[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.53,0:00:05.62,Default,,0000,0000,0000,,Some relationships between two\Nquantities or variables are so
Dialogue: 0,0:00:05.62,0:00:09.70,Default,,0000,0000,0000,,complicated that we sometimes\Nintroduce a third variable or
Dialogue: 0,0:00:09.70,0:00:11.97,Default,,0000,0000,0000,,quantity to make things easier.
Dialogue: 0,0:00:12.59,0:00:17.46,Default,,0000,0000,0000,,In mathematics, this third\Nquantity is called a parameter,
Dialogue: 0,0:00:17.46,0:00:22.87,Default,,0000,0000,0000,,and instead of having one\Nequation, say relating X&Y, we
Dialogue: 0,0:00:22.87,0:00:28.82,Default,,0000,0000,0000,,have two equations, one relating\Nthe parameter with X and one
Dialogue: 0,0:00:28.82,0:00:31.52,Default,,0000,0000,0000,,relating the parameter with Y.
Dialogue: 0,0:00:32.25,0:00:34.08,Default,,0000,0000,0000,,Let's have a look at an example.
Dialogue: 0,0:00:35.62,0:00:39.12,Default,,0000,0000,0000,,X equals Cos T.
Dialogue: 0,0:00:39.83,0:00:43.83,Default,,0000,0000,0000,,And why it was
Dialogue: 0,0:00:43.83,0:00:48.16,Default,,0000,0000,0000,,Scienti? It's our parametric\Nequations we have.
Dialogue: 0,0:00:49.06,0:00:49.80,Default,,0000,0000,0000,,X.
Dialogue: 0,0:00:50.94,0:00:56.96,Default,,0000,0000,0000,,And T the relationship in one\Nequation and Y&T related in the
Dialogue: 0,0:00:56.96,0:01:01.95,Default,,0000,0000,0000,,other equation. Let's have a\Nlook at what the graph looks
Dialogue: 0,0:01:01.95,0:01:06.13,Default,,0000,0000,0000,,like and to do that, we\Nsubstitute some values for T
Dialogue: 0,0:01:06.13,0:01:09.93,Default,,0000,0000,0000,,into both the equations and we\Nworkout values for X&Y.
Dialogue: 0,0:01:11.35,0:01:15.09,Default,,0000,0000,0000,,Let's take some values of tea.
Dialogue: 0,0:01:15.09,0:01:18.13,Default,,0000,0000,0000,,Calculate X&Y.
Dialogue: 0,0:01:20.80,0:01:27.41,Default,,0000,0000,0000,,And will take\Nsome values Zero
Dialogue: 0,0:01:27.41,0:01:32.89,Default,,0000,0000,0000,,Pi 2π? 35 by\N2 and 2π.
Dialogue: 0,0:01:34.26,0:01:35.81,Default,,0000,0000,0000,,And to make it a little bit
Dialogue: 0,0:01:35.81,0:01:39.37,Default,,0000,0000,0000,,easier. Well, draw the curves.
Dialogue: 0,0:01:41.57,0:01:46.70,Default,,0000,0000,0000,,Of cause T and sign\NTA little bit more.
Dialogue: 0,0:01:49.31,0:01:53.57,Default,,0000,0000,0000,,Call\NZetty.
Dialogue: 0,0:01:54.64,0:01:56.79,Default,,0000,0000,0000,,He just put some labels on.
Dialogue: 0,0:02:12.05,0:02:13.18,Default,,0000,0000,0000,,Now we have one.
Dialogue: 0,0:02:13.75,0:02:18.41,Default,,0000,0000,0000,,Minus one. So that's\Nour graph of Costi.
Dialogue: 0,0:02:20.03,0:02:21.41,Default,,0000,0000,0000,,And Scienti.
Dialogue: 0,0:02:42.75,0:02:43.49,Default,,0000,0000,0000,,And.
Dialogue: 0,0:02:44.73,0:02:48.95,Default,,0000,0000,0000,,This
Dialogue: 0,0:02:48.95,0:02:57.59,Default,,0000,0000,0000,,scientist.\NOK, so when T is 0.
Dialogue: 0,0:02:58.58,0:03:00.86,Default,,0000,0000,0000,,X is cause T.
Dialogue: 0,0:03:01.51,0:03:02.58,Default,,0000,0000,0000,,And that's one.
Dialogue: 0,0:03:03.14,0:03:08.82,Default,,0000,0000,0000,,When she is, OY is scienti\Nand that zero.
Dialogue: 0,0:03:10.91,0:03:16.65,Default,,0000,0000,0000,,20 is π by two, X is the cause\Nof Π by two, which is 0.
Dialogue: 0,0:03:17.87,0:03:23.29,Default,,0000,0000,0000,,20 is π by two, Y is the sign of\NΠ by two, which is one.
Dialogue: 0,0:03:25.26,0:03:29.25,Default,,0000,0000,0000,,20 is π the cause\Nof Pi is minus one?
Dialogue: 0,0:03:30.39,0:03:31.77,Default,,0000,0000,0000,,Anne for why?
Dialogue: 0,0:03:32.57,0:03:35.67,Default,,0000,0000,0000,,The sign of Π Zero.
Dialogue: 0,0:03:37.27,0:03:42.84,Default,,0000,0000,0000,,20 is 3 Pi by two the cause of\Nthree Pi by two is 0.
Dialogue: 0,0:03:43.77,0:03:46.92,Default,,0000,0000,0000,,And the sign of three\NPi by two is minus one.
Dialogue: 0,0:03:47.97,0:03:55.22,Default,,0000,0000,0000,,And at 2π. Twenty is 2π the\Ncause of 2π is one and the sign
Dialogue: 0,0:03:55.22,0:03:57.15,Default,,0000,0000,0000,,of 2π is 0.
Dialogue: 0,0:03:57.97,0:04:02.49,Default,,0000,0000,0000,,So we now have X&Y coordinates\Nthat we can plot.
Dialogue: 0,0:04:03.15,0:04:04.80,Default,,0000,0000,0000,,To show the curve.
Dialogue: 0,0:04:06.49,0:04:11.00,Default,,0000,0000,0000,,How\Nabout
Dialogue: 0,0:04:11.00,0:04:13.26,Default,,0000,0000,0000,,X&Y?
Dialogue: 0,0:04:14.90,0:04:22.04,Default,,0000,0000,0000,,Once he was zero,\NX is one, Y
Dialogue: 0,0:04:22.04,0:04:25.60,Default,,0000,0000,0000,,is 0, so 10.
Dialogue: 0,0:04:26.19,0:04:27.50,Default,,0000,0000,0000,,01
Dialogue: 0,0:04:28.75,0:04:30.47,Default,,0000,0000,0000,,Minus 10.
Dialogue: 0,0:04:31.63,0:04:33.17,Default,,0000,0000,0000,,0 - 1.
Dialogue: 0,0:04:34.29,0:04:36.10,Default,,0000,0000,0000,,And back again to 10.
Dialogue: 0,0:04:37.38,0:04:41.29,Default,,0000,0000,0000,,Now, with those points, we've\Nnot actually plotted enough to
Dialogue: 0,0:04:41.29,0:04:45.98,Default,,0000,0000,0000,,be able to see what's happening\Nin between these points, but if
Dialogue: 0,0:04:45.98,0:04:51.85,Default,,0000,0000,0000,,we were to take values for T\Nbetween 0 and Π by two and some
Dialogue: 0,0:04:51.85,0:04:56.93,Default,,0000,0000,0000,,more between pie by two and Π\Nand so on, what we'd actually
Dialogue: 0,0:04:56.93,0:04:59.28,Default,,0000,0000,0000,,find is that these are the
Dialogue: 0,0:04:59.28,0:05:03.87,Default,,0000,0000,0000,,parametric equations. That\Ndescribe a circle.
Dialogue: 0,0:05:04.77,0:05:11.30,Default,,0000,0000,0000,,Sensor. 00 and with\Na radius of 1.
Dialogue: 0,0:05:13.26,0:05:18.36,Default,,0000,0000,0000,,Now what we often want to find\Nout is how to variables are
Dialogue: 0,0:05:18.36,0:05:22.67,Default,,0000,0000,0000,,changing in relationship to each\Nother. So when exchange is, how
Dialogue: 0,0:05:22.67,0:05:28.16,Default,,0000,0000,0000,,is why changing what's the rate\Nof change? So we need to be able
Dialogue: 0,0:05:28.16,0:05:32.07,Default,,0000,0000,0000,,to differentiate. Now what we\Ndon't want to do is to actually.
Dialogue: 0,0:05:32.94,0:05:35.66,Default,,0000,0000,0000,,Eliminate the parameter.
Dialogue: 0,0:05:36.20,0:05:39.56,Default,,0000,0000,0000,,And get back to an equation\Ndirectly relating X&Y, 'cause
Dialogue: 0,0:05:39.56,0:05:43.93,Default,,0000,0000,0000,,the whole point of having it's a\Nparameter is that it makes it
Dialogue: 0,0:05:43.93,0:05:48.97,Default,,0000,0000,0000,,easier for us and simpler, So\Nwhat we need to do is to find a
Dialogue: 0,0:05:48.97,0:05:52.66,Default,,0000,0000,0000,,way of differentiating when we\Ngot them in the parametric form.
Dialogue: 0,0:05:53.46,0:05:56.03,Default,,0000,0000,0000,,And that's what we do.
Dialogue: 0,0:05:56.09,0:06:01.94,Default,,0000,0000,0000,,Thanks, right, the two equations\Nagain X equals Cos T.
Dialogue: 0,0:06:02.45,0:06:08.43,Default,,0000,0000,0000,,Y equals sign T. What we're\Ngoing to do to differentiate?
Dialogue: 0,0:06:09.33,0:06:12.91,Default,,0000,0000,0000,,Is to differentiate each\Nequation with respect to
Dialogue: 0,0:06:12.91,0:06:14.26,Default,,0000,0000,0000,,the parameter T.
Dialogue: 0,0:06:15.84,0:06:23.58,Default,,0000,0000,0000,,So the X5 DT, the derivative\Nof Cos T is minus sign
Dialogue: 0,0:06:23.58,0:06:29.79,Default,,0000,0000,0000,,T. 4 divided by DT.\NThe derivative of Scienti.
Dialogue: 0,0:06:30.29,0:06:31.52,Default,,0000,0000,0000,,Is cause T?
Dialogue: 0,0:06:32.29,0:06:35.62,Default,,0000,0000,0000,,Now using the chain rule.
Dialogue: 0,0:06:36.86,0:06:43.91,Default,,0000,0000,0000,,Which says that DY by\Nthe T is equal to DY
Dialogue: 0,0:06:43.91,0:06:45.19,Default,,0000,0000,0000,,by DX.
Dialogue: 0,0:06:46.23,0:06:49.90,Default,,0000,0000,0000,,Times by DX by DT.
Dialogue: 0,0:06:51.07,0:06:55.04,Default,,0000,0000,0000,,What we have here is DX by DT&EY
Dialogue: 0,0:06:55.04,0:06:58.66,Default,,0000,0000,0000,,by DT. What we wish to find is
Dialogue: 0,0:06:58.66,0:07:04.02,Default,,0000,0000,0000,,divided by DX. So if we\Nrearrange that equation D, why
Dialogue: 0,0:07:04.02,0:07:11.45,Default,,0000,0000,0000,,by DX is multiplied by DX by DT\Nso to get divided by DX on its
Dialogue: 0,0:07:11.45,0:07:15.16,Default,,0000,0000,0000,,own, we divide by the X by DT.
Dialogue: 0,0:07:15.17,0:07:22.32,Default,,0000,0000,0000,,We have divide by DX equals\NDY by DT all divided by
Dialogue: 0,0:07:22.32,0:07:24.11,Default,,0000,0000,0000,,DX by BT.
Dialogue: 0,0:07:25.51,0:07:28.72,Default,,0000,0000,0000,,So if we now
Dialogue: 0,0:07:28.72,0:07:34.35,Default,,0000,0000,0000,,substitute. Ty by DT\Nis cause T.
Dialogue: 0,0:07:36.08,0:07:39.78,Default,,0000,0000,0000,,And The X by the T is minus sign
Dialogue: 0,0:07:39.78,0:07:46.88,Default,,0000,0000,0000,,T. So what we have is\Nthe derivative divided by DX is
Dialogue: 0,0:07:46.88,0:07:48.56,Default,,0000,0000,0000,,Mina Scott T.
Dialogue: 0,0:07:49.41,0:07:56.17,Default,,0000,0000,0000,,Let's look at another\Nexample. One is a little
Dialogue: 0,0:07:56.17,0:07:58.42,Default,,0000,0000,0000,,bit more complicated.
Dialogue: 0,0:08:01.30,0:08:08.17,Default,,0000,0000,0000,,The parametric equations\Nfor this example
Dialogue: 0,0:08:08.17,0:08:15.04,Default,,0000,0000,0000,,RX equals T\Ncubed minus T&Y
Dialogue: 0,0:08:15.04,0:08:20.76,Default,,0000,0000,0000,,equals 4 minus\NT squared.
Dialogue: 0,0:08:21.80,0:08:23.86,Default,,0000,0000,0000,,Again, to find the gradient
Dialogue: 0,0:08:23.86,0:08:30.31,Default,,0000,0000,0000,,function. Of the equation, we're\Ngoing to differentiate each with
Dialogue: 0,0:08:30.31,0:08:37.63,Default,,0000,0000,0000,,respect to the parameter, so DX\Nby DT is 3 T squared
Dialogue: 0,0:08:37.63,0:08:40.77,Default,,0000,0000,0000,,minus one. Until why by duty?
Dialogue: 0,0:08:41.32,0:08:44.23,Default,,0000,0000,0000,,Is equal to minus 2 T?
Dialogue: 0,0:08:45.24,0:08:52.55,Default,,0000,0000,0000,,Again, using the chain rule\ND, why by DX equals
Dialogue: 0,0:08:52.55,0:08:54.74,Default,,0000,0000,0000,,DY by DT?
Dialogue: 0,0:08:55.71,0:09:02.72,Default,,0000,0000,0000,,Divided by the X by DT. That\Nis assuming that DX by DT does
Dialogue: 0,0:09:02.72,0:09:04.23,Default,,0000,0000,0000,,not equal 0.
Dialogue: 0,0:09:04.84,0:09:10.11,Default,,0000,0000,0000,,Let's substituting do why by DT\Nis minus 2 T.
Dialogue: 0,0:09:10.71,0:09:17.89,Default,,0000,0000,0000,,And EX by DT is\N3 two squared minus one.
Dialogue: 0,0:09:17.89,0:09:24.35,Default,,0000,0000,0000,,So again, we found the gradient\Nfunction of the curve.
Dialogue: 0,0:09:25.91,0:09:28.99,Default,,0000,0000,0000,,From The parametric equations.
Dialogue: 0,0:09:30.05,0:09:32.51,Default,,0000,0000,0000,,But it's in terms\Nof the parameter T.
Dialogue: 0,0:09:33.75,0:09:37.07,Default,,0000,0000,0000,,Let's look at
Dialogue: 0,0:09:37.07,0:09:43.26,Default,,0000,0000,0000,,another example. This\Ntime a parametric equations are
Dialogue: 0,0:09:43.26,0:09:45.44,Default,,0000,0000,0000,,X equals T cubed.
Dialogue: 0,0:09:46.22,0:09:48.33,Default,,0000,0000,0000,,And why he cause?
Dialogue: 0,0:09:49.37,0:09:52.26,Default,,0000,0000,0000,,T squared minus T.
Dialogue: 0,0:09:53.55,0:09:57.100,Default,,0000,0000,0000,,So let's have a look at what\Nthis curve looks like before we
Dialogue: 0,0:09:57.100,0:09:59.36,Default,,0000,0000,0000,,differentiate and find the
Dialogue: 0,0:09:59.36,0:10:03.22,Default,,0000,0000,0000,,gradient function. So we're\Ngoing to substitute for some
Dialogue: 0,0:10:03.22,0:10:07.65,Default,,0000,0000,0000,,values of tea again to workout\Nsome values of X&Y so that we
Dialogue: 0,0:10:07.65,0:10:09.01,Default,,0000,0000,0000,,can plot the curve.
Dialogue: 0,0:10:11.16,0:10:17.63,Default,,0000,0000,0000,,Let's take values of tea from\Nminus two through to two.
Dialogue: 0,0:10:18.56,0:10:25.41,Default,,0000,0000,0000,,So when T is minus, 2X is minus\N2 cubed, which is minus 8.
Dialogue: 0,0:10:26.26,0:10:32.60,Default,,0000,0000,0000,,When T is minus two, Y is minus\N2 squared, which is 4.
Dialogue: 0,0:10:33.53,0:10:36.09,Default,,0000,0000,0000,,Takeaway minus 2?
Dialogue: 0,0:10:36.96,0:10:40.28,Default,,0000,0000,0000,,Four takeaway minus two\Ngives us 6.
Dialogue: 0,0:10:41.68,0:10:47.45,Default,,0000,0000,0000,,20 is minus One X is minus 1\Ncubed, which is minus one.
Dialogue: 0,0:10:48.80,0:10:55.24,Default,,0000,0000,0000,,20 is minus one, Y is going to\Nbe minus one squared, which is
Dialogue: 0,0:10:55.24,0:10:58.92,Default,,0000,0000,0000,,one takeaway minus one which\Ngives us 2.
Dialogue: 0,0:10:59.98,0:11:03.79,Default,,0000,0000,0000,,Went to 0, then access 0.
Dialogue: 0,0:11:04.63,0:11:06.44,Default,,0000,0000,0000,,And why is era?
Dialogue: 0,0:11:07.82,0:11:10.71,Default,,0000,0000,0000,,20 is One X is one.
Dialogue: 0,0:11:11.43,0:11:14.92,Default,,0000,0000,0000,,I'm 20 is one, Y is
Dialogue: 0,0:11:14.92,0:11:18.60,Default,,0000,0000,0000,,one. Take away one giving a
Dialogue: 0,0:11:18.60,0:11:22.01,Default,,0000,0000,0000,,0 again. When T is 2.
Dialogue: 0,0:11:22.68,0:11:24.82,Default,,0000,0000,0000,,The next is 8.
Dialogue: 0,0:11:25.36,0:11:30.77,Default,,0000,0000,0000,,And when T is 2, Y is 2\Nsquared, four takeaway, two
Dialogue: 0,0:11:30.77,0:11:32.12,Default,,0000,0000,0000,,giving us 2.
Dialogue: 0,0:11:33.48,0:11:35.15,Default,,0000,0000,0000,,So let's plot curve.
Dialogue: 0,0:11:35.94,0:11:39.17,Default,,0000,0000,0000,,X axis.
Dialogue: 0,0:11:41.81,0:11:43.49,Default,,0000,0000,0000,,And IY axis.
Dialogue: 0,0:11:44.55,0:11:47.51,Default,,0000,0000,0000,,And we've got to go from minus 8
Dialogue: 0,0:11:47.51,0:11:50.87,Default,,0000,0000,0000,,to +8. So would take fairly
Dialogue: 0,0:11:50.87,0:11:53.84,Default,,0000,0000,0000,,large. Steps.
Dialogue: 0,0:11:55.71,0:11:59.58,Default,,0000,0000,0000,,So we plot minus
Dialogue: 0,0:11:59.58,0:12:01.52,Default,,0000,0000,0000,,eight 6.
Dialogue: 0,0:12:02.27,0:12:06.34,Default,,0000,0000,0000,,So. Minus 1\Ntwo.
Dialogue: 0,0:12:08.59,0:12:09.71,Default,,0000,0000,0000,,00
Dialogue: 0,0:12:11.02,0:12:12.51,Default,,0000,0000,0000,,10
Dialogue: 0,0:12:13.52,0:12:17.59,Default,,0000,0000,0000,,And eight\N2.
Dialogue: 0,0:12:24.90,0:12:26.38,Default,,0000,0000,0000,,So those are curve.
Dialogue: 0,0:12:27.70,0:12:32.06,Default,,0000,0000,0000,,And here we're not. Perhaps\Ncertain what happens.
Dialogue: 0,0:12:32.71,0:12:35.28,Default,,0000,0000,0000,,It does look as if that is a
Dialogue: 0,0:12:35.28,0:12:39.18,Default,,0000,0000,0000,,turning point. But let's\Ninvestigate a bit further
Dialogue: 0,0:12:39.18,0:12:41.94,Default,,0000,0000,0000,,and actually differentiate\Nthese parametric equations.
Dialogue: 0,0:12:43.35,0:12:46.78,Default,,0000,0000,0000,,So as before the X5
Dialogue: 0,0:12:46.78,0:12:51.73,Default,,0000,0000,0000,,ET. The derivative of T cubed is\N3 two squared.
Dialogue: 0,0:12:52.96,0:12:59.25,Default,,0000,0000,0000,,And if we look at the why by DT,\Nthe derivative of T squared is 2
Dialogue: 0,0:12:59.25,0:13:02.09,Default,,0000,0000,0000,,T. Minus one.
Dialogue: 0,0:13:03.43,0:13:09.88,Default,,0000,0000,0000,,Again, using the chain rule\Ndivide by DX is equal to
Dialogue: 0,0:13:09.88,0:13:13.39,Default,,0000,0000,0000,,DY by DT divided by DX
Dialogue: 0,0:13:13.39,0:13:17.78,Default,,0000,0000,0000,,by beauty. And again,\Nassuming that the X by
Dialogue: 0,0:13:17.78,0:13:19.83,Default,,0000,0000,0000,,DT does not equal 0.
Dialogue: 0,0:13:21.12,0:13:24.12,Default,,0000,0000,0000,,So if we substitute.
Dialogue: 0,0:13:24.85,0:13:26.57,Default,,0000,0000,0000,,For RDY by DT.
Dialogue: 0,0:13:27.13,0:13:33.73,Default,,0000,0000,0000,,We get 2T minus 1 divided\Nby R DX by DT, which is
Dialogue: 0,0:13:33.73,0:13:35.26,Default,,0000,0000,0000,,3 T squared.
Dialogue: 0,0:13:37.41,0:13:41.48,Default,,0000,0000,0000,,From this we can analyze the\Ncurve further and we can see
Dialogue: 0,0:13:41.48,0:13:43.85,Default,,0000,0000,0000,,that in fact when divided by DX
Dialogue: 0,0:13:43.85,0:13:51.15,Default,,0000,0000,0000,,is 0. Then T must be 1/2,\Nso in this section here we do
Dialogue: 0,0:13:51.15,0:13:53.09,Default,,0000,0000,0000,,have a stationary point.
Dialogue: 0,0:13:54.15,0:13:56.50,Default,,0000,0000,0000,,Also, we can see that when.
Dialogue: 0,0:13:57.06,0:13:58.50,Default,,0000,0000,0000,,T is 0.
Dialogue: 0,0:13:59.99,0:14:02.22,Default,,0000,0000,0000,,DY by DX is Infinity.
Dialogue: 0,0:14:03.15,0:14:08.61,Default,,0000,0000,0000,,So we have got the Y axis\Nhere being a tangent to the
Dialogue: 0,0:14:08.61,0:14:10.29,Default,,0000,0000,0000,,curve at the .00.
Dialogue: 0,0:14:12.08,0:14:17.69,Default,,0000,0000,0000,,Sometimes it is necessary to\Ndifferentiate a second time
Dialogue: 0,0:14:17.69,0:14:23.29,Default,,0000,0000,0000,,and we can do this with\Nour parametric equations.
Dialogue: 0,0:14:24.46,0:14:26.79,Default,,0000,0000,0000,,Let's have a look at a fairly
Dialogue: 0,0:14:26.79,0:14:30.96,Default,,0000,0000,0000,,straightforward example. X\Nequals T squared.
Dialogue: 0,0:14:31.79,0:14:35.28,Default,,0000,0000,0000,,And why equals T cubed?
Dialogue: 0,0:14:36.06,0:14:39.40,Default,,0000,0000,0000,,And what we're going to do is to\Ndifferentiate using the chain
Dialogue: 0,0:14:39.40,0:14:42.73,Default,,0000,0000,0000,,rule, as we've done before, and\Nthen we're going to apply the
Dialogue: 0,0:14:42.73,0:14:46.35,Default,,0000,0000,0000,,chain rule the second time to\Nfind the two. Why by DX squared.
Dialogue: 0,0:14:47.40,0:14:54.25,Default,,0000,0000,0000,,So starting us before DX bite\Nbeauty is equal to T.
Dialogue: 0,0:14:54.87,0:14:57.27,Default,,0000,0000,0000,,And why by DT?
Dialogue: 0,0:14:58.05,0:15:00.55,Default,,0000,0000,0000,,Is equal to three T squared.
Dialogue: 0,0:15:01.81,0:15:03.24,Default,,0000,0000,0000,,Using the chain rule.
Dialogue: 0,0:15:03.81,0:15:07.72,Default,,0000,0000,0000,,Dude, why by DX equals.
Dialogue: 0,0:15:08.39,0:15:10.78,Default,,0000,0000,0000,,Divide by BT.
Dialogue: 0,0:15:11.55,0:15:14.35,Default,,0000,0000,0000,,Divided by DX by DT.
Dialogue: 0,0:15:14.91,0:15:19.54,Default,,0000,0000,0000,,And assuming, of course that the\NX by DT does not equal 0.
Dialogue: 0,0:15:20.27,0:15:25.12,Default,,0000,0000,0000,,So let's substitute for divide\Nby DT. It's 3T squared.
Dialogue: 0,0:15:25.70,0:15:29.38,Default,,0000,0000,0000,,Divided by DX by DT, which is
Dialogue: 0,0:15:29.38,0:15:35.84,Default,,0000,0000,0000,,2 two. And here at TI\Ngoes into 2 squared two times.
Dialogue: 0,0:15:35.84,0:15:39.17,Default,,0000,0000,0000,,So we've got three over 2 times
Dialogue: 0,0:15:39.17,0:15:46.23,Default,,0000,0000,0000,,by teeth. Now applying the\Nchain rule for a second
Dialogue: 0,0:15:46.23,0:15:53.59,Default,,0000,0000,0000,,time. We have the two\NY by DX squared equals D
Dialogue: 0,0:15:53.59,0:16:00.14,Default,,0000,0000,0000,,by DX of divide by DX\N'cause we need to differentiate
Dialogue: 0,0:16:00.14,0:16:02.52,Default,,0000,0000,0000,,divided by DX again.
Dialogue: 0,0:16:03.32,0:16:04.63,Default,,0000,0000,0000,,And that is.
Dialogue: 0,0:16:05.13,0:16:10.70,Default,,0000,0000,0000,,The derivative of divide by DX\Nwith respect to T.
Dialogue: 0,0:16:11.36,0:16:14.84,Default,,0000,0000,0000,,Divided by DX by DT.
Dialogue: 0,0:16:15.41,0:16:19.14,Default,,0000,0000,0000,,Now, just to recap, as YY
Dialogue: 0,0:16:19.14,0:16:24.17,Default,,0000,0000,0000,,by ZX. Was equal to three\Nover 2 T.
Dialogue: 0,0:16:24.83,0:16:27.35,Default,,0000,0000,0000,,And our DX by DT.
Dialogue: 0,0:16:28.33,0:16:29.82,Default,,0000,0000,0000,,Was equal to 2 T.
Dialogue: 0,0:16:30.74,0:16:36.38,Default,,0000,0000,0000,,So now we can do the\Nsubstitution and find D2Y by the
Dialogue: 0,0:16:36.38,0:16:39.16,Default,,0000,0000,0000,,X squared. Is equal to.
Dialogue: 0,0:16:40.29,0:16:44.51,Default,,0000,0000,0000,,The derivative of divide by DX\Nwith respect to T.
Dialogue: 0,0:16:45.81,0:16:47.62,Default,,0000,0000,0000,,So that's three over 2.
Dialogue: 0,0:16:48.64,0:16:52.64,Default,,0000,0000,0000,,Divided by. DX by BT which is
Dialogue: 0,0:16:52.64,0:16:59.74,Default,,0000,0000,0000,,2 T? And that gives\Nus three over 4T.
Dialogue: 0,0:17:00.51,0:17:06.40,Default,,0000,0000,0000,,So do 2 white by the X squared\Nis 3 / 40.
Dialogue: 0,0:17:07.36,0:17:14.47,Default,,0000,0000,0000,,Let's do one more example.\NThis time are parametric
Dialogue: 0,0:17:14.47,0:17:21.58,Default,,0000,0000,0000,,equation is X equals T\Ncubed plus 3T squared.
Dialogue: 0,0:17:22.61,0:17:29.91,Default,,0000,0000,0000,,And why equals T to\Nthe Power 4 - 8
Dialogue: 0,0:17:29.91,0:17:33.77,Default,,0000,0000,0000,,T squared? So we're\Ngoing to
Dialogue: 0,0:17:33.77,0:17:36.47,Default,,0000,0000,0000,,differentiate X with\Nrespect to T.
Dialogue: 0,0:17:38.34,0:17:42.26,Default,,0000,0000,0000,,Which gives us 3T squared
Dialogue: 0,0:17:42.26,0:17:45.93,Default,,0000,0000,0000,,plus 60. And that is why by
Dialogue: 0,0:17:45.93,0:17:52.20,Default,,0000,0000,0000,,duty? Is equal to 40\Ncubed minus 16 T.
Dialogue: 0,0:17:52.94,0:18:00.88,Default,,0000,0000,0000,,Using the chain rule, divide by\NDX equals DY by the T
Dialogue: 0,0:18:00.88,0:18:04.19,Default,,0000,0000,0000,,divided by DX by DT.
Dialogue: 0,0:18:05.03,0:18:08.97,Default,,0000,0000,0000,,Assuming the exploited seat does\Nnot equal 0.
Dialogue: 0,0:18:09.48,0:18:17.08,Default,,0000,0000,0000,,So we get the why by the\NT is 40 cubed minus 16 T.
Dialogue: 0,0:18:17.73,0:18:24.78,Default,,0000,0000,0000,,Divided by DX by DT\Nwhich is 3 T squared
Dialogue: 0,0:18:24.78,0:18:28.35,Default,,0000,0000,0000,,plus 60. Now that let's tidy\Nthis up a bit.
Dialogue: 0,0:18:28.99,0:18:31.39,Default,,0000,0000,0000,,And see if there's things\Nthat we can cancel.
Dialogue: 0,0:18:32.97,0:18:38.05,Default,,0000,0000,0000,,Here at the top we've got 40\Ncubed takeaway 16 T so common to
Dialogue: 0,0:18:38.05,0:18:44.59,Default,,0000,0000,0000,,both parts of this is a four and\Na T, so if we take four and a T
Dialogue: 0,0:18:44.59,0:18:45.68,Default,,0000,0000,0000,,outside the bracket.
Dialogue: 0,0:18:46.73,0:18:52.58,Default,,0000,0000,0000,,Inside will have left\NTI squared that makes
Dialogue: 0,0:18:52.58,0:18:55.50,Default,,0000,0000,0000,,40 cubed takeaway 4.
Dialogue: 0,0:18:56.68,0:19:03.32,Default,,0000,0000,0000,,Underneath common to both these\Nparts is 3 T.
Dialogue: 0,0:19:04.34,0:19:07.47,Default,,0000,0000,0000,,So take 3T outside of bracket.
Dialogue: 0,0:19:08.20,0:19:13.43,Default,,0000,0000,0000,,And inside we're left with TI so\Nthat when it's multiplied out we
Dialogue: 0,0:19:13.43,0:19:14.63,Default,,0000,0000,0000,,get 3T squared.
Dialogue: 0,0:19:15.14,0:19:21.56,Default,,0000,0000,0000,,+2 again three 2 * 2\Ngives us our 60.
Dialogue: 0,0:19:23.24,0:19:28.22,Default,,0000,0000,0000,,Now we can go further here\Nbecause this one here, T squared
Dialogue: 0,0:19:28.22,0:19:32.61,Default,,0000,0000,0000,,minus 4. Is actually a\Ndifference, the minus the
Dialogue: 0,0:19:32.61,0:19:36.29,Default,,0000,0000,0000,,takeaway between 2 square\Nnumbers? It's a difference of
Dialogue: 0,0:19:36.29,0:19:39.75,Default,,0000,0000,0000,,two squares. So we can express
Dialogue: 0,0:19:39.75,0:19:46.77,Default,,0000,0000,0000,,that. As T plus\N2 multiplied by T
Dialogue: 0,0:19:46.77,0:19:52.48,Default,,0000,0000,0000,,minus 2. And that's going to\Nhelp us because we can do some
Dialogue: 0,0:19:52.48,0:19:56.17,Default,,0000,0000,0000,,more. Counseling and make it\Nsimpler for us before we
Dialogue: 0,0:19:56.17,0:19:57.65,Default,,0000,0000,0000,,differentiate a second time.
Dialogue: 0,0:19:58.44,0:20:06.21,Default,,0000,0000,0000,,So here T goes into T once\N2 + 2 goes into 2 +
Dialogue: 0,0:20:06.21,0:20:12.87,Default,,0000,0000,0000,,2 once, so we're left with four\N2 - 2 over 3.
Dialogue: 0,0:20:14.63,0:20:17.29,Default,,0000,0000,0000,,Now differentiating a second
Dialogue: 0,0:20:17.29,0:20:23.92,Default,,0000,0000,0000,,time. The two\NY by the X
Dialogue: 0,0:20:23.92,0:20:31.23,Default,,0000,0000,0000,,squared. Is the differential\Nof DY by DX with
Dialogue: 0,0:20:31.23,0:20:36.74,Default,,0000,0000,0000,,respect to T divided by\NDX by BT.
Dialogue: 0,0:20:38.60,0:20:45.42,Default,,0000,0000,0000,,Now recapping from before, let's\Njust note down the why by
Dialogue: 0,0:20:45.42,0:20:50.62,Default,,0000,0000,0000,,DX. What is 4 thirds of\NT minus 2?
Dialogue: 0,0:20:51.62,0:20:54.21,Default,,0000,0000,0000,,And our DX by DT.
Dialogue: 0,0:20:55.35,0:20:58.94,Default,,0000,0000,0000,,Was 3T squared plus 60.
Dialogue: 0,0:21:00.09,0:21:04.63,Default,,0000,0000,0000,,So differentiating divided by DX\Nwith respect to T.
Dialogue: 0,0:21:05.42,0:21:07.97,Default,,0000,0000,0000,,We get 4 thirds.
Dialogue: 0,0:21:08.55,0:21:11.52,Default,,0000,0000,0000,,And then we divide by DX by BT.
Dialogue: 0,0:21:12.33,0:21:15.77,Default,,0000,0000,0000,,Which is 3 T squared plus 60.
Dialogue: 0,0:21:16.37,0:21:23.32,Default,,0000,0000,0000,,So that gives us 4 over\N3 lots of three 2 squared
Dialogue: 0,0:21:23.32,0:21:30.16,Default,,0000,0000,0000,,plus 60. So do 2 white\Nby DX squared is equal to.
Dialogue: 0,0:21:30.73,0:21:37.45,Default,,0000,0000,0000,,Full And here we can\Ntake another three and a T
Dialogue: 0,0:21:37.45,0:21:41.59,Default,,0000,0000,0000,,outside of a bracket to tidy\Nthis up 90.
Dialogue: 0,0:21:42.35,0:21:45.66,Default,,0000,0000,0000,,Into T +2.
Dialogue: 0,0:21:46.71,0:21:49.94,Default,,0000,0000,0000,,I'm not so there is to it.