0:00:01.530,0:00:05.616
Some relationships between two[br]quantities or variables are so

0:00:05.616,0:00:09.702
complicated that we sometimes[br]introduce a third variable or

0:00:09.702,0:00:11.972
quantity to make things easier.

0:00:12.590,0:00:17.459
In mathematics, this third[br]quantity is called a parameter,

0:00:17.459,0:00:22.869
and instead of having one[br]equation, say relating X&Y, we

0:00:22.869,0:00:28.820
have two equations, one relating[br]the parameter with X and one

0:00:28.820,0:00:31.525
relating the parameter with Y.

0:00:32.250,0:00:34.077
Let's have a look at an example.

0:00:35.620,0:00:39.120
X equals Cos T.

0:00:39.830,0:00:43.830
And why it was

0:00:43.830,0:00:48.160
Scienti? It's our parametric[br]equations we have.

0:00:49.060,0:00:49.800
X.

0:00:50.940,0:00:56.964
And T the relationship in one[br]equation and Y&T related in the

0:00:56.964,0:01:01.950
other equation. Let's have a[br]look at what the graph looks

0:01:01.950,0:01:06.130
like and to do that, we[br]substitute some values for T

0:01:06.130,0:01:09.930
into both the equations and we[br]workout values for X&Y.

0:01:11.350,0:01:15.088
Let's take some values of tea.

0:01:15.090,0:01:18.130
Calculate X&Y.

0:01:20.800,0:01:27.406
And will take[br]some values Zero

0:01:27.406,0:01:32.890
Pi 2π? 35 by[br]2 and 2π.

0:01:34.260,0:01:35.814
And to make it a little bit

0:01:35.814,0:01:39.368
easier. Well, draw the curves.

0:01:41.570,0:01:46.700
Of cause T and sign[br]TA little bit more.

0:01:49.310,0:01:53.570
Call[br]Zetty.

0:01:54.640,0:01:56.788
He just put some labels on.

0:02:12.050,0:02:13.178
Now we have one.

0:02:13.750,0:02:18.408
Minus one. So that's[br]our graph of Costi.

0:02:20.030,0:02:21.410
And Scienti.

0:02:42.750,0:02:43.490
And.

0:02:44.730,0:02:48.950
This

0:02:48.950,0:02:57.586
scientist.[br]OK, so when T is 0.

0:02:58.580,0:03:00.860
X is cause T.

0:03:01.510,0:03:02.578
And that's one.

0:03:03.140,0:03:08.819
When she is, OY is scienti[br]and that zero.

0:03:10.910,0:03:16.654
20 is π by two, X is the cause[br]of Π by two, which is 0.

0:03:17.870,0:03:23.294
20 is π by two, Y is the sign of[br]Π by two, which is one.

0:03:25.260,0:03:29.250
20 is π the cause[br]of Pi is minus one?

0:03:30.390,0:03:31.770
Anne for why?

0:03:32.570,0:03:35.670
The sign of Π Zero.

0:03:37.270,0:03:42.835
20 is 3 Pi by two the cause of[br]three Pi by two is 0.

0:03:43.770,0:03:46.916
And the sign of three[br]Pi by two is minus one.

0:03:47.970,0:03:55.215
And at 2π. Twenty is 2π the[br]cause of 2π is one and the sign

0:03:55.215,0:03:57.147
of 2π is 0.

0:03:57.970,0:04:02.490
So we now have X&Y coordinates[br]that we can plot.

0:04:03.150,0:04:04.798
To show the curve.

0:04:06.490,0:04:11.002
How[br]about

0:04:11.002,0:04:13.258
X&Y?

0:04:14.900,0:04:22.036
Once he was zero,[br]X is one, Y

0:04:22.036,0:04:25.604
is 0, so 10.

0:04:26.190,0:04:27.500
01

0:04:28.750,0:04:30.470
Minus 10.

0:04:31.630,0:04:33.169
0 - 1.

0:04:34.290,0:04:36.100
And back again to 10.

0:04:37.380,0:04:41.290
Now, with those points, we've[br]not actually plotted enough to

0:04:41.290,0:04:45.982
be able to see what's happening[br]in between these points, but if

0:04:45.982,0:04:51.847
we were to take values for T[br]between 0 and Π by two and some

0:04:51.847,0:04:56.930
more between pie by two and Π[br]and so on, what we'd actually

0:04:56.930,0:04:59.276
find is that these are the

0:04:59.276,0:05:03.870
parametric equations. That[br]describe a circle.

0:05:04.770,0:05:11.298
Sensor. 00 and with[br]a radius of 1.

0:05:13.260,0:05:18.356
Now what we often want to find[br]out is how to variables are

0:05:18.356,0:05:22.668
changing in relationship to each[br]other. So when exchange is, how

0:05:22.668,0:05:28.156
is why changing what's the rate[br]of change? So we need to be able

0:05:28.156,0:05:32.070
to differentiate. Now what we[br]don't want to do is to actually.

0:05:32.940,0:05:35.658
Eliminate the parameter.

0:05:36.200,0:05:39.560
And get back to an equation[br]directly relating X&Y, 'cause

0:05:39.560,0:05:43.928
the whole point of having it's a[br]parameter is that it makes it

0:05:43.928,0:05:48.968
easier for us and simpler, So[br]what we need to do is to find a

0:05:48.968,0:05:52.664
way of differentiating when we[br]got them in the parametric form.

0:05:53.460,0:05:56.030
And that's what we do.

0:05:56.090,0:06:01.940
Thanks, right, the two equations[br]again X equals Cos T.

0:06:02.450,0:06:08.434
Y equals sign T. What we're[br]going to do to differentiate?

0:06:09.330,0:06:12.914
Is to differentiate each[br]equation with respect to

0:06:12.914,0:06:14.258
the parameter T.

0:06:15.840,0:06:23.580
So the X5 DT, the derivative[br]of Cos T is minus sign

0:06:23.580,0:06:29.786
T. 4 divided by DT.[br]The derivative of Scienti.

0:06:30.290,0:06:31.520
Is cause T?

0:06:32.290,0:06:35.620
Now using the chain rule.

0:06:36.860,0:06:43.911
Which says that DY by[br]the T is equal to DY

0:06:43.911,0:06:45.193
by DX.

0:06:46.230,0:06:49.900
Times by DX by DT.

0:06:51.070,0:06:55.038
What we have here is DX by DT&EY

0:06:55.038,0:06:58.656
by DT. What we wish to find is

0:06:58.656,0:07:04.022
divided by DX. So if we[br]rearrange that equation D, why

0:07:04.022,0:07:11.446
by DX is multiplied by DX by DT[br]so to get divided by DX on its

0:07:11.446,0:07:15.158
own, we divide by the X by DT.

0:07:15.170,0:07:22.322
We have divide by DX equals[br]DY by DT all divided by

0:07:22.322,0:07:24.110
DX by BT.

0:07:25.510,0:07:28.718
So if we now

0:07:28.718,0:07:34.350
substitute. Ty by DT[br]is cause T.

0:07:36.080,0:07:39.779
And The X by the T is minus sign

0:07:39.779,0:07:46.881
T. So what we have is[br]the derivative divided by DX is

0:07:46.881,0:07:48.564
Mina Scott T.

0:07:49.410,0:07:56.169
Let's look at another[br]example. One is a little

0:07:56.169,0:07:58.422
bit more complicated.

0:08:01.300,0:08:08.170
The parametric equations[br]for this example

0:08:08.170,0:08:15.040
RX equals T[br]cubed minus T&Y

0:08:15.040,0:08:20.765
equals 4 minus[br]T squared.

0:08:21.800,0:08:23.865
Again, to find the gradient

0:08:23.865,0:08:30.310
function. Of the equation, we're[br]going to differentiate each with

0:08:30.310,0:08:37.630
respect to the parameter, so DX[br]by DT is 3 T squared

0:08:37.630,0:08:40.768
minus one. Until why by duty?

0:08:41.320,0:08:44.230
Is equal to minus 2 T?

0:08:45.240,0:08:52.550
Again, using the chain rule[br]D, why by DX equals

0:08:52.550,0:08:54.743
DY by DT?

0:08:55.710,0:09:02.724
Divided by the X by DT. That[br]is assuming that DX by DT does

0:09:02.724,0:09:04.227
not equal 0.

0:09:04.840,0:09:10.110
Let's substituting do why by DT[br]is minus 2 T.

0:09:10.710,0:09:17.890
And EX by DT is[br]3 two squared minus one.

0:09:17.890,0:09:24.350
So again, we found the gradient[br]function of the curve.

0:09:25.910,0:09:28.989
From The parametric equations.

0:09:30.050,0:09:32.514
But it's in terms[br]of the parameter T.

0:09:33.750,0:09:37.068
Let's look at

0:09:37.068,0:09:43.260
another example. This[br]time a parametric equations are

0:09:43.260,0:09:45.440
X equals T cubed.

0:09:46.220,0:09:48.328
And why he cause?

0:09:49.370,0:09:52.258
T squared minus T.

0:09:53.550,0:09:57.996
So let's have a look at what[br]this curve looks like before we

0:09:57.996,0:09:59.364
differentiate and find the

0:09:59.364,0:10:03.217
gradient function. So we're[br]going to substitute for some

0:10:03.217,0:10:07.650
values of tea again to workout[br]some values of X&Y so that we

0:10:07.650,0:10:09.014
can plot the curve.

0:10:11.160,0:10:17.628
Let's take values of tea from[br]minus two through to two.

0:10:18.560,0:10:25.406
So when T is minus, 2X is minus[br]2 cubed, which is minus 8.

0:10:26.260,0:10:32.604
When T is minus two, Y is minus[br]2 squared, which is 4.

0:10:33.530,0:10:36.089
Takeaway minus 2?

0:10:36.960,0:10:40.285
Four takeaway minus two[br]gives us 6.

0:10:41.680,0:10:47.452
20 is minus One X is minus 1[br]cubed, which is minus one.

0:10:48.800,0:10:55.240
20 is minus one, Y is going to[br]be minus one squared, which is

0:10:55.240,0:10:58.920
one takeaway minus one which[br]gives us 2.

0:10:59.980,0:11:03.790
Went to 0, then access 0.

0:11:04.630,0:11:06.438
And why is era?

0:11:07.820,0:11:10.706
20 is One X is one.

0:11:11.430,0:11:14.916
I'm 20 is one, Y is

0:11:14.916,0:11:18.600
one. Take away one giving a

0:11:18.600,0:11:22.010
0 again. When T is 2.

0:11:22.680,0:11:24.820
The next is 8.

0:11:25.360,0:11:30.772
And when T is 2, Y is 2[br]squared, four takeaway, two

0:11:30.772,0:11:32.125
giving us 2.

0:11:33.480,0:11:35.148
So let's plot curve.

0:11:35.940,0:11:39.170
X axis.

0:11:41.810,0:11:43.490
And IY axis.

0:11:44.550,0:11:47.510
And we've got to go from minus 8

0:11:47.510,0:11:50.870
to +8. So would take fairly

0:11:50.870,0:11:53.840
large. Steps.

0:11:55.710,0:11:59.582
So we plot minus

0:11:59.582,0:12:01.518
eight 6.

0:12:02.270,0:12:06.338
So. Minus 1[br]two.

0:12:08.590,0:12:09.710
00

0:12:11.020,0:12:12.510
10

0:12:13.520,0:12:17.588
And eight[br]2.

0:12:24.900,0:12:26.380
So those are curve.

0:12:27.700,0:12:32.060
And here we're not. Perhaps[br]certain what happens.

0:12:32.710,0:12:35.278
It does look as if that is a

0:12:35.278,0:12:39.180
turning point. But let's[br]investigate a bit further

0:12:39.180,0:12:41.940
and actually differentiate[br]these parametric equations.

0:12:43.350,0:12:46.775
So as before the X5

0:12:46.775,0:12:51.726
ET. The derivative of T cubed is[br]3 two squared.

0:12:52.960,0:12:59.248
And if we look at the why by DT,[br]the derivative of T squared is 2

0:12:59.248,0:13:02.090
T. Minus one.

0:13:03.430,0:13:09.876
Again, using the chain rule[br]divide by DX is equal to

0:13:09.876,0:13:13.392
DY by DT divided by DX

0:13:13.392,0:13:17.783
by beauty. And again,[br]assuming that the X by

0:13:17.783,0:13:19.828
DT does not equal 0.

0:13:21.120,0:13:24.120
So if we substitute.

0:13:24.850,0:13:26.570
For RDY by DT.

0:13:27.130,0:13:33.734
We get 2T minus 1 divided[br]by R DX by DT, which is

0:13:33.734,0:13:35.258
3 T squared.

0:13:37.410,0:13:41.478
From this we can analyze the[br]curve further and we can see

0:13:41.478,0:13:43.851
that in fact when divided by DX

0:13:43.851,0:13:51.150
is 0. Then T must be 1/2,[br]so in this section here we do

0:13:51.150,0:13:53.090
have a stationary point.

0:13:54.150,0:13:56.496
Also, we can see that when.

0:13:57.060,0:13:58.500
T is 0.

0:13:59.990,0:14:02.220
DY by DX is Infinity.

0:14:03.150,0:14:08.610
So we have got the Y axis[br]here being a tangent to the

0:14:08.610,0:14:10.290
curve at the .00.

0:14:12.080,0:14:17.687
Sometimes it is necessary to[br]differentiate a second time

0:14:17.687,0:14:23.294
and we can do this with[br]our parametric equations.

0:14:24.460,0:14:26.791
Let's have a look at a fairly

0:14:26.791,0:14:30.960
straightforward example. X[br]equals T squared.

0:14:31.790,0:14:35.280
And why equals T cubed?

0:14:36.060,0:14:39.396
And what we're going to do is to[br]differentiate using the chain

0:14:39.396,0:14:42.732
rule, as we've done before, and[br]then we're going to apply the

0:14:42.732,0:14:46.346
chain rule the second time to[br]find the two. Why by DX squared.

0:14:47.400,0:14:54.253
So starting us before DX bite[br]beauty is equal to T.

0:14:54.870,0:14:57.270
And why by DT?

0:14:58.050,0:15:00.546
Is equal to three T squared.

0:15:01.810,0:15:03.238
Using the chain rule.

0:15:03.810,0:15:07.720
Dude, why by DX equals.

0:15:08.390,0:15:10.778
Divide by BT.

0:15:11.550,0:15:14.350
Divided by DX by DT.

0:15:14.910,0:15:19.538
And assuming, of course that the[br]X by DT does not equal 0.

0:15:20.270,0:15:25.120
So let's substitute for divide[br]by DT. It's 3T squared.

0:15:25.700,0:15:29.375
Divided by DX by DT, which is

0:15:29.375,0:15:35.840
2 two. And here at TI[br]goes into 2 squared two times.

0:15:35.840,0:15:39.172
So we've got three over 2 times

0:15:39.172,0:15:46.234
by teeth. Now applying the[br]chain rule for a second

0:15:46.234,0:15:53.590
time. We have the two[br]Y by DX squared equals D

0:15:53.590,0:16:00.135
by DX of divide by DX[br]'cause we need to differentiate

0:16:00.135,0:16:02.515
divided by DX again.

0:16:03.320,0:16:04.628
And that is.

0:16:05.130,0:16:10.700
The derivative of divide by DX[br]with respect to T.

0:16:11.360,0:16:14.840
Divided by DX by DT.

0:16:15.410,0:16:19.136
Now, just to recap, as YY

0:16:19.136,0:16:24.167
by ZX. Was equal to three[br]over 2 T.

0:16:24.830,0:16:27.350
And our DX by DT.

0:16:28.330,0:16:29.820
Was equal to 2 T.

0:16:30.740,0:16:36.380
So now we can do the[br]substitution and find D2Y by the

0:16:36.380,0:16:39.159
X squared. Is equal to.

0:16:40.290,0:16:44.510
The derivative of divide by DX[br]with respect to T.

0:16:45.810,0:16:47.620
So that's three over 2.

0:16:48.640,0:16:52.640
Divided by. DX by BT which is

0:16:52.640,0:16:59.736
2 T? And that gives[br]us three over 4T.

0:17:00.510,0:17:06.402
So do 2 white by the X squared[br]is 3 / 40.

0:17:07.360,0:17:14.470
Let's do one more example.[br]This time are parametric

0:17:14.470,0:17:21.580
equation is X equals T[br]cubed plus 3T squared.

0:17:22.610,0:17:29.910
And why equals T to[br]the Power 4 - 8

0:17:29.910,0:17:33.770
T squared? So we're[br]going to

0:17:33.770,0:17:36.470
differentiate X with[br]respect to T.

0:17:38.340,0:17:42.265
Which gives us 3T squared

0:17:42.265,0:17:45.930
plus 60. And that is why by

0:17:45.930,0:17:52.204
duty? Is equal to 40[br]cubed minus 16 T.

0:17:52.940,0:18:00.884
Using the chain rule, divide by[br]DX equals DY by the T

0:18:00.884,0:18:04.194
divided by DX by DT.

0:18:05.030,0:18:08.966
Assuming the exploited seat does[br]not equal 0.

0:18:09.480,0:18:17.082
So we get the why by the[br]T is 40 cubed minus 16 T.

0:18:17.730,0:18:24.780
Divided by DX by DT[br]which is 3 T squared

0:18:24.780,0:18:28.350
plus 60. Now that let's tidy[br]this up a bit.

0:18:28.990,0:18:31.393
And see if there's things[br]that we can cancel.

0:18:32.970,0:18:38.052
Here at the top we've got 40[br]cubed takeaway 16 T so common to

0:18:38.052,0:18:44.586
both parts of this is a four and[br]a T, so if we take four and a T

0:18:44.586,0:18:45.675
outside the bracket.

0:18:46.730,0:18:52.578
Inside will have left[br]TI squared that makes

0:18:52.578,0:18:55.502
40 cubed takeaway 4.

0:18:56.680,0:19:03.322
Underneath common to both these[br]parts is 3 T.

0:19:04.340,0:19:07.466
So take 3T outside of bracket.

0:19:08.200,0:19:13.426
And inside we're left with TI so[br]that when it's multiplied out we

0:19:13.426,0:19:14.632
get 3T squared.

0:19:15.140,0:19:21.560
+2 again three 2 * 2[br]gives us our 60.

0:19:23.240,0:19:28.220
Now we can go further here[br]because this one here, T squared

0:19:28.220,0:19:32.613
minus 4. Is actually a[br]difference, the minus the

0:19:32.613,0:19:36.294
takeaway between 2 square[br]numbers? It's a difference of

0:19:36.294,0:19:39.746
two squares. So we can express

0:19:39.746,0:19:46.772
that. As T plus[br]2 multiplied by T

0:19:46.772,0:19:52.482
minus 2. And that's going to[br]help us because we can do some

0:19:52.482,0:19:56.170
more. Counseling and make it[br]simpler for us before we

0:19:56.170,0:19:57.650
differentiate a second time.

0:19:58.440,0:20:06.210
So here T goes into T once[br]2 + 2 goes into 2 +

0:20:06.210,0:20:12.870
2 once, so we're left with four[br]2 - 2 over 3.

0:20:14.630,0:20:17.286
Now differentiating a second

0:20:17.286,0:20:23.922
time. The two[br]Y by the X

0:20:23.922,0:20:31.234
squared. Is the differential[br]of DY by DX with

0:20:31.234,0:20:36.738
respect to T divided by[br]DX by BT.

0:20:38.600,0:20:45.420
Now recapping from before, let's[br]just note down the why by

0:20:45.420,0:20:50.616
DX. What is 4 thirds of[br]T minus 2?

0:20:51.620,0:20:54.210
And our DX by DT.

0:20:55.350,0:20:58.940
Was 3T squared plus 60.

0:21:00.090,0:21:04.626
So differentiating divided by DX[br]with respect to T.

0:21:05.420,0:21:07.968
We get 4 thirds.

0:21:08.550,0:21:11.518
And then we divide by DX by BT.

0:21:12.330,0:21:15.767
Which is 3 T squared plus 60.

0:21:16.370,0:21:23.318
So that gives us 4 over[br]3 lots of three 2 squared

0:21:23.318,0:21:30.160
plus 60. So do 2 white[br]by DX squared is equal to.

0:21:30.730,0:21:37.450
Full And here we can[br]take another three and a T

0:21:37.450,0:21:41.590
outside of a bracket to tidy[br]this up 90.

0:21:42.350,0:21:45.659
Into T +2.

0:21:46.710,0:21:49.944
I'm not so there is to it.