Empirical and Molecular Formulas from Stoichiometry

WEBVTT
Kind: captions
Language: en

00:00:00.630 --> 00:00:05.680
Our problem tells us that when
1.125 grams of a liquid

00:00:05.680 --> 00:00:09.540
hydrocarbon, CxHy-- so we don't
know how many C's and

00:00:09.540 --> 00:00:12.320
H's there are, so it's a mystery
liquid hydrocarbon.

00:00:12.320 --> 00:00:14.520
It was burned in an apparatus
like that shown in figure

00:00:14.520 --> 00:00:17.470
4-7-- this is the apparatus,
this is our mystery

00:00:17.470 --> 00:00:21.470
hydrocarbon getting burned, or
it's combusting with oxygen--

00:00:21.470 --> 00:00:24.090
you need oxygen for the
combustion to occur.

00:00:24.090 --> 00:00:27.480
So when it burns in this
apparatus, and it burns with

00:00:27.480 --> 00:00:31.890
oxygen, it tells us -- and we
have 1.125 of it to start

00:00:31.890 --> 00:00:37.330
with-- it tells us that 3.447
grams of carbon dioxide and

00:00:37.330 --> 00:00:42.360
1.647 grams of water
were produced.

00:00:42.360 --> 00:00:45.080
So let me write that reaction
right there.

00:00:45.080 --> 00:00:48.360
So let me just write this first
sentence right here.

00:00:48.360 --> 00:00:53.100
Write all of the information
that they're giving us in that

00:00:53.100 --> 00:00:53.910
first sentence.

00:00:53.910 --> 00:00:56.000
They're saying that when we
start off with that liquid

00:00:56.000 --> 00:01:02.350
hydrocarbon that has x C's and
y H's in it, and it's in its

00:01:02.350 --> 00:01:06.670
liquid state, they say it's
a liquid hydrocarbon.

00:01:06.670 --> 00:01:13.960
And when we combust it using
oxygen in its gaseous state,

00:01:13.960 --> 00:01:25.340
we will produce carbon
dioxide and water.

00:01:25.340 --> 00:01:28.270
And you can't even really
realistically balance this

00:01:28.270 --> 00:01:30.110
equation because you don't
know how many carbons and

00:01:30.110 --> 00:01:32.060
hydrogens you have on
the left-hand side.

00:01:32.060 --> 00:01:35.450
But what they do tell us is how
many grams of everything,

00:01:35.450 --> 00:01:38.670
of at least this we start with,
and how many grams of

00:01:38.670 --> 00:01:41.270
each of these we end up with,
and we're assuming that we

00:01:41.270 --> 00:01:43.050
have an excess of oxygen.

00:01:43.050 --> 00:01:45.290
That this hydrocarbon is
the limiting reactant.

00:01:50.730 --> 00:01:52.860
We just have an abundance
of oxygen.

00:01:52.860 --> 00:01:56.470
Just enough oxygen to turn all
of this hydrocarbon, to

00:01:56.470 --> 00:01:59.970
combust all of it into carbon
dioxide and water.

00:01:59.970 --> 00:02:04.970
And they tell us that we have
1.125 grams of the mystery

00:02:04.970 --> 00:02:11.510
substance, and then it produces
3.447 grams of carbon

00:02:11.510 --> 00:02:20.470
dioxide, and 1.647
grams of water.

00:02:20.470 --> 00:02:24.180
Now to solve this-- well let me
finish reading the problem.

00:02:24.180 --> 00:02:26.430
It tells in a separate reaction
or in a separate

00:02:26.430 --> 00:02:30.100
experiment, the molar mass of
the compound was found to be

00:02:30.100 --> 00:02:32.740
86.2 grams per mole.

00:02:32.740 --> 00:02:33.970
So I'll write that down here.

00:02:33.970 --> 00:02:44.200
So CxHy molar mass is equal to--
let me write it this way

00:02:44.200 --> 00:02:46.300
instead just to go with the
convention that I've been

00:02:46.300 --> 00:02:58.620
doing-- we have 86.2 grams of
our mystery substance, CxHy

00:02:58.620 --> 00:03:04.140
per mole of our mystery
substance, CxHy.

00:03:04.140 --> 00:03:07.890
And then they're asking us
in this problem, they say

00:03:07.890 --> 00:03:12.220
determine the empirical and
molecular formulas for the

00:03:12.220 --> 00:03:15.200
unknown hydrocarbon CxHy.

00:03:15.200 --> 00:03:17.950
And just as a refresher,
empirical means just the

00:03:17.950 --> 00:03:23.550
simplest whole number ratio of
the atoms in our molecule.

00:03:23.550 --> 00:03:26.120
So the simplest ratio of our
carbons to hydrogens, or

00:03:26.120 --> 00:03:29.220
hydrogens to carbons, will give
us the empirical formula.

00:03:29.220 --> 00:03:33.610
The molecular formula is how
many carbons, exactly how many

00:03:33.610 --> 00:03:36.370
carbons and hydrogens do
we have in each atom.

00:03:36.370 --> 00:03:38.100
And I suspect we're going
to be able to use this

00:03:38.100 --> 00:03:41.420
information to come up, once
we have the empirical

00:03:41.420 --> 00:03:45.090
formulation, the empirical
formula, we then can use this

00:03:45.090 --> 00:03:48.070
molar mass right here to
figure out the actual

00:03:48.070 --> 00:03:48.950
molecular formula.

00:03:48.950 --> 00:03:52.340
But let's figure out the
empirical formula first.

00:03:52.340 --> 00:03:55.300
And the way we can do it is we
could say, well gee, how many

00:03:55.300 --> 00:03:57.590
moles of carbon dioxide
we get produced, how

00:03:57.590 --> 00:03:59.290
many moles of hydrogen?

00:03:59.290 --> 00:04:01.040
And then we could say, well in
the final product, how many

00:04:01.040 --> 00:04:02.530
moles of carbon do
we have then?

00:04:02.530 --> 00:04:06.130
And how many moles of hydrogen
do we have there?

00:04:06.130 --> 00:04:07.950
And then we can figure
out the ratio.

00:04:07.950 --> 00:04:10.620
And the ratio of carbons to
hydrogens or hydrogens to

00:04:10.620 --> 00:04:13.340
carbons in the product, that's
going to have to be the same

00:04:13.340 --> 00:04:15.320
as the ratio in the reactants.

00:04:15.320 --> 00:04:18.010
And they all came from this
reactant right there.

00:04:18.010 --> 00:04:20.120
They didn't come
from the water.

00:04:20.120 --> 00:04:22.570
And then hopefully we'll be
able to figure out the

00:04:22.570 --> 00:04:25.950
simplest whole number ratio
of hydrogens to carbons or

00:04:25.950 --> 00:04:27.180
carbons to hydrogens is.

00:04:27.180 --> 00:04:29.740
That sounds very complicated I
think, but I think if we take

00:04:29.740 --> 00:04:33.440
it step-by-step it
won't be too bad.

00:04:33.440 --> 00:04:35.980
So a good place to start, let's
figure out how many

00:04:35.980 --> 00:04:39.000
moles of each of these
substances we have. I'll start

00:04:39.000 --> 00:04:41.580
off with the carbon dioxide.

00:04:41.580 --> 00:04:44.260
So if I start off with the
carbon dioxide, what is the

00:04:44.260 --> 00:04:47.450
molar mass of carbon dioxide?

00:04:47.450 --> 00:04:53.990
So its atomic weight, carbon is
12 plus you have 2 oxygens,

00:04:53.990 --> 00:04:57.480
so 2 times-- the atomic weight
of oxygen is 16, you could

00:04:57.480 --> 00:04:59.870
look that up if you like--
but that's equal to

00:04:59.870 --> 00:05:02.440
44 atomic mass units.

00:05:02.440 --> 00:05:05.640
That's its atomic weight, which
is the weighted average

00:05:05.640 --> 00:05:08.330
of all of the isotopes and
all of that business.

00:05:08.330 --> 00:05:12.620
But if you have a whole mole of
carbon dioxide, 6.022 times

00:05:12.620 --> 00:05:15.780
10 to the 23rd of these
molecules, then you're going

00:05:15.780 --> 00:05:19.000
to have 44 grams of
the substance.

00:05:19.000 --> 00:05:30.680
So we have 44 grams of
CO2 per mole of CO2.

00:05:30.680 --> 00:05:34.540
And if we want to figure out
how many moles we have with

00:05:34.540 --> 00:05:38.690
us, we know that we're
starting off with

00:05:38.690 --> 00:05:45.840
3.447 grams of CO2.

00:05:45.840 --> 00:05:48.810
And we want to multiply this by
something that has grams of

00:05:48.810 --> 00:05:50.580
CO2 in the denominator.

00:05:50.580 --> 00:05:53.870
So we want to multiply it by
something that looks like

00:05:53.870 --> 00:05:57.270
this-- and we want to turn
it into moles of CO2.

00:05:57.270 --> 00:06:04.420
So it should have moles of
CO2 over grams of CO2.

00:06:04.420 --> 00:06:06.740
Well that's the inverse
of this right here.

00:06:06.740 --> 00:06:09.010
This is grams per mole, we
want moles per gram.

00:06:09.010 --> 00:06:14.920
So instead of 44 grams per mole,
we have 1 over 44 moles

00:06:14.920 --> 00:06:18.190
per gram, or for every 44 grams
we have 1 mole, for

00:06:18.190 --> 00:06:21.550
every 1 mole we have 44 grams.
I just inverted this.

00:06:21.550 --> 00:06:24.890
So let's multiply.

00:06:24.890 --> 00:06:29.530
The grams of CO2 cancel out with
the grams of CO2, and you

00:06:29.530 --> 00:06:35.890
are left with 3.447
divided by 44.

00:06:35.890 --> 00:06:38.040
So let's figure out
what that is.

00:06:40.780 --> 00:06:55.020
So you have 3.447 divided by 44,
which is equal to 0.07834.

00:06:55.020 --> 00:06:56.450
So 7-8-3-4.

00:06:56.450 --> 00:06:57.760
Let me remember that.

00:06:57.760 --> 00:07:04.020
So that is 0.07834 moles
of CO2-- that's

00:07:04.020 --> 00:07:05.270
our only units left.

00:07:09.370 --> 00:07:12.320
Now let's figure out how many
moles of water we have. So I

00:07:12.320 --> 00:07:14.920
could do the exact same thing
over-- let's see,

00:07:14.920 --> 00:07:16.080
where should I do that.

00:07:16.080 --> 00:07:18.900
Let me write it over here.

00:07:18.900 --> 00:07:22.830
So let's do the water in
orange right here.

00:07:22.830 --> 00:07:30.780
So we're starting off with
1.647 grams of H2O.

00:07:30.780 --> 00:07:34.710
And what is water's
molar mass?

00:07:34.710 --> 00:07:39.160
Same argument, its atomic mass
is-- each hydrogen is 1.

00:07:39.160 --> 00:07:42.750
But you have 2 of them so it's
2 times 1 plus 16, which is

00:07:42.750 --> 00:07:45.380
the atomic weight of oxygen.

00:07:45.380 --> 00:07:47.910
So 2 times 1 plus 16 is 18.

00:07:47.910 --> 00:07:49.470
That's also its molar mass.

00:07:49.470 --> 00:07:59.810
You could say 18 grams of
H2O per mole of H2O.

00:07:59.810 --> 00:08:03.330
Now to convert this, the amount
of grams we have to

00:08:03.330 --> 00:08:06.810
moles, what we want to do is
multiply it by something that

00:08:06.810 --> 00:08:11.890
has moles in the numerator and
grams in the denominator.

00:08:11.890 --> 00:08:18.600
We want grams of H2O in the
denominator and moles of H2O

00:08:18.600 --> 00:08:19.930
in the numerator.

00:08:19.930 --> 00:08:21.820
And that's the exact
inverse of this.

00:08:21.820 --> 00:08:24.450
So if we have 18 grams per
mole, we have 1 over

00:08:24.450 --> 00:08:27.310
18 moles per gram.

00:08:27.310 --> 00:08:31.400
And what is this going
to be equal to?

00:08:31.400 --> 00:08:36.549
So this cancels out with that,
so we're left with 1.647

00:08:36.549 --> 00:08:40.006
divided by 18 moles of H2O.

00:08:40.006 --> 00:08:42.549
Let me scroll over
a little bit.

00:08:42.549 --> 00:08:44.780
Get the calculator out.

00:08:44.780 --> 00:08:45.850
Clear it out.

00:08:45.850 --> 00:08:56.360
And so we have 1.647 divided
by 18, which is 0.0915.

00:08:56.360 --> 00:08:58.970
Let me do this in a new color.

00:08:58.970 --> 00:09:06.310
We have a 0.0915 moles of H2O.

00:09:06.310 --> 00:09:10.360
So so far we figured out in our
products how many moles of

00:09:10.360 --> 00:09:17.770
carbon dioxide we have, and
how many moles of water.

00:09:17.770 --> 00:09:21.520
Now, I said we want to figure
out the ratio of hydrogens to

00:09:21.520 --> 00:09:23.700
carbons or carbons to hydrogens
to figure out the

00:09:23.700 --> 00:09:25.630
empirical formula right here.

00:09:25.630 --> 00:09:28.800
And to do that we can say hey
look, all of the carbons and

00:09:28.800 --> 00:09:32.590
the hydrogens in our products,
they all had to come from our

00:09:32.590 --> 00:09:34.420
mystery hydrocarbon.

00:09:34.420 --> 00:09:38.100
So if we can figure out the
ratio of hydrogens to carbons

00:09:38.100 --> 00:09:41.470
in our products, we'll know
the ratio of hydrogens to

00:09:41.470 --> 00:09:42.490
carbons in our reactants.

00:09:42.490 --> 00:09:44.470
Because they all had to come
from this guy right here.

00:09:44.470 --> 00:09:46.890
The oxygen, obviously, had no
hydrogens and carbons in it.

00:09:46.890 --> 00:09:48.950
So all of these hydrogens
and carbons came

00:09:48.950 --> 00:09:51.520
from our mystery substance.

00:09:51.520 --> 00:09:55.450
So how many moles of
carbon do we have?

00:09:55.450 --> 00:09:58.390
How many moles of carbon do we
have in the carbon dioxide, in

00:09:58.390 --> 00:09:59.600
our products?

00:09:59.600 --> 00:10:04.470
Well the only product with
carbon is the carbon dioxide.

00:10:04.470 --> 00:10:05.660
This has no carbon in it.

00:10:05.660 --> 00:10:12.200
So if we have this many carbon
dioxides, we also have the

00:10:12.200 --> 00:10:22.830
same number of carbon atoms,
moles of atomic carbon.

00:10:22.830 --> 00:10:25.780
And you might say hey,
that's just a number.

00:10:25.780 --> 00:10:30.870
This is saying 0.07 times 6.022
times 10 to the 23rd

00:10:30.870 --> 00:10:31.760
times that huge number.

00:10:31.760 --> 00:10:34.690
A mole is just a number-- very
important to remember.

00:10:34.690 --> 00:10:37.120
So this means we have the
same number of carbons.

00:10:37.120 --> 00:10:40.060
And if we have that many
waters, how many

00:10:40.060 --> 00:10:41.670
hydrogens do we have?

00:10:41.670 --> 00:10:45.680
Well for every water molecule
we have 2 hydrogens.

00:10:45.680 --> 00:10:47.960
So we're going to have twice
as many hydrogen atoms.

00:10:47.960 --> 00:10:50.780
So let's multiply
that times 2.

00:10:50.780 --> 00:10:57.670
So times 2 is equal to 0.183.

00:10:57.670 --> 00:11:03.600
So we have 0.183.

00:11:03.600 --> 00:11:06.980
I could even put a trailing 0,
because we actually had-- well

00:11:06.980 --> 00:11:08.680
I think I lost the significant
digit here.

00:11:08.680 --> 00:11:10.890
I could have put a
zero right there.

00:11:10.890 --> 00:11:13.360
But anyway, I won't worry
too much about that.

00:11:13.360 --> 00:11:20.300
0.183 moles of atomic hydrogen,
because for every

00:11:20.300 --> 00:11:21.710
water molecule we have
2 hydrogens.

00:11:21.710 --> 00:11:24.610
So I just took this number
and multiplied by 2.

00:11:24.610 --> 00:11:29.630
And now let's figure out the
ratio of hydrogens to carbons,

00:11:29.630 --> 00:11:31.260
and that's pretty easy.

00:11:31.260 --> 00:11:33.470
We just take a ratio
of that to that.

00:11:33.470 --> 00:11:42.750
So the ratio of hydrogens to
carbons is 0.183 moles of

00:11:42.750 --> 00:11:52.400
hydrogen in our product for
every 0.07834 moles of the

00:11:52.400 --> 00:11:54.330
carbon in our product.

00:11:54.330 --> 00:11:57.590
And somehow we have to make this
into some type of a whole

00:11:57.590 --> 00:11:58.400
number ratio.

00:11:58.400 --> 00:12:00.780
Let's see whether we can
do anything here.

00:12:00.780 --> 00:12:04.560
Let's just divide this right
here and see what we get.

00:12:04.560 --> 00:12:07.690
So if we just take-- well we
have the 0.183 there, so let's

00:12:07.690 --> 00:12:20.600
divide it by 0.07834,
and you get 2.336.

00:12:20.600 --> 00:12:26.410
So this is approximately equal
to-- I'll do a new color-- is

00:12:26.410 --> 00:12:36.590
approximately equal to 2.336
moles of hydrogen for every 1

00:12:36.590 --> 00:12:39.530
mole of carbon.

00:12:39.530 --> 00:12:42.520
This doesn't look like that neat
of a ratio, but this is

00:12:42.520 --> 00:12:44.790
pretty close to 1/3.

00:12:44.790 --> 00:12:47.130
So maybe if we multiply the
numerator and the denominator

00:12:47.130 --> 00:12:50.680
by 3 what do we get?

00:12:50.680 --> 00:12:55.060
Well 3 times 2 is 6 and 3
times 1/3 is another 1.

00:12:55.060 --> 00:12:56.670
So this will be 7.

00:12:56.670 --> 00:12:59.590
7 moles of hydrogen.

00:12:59.590 --> 00:13:00.850
I'm approximating it.

00:13:00.850 --> 00:13:03.640
I'm trying to get to some
reasonable whole number.

00:13:03.640 --> 00:13:11.350
7 moles of hydrogen for every
3 moles of carbon.

00:13:11.350 --> 00:13:12.510
And this was a little
bit of an art to it.

00:13:12.510 --> 00:13:18.890
I said OK, 2.33:1 or 2.336:1,
well gee, this looks like,

00:13:18.890 --> 00:13:22.010
over here, this part of it looks
like if we multiplied

00:13:22.010 --> 00:13:24.140
that by 3 we're going
to get pretty

00:13:24.140 --> 00:13:25.330
close to a whole number.

00:13:25.330 --> 00:13:26.670
So let me multiply the
numerator and the

00:13:26.670 --> 00:13:28.040
denominator by 3.

00:13:28.040 --> 00:13:29.990
That's kind of the art there to
recognize what do you have

00:13:29.990 --> 00:13:33.590
to multiply this to turn
it into a whole number.

00:13:33.590 --> 00:13:35.460
And if you don't believe me, we
could take that number and

00:13:35.460 --> 00:13:39.720
multiply it by 3 and you get
something very, very, very,

00:13:39.720 --> 00:13:40.820
very close to 7.

00:13:40.820 --> 00:13:43.200
So that's enough for me.

00:13:43.200 --> 00:13:46.100
So you're multiplying both by 3,
you get 7 moles of hydrogen

00:13:46.100 --> 00:13:47.950
for every 3 moles of carbon.

00:13:47.950 --> 00:13:50.610
And we've figured out our
empirical formula.

00:13:50.610 --> 00:13:55.420
We have 7 hydrogens for
every 3 carbons.

00:13:55.420 --> 00:13:55.640
Right?

00:13:55.640 --> 00:13:58.710
If we have 7 moles, for every
3 moles that means we have 7

00:13:58.710 --> 00:14:00.300
hydrogens for every 3 carbons.

00:14:00.300 --> 00:14:13.430
And so the empirical formula of
our mystery hydrocarbon is

00:14:13.430 --> 00:14:21.590
for every 3 carbons we
have 7 hydrogens.

00:14:21.590 --> 00:14:22.960
Now the next thing we want
to figure is the

00:14:22.960 --> 00:14:24.280
exact molecular formula.

00:14:24.280 --> 00:14:26.740
The exact number of-- this is
just the minimum ratio.

00:14:26.740 --> 00:14:30.400
But maybe we have 9 carbons
for every 21 hydrogens.

00:14:30.400 --> 00:14:31.310
We don't know.

00:14:31.310 --> 00:14:34.070
So let's use the other
information they gave us in

00:14:34.070 --> 00:14:41.210
the problem where they told us
that the molar mass is 86.2

00:14:41.210 --> 00:14:47.310
grams per mole-- this
information right here.

00:14:47.310 --> 00:14:49.510
Let me copy that.

00:14:49.510 --> 00:14:51.350
86.2-- that looks like
an 80, but we'll

00:14:51.350 --> 00:14:54.060
remember it says 86.2.

00:14:54.060 --> 00:14:58.540
So let me paste it over here.

00:14:58.540 --> 00:15:00.250
That's what they gave
us in the problem.

00:15:00.250 --> 00:15:06.190
So just for kicks, what would be
the molar mass if this was

00:15:06.190 --> 00:15:08.120
the molecular formula?

00:15:08.120 --> 00:15:13.910
So as it's written right now--
well there's a couple of ways.

00:15:13.910 --> 00:15:15.440
We could do it algebraically
and all of that.

00:15:15.440 --> 00:15:17.520
But the simplest way is to say
as it's written right now what

00:15:17.520 --> 00:15:21.780
would be the molecular
mass of this?

00:15:21.780 --> 00:15:23.140
Or the molar mass,
I should say.

00:15:23.140 --> 00:15:29.580
So it would be 3 times
12, plus 7, times 1.

00:15:29.580 --> 00:15:36.120
3 times 12 is 36 plus
7 is equal to 43.

00:15:36.120 --> 00:15:39.710
So if this was the molecular
formula, its molecular mass

00:15:39.710 --> 00:15:42.510
would be 43 grams per mole.

00:15:42.510 --> 00:15:46.060
But they tell us in the problem
that the molecular

00:15:46.060 --> 00:15:49.900
mass is 86 grams per mole.

00:15:49.900 --> 00:15:52.110
So they're saying that the mass
is twice as much as this.

00:15:52.110 --> 00:15:55.810
We must have twice as many
atoms in the same ratio.

00:15:55.810 --> 00:16:00.660
So the molecular formula for
this-- we're in the home

00:16:00.660 --> 00:16:09.560
stretch now-- the molecular
formula must have twice as

00:16:09.560 --> 00:16:12.740
many atoms in the exact
same ratio.

00:16:12.740 --> 00:16:27.180
So it must be C3H7 times 2, or
we could write that as C6H14.

00:16:27.180 --> 00:16:32.590
So this is still a ratio
of 14:6, which is 7:3.

00:16:32.590 --> 00:16:35.030
But using this information, we
were able to figure out the

00:16:35.030 --> 00:16:38.320
exact number of carbons and the
exact number of hydrogens

00:16:38.320 --> 00:16:41.230
in this molecule.

00:16:41.230 --> 00:16:44.040
Anyway, hopefully you
found that fun.


Absolute value equation example 2

WEBVTT
Kind: captions
Language: en

00:00:00.510 --> 00:00:04.330
So we're given the following
equation, that 5 times the

00:00:04.330 --> 00:00:09.860
absolute value of x plus 3
minus 3 is equal to 7.

00:00:09.860 --> 00:00:12.020
It's always a little daunting
to see an equation with an

00:00:12.020 --> 00:00:13.310
absolute value sign.

00:00:13.310 --> 00:00:15.620
And when you only have one of
them, like this, what I like

00:00:15.620 --> 00:00:18.710
to do is isolate it, and just
kind of think it through from

00:00:18.710 --> 00:00:19.290
that point.

00:00:19.290 --> 00:00:21.570
So let's try to isolate
the absolute

00:00:21.570 --> 00:00:23.140
value of x plus 3 here.

00:00:23.140 --> 00:00:27.250
Let's try to isolate that
part of our equation.

00:00:27.250 --> 00:00:29.960
So the first thing we might want
to do is add 3 to both

00:00:29.960 --> 00:00:31.000
sides of the equation.

00:00:31.000 --> 00:00:34.270
That'll get rid of this minus 3,
or this negative 3, on the

00:00:34.270 --> 00:00:35.520
left-hand side.

00:00:35.520 --> 00:00:40.190
So let's add 3 to both sides
of this equation.

00:00:40.190 --> 00:00:44.690
And this'll turn the equation
into-- do it in that same pink

00:00:44.690 --> 00:00:48.270
color --the left-hand side
will still be 5 times the

00:00:48.270 --> 00:00:51.810
absolute value of x plus 3.

00:00:51.810 --> 00:00:54.280
The minus 3, or the negative
3 plus the 3,

00:00:54.280 --> 00:00:55.440
those will cancel out.

00:00:55.440 --> 00:00:56.770
That'll just be 0.

00:00:56.770 --> 00:01:02.230
And then that will be equal
to 7 plus 3, which is 10.

00:01:02.230 --> 00:01:05.610
Now, we have 5 times the thing
we want to isolate.

00:01:05.610 --> 00:01:08.380
The best way to isolate it
completely is to divide both

00:01:08.380 --> 00:01:11.030
sides of this equation by 5.

00:01:11.030 --> 00:01:13.800
So if you divide that side
by 5, and then the

00:01:13.800 --> 00:01:15.630
right-hand side by 5.

00:01:15.630 --> 00:01:18.390
We divide it by 5 so that these
guys will cancel out.

00:01:18.390 --> 00:01:23.050
5 times something divided by
5 is just that something.

00:01:23.050 --> 00:01:24.930
So these cancel out.

00:01:24.930 --> 00:01:25.810
That's 2.

00:01:25.810 --> 00:01:31.590
So we're left with the absolute
value of x plus 3 is

00:01:31.590 --> 00:01:35.860
equal to 10 divided
by 5, which is 2.

00:01:35.860 --> 00:01:38.350
So we've simplified the equation
a good bit, now we

00:01:38.350 --> 00:01:41.580
just have to put our thinking
caps on a little bit.

00:01:41.580 --> 00:01:45.180
If the absolute value of
something is 2, what does it

00:01:45.180 --> 00:01:47.060
mean that that something is?

00:01:47.060 --> 00:01:49.620
What are the two numbers that,
if I were take it's absolute

00:01:49.620 --> 00:01:51.805
value, I could get 2.

00:01:51.805 --> 00:01:53.450
I'll do a little thing
on the side here.

00:01:53.450 --> 00:01:56.740
We know that the absolute value
of 2 is equal to 2.

00:01:56.740 --> 00:01:59.160
So maybe this thing
was equal to 2.

00:01:59.160 --> 00:02:02.300
So maybe x plus 3
is equal to 2.

00:02:02.300 --> 00:02:05.600
If x plus 3 is equal to 2, and
you take its absolute value,

00:02:05.600 --> 00:02:07.150
you're going to get 2 again.

00:02:07.150 --> 00:02:12.960
But we also know that the
absolute value of negative 2

00:02:12.960 --> 00:02:15.700
is also equal to 2.

00:02:15.700 --> 00:02:18.560
So maybe x plus 3 is equal
to negative 2.

00:02:18.560 --> 00:02:21.480
Because if x plus 3 is equal to
negative 2, and we take its

00:02:21.480 --> 00:02:24.300
absolute value, then we're
going to get 2 again.

00:02:24.300 --> 00:02:29.990
So we could also write or x
plus 3 could be equal to

00:02:29.990 --> 00:02:30.900
negative 2.

00:02:30.900 --> 00:02:32.570
That's what I mean
about thinking

00:02:32.570 --> 00:02:33.410
about it a little bit.

00:02:33.410 --> 00:02:36.280
Another way to think about it--
I've said this in other

00:02:36.280 --> 00:02:40.500
videos --is absolute value
means distance from 0.

00:02:40.500 --> 00:02:48.420
So if we were to draw a number
line here, that is 0, this is

00:02:48.420 --> 00:02:51.420
saying that, whatever this
quantity is inside the

00:02:51.420 --> 00:02:55.550
absolute value sign, its
distance from 0 is 2.

00:02:55.550 --> 00:02:58.140
So what numbers are
2 away from 0?

00:02:58.140 --> 00:03:00.690
Well, you have positive 2.

00:03:00.690 --> 00:03:02.640
I'll write a positive
there explicitly.

00:03:02.640 --> 00:03:05.500
And you also have
a negative 2.

00:03:05.500 --> 00:03:09.170
So this thing here could be a
positive 2, or it could be a

00:03:09.170 --> 00:03:09.980
negative 2.

00:03:09.980 --> 00:03:12.120
Either way you take the absolute
value of a positive 2

00:03:12.120 --> 00:03:15.020
or a negative 2, you're
going to get a 2.

00:03:15.020 --> 00:03:16.980
So let's solve these.

00:03:16.980 --> 00:03:19.680
So over here we can subtract
3 from both

00:03:19.680 --> 00:03:21.210
sides of this equation.

00:03:21.210 --> 00:03:25.670
So if you subtract 3 from both
sides, you get-- The left-hand

00:03:25.670 --> 00:03:27.250
side, you're just
left with an x.

00:03:27.250 --> 00:03:28.650
These 3's cancel out.

00:03:28.650 --> 00:03:31.105
That's the whole point
of subtracting the 3.

00:03:31.105 --> 00:03:36.060
x will be equal to 2 minus
3 is negative 1.

00:03:36.060 --> 00:03:40.580
So that is one solution to our
absolute value equation.

00:03:40.580 --> 00:03:42.560
And what's our other solution?

00:03:42.560 --> 00:03:45.450
Well here, once again, let's
subtract 3 from both sides.

00:03:45.450 --> 00:03:48.540
So you subtract 3
from both sides.

00:03:48.540 --> 00:03:51.520
The left-hand side just
becomes an x.

00:03:51.520 --> 00:03:56.200
The right-hand side, negative
2 minus 3 is negative 5.

00:03:56.200 --> 00:04:01.000
So this is our other solution.

00:04:01.000 --> 00:04:03.570
And let's verify that
they both work.

00:04:03.570 --> 00:04:06.640
So if x is equal to negative
1, what does

00:04:06.640 --> 00:04:08.370
this equation become?

00:04:08.370 --> 00:04:12.320
We have 5 times the absolute
value of negative 1

00:04:12.320 --> 00:04:16.060
plus 3 minus 3.

00:04:16.060 --> 00:04:19.790
And if this really is a
solution, then this should be

00:04:19.790 --> 00:04:21.600
equal to positive 7.

00:04:21.600 --> 00:04:21.899
So let's see.

00:04:21.899 --> 00:04:26.700
This is 5 times negative
1 plus 3 is 2.

00:04:26.700 --> 00:04:31.510
So it's 5 times the absolute
value of 2 minus 3.

00:04:31.510 --> 00:04:32.920
The absolute value
of 2 is just 2.

00:04:32.920 --> 00:04:36.670
So this is 5 times 2 minus 3.

00:04:36.670 --> 00:04:43.740
5 times 2 is 10 minus 3, which
is, indeed, equal to 7.

00:04:43.740 --> 00:04:46.030
So this is definitely
a solution.

00:04:46.030 --> 00:04:50.110
Let's try the other one out.
x is equal to negative 5.

00:04:50.110 --> 00:04:52.320
Running out of some real estate,
but I could clear some

00:04:52.320 --> 00:04:53.090
up right here.

00:04:53.090 --> 00:04:55.570
If x is equal to negative
5, same drill.

00:04:55.570 --> 00:05:03.900
5 times the absolute value of
negative 5 plus 3 minus 3.

00:05:03.900 --> 00:05:06.300
This again should
be equal to 7.

00:05:06.300 --> 00:05:08.480
Negative 5 plus 3
is negative 2.

00:05:08.480 --> 00:05:12.110
Since there's 5 times the
absolute value of negative 2

00:05:12.110 --> 00:05:15.970
minus 3, the absolute value of
negative 2 is positive 2.

00:05:15.970 --> 00:05:24.840
So this is 5 times 2 minus 3,
which is 10 minus 3, which is

00:05:24.840 --> 00:05:27.110
again equal to 7.

00:05:27.110 --> 00:05:29.580
And notice, in this situation,
the thing in the absolute

00:05:29.580 --> 00:05:32.260
value sign became a negative 2,
because we said, oh, x plus

00:05:32.260 --> 00:05:33.960
3 could be equal
to negative 2.

00:05:33.960 --> 00:05:36.940
In the magenta situation, the
thing in the absolute value

00:05:36.940 --> 00:05:39.690
sign was positive 2, because
this was a situation where we

00:05:39.690 --> 00:05:42.760
assumed that the thing in the
absolute value sign was a

00:05:42.760 --> 00:05:43.950
positive 2.

00:05:43.950 --> 00:05:47.910
Now if we were to graph these
solutions on the number line,

00:05:47.910 --> 00:05:50.380
if we were to graph the two
solutions on a number line--

00:05:50.380 --> 00:05:53.580
I'll draw a new number
line here.

00:05:53.580 --> 00:05:57.510
Let me draw a new number
line, just like that.

00:05:57.510 --> 00:06:00.910
Both of these are less than
0, so I'll put 0 all

00:06:00.910 --> 00:06:01.520
the way over here.

00:06:01.520 --> 00:06:04.050
Maybe 0, 1, 2.

00:06:04.050 --> 00:06:09.820
Let's go negative, negative
1, negative 2, negative 3,

00:06:09.820 --> 00:06:15.320
negative 4, negative
5, negative 6.

00:06:15.320 --> 00:06:18.100
This solution right here-- x
is equal to negative 1 --we

00:06:18.100 --> 00:06:21.450
could plot on the number
line right over there.

00:06:21.450 --> 00:06:25.650
And then the other solution-- x
is equal to negative 5 --we

00:06:25.650 --> 00:06:28.430
can plot it right over there.

00:06:28.430 --> 00:06:30.980
Hopefully you found
that enjoyable.


Absolute value and number lines

WEBVTT
Kind: captions
Language: en

00:00:00.370 --> 00:00:04.230
We're told to plot these values
on a number line.

00:00:04.230 --> 00:00:06.700
And you see every one of
these values have an

00:00:06.700 --> 00:00:08.290
absolute value sign.

00:00:08.290 --> 00:00:11.020
So let's take a little bit of
a review of what absolute

00:00:11.020 --> 00:00:12.300
value even is.

00:00:18.100 --> 00:00:19.990
The way I think about
it, there's two ways

00:00:19.990 --> 00:00:21.000
to think about it.

00:00:21.000 --> 00:00:23.200
The first way to think about
it is, how far is

00:00:23.200 --> 00:00:25.270
something from 0?

00:00:25.270 --> 00:00:28.340
So let me plot this
negative 3 here.

00:00:28.340 --> 00:00:29.890
So let me do a number line.

00:00:29.890 --> 00:00:33.170
This isn't the number line for
our actual answer, or to this

00:00:33.170 --> 00:00:36.070
command-- plot these values
on a number line.

00:00:36.070 --> 00:00:38.470
I'm just first going to plot
the numbers inside the

00:00:38.470 --> 00:00:40.250
absolute value sign, and then
we're going to take the

00:00:40.250 --> 00:00:42.790
absolute value and plot those,
just like they're

00:00:42.790 --> 00:00:44.360
asking us to do.

00:00:44.360 --> 00:00:47.690
So on this number line, if this
is 0, if we go to the

00:00:47.690 --> 00:00:49.440
negative, we're going to
go to the left of 0.

00:00:49.440 --> 00:00:55.680
So this is negative 1, negative
2, negative 3.

00:00:55.680 --> 00:00:58.880
Negative 3 sits right over
there, so this is negative 3

00:00:58.880 --> 00:01:01.490
right there.

00:01:01.490 --> 00:01:04.900
The absolute value of negative
3 is essentially saying, how

00:01:04.900 --> 00:01:07.400
far are you away from 0?

00:01:07.400 --> 00:01:10.590
How far is negative 3 from 0?

00:01:10.590 --> 00:01:15.750
And you say, well, it's
1, 2, 3 away from 0.

00:01:15.750 --> 00:01:19.830
So you'd say that the absolute
value of negative 3 is equal

00:01:19.830 --> 00:01:22.470
to positive 3.

00:01:22.470 --> 00:01:25.490
Now that's really the conceptual
way to imagine

00:01:25.490 --> 00:01:26.360
absolute value.

00:01:26.360 --> 00:01:31.240
How far are you away from 0?

00:01:31.240 --> 00:01:34.110
But the easy way to calculate
absolute value signs, if you

00:01:34.110 --> 00:01:37.480
don't care too much about the
concept, is whether it's

00:01:37.480 --> 00:01:40.180
negative or positive, the
absolute value of it's always

00:01:40.180 --> 00:01:41.180
going to be positive.

00:01:41.180 --> 00:01:44.210
Absolute value of negative
3 is positive 3.

00:01:44.210 --> 00:01:47.290
Absolute value of positive
3 is still positive 3.

00:01:47.290 --> 00:01:49.880
So you're always going to get
the positive version of the

00:01:49.880 --> 00:01:50.820
number, so to speak.

00:01:50.820 --> 00:01:53.920
But conceptualy, you're just
saying how far away

00:01:53.920 --> 00:01:55.330
are you from 0.

00:01:55.330 --> 00:01:56.980
So let's do what they asked.

00:01:56.980 --> 00:02:01.150
So that first value, on this
number line, so all of these

00:02:01.150 --> 00:02:02.090
are absolute values.

00:02:02.090 --> 00:02:03.750
So they're all going to
be positive values.

00:02:03.750 --> 00:02:05.220
So they're all going to
be greater than 0.

00:02:05.220 --> 00:02:08.320
So let me draw my number
line, like this.

00:02:08.320 --> 00:02:10.789
I can do a straighter number
line than that.

00:02:10.789 --> 00:02:11.500
Let's see.

00:02:11.500 --> 00:02:14.500
Well, that's a little
bit straighter.

00:02:14.500 --> 00:02:18.210
And let's say, if this is 0,
this would be negative 1, then

00:02:18.210 --> 00:02:25.580
you'd have 1, 2,
3, 4, 5, 6, 7.

00:02:25.580 --> 00:02:26.910
I think that'll do the trick.

00:02:26.910 --> 00:02:30.150
So this first quantity here--
I'll do it in orange --the

00:02:30.150 --> 00:02:33.930
absolute value of negative 3,
we just figured out, that is

00:02:33.930 --> 00:02:34.765
positive 3.

00:02:34.765 --> 00:02:38.810
So I'll plot it right over
there, positive 3.

00:02:38.810 --> 00:02:43.750
Then this next value, right
here, the absolute value of 7.

00:02:43.750 --> 00:02:48.010
If we look over here,
1, 2, 3, 4, 5, 6, 7.

00:02:48.010 --> 00:02:50.760
7 is how far away from 0?

00:02:50.760 --> 00:02:53.430
It is 7 away from 0.

00:02:53.430 --> 00:02:58.830
So the absolute value
of 7 is equal to 7.

00:02:58.830 --> 00:03:00.190
So you already see the
pattern there.

00:03:00.190 --> 00:03:01.755
If it's negative, it just
becomes positive.

00:03:01.755 --> 00:03:05.020
If it's already positive,
it just equals itself.

00:03:05.020 --> 00:03:07.270
So plotting this value,
I'll just place

00:03:07.270 --> 00:03:10.170
it right over there.

00:03:10.170 --> 00:03:12.830
So the absolute value
of 7 is 7.

00:03:12.830 --> 00:03:16.510
Absolute value of negative
3 is positive 3.

00:03:16.510 --> 00:03:20.170
Let me mark out the 0 a little
bit better, so you see

00:03:20.170 --> 00:03:22.830
relative to 0.

00:03:22.830 --> 00:03:29.620
Now we have the absolute
value of 8 minus 12.

00:03:29.620 --> 00:03:32.670
Well, first of all, let's figure
out what 8 minus 12 is.

00:03:32.670 --> 00:03:37.080
So if you take 12 away from
8, you're at negative 4.

00:03:37.080 --> 00:03:39.050
12 less than 8 is negative 4.

00:03:39.050 --> 00:03:44.530
And you can do that on a number
line if you don't quite

00:03:44.530 --> 00:03:45.500
remember how to do this.

00:03:45.500 --> 00:03:49.070
But if you, you know, if you
take 8 away from 8, you're at

00:03:49.070 --> 00:03:51.310
0, and then you take another
one, you're at negative 1,

00:03:51.310 --> 00:03:55.220
negative 2, negative 3, all
the way to negative 4.

00:03:55.220 --> 00:03:59.580
So this is equal to the absolute
value of negative 4.

00:03:59.580 --> 00:04:05.440
If we just plot negative 4, we
go 1, 2, 3, negative 4 is

00:04:05.440 --> 00:04:06.550
right over there.

00:04:06.550 --> 00:04:09.120
But if we're taking its absolute
value, we're saying

00:04:09.120 --> 00:04:11.950
how far is negative 4 from 0?

00:04:11.950 --> 00:04:13.910
Well it's 4 away from 0.

00:04:13.910 --> 00:04:16.220
1, 2, 3, 4.

00:04:16.220 --> 00:04:18.779
So this is equal
to positive 4.

00:04:18.779 --> 00:04:20.100
So we'll plot it right here.

00:04:20.100 --> 00:04:23.980
This number line is the answer
to this command up here.

00:04:23.980 --> 00:04:29.240
So the absolute value of 8 minus
12, which is negative 4,

00:04:29.240 --> 00:04:31.690
is positive 4.

00:04:31.690 --> 00:04:34.960
Then we have the absolute
value of 0.

00:04:37.660 --> 00:04:40.650
So how far is 0 from 0?

00:04:40.650 --> 00:04:42.590
Well, it's 0 away from 0.

00:04:42.590 --> 00:04:47.860
The absolute value of 0 is 0, so
you can just plot it right

00:04:47.860 --> 00:04:48.780
over there.

00:04:48.780 --> 00:04:49.960
And we have one left.

00:04:49.960 --> 00:04:53.190
Let me pick a suitable
color here.

00:04:53.190 --> 00:04:56.050
The absolute value
of 7 minus 2.

00:04:59.540 --> 00:05:02.170
Well, 7 minus 2 is 5, so this
is the same thing as the

00:05:02.170 --> 00:05:04.380
absolute value of 5.

00:05:04.380 --> 00:05:06.655
How far is 5 away from 0?

00:05:06.655 --> 00:05:08.870
Well, it's just 5 away.

00:05:08.870 --> 00:05:10.350
It's almost, you
know, too easy.

00:05:10.350 --> 00:05:11.400
That's what makes
it confusing.

00:05:11.400 --> 00:05:15.330
If I were to plot 5,
it's 1, 2, 3, 4, 5.

00:05:15.330 --> 00:05:18.710
It is 1, 2, 3, 4,
5 spaces from 0.

00:05:18.710 --> 00:05:21.310
So the absolute value
of 5 is 5.

00:05:21.310 --> 00:05:24.420
So you plot it just like that.

00:05:24.420 --> 00:05:27.300
So conceptually, it's how
far you are away from 0.

00:05:27.300 --> 00:05:29.620
But if you think about it in
kind of just very simple

00:05:29.620 --> 00:05:32.570
terms, if it's a negative
number, it becomes a positive

00:05:32.570 --> 00:05:33.160
version of it.

00:05:33.160 --> 00:05:35.710
If it's a positive number
already, it just equals itself

00:05:35.710 --> 00:05:38.100
when you take the
absolute value.


Example: Solving for a variable

WEBVTT
Kind: captions
Language: en

00:00:00.320 --> 00:00:05.970
We are told that the formula for
finding the perimeter of a

00:00:05.970 --> 00:00:12.180
rectangle is P is equal to 2l
plus 2w, where P is the

00:00:12.180 --> 00:00:16.520
perimeter, l is the length,
and w is the width.

00:00:16.520 --> 00:00:18.840
And just to visualize what
they're saying, and you might

00:00:18.840 --> 00:00:24.080
already be familiar with this,
let me draw a rectangle.

00:00:24.080 --> 00:00:27.080
That looks like a rectangle.

00:00:27.080 --> 00:00:31.780
And if this side's length is l,
then this side's length is

00:00:31.780 --> 00:00:33.230
also going to be l.

00:00:33.230 --> 00:00:38.140
And if this width is w, then
this width up here is w.

00:00:38.140 --> 00:00:40.690
And the perimeter is just how,
what is the distance if you

00:00:40.690 --> 00:00:43.460
were to go around
this rectangle?

00:00:43.460 --> 00:00:49.310
And so, that distance is going
to be this w plus this l, plus

00:00:49.310 --> 00:00:53.350
this w-- or that width--
plus this length.

00:00:53.350 --> 00:00:59.070
And if you have 1 w and you add
it to another w, that's

00:00:59.070 --> 00:01:00.320
going to give you 2 w's.

00:01:00.320 --> 00:01:01.940
So that's 2 w's.

00:01:01.940 --> 00:01:06.700
And then if you have 1 l, and
then you have another l,

00:01:06.700 --> 00:01:08.260
that's going to give you, if you
add them together, that's

00:01:08.260 --> 00:01:09.970
going to give you 2 l's.

00:01:09.970 --> 00:01:13.150
So the perimeter is going
to be 2 l's plus 2 w's.

00:01:13.150 --> 00:01:14.720
They just wrote it in a
different order than

00:01:14.720 --> 00:01:16.300
the way I wrote it.

00:01:16.300 --> 00:01:18.650
But the same thing, so hopefully
that makes sense.

00:01:18.650 --> 00:01:22.040
Now, their question is, rewrite
the formula so that it

00:01:22.040 --> 00:01:25.630
solves for width.

00:01:25.630 --> 00:01:29.050
So the formula, the way it's
written now, it says P is

00:01:29.050 --> 00:01:30.120
equal to something.

00:01:30.120 --> 00:01:33.770
They want us to write it so
it's, this w, right here, they

00:01:33.770 --> 00:01:38.410
want it to be w is equal to a
bunch of stuff with l's and

00:01:38.410 --> 00:01:40.210
P's in it, and maybe
some numbers there.

00:01:40.210 --> 00:01:43.240
So let's think about
how we can do this.

00:01:43.240 --> 00:01:48.260
So they tell us that P is
equal to 2 times l,

00:01:48.260 --> 00:01:50.380
plus 2 times w.

00:01:50.380 --> 00:01:53.300
We want to solve for w.

00:01:53.300 --> 00:01:56.750
Well, a good starting point
might be to get rid of the l

00:01:56.750 --> 00:01:58.630
on this side of the equation.

00:01:58.630 --> 00:02:00.790
And to get rid of it on that
side of the equation, we could

00:02:00.790 --> 00:02:05.050
subtract the 2l from both
sides of the equation.

00:02:05.050 --> 00:02:06.040
So let's do it this way.

00:02:06.040 --> 00:02:08.259
So you subtract 2l over here.

00:02:08.259 --> 00:02:09.870
Minus 2l.

00:02:09.870 --> 00:02:12.070
You're also going to have to do
that on the left-hand side.

00:02:12.070 --> 00:02:15.400
So you're going to
have minus 2l.

00:02:15.400 --> 00:02:18.060
We're doing it on both sides
of the equation.

00:02:18.060 --> 00:02:21.180
And remember, an equation says
P is equal to that, so if you

00:02:21.180 --> 00:02:23.570
do anything to that, you
have to do it to P.

00:02:23.570 --> 00:02:25.780
So if you subtract 2l from this,
you're going to have to

00:02:25.780 --> 00:02:28.950
subtract 2l from P in order
for the equality

00:02:28.950 --> 00:02:31.260
to keep being true.

00:02:31.260 --> 00:02:39.450
So the left-hand side is going
to be P minus 2l, and then

00:02:39.450 --> 00:02:43.250
that is going to be equal to--
well, 2l minus 2l, the whole

00:02:43.250 --> 00:02:46.150
reason why we subtracted 2l is
because these are going to

00:02:46.150 --> 00:02:47.280
cancel out.

00:02:47.280 --> 00:02:49.350
So these cancel out,
and you're just

00:02:49.350 --> 00:02:51.870
left with a 2w here.

00:02:51.870 --> 00:02:54.170
You're just left with a 2w.

00:02:54.170 --> 00:02:55.260
We're almost there.

00:02:55.260 --> 00:02:58.250
We've almost solved for w.

00:02:58.250 --> 00:03:00.990
To finish it up, we just have
to divide both sides of this

00:03:00.990 --> 00:03:02.350
equation by 2.

00:03:02.350 --> 00:03:05.510
And the whole reason why I'm
dividing both sides of this

00:03:05.510 --> 00:03:09.400
equation by 2 is to get rid of
this 2 coefficient, this 2

00:03:09.400 --> 00:03:11.830
that's multiplying w.

00:03:11.830 --> 00:03:15.770
So if you divide both sides of
this equation by 2, once

00:03:15.770 --> 00:03:17.780
again, if you do something to
one side of the equation, you

00:03:17.780 --> 00:03:19.260
do it to the other side.

00:03:19.260 --> 00:03:22.330
The whole reason why I divided
the right-hand side by 2 is 2

00:03:22.330 --> 00:03:25.120
times anything divided by 2
is just going to be that

00:03:25.120 --> 00:03:27.240
anything, so this is
going to be a w.

00:03:27.240 --> 00:03:29.240
And then we have our
left-hand side.

00:03:29.240 --> 00:03:31.300
So we're done.

00:03:31.300 --> 00:03:38.240
If we flip these two sides, we
have our w will be equal to

00:03:38.240 --> 00:03:45.830
this thing over here-- equals
P minus 2l, all

00:03:45.830 --> 00:03:50.700
of that over 2.

00:03:50.700 --> 00:03:52.830
Now, this is the
correct answer.

00:03:52.830 --> 00:03:54.990
There's other ways to
write it, though.

00:03:54.990 --> 00:03:58.610
You might want to rewrite this,
so let me square this

00:03:58.610 --> 00:04:02.840
off, because this is completely
the correct answer.

00:04:02.840 --> 00:04:05.590
This is the correct answer, but
there's other ways that

00:04:05.590 --> 00:04:07.650
you might have been able to
get this answer, other

00:04:07.650 --> 00:04:09.360
expressions for this answer.

00:04:09.360 --> 00:04:12.630
You might have also, you know,
another completely legitimate

00:04:12.630 --> 00:04:15.720
way to do this problem-- let me
write it this way-- so our

00:04:15.720 --> 00:04:21.779
original problem is P is equal
to 2l plus 2w-- is on this

00:04:21.779 --> 00:04:25.960
right-hand side, what if
we factor out a 2?

00:04:25.960 --> 00:04:27.010
So let me make this clear.

00:04:27.010 --> 00:04:30.280
You have a 2 here, and
you have a 2 here.

00:04:30.280 --> 00:04:33.090
So you could imagine
undistributing the 2.

00:04:33.090 --> 00:04:41.650
So we would get P is equal
to 2 times l plus w.

00:04:41.650 --> 00:04:45.680
This is an equally legitimate
way to do this problem.

00:04:45.680 --> 00:04:49.780
Now, we can divide both sides of
this equation by 2, so that

00:04:49.780 --> 00:04:52.710
we get rid of this 2 on
the right-hand side.

00:04:52.710 --> 00:04:58.170
So if you divide both sides of
this equation by 2, these 2's

00:04:58.170 --> 00:05:00.760
are going to cancel out-- 2
times anything divided by 2 is

00:05:00.760 --> 00:05:02.220
just going to be the
anything-- is

00:05:02.220 --> 00:05:04.440
equal to P over 2.

00:05:04.440 --> 00:05:07.280
So let me just rewrite
this over here.

00:05:07.280 --> 00:05:08.350
Let me just rewrite this.

00:05:08.350 --> 00:05:18.310
So we will get P over 2 is going
to be equal to l plus w.

00:05:18.310 --> 00:05:21.460
And then if we want to solve for
w, we just subtract l from

00:05:21.460 --> 00:05:22.500
both sides.

00:05:22.500 --> 00:05:24.320
And sometimes, you know, you
could write it in a separate

00:05:24.320 --> 00:05:25.030
line like this.

00:05:25.030 --> 00:05:27.840
Sometimes you could just
write it like this.

00:05:27.840 --> 00:05:30.560
You could say, I'm going to
subtract an l on that side.

00:05:30.560 --> 00:05:31.790
If I do it on that side,
I have to do it

00:05:31.790 --> 00:05:32.830
on this side, too.

00:05:32.830 --> 00:05:35.470
That's the same thing as
adding a negative l.

00:05:35.470 --> 00:05:39.160
And so the right-hand side,
you're just left with a w.

00:05:42.570 --> 00:05:45.230
And then the left-hand side,
you're going to have, it could

00:05:45.230 --> 00:05:49.080
be a negative l plus P over 2,
or you could just change the

00:05:49.080 --> 00:05:58.190
order, and you can write this
as P over 2 minus l.

00:05:58.190 --> 00:06:01.650
And this is also an equally
legitimate answer.

00:06:01.650 --> 00:06:02.750
And you're probably saying,
hey Sal, wait.

00:06:02.750 --> 00:06:04.410
These things look different.

00:06:04.410 --> 00:06:07.850
P minus 2l over 2, that
looks different than P

00:06:07.850 --> 00:06:09.530
over 2 minus l.

00:06:09.530 --> 00:06:11.880
And they're not.

00:06:11.880 --> 00:06:12.580
Think about this.

00:06:12.580 --> 00:06:16.780
We could rewrite this as P-- let
me do this with the same

00:06:16.780 --> 00:06:26.690
colors-- this over here is the
same thing as P over 2

00:06:26.690 --> 00:06:32.570
minus 2l over 2.

00:06:32.570 --> 00:06:32.825
Right?

00:06:32.825 --> 00:06:35.450
If I have a minus b and they're
both being divided by

00:06:35.450 --> 00:06:38.360
2, I can just separate-- you can
imagine I'm distributing

00:06:38.360 --> 00:06:40.850
the division by 2
right over here.

00:06:40.850 --> 00:06:42.640
And over here, 2 times
l divided by 2,

00:06:42.640 --> 00:06:44.040
that's just an l.

00:06:44.040 --> 00:06:53.710
So this is going to be equal to
P over 2 minus l, which is

00:06:53.710 --> 00:06:57.440
the exact same thing as
this right there.


Introduction to the surface integral

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:04.140
In the last video, we finished
off with these two results.

00:00:04.140 --> 00:00:06.850
We started off just thinking
about what it means to take

00:00:06.850 --> 00:00:10.000
the partial derivative of
vector-valued function, and I

00:00:10.000 --> 00:00:13.000
got to these kind of, you might
call them, bizarre results.

00:00:13.000 --> 00:00:15.230
You know, what was the whole
point in getting here, Sal?

00:00:15.230 --> 00:00:18.820
And the whole point is so that
I can give you the tools you

00:00:18.820 --> 00:00:21.840
need to understand what
a surface integral is.

00:00:21.840 --> 00:00:27.320
So let's just think about,
let's draw the st plane,

00:00:27.320 --> 00:00:29.640
and then see how it gets
transformed into

00:00:29.640 --> 00:00:31.720
this surface r.

00:00:31.720 --> 00:00:33.620
So let's do that.

00:00:36.490 --> 00:00:43.760
Let's say that is the t-axis,
and let's say that that is the

00:00:43.760 --> 00:00:49.240
s-axis, and let's say that our
vector-valued function, our

00:00:49.240 --> 00:00:52.550
positioned vector-valued
function, is defined from s's

00:00:52.550 --> 00:00:57.590
between a and b, I'm just
picking arbitrary boundaries,

00:00:57.590 --> 00:01:01.550
and between t being
equal to c and d.

00:01:04.250 --> 00:01:09.320
So the area under question, if
you take any t and any s in

00:01:09.320 --> 00:01:15.710
this rectangle right here, it
will be mapped to part

00:01:15.710 --> 00:01:16.890
of that surface.

00:01:16.890 --> 00:01:19.100
And if you map each of these
points, you will eventually

00:01:19.100 --> 00:01:21.130
get the surface r.

00:01:21.130 --> 00:01:25.250
So let me draw r
in 3 dimensions.

00:01:25.250 --> 00:01:26.770
A surface in 3D.

00:01:26.770 --> 00:01:33.090
So that is our x-axis, that
is our y-axis, and then

00:01:33.090 --> 00:01:36.930
that is the z-axis.

00:01:36.930 --> 00:01:40.780
And just as a bit of a
reminder, it might look

00:01:40.780 --> 00:01:42.370
something like this.

00:01:42.370 --> 00:01:46.570
If we were to, this point right
here, where s is equal to a,

00:01:46.570 --> 00:01:51.560
and t is equal to c, remember,
we're going to draw the surface

00:01:51.560 --> 00:01:54.540
indicated by the position
vector function

00:01:54.540 --> 00:01:57.380
s, r of s and t.

00:01:57.380 --> 00:02:01.230
So this point right here, when
s is a and t is c, maybe it

00:02:01.230 --> 00:02:04.120
maps to, I'm just, you
know, that point!

00:02:04.120 --> 00:02:05.070
Right there.

00:02:05.070 --> 00:02:09.540
When you take a and c, and you
put it into this thing over

00:02:09.540 --> 00:02:12.870
here, you're just going to get
the vector that points at that.

00:02:12.870 --> 00:02:15.840
So you could say, it'll
give you a position vector

00:02:15.840 --> 00:02:19.170
that'll point right at that
position, right there.

00:02:19.170 --> 00:02:23.407
And then, let's say that this
line, right here, if we were to

00:02:23.407 --> 00:02:31.570
hold s constant at a, and if we
were to just vary t from c

00:02:31.570 --> 00:02:36.810
to d, maybe that looks
something like this.

00:02:36.810 --> 00:02:40.080
I'm just drawing some
arbitrary contour there.

00:02:40.080 --> 00:02:45.980
Maybe if we hold t constant at
c, and vary s from a to b,

00:02:45.980 --> 00:02:50.210
maybe that'll look at
something like that.

00:02:50.210 --> 00:02:50.980
I don't know.

00:02:50.980 --> 00:02:53.980
I'm just trying to
show you an example.

00:02:53.980 --> 00:02:58.150
So this point right here would
correspond to that point right

00:02:58.150 --> 00:03:03.380
there, when you put it into the
vector-valued function r, you

00:03:03.380 --> 00:03:07.420
would get a vector that points
to that point, just like that.

00:03:07.420 --> 00:03:11.710
And this point right here in
purple, when you evaluate r of

00:03:11.710 --> 00:03:16.660
s and t, it'll give you a
vector that points right there,

00:03:16.660 --> 00:03:18.850
to that point over there, and
we could do a couple of other

00:03:18.850 --> 00:03:21.430
points, just to get an idea of
what the surface looks like,

00:03:21.430 --> 00:03:24.390
although I'm trying to keep
things as general as possible.

00:03:24.390 --> 00:03:30.730
So maybe, let me do it
in this bluish color.

00:03:30.730 --> 00:03:37.350
This, if we hold p at d and
vary s from a to b, we're

00:03:37.350 --> 00:03:38.230
going to start here.

00:03:38.230 --> 00:03:41.540
This is when t is d and s is a.

00:03:41.540 --> 00:03:44.880
And when you vary it, maybe
you get something like that.

00:03:44.880 --> 00:03:45.320
I don't know.

00:03:45.320 --> 00:03:50.900
So this point right here would
correspond to a vector that

00:03:50.900 --> 00:03:56.430
points to that point, right
there, and then finally this

00:03:56.430 --> 00:04:02.600
line, or this, if we hold s at
b and vary t between c and d,

00:04:02.600 --> 00:04:08.480
we're going to go from that
point, to that point.

00:04:08.480 --> 00:04:12.140
So it's going to look something
like this-- oh, sorry, we're

00:04:12.140 --> 00:04:14.630
going to go from this
point to that point.

00:04:14.630 --> 00:04:18.580
We're holding s at b, varying
t from c to d, maybe it

00:04:18.580 --> 00:04:21.290
looks something like that.

00:04:21.290 --> 00:04:26.200
So our surface, we went from
this nice rectangular area in

00:04:26.200 --> 00:04:30.250
the ts plane, and it gets
transformed into this

00:04:30.250 --> 00:04:31.880
wacky-looking surface.

00:04:31.880 --> 00:04:34.920
We could even draw some
other things right here.

00:04:34.920 --> 00:04:40.150
Let's say we get some
arbitrary value.

00:04:40.150 --> 00:04:42.720
Let me pick a new color,
I'll do it in white.

00:04:42.720 --> 00:04:44.430
Or a new noncolor.

00:04:44.430 --> 00:04:46.860
And let's say if we hold s at
that constant value and we

00:04:46.860 --> 00:04:51.220
vary t, maybe that will look
for something like this.

00:04:51.220 --> 00:04:55.690
Maybe that we'll look something
like, well, I don't know.

00:04:55.690 --> 00:04:58.180
Maybe it'll look
something like that.

00:04:58.180 --> 00:05:01.520
So you get an idea of what
the surface might look like.

00:05:01.520 --> 00:05:04.140
Now, given this, I want
to think about what

00:05:04.140 --> 00:05:06.350
these quantities are.

00:05:06.350 --> 00:05:08.830
And then when we visualize what
these quantities are, we'll be

00:05:08.830 --> 00:05:11.630
able to kind of use these
results of the last videos

00:05:11.630 --> 00:05:15.160
to do something that I
think will be useful.

00:05:15.160 --> 00:05:19.330
So let's say that we
pick arbitrary s and t.

00:05:19.330 --> 00:05:23.860
So this is the point, let me
just we pick it right here.

00:05:27.140 --> 00:05:30.020
That is the point s, t.

00:05:30.020 --> 00:05:31.090
s comma t.

00:05:31.090 --> 00:05:36.390
If you were to put those values
in here maybe it maps to, and I

00:05:36.390 --> 00:05:38.850
want to make sure I'm
consistent with everything I've

00:05:38.850 --> 00:05:43.720
drawn, maybe it maps to
this point right here.

00:05:43.720 --> 00:05:46.430
Maybe it maps to that
point right there.

00:05:46.430 --> 00:05:52.620
So this point right here,
that is r of s and t.

00:05:52.620 --> 00:05:53.940
For a particular s and t.

00:05:53.940 --> 00:05:55.650
I mean, I could put little
subscripts here, but

00:05:55.650 --> 00:05:56.790
I want to be general.

00:05:56.790 --> 00:05:59.070
I could call this a, well,
I already used a and b.

00:05:59.070 --> 00:06:02.010
I could call this x and y,
this would be r of x and y.

00:06:02.010 --> 00:06:05.820
It would map to that
point, right there.

00:06:05.820 --> 00:06:10.090
So that right there,
or that right there.

00:06:10.090 --> 00:06:13.500
Now let's see what happens
if we take, if we move

00:06:13.500 --> 00:06:15.790
just in the s direction.

00:06:15.790 --> 00:06:20.270
So this is we could
do that as s.

00:06:20.270 --> 00:06:24.310
Now let us move forward
by a differential, by a

00:06:24.310 --> 00:06:26.310
super small amount of s.

00:06:26.310 --> 00:06:30.830
So this right here, let's
call that a s plus a super

00:06:30.830 --> 00:06:33.150
small differential in s.

00:06:33.150 --> 00:06:33.910
That's right there.

00:06:33.910 --> 00:06:35.850
So that point.

00:06:35.850 --> 00:06:39.430
Let me do that in a better
color, in this yellow.

00:06:39.430 --> 00:06:42.320
So that point right there
is the point s plus

00:06:42.320 --> 00:06:44.080
my differential of s.

00:06:44.080 --> 00:06:46.550
I could write delta s, but
I wanted a super small

00:06:46.550 --> 00:06:49.560
change in s, comma t.

00:06:49.560 --> 00:06:52.490
And what is that going
to get mapped to?

00:06:52.490 --> 00:06:55.380
Well, if we apply these two
point in r, we're going to

00:06:55.380 --> 00:07:01.660
get something that maybe
is right over there.

00:07:01.660 --> 00:07:02.570
And I want to be very clear.

00:07:02.570 --> 00:07:11.190
This right here, that is
r of s plus ds comma t.

00:07:11.190 --> 00:07:11.850
That's what that is.

00:07:11.850 --> 00:07:15.520
That's the point when we just
shift s by a super small

00:07:15.520 --> 00:07:19.000
differential, this distance
here, you can view as ds.

00:07:19.000 --> 00:07:21.400
It's a super small change in s.

00:07:21.400 --> 00:07:24.930
And then when we map it or
transform it, or put that

00:07:24.930 --> 00:07:29.180
point into our vector-valued
function, let me copy and paste

00:07:29.180 --> 00:07:32.600
the original vector-valued
function, just so we have a

00:07:32.600 --> 00:07:34.900
good image of what we're
talking about this whole time.

00:07:41.080 --> 00:07:44.620
Let me put it right down there.

00:07:44.620 --> 00:07:51.000
So just to be clear what's
going on, when we took this

00:07:51.000 --> 00:07:56.160
little blue point right here,
this s and t, and we put the s

00:07:56.160 --> 00:08:00.480
and t values here, we get a
vector that points to that

00:08:00.480 --> 00:08:02.560
point on the surface,
right there.

00:08:02.560 --> 00:08:06.770
When you add a ds to your
s-values, you get a vector

00:08:06.770 --> 00:08:09.660
that points at that
yellow point right there.

00:08:12.570 --> 00:08:16.695
So going back to the results we
got in the last presentation,

00:08:16.695 --> 00:08:18.640
or the last video,
what is this?

00:08:18.640 --> 00:08:22.967
r of s plus delta s, or r of s
plus ds, the differential of

00:08:22.967 --> 00:08:27.250
s, t, well, that is
that, right there.

00:08:27.250 --> 00:08:31.020
That is the vector that
points to that position.

00:08:31.020 --> 00:08:33.190
This right here is a vector
that points to this

00:08:33.190 --> 00:08:34.480
blue position.

00:08:34.480 --> 00:08:36.760
So what is the difference
of those two vectors?

00:08:36.760 --> 00:08:40.430
And this is a bit of
basic vector math,

00:08:40.430 --> 00:08:41.060
you might remember.

00:08:41.060 --> 00:08:43.910
The difference of these two
vectors, head to tails.

00:08:43.910 --> 00:08:45.830
The difference of these
two vectors is going

00:08:45.830 --> 00:08:47.020
to be this vector.

00:08:47.020 --> 00:08:49.450
If you subtract this vector
that vector, you're going to

00:08:49.450 --> 00:08:51.190
get that vector, right there.

00:08:54.695 --> 00:08:57.650
A vector that looks
just like that.

00:08:57.650 --> 00:09:01.380
So that's what this is
equal to, that vector.

00:09:01.380 --> 00:09:02.340
And it makes sense.

00:09:02.340 --> 00:09:06.450
This blue vector plus the
orange vector, this vector,

00:09:06.450 --> 00:09:08.120
right here, plus the orange
vector, is equal

00:09:08.120 --> 00:09:08.755
to this vector.

00:09:08.755 --> 00:09:09.860
It makes complete sense.

00:09:09.860 --> 00:09:11.080
Heads to tails.

00:09:11.080 --> 00:09:12.650
So that's what that represents.

00:09:12.650 --> 00:09:14.625
Now let's do the same
thing in the t direction.

00:09:17.880 --> 00:09:19.020
I'm running out of color.

00:09:21.560 --> 00:09:24.800
I'll go back to the pink,
or maybe the magenta.

00:09:24.800 --> 00:09:26.610
So we had that s and t.

00:09:26.610 --> 00:09:32.960
Now if we go up a little bit,
in that direction, let's say

00:09:32.960 --> 00:09:38.360
that that is t, so this is the
point s, t plus a super small

00:09:38.360 --> 00:09:40.910
change in t, that's that
point right there.

00:09:40.910 --> 00:09:46.180
This distance right there is
dt, you can view it that way.

00:09:46.180 --> 00:09:51.280
If you put s and t plus dt into
our vector value function,

00:09:51.280 --> 00:09:52.180
what are you to get?

00:09:52.180 --> 00:09:54.700
You're going to get a vector
that maybe points to

00:09:54.700 --> 00:09:57.390
this point, right here.

00:09:57.390 --> 00:09:58.300
Maybe I'll draw it right here.

00:09:58.300 --> 00:10:00.690
Maybe it points to this
point, right here.

00:10:00.690 --> 00:10:02.660
A vector that points
right there.

00:10:02.660 --> 00:10:05.910
So that will be mapped to a
vector that points to that

00:10:05.910 --> 00:10:07.550
position, right over there.

00:10:07.550 --> 00:10:11.860
Now by the same argument that
we did on the s-side, this

00:10:11.860 --> 00:10:14.910
point, or the vector that
points to that, that

00:10:14.910 --> 00:10:19.430
is r of st plus dt.

00:10:19.430 --> 00:10:24.250
That is the exact same thing
as that right there, and of

00:10:24.250 --> 00:10:26.940
course, this, we already saw.

00:10:26.940 --> 00:10:29.800
This is the same thing
as that over there.

00:10:29.800 --> 00:10:33.370
So what is that vector
minus this blue vector?

00:10:33.370 --> 00:10:36.830
The magenta vector
minus the blue vector?

00:10:36.830 --> 00:10:40.100
Well, once again, this
is hopefully a bit of a

00:10:40.100 --> 00:10:41.960
review of adding vectors.

00:10:41.960 --> 00:10:44.600
It's going to be a vector
that looks like this.

00:10:44.600 --> 00:10:45.630
I'll do it in white.

00:10:45.630 --> 00:10:47.520
It's going to be a vector
that looks like that.

00:10:52.310 --> 00:10:54.740
And you can imagine, if you
take the blue vector plus the

00:10:54.740 --> 00:10:57.210
white vector, the blue vector
plus this white vector is going

00:10:57.210 --> 00:10:58.320
to equal this purple vector.

00:10:58.320 --> 00:11:00.910
So it makes sense, if the
purple vector minus the blue

00:11:00.910 --> 00:11:04.190
vector is going to be equal
to this white vector.

00:11:04.190 --> 00:11:06.330
So something interesting
is going on here.

00:11:06.330 --> 00:11:10.020
I have these two, this is a
vector that is kind of going

00:11:10.020 --> 00:11:15.760
along this parameterized
surface, as we changed our

00:11:15.760 --> 00:11:17.590
s by a super small amount.

00:11:17.590 --> 00:11:20.700
And then this is a vector that
is going along our surface if

00:11:20.700 --> 00:11:25.030
we change our t by a
super small amount.

00:11:25.030 --> 00:11:28.190
Now, you may or may not
remember this, and I've

00:11:28.190 --> 00:11:31.280
done several videos where
I showed this to you.

00:11:31.280 --> 00:11:35.500
But the magnitude, if I take 2
vectors, and I take their cross

00:11:35.500 --> 00:11:40.370
product, so if I take the cross
product of a and b, and I take

00:11:40.370 --> 00:11:42.870
the magnitude of the resulting
vector-- remember, when you

00:11:42.870 --> 00:11:45.870
take the cross product, you
get a third vector that is

00:11:45.870 --> 00:11:47.530
perpendicular to both of these.

00:11:47.530 --> 00:11:51.030
But if you were just to take
the magnitude of that vector,

00:11:51.030 --> 00:12:04.280
that is equal to the area
of a parallelogram,

00:12:04.280 --> 00:12:09.510
defined by a and b.

00:12:09.510 --> 00:12:11.290
What do I mean by that?

00:12:11.290 --> 00:12:20.940
Well, if that is vector a and
that is vector b, that's a and

00:12:20.940 --> 00:12:24.400
that is b, if you were to just
take the cross product of those

00:12:24.400 --> 00:12:26.290
two, you're going to
get a third vector.

00:12:26.290 --> 00:12:27.880
You're going to get a third
vector that's perpendicular

00:12:27.880 --> 00:12:30.660
to both of them, it kind
of popped out of the page.

00:12:30.660 --> 00:12:33.560
That would be a cross b.

00:12:33.560 --> 00:12:37.050
But the magnitude of this, so
if you just take a cross

00:12:37.050 --> 00:12:38.640
product, you're going
to get a vector.

00:12:38.640 --> 00:12:41.210
But then if you take the
magnitude of that vector,

00:12:41.210 --> 00:12:43.300
you're just saying, how
big is that vector, how

00:12:43.300 --> 00:12:44.690
long is that vector?

00:12:44.690 --> 00:12:47.160
That's going to be the
area of the parallelogram

00:12:47.160 --> 00:12:48.690
defined by a and b.

00:12:48.690 --> 00:12:50.650
And I've proved that in the
linear algebra videos, maybe

00:12:50.650 --> 00:12:51.900
I'll prove it again in this.

00:12:51.900 --> 00:12:57.690
I mean it's because-- well, I
won't go into that in detail.

00:12:57.690 --> 00:13:00.290
I've done it before, don't want
to make this video too long.

00:13:00.290 --> 00:13:03.900
So the parallelogram defined by
a and b, you just imagine a,

00:13:03.900 --> 00:13:07.400
and then take another kind of
parallel version of a, is right

00:13:07.400 --> 00:13:10.230
over there, and another
parallel version of b

00:13:10.230 --> 00:13:11.950
is right over there.

00:13:11.950 --> 00:13:16.980
So this is the parallelogram
defined by a and b.

00:13:16.980 --> 00:13:20.910
So, going back to our surface
example, if we were to take the

00:13:20.910 --> 00:13:25.070
cross product of this orange
vector and this white vector,

00:13:25.070 --> 00:13:28.720
I'm going to get the surface
area, I'm going to get the area

00:13:28.720 --> 00:13:32.580
of the parallelogram, defined
by these two vectors.

00:13:32.580 --> 00:13:34.760
So if I take the parallel
to that one, it will look

00:13:34.760 --> 00:13:38.800
something like this, and then a
parallel to the orange one it

00:13:38.800 --> 00:13:41.570
will look something like that.

00:13:41.570 --> 00:13:45.360
So if I take the cross product
of that and that, I'm going to

00:13:45.360 --> 00:13:49.880
get the area of that
parallelogram.

00:13:49.880 --> 00:13:51.620
Now you might say, this is a
surface, you're taking a

00:13:51.620 --> 00:13:53.250
straight-up parallelogram,
but remember, these are

00:13:53.250 --> 00:13:55.170
super small changes.

00:13:55.170 --> 00:13:58.410
So you can imagine, a surface
can be broken up into

00:13:58.410 --> 00:14:03.240
super small changes in
parallelograms, or infinitely

00:14:03.240 --> 00:14:04.750
many parallelograms.

00:14:04.750 --> 00:14:08.130
And the more parallelograms you
have, the better approximation

00:14:08.130 --> 00:14:10.160
of the surface you're
going to have.

00:14:10.160 --> 00:14:13.560
And this is no different than
when we first took integrals.

00:14:13.560 --> 00:14:15.690
We approximated the area
under the curve with a

00:14:15.690 --> 00:14:17.600
bunch of rectangles.

00:14:17.600 --> 00:14:20.050
The more rectangles
we had, the better.

00:14:20.050 --> 00:14:28.190
So let's call this little
change in our surface d sigma,

00:14:28.190 --> 00:14:31.660
for a little change, for a
little amount of our surface.

00:14:31.660 --> 00:14:33.820
And we could even say that, you
know, the surface area of the

00:14:33.820 --> 00:14:36.580
surface will be the infinite
sum of all of these

00:14:36.580 --> 00:14:39.070
infinitely small d sigmas.

00:14:39.070 --> 00:14:40.700
And there's actually a
little notation for that.

00:14:40.700 --> 00:14:48.750
So surface area is equal to, we
could integrate over the

00:14:48.750 --> 00:14:52.950
surface, and the notation
usually is a capital sigma for

00:14:52.950 --> 00:14:56.080
a surface as opposed to a
region or-- so you're

00:14:56.080 --> 00:14:57.970
integrating over the surface,
and you do a double integral,

00:14:57.970 --> 00:14:59.490
because you're going in
two directions, right?

00:14:59.490 --> 00:15:02.450
A surface is a kind of a
folded, two-dimensional

00:15:02.450 --> 00:15:03.810
structure.

00:15:03.810 --> 00:15:05.310
And you're going to take
the infinite sum of

00:15:05.310 --> 00:15:06.380
all of the d sigmas.

00:15:09.690 --> 00:15:11.460
So this would be
the surface area.

00:15:11.460 --> 00:15:13.110
So that's what a d sigma is.

00:15:13.110 --> 00:15:15.950
Now we just figured out, we
just said, well, that d sigma

00:15:15.950 --> 00:15:20.050
can be represented, that value,
that area, of that little part

00:15:20.050 --> 00:15:22.210
of the surface, of that
parallelogram, can be

00:15:22.210 --> 00:15:25.800
represented as a cross product
of those two vectors.

00:15:25.800 --> 00:15:26.550
So let me write here.

00:15:26.550 --> 00:15:28.840
And this is not
rigorous mathematics.

00:15:28.840 --> 00:15:30.930
The whole point here is to give
you the intuition of what a

00:15:30.930 --> 00:15:32.710
surface integral is all about.

00:15:32.710 --> 00:15:40.670
So we can write that d sigma is
equal to the cross product of

00:15:40.670 --> 00:15:42.940
the orange vector and
the white vector.

00:15:42.940 --> 00:15:45.270
The orange vector is
this, but we could also

00:15:45.270 --> 00:15:45.860
write it like this.

00:15:45.860 --> 00:15:47.840
This was the result
from the last video.

00:15:47.840 --> 00:15:49.460
I'll write it in orange.

00:15:49.460 --> 00:15:57.960
So the partial of r with
respect to, I'm running out of

00:15:57.960 --> 00:16:04.400
space, with respect to s, ds,
and it's going-- well, d sigma

00:16:04.400 --> 00:16:07.120
is going to be the magnitude
of the cross product, not

00:16:07.120 --> 00:16:07.860
just the cross product.

00:16:07.860 --> 00:16:09.705
The cross product by itself
will just give you a vector,

00:16:09.705 --> 00:16:11.930
and that's going to be useful
when we start doing

00:16:11.930 --> 00:16:13.800
vector-valued surface
integrals, but just think

00:16:13.800 --> 00:16:14.360
about it this way.

00:16:14.360 --> 00:16:17.400
So this orange vector is
the same thing is that.

00:16:17.400 --> 00:16:19.680
And we're going to take
the cross product of that

00:16:19.680 --> 00:16:21.280
with this white vector.

00:16:21.280 --> 00:16:24.830
This white vector is the same
thing as that, which we saw,

00:16:24.830 --> 00:16:26.870
which was the same
thing as this.

00:16:26.870 --> 00:16:28.900
The partial of r with
respect to t, dt.

00:16:35.980 --> 00:16:38.430
And we saw, if we take the
magnitude of that, that's going

00:16:38.430 --> 00:16:43.630
to be equal to our little small
change in area, or the

00:16:43.630 --> 00:16:47.500
area of this little
parallelogram over here

00:16:47.500 --> 00:16:50.280
Now, you may or may not
remember that if you take

00:16:50.280 --> 00:16:52.390
these, so let's just be clear.

00:16:52.390 --> 00:16:55.490
This and this, these
are vectors, right?

00:16:55.490 --> 00:16:57.480
When you take the partial
derivative of a vector-valued

00:16:57.480 --> 00:16:59.560
function, you're still
getting a vector.

00:16:59.560 --> 00:17:01.400
This ds, this is a number.

00:17:01.400 --> 00:17:02.810
That's a number and
that's a number.

00:17:02.810 --> 00:17:05.600
And you might remember, when
we, in the linear algebra or

00:17:05.600 --> 00:17:08.960
whenever you first saw taking
cross products, taking the

00:17:08.960 --> 00:17:10.850
cross product of some
scalar multiple.

00:17:10.850 --> 00:17:12.670
You can take the scalars out.

00:17:12.670 --> 00:17:14.870
So if we take this number and
that number, we essentially

00:17:14.870 --> 00:17:17.720
factor them out of
the cross product.

00:17:17.720 --> 00:17:22.190
This is going to be equal to
the magnitude of the cross

00:17:22.190 --> 00:17:30.280
product of the partial of r,
with respect to s, crossed with

00:17:30.280 --> 00:17:35.750
the partial of r with respect
to t, and then all of

00:17:35.750 --> 00:17:38.370
that times these two
guys, over here.

00:17:38.370 --> 00:17:42.820
Times ds and dt.

00:17:42.820 --> 00:17:44.920
So I wrote this here, hey,
maybe our surface area, if we

00:17:44.920 --> 00:17:47.830
were to take the sum of all of
these little d sigmas, but

00:17:47.830 --> 00:17:49.880
there's no obvious way
to evaluate that.

00:17:49.880 --> 00:17:53.980
But we know that all of the d
sigmas, they're the same thing

00:17:53.980 --> 00:17:56.500
as if you take all of the
ds's and all of the dt's.

00:17:56.500 --> 00:17:58.890
So you take all of the
ds's, all of the dt's.

00:17:58.890 --> 00:18:01.240
So this is a ds
times a dt, right?

00:18:01.240 --> 00:18:04.570
A ds times a dt, ds times
a dt is right there.

00:18:04.570 --> 00:18:12.420
If we multiply this times the
cross product of the partial

00:18:12.420 --> 00:18:17.740
derivative, this times this is
going to give us this area.

00:18:17.740 --> 00:18:22.940
So if we summed up all of this
times this, or this times this,

00:18:22.940 --> 00:18:26.490
if we summed them up over this
entire region, we will get all

00:18:26.490 --> 00:18:28.210
of the parallelograms
in this region.

00:18:28.210 --> 00:18:30.280
We will get the surface area.

00:18:30.280 --> 00:18:33.870
So we can write-- I know this
is all a little bit convoluted.

00:18:33.870 --> 00:18:35.810
And you need to kind of
ponder a little bit.

00:18:35.810 --> 00:18:37.830
Surface intervals, at least in
my head, are one of the hardest

00:18:37.830 --> 00:18:40.910
things to really visualize, but
it all hopefully makes sense.

00:18:40.910 --> 00:18:44.380
So we can say that this thing
right over here, the sum of all

00:18:44.380 --> 00:18:49.400
of the little parallelograms on
our surface, or the surface

00:18:49.400 --> 00:18:53.250
area, is going to be equal to,
instead of taking the sum over

00:18:53.250 --> 00:18:58.280
the surface, let's take the sum
of all the ds times dt's over

00:18:58.280 --> 00:19:00.430
this region right here.

00:19:00.430 --> 00:19:01.740
And of course, we're also
going to have to take

00:19:01.740 --> 00:19:02.750
this cross product here.

00:19:02.750 --> 00:19:04.370
And we know how to do that.

00:19:04.370 --> 00:19:06.340
That's a double integral.

00:19:06.340 --> 00:19:10.440
for going to take the double
integral over this, we could

00:19:10.440 --> 00:19:14.400
call it this region, or
this area, right here.

00:19:14.400 --> 00:19:18.460
That area is the same thing
as that whole area, right

00:19:18.460 --> 00:19:22.230
over there, of this thing.

00:19:22.230 --> 00:19:25.710
I'll just write it in yellow.

00:19:25.710 --> 00:19:30.080
Of the cross product of the
partial of r with respect to

00:19:30.080 --> 00:19:35.550
s, and the partial of
r with respect to t.

00:19:35.550 --> 00:19:38.100
ds and dt.

00:19:38.100 --> 00:19:41.600
And so you literally just take,
and it seems very convoluted in

00:19:41.600 --> 00:19:43.800
how you're going to actually
evaluate it, but we were able

00:19:43.800 --> 00:19:47.610
to express this thing called a
surface-- well, this is a very

00:19:47.610 --> 00:19:49.710
simple surface integral-- in
something that we can

00:19:49.710 --> 00:19:50.505
actually calculate.

00:19:50.505 --> 00:19:52.720
And in the next few videos, I'm
going to show you examples

00:19:52.720 --> 00:19:54.450
of actually calculating it.

00:19:54.450 --> 00:19:57.080
Now, this right here will only
give you the surface area.

00:19:57.080 --> 00:20:00.560
But what if, at every point
here-- so over here what

00:20:00.560 --> 00:20:03.970
we've done in both of these
expressions, is we're just

00:20:03.970 --> 00:20:06.780
figuring out the surface area
of each of these parallelograms

00:20:06.780 --> 00:20:08.380
and then adding them all up.

00:20:08.380 --> 00:20:09.430
That's what we're doing.

00:20:09.430 --> 00:20:12.480
But what if, associated with
each of those little

00:20:12.480 --> 00:20:16.330
parallelograms, we had some
value, where that value is

00:20:16.330 --> 00:20:24.030
defined by some third
function f of x, y, z?

00:20:24.030 --> 00:20:27.240
So every parallelogram, it's
super small, it's around a

00:20:27.240 --> 00:20:30.130
point, you can say it's maybe
the center of it, doesn't

00:20:30.130 --> 00:20:30.880
have to be the center.

00:20:30.880 --> 00:20:33.350
But maybe the center of it
is at some point in

00:20:33.350 --> 00:20:36.970
three-dimensional space, and if
you use some other function, f

00:20:36.970 --> 00:20:40.080
of x, y, and z, you'll get the
value of that point.

00:20:40.080 --> 00:20:44.780
And what we want to do is
figure out what happens if

00:20:44.780 --> 00:20:47.610
for every one of those
parallelograms, we were to

00:20:47.610 --> 00:20:51.250
multiply it times the value of
the function at that point?

00:20:51.250 --> 00:20:53.450
So we could write it this way.

00:20:53.450 --> 00:20:55.230
So this is where, you can
imagine, the function

00:20:55.230 --> 00:20:56.570
is just one.

00:20:56.570 --> 00:20:59.960
We're just multiplying each of
the parallelograms by one.

00:20:59.960 --> 00:21:03.640
But we could imagine we're
multiplying each of the little

00:21:03.640 --> 00:21:09.360
parallelograms by f of x, y,
and z, d sigma, and it's going

00:21:09.360 --> 00:21:12.130
to be the exact same thing,
where this is each of the

00:21:12.130 --> 00:21:14.080
little parallelograms, we're
just going to multiply it

00:21:14.080 --> 00:21:16.510
by f of x, y, and z there.

00:21:16.510 --> 00:21:20.880
So we're going to integrate
it over the area, over that

00:21:20.880 --> 00:21:27.120
region, of f of x, y, and z,
and then times the magnitude of

00:21:27.120 --> 00:21:31.660
the partial of r with respect
to f, crossed with the partial

00:21:31.660 --> 00:21:36.900
of r with respect to t, ds, dt.

00:21:36.900 --> 00:21:38.390
And of course, we're
integrating with

00:21:38.390 --> 00:21:39.340
respect to s and t.

00:21:39.340 --> 00:21:42.120
Hopefully we can express this
function in terms of s and

00:21:42.120 --> 00:21:43.590
t, and we should be able
to, because we have a

00:21:43.590 --> 00:21:44.980
parameterization there.

00:21:44.980 --> 00:21:47.090
Wherever we see an x
there, it's really x is

00:21:47.090 --> 00:21:48.510
a function of s and t.

00:21:48.510 --> 00:21:50.510
y is a function of s and t.

00:21:50.510 --> 00:21:52.530
z is a function of s and t.

00:21:52.530 --> 00:21:55.040
And this might look super
convoluted and hard.

00:21:55.040 --> 00:21:57.580
And the visualizations for
this, of why you'd want

00:21:57.580 --> 00:21:59.410
to do this, it has
applications in physics.

00:21:59.410 --> 00:22:00.510
It's a little hard
to visualize.

00:22:00.510 --> 00:22:04.780
It's easier just to visualize
the straight-up surface area.

00:22:04.780 --> 00:22:07.140
But we're going to see the next
few videos that it's a little

00:22:07.140 --> 00:22:09.650
hairy to calculate these
problems, but they're

00:22:09.650 --> 00:22:10.740
not too hard to do.

00:22:10.740 --> 00:22:13.950
That you just kind of
have to stick with them.


Partial derivatives of vector-valued functions

WEBVTT
Kind: captions
Language: en

00:00:00.550 --> 00:00:07.160
Let's have the vector valued
function r of s and t is equal

00:00:07.160 --> 00:00:09.980
to-- well, x is going to
be a function of s and t.

00:00:09.980 --> 00:00:14.090
So we'll just write it as x of
s and t times the x unit

00:00:14.090 --> 00:00:22.180
vector, or i, plus y of s and t
times the y unit factor, or j,

00:00:22.180 --> 00:00:28.240
plus x of s and t times
the z unit vector, k.

00:00:28.240 --> 00:00:32.280
So given that we have this
vector valued function, let's

00:00:32.280 --> 00:00:34.770
define or let's think about
what it means to take the

00:00:34.770 --> 00:00:38.800
partial derivative of this
vector valued function with

00:00:38.800 --> 00:00:41.620
respect to one of the
parameters, s or t.

00:00:41.620 --> 00:00:44.040
I think it's going to be
pretty natural, nothing

00:00:44.040 --> 00:00:45.420
completely bizarre here.

00:00:45.420 --> 00:00:49.030
We've taken partial derivatives
of non-vector valued functions

00:00:49.030 --> 00:00:52.850
before, where we only vary
one of the variables.

00:00:52.850 --> 00:00:54.280
We only take it with
respect to one variable.

00:00:54.280 --> 00:00:55.770
You hold the other
one constant.

00:00:55.770 --> 00:00:57.750
We're going to do the
exact same thing here.

00:00:57.750 --> 00:01:00.920
And we've taken regular
derivatives of vector

00:01:00.920 --> 00:01:01.720
valued functions.

00:01:01.720 --> 00:01:04.120
The path in those just ended up
being the regular derivative

00:01:04.120 --> 00:01:05.490
of each of the terms.

00:01:05.490 --> 00:01:07.180
And we're going to see, it's
going to be the same thing here

00:01:07.180 --> 00:01:08.950
with the partial derivative.

00:01:08.950 --> 00:01:14.670
So let's define the
partial derivative of

00:01:14.670 --> 00:01:16.970
r with respect to s.

00:01:16.970 --> 00:01:19.720
And everything I do with
respect to s, you can just swap

00:01:19.720 --> 00:01:21.930
it with t, and you're going to
get the same exact result.

00:01:21.930 --> 00:01:28.800
I'm going to define it as being
equal to the limit as delta

00:01:28.800 --> 00:01:36.240
s approaches 0 of r
of s plus delta s.

00:01:36.240 --> 00:01:38.250
Only finding the limit
with respect to a

00:01:38.250 --> 00:01:41.570
change in s comma t.

00:01:41.570 --> 00:01:44.340
We're holding t, as you can
imagine, constant for given

00:01:44.340 --> 00:01:50.190
t, minus r of s and t.

00:01:50.190 --> 00:01:56.630
All of that over delta s.

00:01:56.630 --> 00:01:59.270
Now, if you do a little bit of
algebra here, you literally,

00:01:59.270 --> 00:02:01.980
you know-- r of s plus delta s
comma t, that's the same thing

00:02:01.980 --> 00:02:06.310
as x of s plus delta s
t i, plus y of s plus

00:02:06.310 --> 00:02:07.900
delta s t j, plus z.

00:02:07.900 --> 00:02:09.740
All that minus this thing.

00:02:09.740 --> 00:02:11.370
If you do a little bit of
algebra with that, and if you

00:02:11.370 --> 00:02:13.070
don't believe me, try it out.

00:02:13.070 --> 00:02:22.950
This is going to be equal to
the limit of delta s

00:02:22.950 --> 00:02:25.620
approaching 0-- and I'm going
to write it small because it'd

00:02:25.620 --> 00:02:33.120
take up a lot of space-- of x
of s plus delta s comma t minus

00:02:33.120 --> 00:02:38.340
x of s and t, I think you
know where I'm going.

00:02:38.340 --> 00:02:40.280
This is all a little bit
monotonous to write it all

00:02:40.280 --> 00:02:43.440
out, but never hurts.

00:02:43.440 --> 00:02:47.750
Times s or divided by delta s
times i-- and then I'll do it

00:02:47.750 --> 00:02:51.100
in different colors, so it's
less monotonous-- plus y.

00:02:51.100 --> 00:02:53.290
Where every-- those limited
delta s [? approaches ?]

00:02:53.290 --> 00:02:55.720
0 applies to every term
I'm writing out here.

00:02:55.720 --> 00:03:04.590
y of s plus delta s comma t
minus y of s comma t, all of

00:03:04.590 --> 00:03:09.560
that over delta s times j.

00:03:09.560 --> 00:03:19.240
And then finally, plus z of s
plus delta s comma t minus z of

00:03:19.240 --> 00:03:27.910
s and t, all of that over delta
s times the z unit vector, k.

00:03:27.910 --> 00:03:29.890
And this all comes out
of this definition.

00:03:29.890 --> 00:03:32.990
If you literally just put s
plus delta s in place for s--

00:03:32.990 --> 00:03:34.600
you evaluate all this, do a
little algebra-- you're going

00:03:34.600 --> 00:03:36.500
to get the exact same thing.

00:03:36.500 --> 00:03:40.110
And this, hopefully, pops out
at you as, gee, we're just

00:03:40.110 --> 00:03:42.590
taking the partial derivative
of each of these functions

00:03:42.590 --> 00:03:44.480
with respect to s.

00:03:44.480 --> 00:03:49.040
And these functions right here,
this x of s and t, this is a

00:03:49.040 --> 00:03:50.670
non-vector valued function.

00:03:50.670 --> 00:03:53.450
This y, this is also a
non-vector valued function.

00:03:53.450 --> 00:03:56.240
z is also a non-vector
valued function.

00:03:56.240 --> 00:03:59.540
When you put them all together,
it becomes a vector valued

00:03:59.540 --> 00:04:01.410
function, because we're
multiplying the first

00:04:01.410 --> 00:04:02.270
one times a vector.

00:04:02.270 --> 00:04:03.620
The second one times
another vector.

00:04:03.620 --> 00:04:05.210
The third one times
another vector.

00:04:05.210 --> 00:04:06.570
But independently,
these functions are

00:04:06.570 --> 00:04:07.570
non-vector valued.

00:04:07.570 --> 00:04:10.140
So this is just the
definition of the regular

00:04:10.140 --> 00:04:11.110
partial derivatives.

00:04:11.110 --> 00:04:13.170
Where we're taking the limit
as delta s approaches 0

00:04:13.170 --> 00:04:15.020
in each of these cases.

00:04:15.020 --> 00:04:17.380
So this is the
exact same thing.

00:04:17.380 --> 00:04:21.530
This is equal to-- this is the
exact same thing as the partial

00:04:21.530 --> 00:04:30.060
derivative of x with respect to
s times i plus the partial

00:04:30.060 --> 00:04:38.000
derivative y with respect to s
times j plus the partial

00:04:38.000 --> 00:04:43.700
derivative of z with
respect to s times k.

00:04:43.700 --> 00:04:47.160
I'm going to do one more thing
here and this is pseudo mathy,

00:04:47.160 --> 00:04:49.620
but it's going to come out--
the whole reason I'm even doing

00:04:49.620 --> 00:04:52.650
this video, is it's going to
give us some good tools in our

00:04:52.650 --> 00:04:55.470
tool kit for the videos that
I'm about to do on

00:04:55.470 --> 00:04:57.190
surface integrals.

00:04:57.190 --> 00:04:59.510
So I'm going to do one thing
here that's a little pseudo

00:04:59.510 --> 00:05:01.990
mathy, and that's really
because differentials are these

00:05:01.990 --> 00:05:04.600
things that are very hard to
define rigorously, but I think

00:05:04.600 --> 00:05:07.510
it'll give you the intuition
of what's going on.

00:05:07.510 --> 00:05:10.740
So this thing right here, I'm
going to say this is also equal

00:05:10.740 --> 00:05:13.820
to-- and you're not going to
see this in any math textbook,

00:05:13.820 --> 00:05:16.520
and hard core mathematicians
are going to kind of cringe

00:05:16.520 --> 00:05:17.950
when they see me do this.

00:05:17.950 --> 00:05:20.160
But I like to do it because
I think it'll give you the

00:05:20.160 --> 00:05:23.090
intuition on what's going
on when we take our

00:05:23.090 --> 00:05:23.990
surface integrals.

00:05:23.990 --> 00:05:28.340
So I'm going to say that this
whole thing right here, that

00:05:28.340 --> 00:05:36.910
that is equal to r of s plus
the differential of s-- a super

00:05:36.910 --> 00:05:44.130
small change in s-- t minus r
of s and t, all of that

00:05:44.130 --> 00:05:48.900
over that same super
small change in s.

00:05:48.900 --> 00:05:51.440
So hopefully you understand
at least why I view

00:05:51.440 --> 00:05:52.420
things this way.

00:05:52.420 --> 00:05:56.930
When I take the limit as delta
s approaches 0, these delta s's

00:05:56.930 --> 00:05:58.810
are going to get super
duper duper small.

00:05:58.810 --> 00:06:03.370
And in my head, that's how
I imagine differentials.

00:06:03.370 --> 00:06:07.365
When someone writes the
derivative of y with respect to

00:06:07.365 --> 00:06:12.870
x-- and let's say that they say
that that is 2-- and we've done

00:06:12.870 --> 00:06:15.290
a little bit of math with
differentials before.

00:06:15.290 --> 00:06:19.400
You can imagine multiplying
both sides by dx, and you

00:06:19.400 --> 00:06:22.210
could get dy is equal to 2dx.

00:06:22.210 --> 00:06:24.670
We've done this
throughout calculus.

00:06:24.670 --> 00:06:27.870
The way I imagine it is super
small change in y-- infinitely

00:06:27.870 --> 00:06:31.830
small change in y-- is equal to
2 times-- though, you can

00:06:31.830 --> 00:06:35.420
imagine an equally
small change in x.

00:06:35.420 --> 00:06:39.880
So it's a-- well, if you have a
super small change in x, your

00:06:39.880 --> 00:06:42.290
change in y is going to be
still super small, but it's

00:06:42.290 --> 00:06:43.510
going to be 2 times that.

00:06:43.510 --> 00:06:44.710
I guess that's the
best way to view it.

00:06:44.710 --> 00:06:48.310
But in general, I view
differentials as super small

00:06:48.310 --> 00:06:50.420
changes in a variable.

00:06:50.420 --> 00:06:53.600
So with that out of the way,
and me explaining to you that

00:06:53.600 --> 00:06:56.290
many mathematicians would
cringe at what I just wrote,

00:06:56.290 --> 00:06:59.290
hopefully this gives you a
little-- this isn't like

00:06:59.290 --> 00:07:01.080
some crazy thing I did.

00:07:01.080 --> 00:07:04.240
I'm just saying, oh, delta
s as delta approaches 0, I

00:07:04.240 --> 00:07:06.640
kind of imagine that as ds.

00:07:06.640 --> 00:07:09.830
And the whole reason I did
that, is if you take this side

00:07:09.830 --> 00:07:12.610
and that side, and multiply
both sides times this

00:07:12.610 --> 00:07:16.690
differential ds,
then what happens?

00:07:16.690 --> 00:07:23.250
The left hand side, you get the
partial of r with respect to

00:07:23.250 --> 00:07:28.120
s is equal to this times ds.

00:07:28.120 --> 00:07:30.190
I'll do ds in maybe pink.

00:07:30.190 --> 00:07:32.770
Times ds-- this is just
a regular differential,

00:07:32.770 --> 00:07:33.890
super small change in s.

00:07:33.890 --> 00:07:37.220
This is a kind of a partial,
with respect to s.

00:07:37.220 --> 00:07:42.740
That's going to be equal to--
well, if you multiply this side

00:07:42.740 --> 00:07:45.230
of the equation times ds, this
guy's going to disappear.

00:07:45.230 --> 00:07:49.700
So it's going to be r of s,
plus our super small change

00:07:49.700 --> 00:07:55.080
in s, t minus r of s and t.

00:07:55.080 --> 00:07:57.020
Now let me put a little
square around this.

00:07:57.020 --> 00:07:59.650
This is going to be valuable
for us in the next video.

00:07:59.650 --> 00:08:01.980
We're going to actually think
about what this means and how

00:08:01.980 --> 00:08:05.210
to visualize this on a surface.

00:08:05.210 --> 00:08:07.380
As you can imagine, this
is a vector right here.

00:08:07.380 --> 00:08:09.260
You have 2 vector valued
functions and you're

00:08:09.260 --> 00:08:09.990
taking the difference.

00:08:09.990 --> 00:08:11.440
And we're going to visualize
it in the next video.

00:08:11.440 --> 00:08:13.760
It's going to really help
us with surface integrals.

00:08:13.760 --> 00:08:17.270
By the same exact logic, we
can do everything we did

00:08:17.270 --> 00:08:20.140
here with s, we can do
it with t, as well.

00:08:20.140 --> 00:08:24.960
So we can define the partial--
I'll draw a little-- I can

00:08:24.960 --> 00:08:28.650
define the partial of r with
respect-- let me do it in a

00:08:28.650 --> 00:08:31.080
different color, completely
different color.

00:08:31.080 --> 00:08:32.270
It's orange.

00:08:32.270 --> 00:08:36.580
The partial of r with respect
to t-- the definition

00:08:36.580 --> 00:08:37.370
is just right here.

00:08:37.370 --> 00:08:46.750
The limit as delta t approaches
0 of r of s t plus delta

00:08:46.750 --> 00:08:51.100
t minus r of s and t.

00:08:51.100 --> 00:08:52.770
In this situation we're
holding the s, you can

00:08:52.770 --> 00:08:53.880
imagine, in constant.

00:08:53.880 --> 00:08:59.220
We're finding its change in
t, all of that over delta t.

00:08:59.220 --> 00:09:00.380
And the same thing falls out.

00:09:00.380 --> 00:09:04.270
This is equal to the partial of
x with respect to ti plus y

00:09:04.270 --> 00:09:07.580
with respect to tj, plus
z with respect to tk.

00:09:07.580 --> 00:09:11.190
Same exact thing, you just kind
of swap the s's and the t's.

00:09:11.190 --> 00:09:13.740
And by that same logic,
you'd have the same

00:09:13.740 --> 00:09:15.990
result but in terms of t.

00:09:15.990 --> 00:09:18.970
If you do this pseudo mathy
thing that I did up here, then

00:09:18.970 --> 00:09:25.680
you would get the partial of r
with respect to t times a super

00:09:25.680 --> 00:09:28.800
small change in t. dt, our
t differential, you could

00:09:28.800 --> 00:09:40.360
imagine, is equal to r of st
plus dt minus r of s and t.

00:09:40.360 --> 00:09:43.010
So let's box these
two guys away.

00:09:43.010 --> 00:09:44.340
And in the next video,
we're going to actually

00:09:44.340 --> 00:09:46.790
visualize what these mean.

00:09:46.790 --> 00:09:49.640
And sometimes, when you kind of
do a bunch of like, silly math

00:09:49.640 --> 00:09:50.670
like this, you're always like,
all right, what is

00:09:50.670 --> 00:09:51.240
this all about?

00:09:51.240 --> 00:09:53.110
Remember, all I did is I said,
what does it mean to take the

00:09:53.110 --> 00:09:55.540
derivative of this with
respect to s or t?

00:09:55.540 --> 00:09:58.970
Played around with it a little
bit, I got this result.

00:09:58.970 --> 00:10:01.220
These 2 are going to be very
valuable for us, I think, in

00:10:01.220 --> 00:10:03.870
getting the intuition for
why surface integrals

00:10:03.870 --> 00:10:06.450
look the way they do.