So we know by the mean value there. Or we could put in the distance formula. So we could actually subtract the two points and square them and subtract the two points and square them at them together. By the mean value theorem, there must exist some TK star and some TK ** between K1 and K. So basically what we're saying is there's got to be some value where the change of the axis just the difference of the T sub K's and T sub K -, 1, which is F prime of the TK star delta TK delta YK. The same concept where we're letting the X be defined by the F function and the Y be defined by the G function. So if we assume the path is only traveled once as T increases from A to B, then the summation of K equal 1 to north of L sub K is just all of those line segments added up. So K equal 1 to north summation of square root. That's our Pythagorean theorem. Instead of delta X sub K, we're going to have it be F prime TK, star delta TK, and G prime TK double star delta TK. Now what we're eventually doing is we're going to build this to a Riemann sum. So we realized that there's a delta T sub K in both of them, and we could factor that out. And when we have a square root of a square, that gives us just this delta TK at the end. Now we have infinitely many of these. So we're going to figure out what happens when we take the limit as N goes to Infinity when we're adding up all of these. So then we get the summation K equal 1 to square root F prime TK star squared plus G prime TK double prime squared delta TK. Well, that's really by our Riemann sum, just an integral from A to B of the square root of F prime t ^2 + g prime t ^2 DT. So the arc length, the generic formula. If we think about the integral of DS, that's really just the integral of the square root DX squared plus dy squared. We could think about taking each of these terms and dividing by a DT squared. But if we divide by DT squared, we also have to multiply by DT squared. So we get the integral square root DX DT squared plus dy DT squared DT. Or we could think about taking that and dividing everything through by a DX squared, and if we divide by DX squared, we have to multiply by DX squared so we get the integral square root 1 + d YD X ^2 DX. Or we could do the same thing, dividing everything through by a D y ^2, which means we have to also multiply by D y ^2, and that would give us the integral square root DXD y ^2 + 1 DY. So having all three formulas is very nice. But if we just knew this top one and understood how to divide by DT squareds, DX squareds, or DY squareds, we'd come up with all three of the equivalent equations for arc lengths. Example, for circumference of a circle, This is not going to be a unit circle. If we look at a triangle on a circle, we can see that we've gone over X and up Y with a radius of R, so XY. The point could actually be thought of as R cosine Theta and R sine Theta because this is a right triangle here. So we know that sine Theta could be thought of as y / r and cosine Theta could be thought of as X / r. If we multiply across, we get Y equaling R sine Theta and X equaling R cosine Theta. So then if we want to take dyd Theta, we'd get R cosine Theta. R is just some constant. If we want to take DXD Theta, we'd get negative R sine Theta. So now if we're talking about the length, we're going from zero to 2π of the square root DXD Theta squared plus dyd Theta squared D Theta. So zero to 2π, the square root negative R sine Theta squared plus R cosine Theta squared D Theta. They each have an R-squared in them, so the square root of the R-squared is going to leave us R. We also get sine squared Theta plus cosine squared Theta which we know is one and sqrt 1 is just one. Now remember R is just a constant so it can come in and out of the integral. So we have R0 to 2π of D Theta. We know that the integral of D Theta is just Theta. We put in our upper bound 2π minus our lower bound 0 for our Theta. So we get 2π R for the arc length, which is really the circumference of a circle. In this case, because we went from zero to 2π. It's going to be units because it's a length, the area of surface of revolution. For parameterized curves. If a smooth curve X equal F of Ty equal G of TA less than or equal to T less than or equal to B is traversed exactly once as T increases from A to B, then the areas of the surfaces generated by revolving the curve about the coordinate axis are as follows. Revolution about the X axis 2π RH. Well, the H is just the length of the arc that we found a moment ago. If we're going about the X axis 2π the radius times the height again. Once again, the height is just what we found as far as the arc length goes. Thank you and have a wonderful day.