_-substitution: multiplying by a constant

WEBVTT
Kind: captions
Language: en

00:00:00.580 --> 00:00:02.260
Let's take the
indefinite integral

00:00:02.260 --> 00:00:07.990
of the square root
of 7x plus 9 dx.

00:00:07.990 --> 00:00:10.290
So my first question
to you is, is this

00:00:10.290 --> 00:00:14.060
going to be a good case
for u-substitution?

00:00:14.060 --> 00:00:16.830
Well, when you look here,
maybe the natural thing

00:00:16.830 --> 00:00:20.880
to set to be equal
to u is 7x plus 9.

00:00:20.880 --> 00:00:24.204
But do I see its derivative
anywhere over here?

00:00:24.204 --> 00:00:24.870
Well, let's see.

00:00:24.870 --> 00:00:30.120
If we set u to be
equal to 7x plus 9,

00:00:30.120 --> 00:00:33.410
what is the derivative of u
with respect to x going to be?

00:00:33.410 --> 00:00:35.310
Derivative of u
with respect to x

00:00:35.310 --> 00:00:37.080
is just going to be equal to 7.

00:00:37.080 --> 00:00:38.470
Derivative of 7x is 7.

00:00:38.470 --> 00:00:40.650
Derivative of 9 is 0.

00:00:40.650 --> 00:00:44.190
So do we see a 7 lying
around anywhere over here?

00:00:44.190 --> 00:00:45.580
Well, we don't.

00:00:45.580 --> 00:00:49.340
But what could we do in order
to have a 7 lying around,

00:00:49.340 --> 00:00:53.280
but not change the
value of the integral?

00:00:53.280 --> 00:00:55.956
Well, the neat thing-- and we've
seen this multiple times-- is

00:00:55.956 --> 00:00:57.330
when you're
evaluating integrals,

00:00:57.330 --> 00:01:01.240
scalars can go in and outside
of the integral very easily.

00:01:01.240 --> 00:01:05.920
Just to remind ourselves, if I
have the integral of let's say

00:01:05.920 --> 00:01:11.690
some scalar a times
f of x dx, this

00:01:11.690 --> 00:01:17.420
is the same thing as a times
the integral of f of x dx.

00:01:17.420 --> 00:01:19.340
The integral of the
scalar times a function

00:01:19.340 --> 00:01:22.650
is equal to the scalar times
the integral of the functions.

00:01:22.650 --> 00:01:25.280
So let me put this
aside right over here.

00:01:25.280 --> 00:01:28.630
So with that in mind, can
we multiply and divide

00:01:28.630 --> 00:01:31.710
by something that will
have a 7 showing up?

00:01:31.710 --> 00:01:34.330
Well, we can multiply
and divide by 7.

00:01:34.330 --> 00:01:35.830
So imagine doing this.

00:01:35.830 --> 00:01:38.580
Let's rewrite our
original integral.

00:01:38.580 --> 00:01:40.960
So let me draw a
little arrow here just

00:01:40.960 --> 00:01:42.420
to go around that aside.

00:01:42.420 --> 00:01:44.100
We could rewrite our
original integral

00:01:44.100 --> 00:01:51.340
as being 9 to the integral
of times 1/7 times 7 times

00:01:51.340 --> 00:01:57.980
the square root of 7x plus 9 dx.

00:01:57.980 --> 00:02:00.064
And if we want to, we
could take the 1/7 outside

00:02:00.064 --> 00:02:00.730
of the integral.

00:02:00.730 --> 00:02:02.270
We don't have to,
but we can rewrite

00:02:02.270 --> 00:02:06.620
this as 1/7 times
the integral of 7,

00:02:06.620 --> 00:02:12.130
times the square
root of 7x plus 9 dx.

00:02:12.130 --> 00:02:14.840
So now if we set u
equal to 7x plus 9,

00:02:14.840 --> 00:02:16.860
do we have its
derivative laying around?

00:02:16.860 --> 00:02:17.410
Well, sure.

00:02:17.410 --> 00:02:20.350
The 7 is right over here.

00:02:20.350 --> 00:02:23.000
We know that du-- if we want to
write it in differential form--

00:02:23.000 --> 00:02:27.060
du is equal to 7 times dx.

00:02:27.060 --> 00:02:31.460
So du is equal to 7 times dx.

00:02:31.460 --> 00:02:35.320
That part right over
there is equal to du.

00:02:35.320 --> 00:02:37.320
And if we want to care
about u, well, that's

00:02:37.320 --> 00:02:40.050
just going to be the 7x plus 9.

00:02:40.050 --> 00:02:41.520
That is are u.

00:02:41.520 --> 00:02:45.272
So let's rewrite this indefinite
integral in terms of u.

00:02:45.272 --> 00:02:53.120
It's going to be equal to
1/7 times the integral of--

00:02:53.120 --> 00:02:55.160
and I'll just take the 7
and put it in the back.

00:02:55.160 --> 00:02:57.710
So we could just
write the square root

00:02:57.710 --> 00:03:06.450
of u du, 7 times dx is du.

00:03:06.450 --> 00:03:10.110
And we can rewrite this if we
want as u to the 1/2 power.

00:03:10.110 --> 00:03:12.320
It makes it a little bit
easier for us to kind of do

00:03:12.320 --> 00:03:14.490
the reverse power rule here.

00:03:14.490 --> 00:03:20.940
So we can rewrite this as equal
to 1/7 times the integral of u

00:03:20.940 --> 00:03:23.751
to the 1/2 power du.

00:03:23.751 --> 00:03:25.000
And let me just make it clear.

00:03:25.000 --> 00:03:26.680
This u I could have
written in white

00:03:26.680 --> 00:03:27.960
if I want it the same color.

00:03:27.960 --> 00:03:31.100
And this du is the same
du right over here.

00:03:31.100 --> 00:03:35.790
So what is the antiderivative
of u to the 1/2 power?

00:03:35.790 --> 00:03:39.050
Well, we increment
u's power by 1.

00:03:39.050 --> 00:03:41.010
So this is going to be
equal to-- let me not

00:03:41.010 --> 00:03:43.260
forget this 1/7 out front.

00:03:43.260 --> 00:03:49.110
So it's going to be 1/7 times--
if we increment the power here,

00:03:49.110 --> 00:03:55.520
it's going to be u to the 3/2,
1/2 plus 1 is 1 and 1/2 or 3/2.

00:03:55.520 --> 00:03:57.280
So it's going to
be u to the 3/2.

00:04:01.030 --> 00:04:04.140
And then we're going to
multiply this new thing

00:04:04.140 --> 00:04:07.980
times the reciprocal
of 3/2, which is 2/3.

00:04:07.980 --> 00:04:10.840
And I encourage you to verify
the derivative of 2/3 u

00:04:10.840 --> 00:04:14.630
to the 3/2 is
indeed u to the 1/2.

00:04:14.630 --> 00:04:15.750
And so we have that.

00:04:15.750 --> 00:04:17.450
And since we're
multiplying 1/7 times

00:04:17.450 --> 00:04:19.440
this entire indefinite
integral, we

00:04:19.440 --> 00:04:21.560
could also throw in a
plus c right over here.

00:04:21.560 --> 00:04:23.430
There might have
been a constant.

00:04:23.430 --> 00:04:25.990
And if we want, we can
distribute the 1/7.

00:04:25.990 --> 00:04:36.170
So it would get 1/7 times
2/3 is 2/21 u to the 3/2.

00:04:36.170 --> 00:04:38.750
And 1/7 times some
constant, well, that's

00:04:38.750 --> 00:04:40.210
just going to be some constant.

00:04:40.210 --> 00:04:41.959
And so I could write
a constant like that.

00:04:41.959 --> 00:04:44.510
I could call that c1 and
then I could call this c2,

00:04:44.510 --> 00:04:47.140
but it's really just
some arbitrary constant.

00:04:47.140 --> 00:04:47.850
And we're done.

00:04:47.850 --> 00:04:49.183
Oh, actually, no we aren't done.

00:04:49.183 --> 00:04:51.520
We still just have our
entire thing in terms of u.

00:04:51.520 --> 00:04:54.290
So now let's unsubstitute it.

00:04:54.290 --> 00:05:01.290
So this is going to be equal
to 2/21 times u to the 3/2.

00:05:01.290 --> 00:05:03.710
And we already know
what u is equal to.

00:05:03.710 --> 00:05:06.060
u is equal to 7x plus 9.

00:05:06.060 --> 00:05:08.720
Let me put a new color here
just to ease the monotony.

00:05:08.720 --> 00:05:13.460
So it's going to be
2/21 times 7x plus 9

00:05:13.460 --> 00:05:18.770
to the 3/2 power plus c.

00:05:21.870 --> 00:05:23.030
And we are done.

00:05:23.030 --> 00:05:25.500
We were able to take a kind
of hairy looking integral

00:05:25.500 --> 00:05:28.120
and realize that even
though it wasn't completely

00:05:28.120 --> 00:05:31.869
obvious at first, that
u-substitution is applicable.


_-substitution: logarithmic function

WEBVTT
Kind: captions
Language: en

00:00:00.570 --> 00:00:03.040
We are faced with a
fairly daunting-looking

00:00:03.040 --> 00:00:08.870
indefinite integral of pi
over x natural log of x dx.

00:00:08.870 --> 00:00:11.340
Now, what can we
do to address this?

00:00:11.340 --> 00:00:14.730
Is u-substitution
a possibility here?

00:00:14.730 --> 00:00:16.500
Well for u-substitution,
we want to look

00:00:16.500 --> 00:00:19.660
for an expression
and its derivative.

00:00:19.660 --> 00:00:25.860
Well, what happens if we set u
equal to the natural log of x?

00:00:25.860 --> 00:00:29.280
Now what would du be
equal to in that scenario?

00:00:29.280 --> 00:00:32.130
du is going to be the derivative
of the natural log of x

00:00:32.130 --> 00:00:37.920
with respect to x,
which is just 1/x dx.

00:00:37.920 --> 00:00:40.360
This is an equivalent
statement to saying

00:00:40.360 --> 00:00:45.330
that du dx is equal to 1/x.

00:00:45.330 --> 00:00:50.020
So do we see a 1/x dx anywhere
in this original expression?

00:00:50.020 --> 00:00:51.120
Well, it's kind of hiding.

00:00:51.120 --> 00:00:53.520
It's not so obvious, but
this x in the denominator

00:00:53.520 --> 00:00:55.250
is essentially a 1/x.

00:00:55.250 --> 00:00:57.210
And then that's being
multiplied by a dx.

00:00:57.210 --> 00:00:59.010
Let me rewrite this
original expression

00:00:59.010 --> 00:01:00.485
to make a little bit more sense.

00:01:00.485 --> 00:01:03.110
So the first thing I'm going to
do is I'm going to take the pi.

00:01:03.110 --> 00:01:04.651
I should do that in
a different color

00:01:04.651 --> 00:01:06.940
since I've already
used-- let me take the pi

00:01:06.940 --> 00:01:08.510
and just stick it out front.

00:01:08.510 --> 00:01:11.640
So I'm going to stick the pi
out in front of the integral.

00:01:11.640 --> 00:01:13.970
And so this becomes
the integral of--

00:01:13.970 --> 00:01:17.010
and let me write the 1 over
natural log of x first.

00:01:17.010 --> 00:01:22.891
1 over the natural
log of x times 1/x dx.

00:01:22.891 --> 00:01:24.390
Now it becomes a
little bit clearer.

00:01:24.390 --> 00:01:29.430
These are completely
equivalent statements.

00:01:29.430 --> 00:01:31.230
But this makes it
clear that, yes,

00:01:31.230 --> 00:01:33.540
u-substitution will
work over here.

00:01:33.540 --> 00:01:37.170
If we set our u equal to
natural log of x, then

00:01:37.170 --> 00:01:38.980
our du is 1/x dx.

00:01:44.650 --> 00:01:46.320
Let's rewrite this integral.

00:01:46.320 --> 00:01:52.280
It's going to be equal to pi
times the indefinite integral

00:01:52.280 --> 00:01:52.805
of 1/u.

00:01:55.310 --> 00:01:59.220
Natural log of x is u-- we
set that equal to natural log

00:01:59.220 --> 00:02:01.130
of x-- times du.

00:02:04.664 --> 00:02:06.330
Now this becomes
pretty straightforward.

00:02:06.330 --> 00:02:08.630
What is the antiderivative
of all of this business?

00:02:08.630 --> 00:02:11.210
And we've done very similar
things like this multiple times

00:02:11.210 --> 00:02:12.250
already.

00:02:12.250 --> 00:02:19.460
This is going to be equal
to pi times the natural log

00:02:19.460 --> 00:02:23.580
of the absolute
value of u so that we

00:02:23.580 --> 00:02:26.200
can handle even
negative values of u.

00:02:26.200 --> 00:02:30.040
The natural log of the
absolute value of u plus c,

00:02:30.040 --> 00:02:33.870
just in case we had a
constant factor out here.

00:02:33.870 --> 00:02:34.830
And we're almost done.

00:02:34.830 --> 00:02:36.860
We just have to
unsubstitute for the u.

00:02:36.860 --> 00:02:38.630
u is equal to natural log of x.

00:02:38.630 --> 00:02:41.220
So we end up with this kind
of neat-looking expression.

00:02:41.220 --> 00:02:42.690
The anti of this
entire indefinite

00:02:42.690 --> 00:02:44.050
integral we have simplified.

00:02:44.050 --> 00:02:46.110
We have evaluated
it, and it is now

00:02:46.110 --> 00:02:53.050
equal to pi times the natural
log of the absolute value of u.

00:02:53.050 --> 00:02:55.100
But u is just the
natural log of x.

00:02:57.860 --> 00:03:03.220
And then we have this
plus c right over here.

00:03:03.220 --> 00:03:05.220
And we could have assumed
that, from the get go,

00:03:05.220 --> 00:03:07.680
this original
expression was only

00:03:07.680 --> 00:03:10.441
defined for positive
values of x because you had

00:03:10.441 --> 00:03:11.940
to take the natural
log here, and it

00:03:11.940 --> 00:03:15.070
wasn't an absolute value.

00:03:15.070 --> 00:03:17.510
So we can leave this as
just a natural log of x,

00:03:17.510 --> 00:03:19.377
but this also works
for the situations

00:03:19.377 --> 00:03:21.710
now because we're doing the
absolute value of that where

00:03:21.710 --> 00:03:25.220
the natural log of x might
have been a negative number.

00:03:25.220 --> 00:03:30.360
For example, if it was a natural
log of 0.5 or, who knows,

00:03:30.360 --> 00:03:31.660
whatever it might be.

00:03:31.660 --> 00:03:32.980
But then we are all done.

00:03:32.980 --> 00:03:36.520
We have simplified
what seemed like a kind

00:03:36.520 --> 00:03:38.620
of daunting expression.


_-substitution: rational function

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:08.501
So we want to take the indefinite integral of 4x^3 over x^4 plus 7 dx.

00:00:08.501 --> 00:00:11.573
So how can we tackle this? It seems like a hairy integral.

00:00:11.573 --> 00:00:18.184
Now the key inside here is to realize you have this expression x^4 + 7

00:00:18.184 --> 00:00:26.171
and you also have its derivative up here. The derivative of x^4 plus 7 is equal to 4x^3.

00:00:26.171 --> 00:00:30.843
Derivative of x^4 is 4x^3; derivative of 7 is just 0.

00:00:30.843 --> 00:00:36.903
So that's a big clue that u-substitution might be the tool of choice here.

00:00:36.903 --> 00:00:39.575
U-sub -- I'll just write u- -- I'll write the whole thing.

00:00:39.575 --> 00:00:47.180
U-Substitution could be the tool of choice. So given that, what would you want to set your u equal to?

00:00:47.180 --> 00:00:49.486
And I'll let you think about that 'cause it can figure out this part

00:00:49.486 --> 00:00:53.982
and the rest will just boil down to a fairly straightforward integral.

00:00:53.982 --> 00:00:58.737
Well, you want to set u be equal to the expression that you have its derivative laying around.

00:00:58.737 --> 00:01:09.745
So we could set u equal to x^4 plus 7. Now, what is du going to be equal to?

00:01:09.745 --> 00:01:18.406
du, I'm doing it in magenta. du, well it's just going to be the derivative of x^4 plus 7 with respect to x,

00:01:18.406 --> 00:01:27.278
so 4x^3 plus 0 times dx. I wrote it in differential form right over here,

00:01:27.278 --> 00:01:32.929
but it's a completely equivalent statement to saying that du, the derivative of u with respect to x,

00:01:32.929 --> 00:01:40.848
is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say

00:01:40.848 --> 00:01:46.512
the derivative of u with respect to x. It really isn't a fraction in a very formal way,

00:01:46.512 --> 00:01:50.421
but often times, you can kind of pseudo-manipulate them like fractions.

00:01:50.421 --> 00:01:56.378
So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx.

00:01:56.378 --> 00:01:59.377
But these are equivalent statements and we want to get it in differential form

00:01:59.377 --> 00:02:03.911
in order to do proper use of u-substitution. And the reason why this is useful --

00:02:03.911 --> 00:02:07.277
and I'll just rewrite it up here so that it becomes very obvious;

00:02:07.277 --> 00:02:15.410
our original integral we can rewrite as 4x^3 dx over x^4 plus 7.

00:02:15.410 --> 00:02:18.580
And then it's pretty clear what's du and what's u.

00:02:18.580 --> 00:02:26.110
U, which we set to be equal to x^4 plus 7. And then du is equal to this.

00:02:26.110 --> 00:02:32.580
It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral --

00:02:32.580 --> 00:02:37.244
I'll try to stay consistent with the colors -- as the indefinite integral,

00:02:37.244 --> 00:02:44.722
well we have in magenta right over here, that's du over -- try to stay true to the colors --

00:02:44.722 --> 00:02:52.978
over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as

00:02:52.978 --> 00:03:09.961
the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du?

00:03:09.961 --> 00:03:16.120
Well that's just going to be equal to the natural log of the absolute value --

00:03:16.120 --> 00:03:20.060
and we use the absolute value so it'll be defined even for negative U's --

00:03:20.060 --> 00:03:23.881
and it actually does work out. I'll do another video where I'll show you it definitely does.

00:03:23.881 --> 00:03:32.564
The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative.

00:03:32.564 --> 00:03:38.631
So that's essentially our answer in terms of u. But now we need to un-substitute the u.

00:03:38.631 --> 00:03:42.301
So what happens when we un-substitute the u?

00:03:42.301 --> 00:03:45.132
Well, then we are left with -- this is going to be equal to --

00:03:45.132 --> 00:03:55.392
the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 --

00:03:55.392 --> 00:03:59.392
and then we can't forget our plus C out here. And we are done!


_-substitution intro

WEBVTT
Kind: captions
Language: en

00:00:00.630 --> 00:00:03.500
Let's say that we have
the indefinite integral,

00:00:03.500 --> 00:00:11.700
and the function is 3x
squared plus 2x times e

00:00:11.700 --> 00:00:18.560
to x to the third
plus x squared dx.

00:00:18.560 --> 00:00:21.457
So how would we go
about solving this?

00:00:21.457 --> 00:00:23.040
So first when you
look at it, it seems

00:00:23.040 --> 00:00:24.720
like a really
complicated integral.

00:00:24.720 --> 00:00:28.340
We have this polynomial right
over here being multiplied

00:00:28.340 --> 00:00:31.880
by this exponential expression,
and over here in the exponent,

00:00:31.880 --> 00:00:33.950
we essentially have
another polynomial.

00:00:33.950 --> 00:00:36.130
It seems kind of crazy.

00:00:36.130 --> 00:00:39.420
And the key intuition
here, the key insight

00:00:39.420 --> 00:00:41.514
is that you might want
to use a technique here

00:00:41.514 --> 00:00:42.430
called u-substitution.

00:00:46.525 --> 00:00:47.900
And I'll tell you
in a second how

00:00:47.900 --> 00:00:49.910
I would recognize that we
have to use u-substitution.

00:00:49.910 --> 00:00:51.450
And then over time,
you might even

00:00:51.450 --> 00:00:53.800
be able to do this type
of thing in your head.

00:00:53.800 --> 00:00:58.750
u-substitution is essentially
unwinding the chain rule.

00:00:58.750 --> 00:01:01.690
And the chain rule-- I'll go
in more depth in another video,

00:01:01.690 --> 00:01:03.624
where I really talk
about that intuition.

00:01:03.624 --> 00:01:05.040
But the way I would
think about it

00:01:05.040 --> 00:01:09.250
is, well, I have this crazy
exponent right over here.

00:01:09.250 --> 00:01:12.920
I have the x to the
third plus x squared,

00:01:12.920 --> 00:01:15.220
and this thing right
over here happens

00:01:15.220 --> 00:01:18.520
to be the derivative of x
to the third plus x squared.

00:01:18.520 --> 00:01:20.960
The derivative of x to
the third is 3x squared,

00:01:20.960 --> 00:01:25.000
derivative of x squared is
2x, which is a huge clue to me

00:01:25.000 --> 00:01:27.920
that I could use u-substitution.

00:01:27.920 --> 00:01:31.580
So what I do here is this
thing, or this little expression

00:01:31.580 --> 00:01:34.590
here, where I also see its
derivative being multiplied,

00:01:34.590 --> 00:01:36.950
I can set that equal to u.

00:01:36.950 --> 00:01:43.370
So I can say u is equal to x
to the third plus x squared.

00:01:43.370 --> 00:01:45.360
Now, what is going to
be the derivative of u

00:01:45.360 --> 00:01:46.900
with respect to x?

00:01:46.900 --> 00:01:48.930
du dx.

00:01:48.930 --> 00:01:51.430
Well, we've done
this multiple times.

00:01:51.430 --> 00:01:56.900
It's going to be
3x squared plus 2x.

00:01:56.900 --> 00:01:59.480
And now we can write this
in differential form.

00:01:59.480 --> 00:02:04.550
And du dx, this isn't really
a fraction of the differential

00:02:04.550 --> 00:02:06.780
of du divided by
differential of dx.

00:02:06.780 --> 00:02:09.000
It really is a form
of notation, but it

00:02:09.000 --> 00:02:12.420
is often useful to kind of
pretend that it is a fraction,

00:02:12.420 --> 00:02:16.357
and you could kind of view this
if you wanted to just get a du,

00:02:16.357 --> 00:02:18.690
if you just wanted to get a
differential form over here,

00:02:18.690 --> 00:02:22.980
how much does u change
for a given change in x?

00:02:22.980 --> 00:02:27.290
You could multiply
both sides times a dx.

00:02:27.290 --> 00:02:30.670
So both sides times a dx.

00:02:30.670 --> 00:02:32.920
And so if we were to pretend
that they were fractions,

00:02:32.920 --> 00:02:35.990
and it will give you the
correct differential form,

00:02:35.990 --> 00:02:42.720
you're going to
be left with du is

00:02:42.720 --> 00:02:49.230
equal to 3x squared plus 2x dx.

00:02:49.230 --> 00:02:51.700
Now why is this over here?

00:02:51.700 --> 00:02:54.490
Why did I go to the
trouble of doing that?

00:02:54.490 --> 00:02:57.960
Well we see we have
a 3x squared plus 2x,

00:02:57.960 --> 00:03:01.270
and then it's being multiplied
by a dx right over here.

00:03:01.270 --> 00:03:03.740
I could rewrite this
original integral.

00:03:03.740 --> 00:03:06.770
I could rewrite this
as the integral of--

00:03:06.770 --> 00:03:13.040
and let me do it in that color--
of 3x squared plus 2x times

00:03:13.040 --> 00:03:21.210
dx times e-- let me do that
in that other color-- times

00:03:21.210 --> 00:03:26.080
e to the x to the
third plus x squared.

00:03:26.080 --> 00:03:28.769
Now what's interesting
about this?

00:03:28.769 --> 00:03:30.310
Well the stuff that
I have in magenta

00:03:30.310 --> 00:03:33.070
here is exactly equal to du.

00:03:33.070 --> 00:03:35.840
This is exactly equal to du.

00:03:35.840 --> 00:03:39.000
And then this stuff I have up
here, x to the third plus x

00:03:39.000 --> 00:03:41.690
squared, that is what
I set u equal to.

00:03:41.690 --> 00:03:43.850
That is going to be equal to u.

00:03:43.850 --> 00:03:46.090
So I can rewrite
my entire integral,

00:03:46.090 --> 00:03:47.590
and now you might
recognize why this

00:03:47.590 --> 00:03:49.730
is going to simplify
things a good bit,

00:03:49.730 --> 00:03:51.897
it's going to be equal to--
and what I'm going to do

00:03:51.897 --> 00:03:53.271
is I'm going to
change the order.

00:03:53.271 --> 00:03:55.210
I'm going to put the
du, this entire du,

00:03:55.210 --> 00:03:57.440
I'm gonna stick it on
the other side here,

00:03:57.440 --> 00:03:59.230
so it looks like more
of the standard form

00:03:59.230 --> 00:04:02.595
that we're used to seeing
our indefinite integrals in.

00:04:02.595 --> 00:04:11.380
So we're going to have
our du times e to the u.

00:04:15.320 --> 00:04:17.810
And so what would the
antiderivative of this

00:04:17.810 --> 00:04:20.645
be in terms of u?

00:04:20.645 --> 00:04:23.030
Well, the derivative of
e to the u is e to the u.

00:04:23.030 --> 00:04:25.930
The antiderivative of e
to the u is e to the u.

00:04:25.930 --> 00:04:29.507
So it's going to be
equal to e to the u.

00:04:29.507 --> 00:04:31.090
Now, there is a
possibility that there

00:04:31.090 --> 00:04:32.840
was some type of a
constant factor here,

00:04:32.840 --> 00:04:33.820
so let me write that.

00:04:33.820 --> 00:04:35.980
So plus c.

00:04:35.980 --> 00:04:38.220
And now, to get
it in terms of x,

00:04:38.220 --> 00:04:41.440
we just have to
unsubstitute the u.

00:04:41.440 --> 00:04:43.410
We know what u is
equal to, so we

00:04:43.410 --> 00:04:45.960
could say that this is
going to be equal to e.

00:04:45.960 --> 00:04:49.010
Instead of writing u, we
could say u is x to the third

00:04:49.010 --> 00:04:49.865
plus x squared.

00:04:55.640 --> 00:04:58.780
And then we have our plus c.

00:04:58.780 --> 00:05:00.330
And we are done.

00:05:00.330 --> 00:05:02.130
We have found the
antiderivative.

00:05:02.130 --> 00:05:04.460
And I encourage you to take
the derivative of this,

00:05:04.460 --> 00:05:06.940
and I think you will find
yourself using the chain rule,

00:05:06.940 --> 00:05:11.050
and getting right back
to what we had over here.


Triangle inequality theorem

WEBVTT
Kind: captions
Language: en

00:00:00.750 --> 00:00:03.420
Let's draw ourselves a triangle.

00:00:03.420 --> 00:00:06.870
Let's say this
side has length 6.

00:00:06.870 --> 00:00:11.340
Let's say this side right
over here has length 10.

00:00:11.340 --> 00:00:15.580
And let's say that this side
right over here has length x.

00:00:15.580 --> 00:00:18.040
And what I'm going to
think about is how large

00:00:18.040 --> 00:00:21.660
or how small that
value x can be.

00:00:21.660 --> 00:00:24.440
How large or small
can this side be?

00:00:24.440 --> 00:00:29.080
So the first question
is how small can it get?

00:00:29.080 --> 00:00:30.900
Well, if we want
to make this small,

00:00:30.900 --> 00:00:32.479
we would just
literally have to look

00:00:32.479 --> 00:00:34.380
at this angle right over here.

00:00:34.380 --> 00:00:37.560
So let me take a look at this
angle and make it smaller.

00:00:37.560 --> 00:00:40.050
So let's try to make that
angle as small as possible.

00:00:40.050 --> 00:00:41.280
So we have our 10 side.

00:00:43.950 --> 00:00:46.150
Actually let me do it down here.

00:00:46.150 --> 00:00:51.407
So you have your 10 side,
the side of length 10,

00:00:51.407 --> 00:00:53.990
and I'm going to make this angle
really, really, really small,

00:00:53.990 --> 00:00:55.250
approaching 0.

00:00:55.250 --> 00:00:59.795
If that angle becomes 0, we end
up with a degenerate triangle.

00:00:59.795 --> 00:01:01.510
It essentially
becomes one dimension.

00:01:01.510 --> 00:01:03.770
We lose our
two-dimensionality there.

00:01:03.770 --> 00:01:06.690
But as we approach
0, this side starts

00:01:06.690 --> 00:01:09.461
to coincide or get closer
and closer to the 10 side.

00:01:09.461 --> 00:01:11.460
And you could imagine the
case where it actually

00:01:11.460 --> 00:01:14.689
coincides with it and you
actually get the degenerate.

00:01:14.689 --> 00:01:16.230
So if want this
point right over here

00:01:16.230 --> 00:01:19.210
to get as close as possible
to that point over there,

00:01:19.210 --> 00:01:21.520
essentially minimizing
your distance x,

00:01:21.520 --> 00:01:23.940
the closest way is if you
make the angle the way

00:01:23.940 --> 00:01:26.184
equal to 0, all the way.

00:01:26.184 --> 00:01:28.100
So let's actually-- let
me draw a progression.

00:01:28.100 --> 00:01:29.645
So now the angle
is getting smaller.

00:01:32.230 --> 00:01:34.180
This is length 6.

00:01:34.180 --> 00:01:37.277
x is getting smaller.

00:01:37.277 --> 00:01:39.860
Then we keep making that angle
smaller and smaller and smaller

00:01:39.860 --> 00:01:42.520
all the way until we get
a degenerate triangle.

00:01:42.520 --> 00:01:45.510
So let me draw that pink side.

00:01:45.510 --> 00:01:48.380
So you have the
side of length 10.

00:01:48.380 --> 00:01:51.940
Now the angle is essentially 0,
this angle that we care about.

00:01:51.940 --> 00:01:56.400
So this side is length 6.

00:01:56.400 --> 00:01:58.980
And so what is the
distance between this point

00:01:58.980 --> 00:01:59.830
and this point?

00:01:59.830 --> 00:02:01.730
And that distance is length x.

00:02:01.730 --> 00:02:06.060
So in the degenerate case, this
length right over here is x.

00:02:06.060 --> 00:02:09.419
We know that 6 plus x is
going to be equal to 10.

00:02:09.419 --> 00:02:14.750
So in this degenerate case,
x is going to be equal to 4.

00:02:14.750 --> 00:02:18.460
So if you want this to be a
real triangle, at x equals 4

00:02:18.460 --> 00:02:20.460
you've got these points
as close as possible.

00:02:20.460 --> 00:02:22.964
It's degenerated into a
line, into a line segment.

00:02:22.964 --> 00:02:24.380
If you want this
to be a triangle,

00:02:24.380 --> 00:02:28.400
x has to be greater than 4.

00:02:28.400 --> 00:02:30.130
Now let's think about
it the other way.

00:02:30.130 --> 00:02:32.710
How large can x be?

00:02:32.710 --> 00:02:34.700
Well to think about
larger and larger x's, we

00:02:34.700 --> 00:02:37.670
need to make this angle bigger.

00:02:37.670 --> 00:02:39.580
So let's try to do that.

00:02:39.580 --> 00:02:42.370
So let's draw my 10 side again.

00:02:42.370 --> 00:02:44.530
So this is my 10 side.

00:02:46.922 --> 00:02:48.880
I'm going to make that
angle bigger and bigger.

00:02:48.880 --> 00:02:52.959
So now let me take my 6
side and put it like that.

00:02:52.959 --> 00:02:55.500
And so now our angle is getting
bigger and bigger and bigger.

00:02:55.500 --> 00:02:58.800
It's approaching 180 degrees.

00:02:58.800 --> 00:03:01.802
At 180 degrees, our
triangle once again

00:03:01.802 --> 00:03:03.260
will be turned into
a line segment.

00:03:03.260 --> 00:03:04.860
It'll become a
degenerate triangle.

00:03:04.860 --> 00:03:06.800
So let me draw the
side of length x,

00:03:06.800 --> 00:03:08.009
try to draw it straight.

00:03:08.009 --> 00:03:09.675
So we're trying to
maximize the distance

00:03:09.675 --> 00:03:11.190
between that point
and that point.

00:03:11.190 --> 00:03:13.080
So this is side of
length x and let's

00:03:13.080 --> 00:03:15.150
go all the way to
the degenerate case.

00:03:15.150 --> 00:03:20.250
In the degenerate case, at 180
degrees, the side of length 6

00:03:20.250 --> 00:03:24.190
forms a straight line with
the side of length 10.

00:03:24.190 --> 00:03:26.960
And this is how you can get
this point and that point

00:03:26.960 --> 00:03:29.116
as far apart as possible.

00:03:29.116 --> 00:03:30.990
Well, in this situation,
what is the distance

00:03:30.990 --> 00:03:32.940
between that point
and that point, which

00:03:32.940 --> 00:03:36.470
is the distance which
is going to be our x?

00:03:36.470 --> 00:03:40.350
Well in this situation, x is
going to be 6 plus 10 is 16.

00:03:40.350 --> 00:03:42.756
If x is 16, we have a
degenerate triangle.

00:03:42.756 --> 00:03:44.380
If we don't want a
degenerate triangle,

00:03:44.380 --> 00:03:46.920
if we want to have two
dimensions to the triangle,

00:03:46.920 --> 00:03:51.410
then x is going to have
to be less than 16.

00:03:51.410 --> 00:03:54.280
Now the whole principle that
we're working on right over

00:03:54.280 --> 00:03:58.450
here is called the
triangle inequality theorem

00:03:58.450 --> 00:04:00.440
and it's a pretty basic idea.

00:04:00.440 --> 00:04:02.979
That any one side of a
triangle has to be less,

00:04:02.979 --> 00:04:04.645
if you don't want a
degenerate triangle,

00:04:04.645 --> 00:04:07.860
than the sum of the
other two sides.

00:04:07.860 --> 00:04:14.020
So length of a
side has to be less

00:04:14.020 --> 00:04:19.405
than the sum of the
lengths of other two sides.

00:04:23.090 --> 00:04:26.000
If you're willing to deal
with degenerate triangles--

00:04:26.000 --> 00:04:27.750
where you essentially
form a line segment,

00:04:27.750 --> 00:04:29.124
you lose all your
dimensionality,

00:04:29.124 --> 00:04:30.860
you turn to a
one-dimensional figure--

00:04:30.860 --> 00:04:33.510
then you could say
less than or equal,

00:04:33.510 --> 00:04:36.607
but we're just going to stick
to non-degenerate triangles.

00:04:36.607 --> 00:04:38.190
So the length of a
side has to be less

00:04:38.190 --> 00:04:40.990
than the sum of the lengths
of the other two sides.

00:04:40.990 --> 00:04:42.800
And just using
this principle, we

00:04:42.800 --> 00:04:45.570
could have come up with
the same exact conclusion.

00:04:45.570 --> 00:04:48.210
You could say, well look,
x is one of the sides.

00:04:48.210 --> 00:04:51.420
It has to be less than the sum
of the lengths of the other two

00:04:51.420 --> 00:04:51.980
sides.

00:04:51.980 --> 00:04:58.890
So it has to be
less than 6 plus 10,

00:04:58.890 --> 00:05:04.530
or x has to be less than
16-- the exact same result

00:05:04.530 --> 00:05:07.250
we got by visualizing
it like this.

00:05:07.250 --> 00:05:09.600
You want to say
how large can x be?

00:05:09.600 --> 00:05:16.650
Well you could say, well,
10 has to be less than-- Or

00:05:16.650 --> 00:05:18.460
how small can x be?

00:05:18.460 --> 00:05:23.810
You have to say 10
has to be less than 6

00:05:23.810 --> 00:05:26.840
plus x, the sum of the lengths
of the other two sides.

00:05:26.840 --> 00:05:29.950
If you subtract 6 from
both sides right over here,

00:05:29.950 --> 00:05:37.310
you get 4 is less than x,
or x is greater than 4.

00:05:37.310 --> 00:05:40.884
So this is a, in some level,
it's a kind of a basic idea,

00:05:40.884 --> 00:05:43.300
but it's something that you'll
see definitely in geometry.

00:05:43.300 --> 00:05:46.040
And then you'll go far into
other types of mathematics

00:05:46.040 --> 00:05:48.590
and you'll see other versions
of what's essentially

00:05:48.590 --> 00:05:51.930
this triangle
inequality theorem.


Fancier logarithm expressions

WEBVTT
Kind: captions
Language: en

00:00:00.620 --> 00:00:02.770
Let's give ourselves a
little bit more practice

00:00:02.770 --> 00:00:04.290
with logarithms.

00:00:04.290 --> 00:00:08.880
So just as a little bit of
review, let's evaluate log base

00:00:08.880 --> 00:00:11.600
2 of 8.

00:00:11.600 --> 00:00:14.740
What does this evaluate to?

00:00:14.740 --> 00:00:17.020
Well, it's asking us
or it will evaluate

00:00:17.020 --> 00:00:19.800
to the power or
the exponent that I

00:00:19.800 --> 00:00:23.360
have to raise our base to,
that I have to raise 2 to,

00:00:23.360 --> 00:00:25.450
to get to 8.

00:00:25.450 --> 00:00:26.810
So 2 to the first power is 2.

00:00:26.810 --> 00:00:28.080
2 to the second power is 4.

00:00:28.080 --> 00:00:30.580
2 to the third power is 8.

00:00:30.580 --> 00:00:34.430
So this right over here
is going to be equal to 3.

00:00:34.430 --> 00:00:35.060
Fair enough.

00:00:35.060 --> 00:00:37.640
We did examples like
that in the last video.

00:00:37.640 --> 00:00:39.681
Let's do something a little
bit more interesting.

00:00:41.960 --> 00:00:42.940
And I'll color-code it.

00:00:42.940 --> 00:00:51.240
What is log base 8 of 2?

00:00:51.240 --> 00:00:53.460
Now, this is interesting.

00:00:53.460 --> 00:00:56.680
I'll give you a few
seconds to think about it.

00:00:56.680 --> 00:00:59.380
Well, we're asking ourselves,
or this will evaluate to,

00:00:59.380 --> 00:01:03.940
the exponent that I have
to raise 8 to to get to 2.

00:01:03.940 --> 00:01:06.140
So let's think about
that in another way.

00:01:06.140 --> 00:01:12.540
So we could say 8 to some
power-- and that exponent

00:01:12.540 --> 00:01:14.290
that I'm raising 8 to
is essentially what

00:01:14.290 --> 00:01:16.100
this logarithm
would evaluate to.

00:01:16.100 --> 00:01:20.870
8 to some power is
going to be equal to 2.

00:01:24.580 --> 00:01:28.470
Well, if 2 to the third power
is 8, 8 to the one-third power

00:01:28.470 --> 00:01:30.030
is equal to 2.

00:01:30.030 --> 00:01:32.890
So x is equal to 1/3.

00:01:32.890 --> 00:01:35.480
8 to the one-third
power is equal to 2,

00:01:35.480 --> 00:01:38.760
or you could say the
cube root of 8 is 2.

00:01:38.760 --> 00:01:40.610
So in this case, x is 1/3.

00:01:40.610 --> 00:01:47.260
This logarithm right over
here will evaluate to 1/3.

00:01:47.260 --> 00:01:48.980
Fascinating.

00:01:48.980 --> 00:01:51.080
Let's mix it up a
little bit more.

00:01:51.080 --> 00:01:55.190
Let's say we had the log base 2.

00:01:55.190 --> 00:02:01.590
Instead of 8, let's put
a 1/8 right over here.

00:02:01.590 --> 00:02:05.270
So I'll give you a few
seconds to think about that.

00:02:05.270 --> 00:02:07.850
Well, it's asking us, or
this will evaluate to,

00:02:07.850 --> 00:02:11.610
the exponent that I have to
raise 2 to to get to 1/8.

00:02:11.610 --> 00:02:13.940
So if we said that
this is equal to x,

00:02:13.940 --> 00:02:21.500
we're essentially saying 2 to
the x power is equal to 1/8.

00:02:25.180 --> 00:02:28.270
Well, we know 2 to the third
power-- let me write this down.

00:02:28.270 --> 00:02:31.470
We already know that 2 to the
third power is equal to 8.

00:02:31.470 --> 00:02:35.150
If we want to get to 1/8,
which is the reciprocal of 8,

00:02:35.150 --> 00:02:37.580
we just have to raise 2
to the negative 3 power.

00:02:37.580 --> 00:02:41.500
2 to the negative
3 power is 1 over 2

00:02:41.500 --> 00:02:46.500
to the third power, which
is the same thing as 1/8.

00:02:46.500 --> 00:02:48.885
So if we're asking
ourselves, what exponent

00:02:48.885 --> 00:02:51.390
do we have to raise
2 to to get to 1/8?

00:02:51.390 --> 00:02:54.660
Well, we have to raise it
to the negative 3 power.

00:02:54.660 --> 00:02:56.820
So x is equal to negative 3.

00:02:56.820 --> 00:03:01.910
This logarithm
evaluates to negative 3.

00:03:01.910 --> 00:03:05.580
Now let's really,
really mix it up.

00:03:05.580 --> 00:03:18.160
What would be the
log base 8 of 1/2?

00:03:18.160 --> 00:03:20.910
What does this evaluate to?

00:03:20.910 --> 00:03:23.710
Let me clean this up so that we
have some space to work with.

00:03:26.630 --> 00:03:30.075
So as always, we're saying, what
power do I have to raise 8 to

00:03:30.075 --> 00:03:32.560
to get to 1/2?

00:03:32.560 --> 00:03:34.720
So let's think about
that a little bit.

00:03:34.720 --> 00:03:38.970
We already know that 8 to the
one-third power is equal to 2.

00:03:38.970 --> 00:03:42.080
If we want the reciprocal
of 2 right over here,

00:03:42.080 --> 00:03:45.570
we have to just raise 8
to the negative one-third.

00:03:45.570 --> 00:03:47.840
So let me write that down.

00:03:47.840 --> 00:03:51.460
8 to the negative
one-third power

00:03:51.460 --> 00:03:56.546
is going to be equal to 1
over 8 to the one-third power.

00:03:56.546 --> 00:03:58.170
And we already know
the cube root of 8,

00:03:58.170 --> 00:04:00.100
or 8 to the one-third
power, is equal to 2.

00:04:00.100 --> 00:04:02.310
This is equal to 1/2.

00:04:02.310 --> 00:04:08.800
So the log base 8 of 1/2 is
equal to-- well, what power

00:04:08.800 --> 00:04:12.770
do I have to raise 8 to to
get to 1/2-- is negative 1/3.

00:04:12.770 --> 00:04:15.510
Equal to negative 1/3.

00:04:15.510 --> 00:04:19.110
I hope you enjoyed
that as much as I did.


Logarithms

WEBVTT
Kind: captions
Language: en

00:00:00.422 --> 00:00:02.512
Let's learn a little bit about the wonderful

00:00:02.512 --> 00:00:06.209
world of logarithms.

00:00:06.209 --> 00:00:09.519
So we already know how to take exponents.

00:00:09.519 --> 00:00:13.871
If I were to say 2 to the fourth power, what does that mean?

00:00:13.871 --> 00:00:18.846
Well that means 2 times 2 times 2 times 2.

00:00:18.846 --> 00:00:24.349
2 multiplied or repeatedly multiplied 4 times, and so

00:00:24.349 --> 00:00:26.499
this is going to be

00:00:26.499 --> 00:00:27.630
2 times 2 is 4

00:00:27.630 --> 00:00:32.887
times 2 is 8, times 2 is 16.

00:00:32.887 --> 00:00:35.346
But what if we think about things in another way.

00:00:35.346 --> 00:00:40.496
We know that we get to 16 when we raise 2 to some power

00:00:40.496 --> 00:00:42.908
but we want to know what that power is.

00:00:42.908 --> 00:00:46.651
So for example, let's say that I start with 2, and I say

00:00:46.651 --> 00:00:52.832
I'm raising it to some power, what does that power have to be to get 16?

00:00:52.832 --> 00:01:00.369
Well we just figured that out. 'X' would have to be 4.

00:01:00.369 --> 00:01:04.762
And this is what logarithms are fundamentally about, figuring out what power

00:01:04.762 --> 00:01:08.514
you have to raise to, to get another number.

00:01:08.514 --> 00:01:12.288
Now the way that we would denote this with logarithm notation

00:01:12.288 --> 00:01:15.269
is we would say, log, base--

00:01:15.269 --> 00:01:18.182
actually let me make it a little bit more colourful.

00:01:18.182 --> 00:01:21.656
Log, base 2--

00:01:21.656 --> 00:01:23.592
I'll do this 2 in blue...

00:01:23.592 --> 00:01:31.777
Log, base 2, of 16

00:01:31.777 --> 00:01:36.118
is equal to what, or is equal in this case since we have the 'x' there,

00:01:36.118 --> 00:01:39.975
is equal to 'x'?

00:01:39.975 --> 00:01:45.063
This and this are completely equivalent statements.

00:01:45.063 --> 00:01:49.078
This is saying "hey well if I take 2 to some 'x' power I get 16'."

00:01:49.078 --> 00:01:57.388
This is saying, "what power do I need to raise 2 to, to get 16 and I'm going to set that to be equal to 'x'."

00:01:57.388 --> 00:01:59.927
And you'll say, "well you have to raise it to the fourth power and once again

00:01:59.927 --> 00:02:01.583
'x' is equal to 4.

00:02:01.583 --> 00:02:07.670
So with that out of the way let's try more examples of evaluating logarithmic expressions.

00:02:07.670 --> 00:02:12.051
Let's say you had...

00:02:12.051 --> 00:02:25.127
log, base 3, of 81. What would this evaluate to?

00:02:25.127 --> 00:02:31.319
Well this is a reminder, this evaluates to the power we have to raise 3 to, to get

00:02:31.319 --> 00:02:33.053
to 81.

00:02:33.053 --> 00:02:37.764
So if you want to, you can set this to be equal to an 'x', and you can

00:02:37.764 --> 00:02:48.779
restate this equation as, 3 to the 'x' power, is equal to 81.

00:02:48.779 --> 00:02:52.287
Why is a logarithm useful? And you'll see that it has very interesting properties

00:02:52.287 --> 00:02:53.112
later on.

00:02:53.112 --> 00:02:57.797
But you didn't necessarily have to use algebra. To do it this way, to say that 'x' is the power

00:02:57.797 --> 00:03:02.151
you raise 3 to to get to 81, you had to use algebra here, while with just

00:03:02.151 --> 00:03:06.257
a straight up logarithmic expression, you didn't really have to use any algebra, we didn't have to

00:03:06.257 --> 00:03:09.248
say that it was equal to 'x', we could just say that this evaluates to

00:03:09.248 --> 00:03:12.614
the power I need to raise 3 to to get to 81.

00:03:12.614 --> 00:03:15.895
The power I need to raise 3 to to get to 81.

00:03:15.895 --> 00:03:19.398
Well what power do you have to raise 3 to to get to 81?

00:03:19.398 --> 00:03:21.244
Well let's experiment a little bit,

00:03:21.244 --> 00:03:25.799
so 3 to the first power is just 3, 3 to the second power is 9,

00:03:25.799 --> 00:03:34.383
3 to the third power is 27, 3 to the fourth power, 27 times 3 is equal to 81.

00:03:34.383 --> 00:03:36.967
3 to the fourth power is equal to 81.

00:03:36.967 --> 00:03:39.388
'X' is equal to four.

00:03:39.388 --> 00:03:41.169
So we could say...

00:03:41.169 --> 00:03:50.953
Log, base 3, of 81, is equal to--

00:03:50.953 --> 00:03:53.993
I'll do this in a different colour.

00:03:53.993 --> 00:03:57.453
Is equal to 4.

00:03:57.453 --> 00:04:00.730
Let's do several more of these examples and I really encourage you to give a shot

00:04:00.730 --> 00:04:04.548
on your own and hopefully you'll get the hang of it.

00:04:04.548 --> 00:04:09.647
So let's try a larger number, let's say we want to take

00:04:09.647 --> 00:04:17.511
log, base 6, of 216. What will this evaluate to?

00:04:17.511 --> 00:04:22.768
Well we're asking ourselves, "what power do we have to raise 6 to, to get 216?"

00:04:22.768 --> 00:04:27.237
6 to the first power is 6, 6 to the second power is 36,

00:04:27.237 --> 00:04:31.411
36 times 6 is 216.

00:04:31.411 --> 00:04:34.330
This is equal to 216.

00:04:34.330 --> 00:04:37.451
So this is 6 to the third power is equal to 216.

00:04:37.451 --> 00:04:40.637
So if someone says "what power do I have to raise 6 to--

00:04:40.637 --> 00:04:41.851
this base here,

00:04:41.851 --> 00:04:44.328
to get to 216?"

00:04:44.328 --> 00:04:48.096
Well that's just going to be equal to 3.

00:04:48.096 --> 00:04:52.063
6 to the third power is equal to 216.

00:04:52.063 --> 00:04:56.891
Let's try another one. Let's say I had, I dunno,

00:04:56.891 --> 00:05:02.599
log, base 2, of 64.

00:05:02.599 --> 00:05:06.056
So what does this evaluate to?

00:05:06.056 --> 00:05:11.749
Well once again we're asking ourselves, "well this will evaluate to the exponent that I have to raise this

00:05:11.749 --> 00:05:14.545
base 2, and you do this as a little subscript right here.

00:05:14.545 --> 00:05:18.748
The exponent that I have to raise 2 to, to get to 64."

00:05:18.748 --> 00:05:22.217
So 2 to the first power is 2, 2 to the second power is 4,

00:05:22.217 --> 00:05:27.369
8, 16, 32, 64.

00:05:27.369 --> 00:05:34.663
So this right over here is 2 to the sixth power, is equal to 64. So when you evaluate this expression

00:05:34.663 --> 00:05:38.518
you say "what power do I have to raise 2 to, to get to 64?"

00:05:38.518 --> 00:05:43.812
Well I have to raise to the sixth power.

00:05:43.812 --> 00:05:49.097
Let's do a slightly more straightforward one, or maybe this will be less

00:05:49.097 --> 00:05:51.349
straightforward depending on how you view it.

00:05:51.349 --> 00:05:58.245
What is log, base 100, of 1?

00:05:58.245 --> 00:06:00.626
I'll let you think about that for a second.

00:06:00.626 --> 00:06:05.101
100 is a subscript so it's, log, base 100, of 1.

00:06:05.101 --> 00:06:08.091
That's one way to think about it, I'll put parentheses around the 1.

00:06:08.091 --> 00:06:10.991
What does this evaluate to?

00:06:10.991 --> 00:06:15.044
Well this is asking ourselves, or we would evaluate this as,

00:06:15.044 --> 00:06:22.197
"what power do I have to raise 100 to, to get to 1?"

00:06:22.197 --> 00:06:24.241
So let me write this down as an equation.

00:06:24.241 --> 00:06:26.909
If I set this to be equal to 'x', this is literally saying

00:06:26.909 --> 00:06:31.598
100, to what power, is equal to 1?

00:06:31.598 --> 00:06:34.424
Well anything that a 0 power is equal to 1.

00:06:34.424 --> 00:06:36.707
So in this case 'x' is equal to 0.

00:06:36.707 --> 00:06:44.114
So log, base 100, of 1, is going to be equal to 0.

00:06:44.114 --> 00:06:50.933
Log base anything of 1, is going to be equal to 0 because anything to the 0

00:06:50.933 --> 00:06:55.905
power and we're not talking about 0 here. Anything that is to the power of 0

00:06:55.905 --> 00:07:01.328
that is not 0, is going to be equal to 1.


Ex 2: Multi-step equation

WEBVTT
Kind: captions
Language: en

00:00:00.750 --> 00:00:07.940
So I have the equation 7 minus
10/x is equal to 2 plus 15/x.

00:00:07.940 --> 00:00:09.550
And so this isn't
the type of equation

00:00:09.550 --> 00:00:11.780
that you might think that
you're used to solving.

00:00:11.780 --> 00:00:13.640
But I'll give you a
few moments to see

00:00:13.640 --> 00:00:17.210
if you can solve it on your own.

00:00:17.210 --> 00:00:20.260
Well, what we'll see is we
can do a quick multiplication

00:00:20.260 --> 00:00:22.740
of both sides to actually
simplify this to a form

00:00:22.740 --> 00:00:24.990
that we are more
used to looking at.

00:00:24.990 --> 00:00:28.630
So what's probably bothering
you, because it's bothering me,

00:00:28.630 --> 00:00:32.220
is these x's that we have in the
denominators right over here.

00:00:32.220 --> 00:00:34.144
We're like, well, how
do we deal with that?

00:00:34.144 --> 00:00:36.060
Well, whenever we see
an x in the denominator,

00:00:36.060 --> 00:00:38.400
the temptation is
to multiply it by x.

00:00:38.400 --> 00:00:40.620
But we can't just multiply
one of the terms by x.

00:00:40.620 --> 00:00:43.320
We have to multiply
the entire side by x.

00:00:43.320 --> 00:00:45.850
So we could multiply
this entire side by x.

00:00:45.850 --> 00:00:48.280
But we can't just multiply
the left-hand side by x.

00:00:48.280 --> 00:00:52.130
We'd also want to multiply
the right-hand side by x.

00:00:52.130 --> 00:00:53.790
And so what will that give us?

00:00:53.790 --> 00:00:55.800
Well, we distribute the x.

00:00:55.800 --> 00:00:59.190
We get x times 7 is 7x.

00:00:59.190 --> 00:01:02.550
And then x times
negative 10/x, well,

00:01:02.550 --> 00:01:04.750
that's just going
to be negative 10.

00:01:04.750 --> 00:01:06.990
So you get negative
10 right over there.

00:01:06.990 --> 00:01:10.750
So the left-hand side
simplifies to 7x minus 10.

00:01:10.750 --> 00:01:12.810
And then your right-hand
side, once again,

00:01:12.810 --> 00:01:17.512
distribute the x.
x times 2 is 2x.

00:01:17.512 --> 00:01:21.890
x times 15/x, well,
x times something

00:01:21.890 --> 00:01:24.910
over x is just going to be
the something. x times 15/x

00:01:24.910 --> 00:01:28.270
is just going to
be 15-- plus 15.

00:01:28.270 --> 00:01:31.180
So now we've simplified
this to a linear equation.

00:01:31.180 --> 00:01:33.220
We have the variable
on both sides.

00:01:33.220 --> 00:01:35.210
So we just have to do
some of the techniques

00:01:35.210 --> 00:01:37.001
that we already know.

00:01:37.001 --> 00:01:38.500
So the first thing
that I like to do

00:01:38.500 --> 00:01:41.210
is maybe get all my x's
on the left-hand side.

00:01:41.210 --> 00:01:43.490
So I want to get rid of
this 2x right over here.

00:01:43.490 --> 00:01:46.510
So I subtract 2x from
the right-hand side.

00:01:46.510 --> 00:01:50.170
Now, and I always remind
you, I can't do that just

00:01:50.170 --> 00:01:51.140
to the right-hand side.

00:01:51.140 --> 00:01:52.420
If I did it just to
the right-hand side,

00:01:52.420 --> 00:01:54.190
it wouldn't be an
equality anymore.

00:01:54.190 --> 00:01:57.430
You have to do that to the
left-hand side as well.

00:01:57.430 --> 00:02:02.470
And so we are left with-- let
me get that pink color again.

00:02:02.470 --> 00:02:06.590
On the left-hand side, 7x, 7 of
something minus 2 of something,

00:02:06.590 --> 00:02:10.340
well, you're going to have 5
of that something, minus 10.

00:02:10.340 --> 00:02:12.720
These two x's negate each other.

00:02:12.720 --> 00:02:16.060
And you're left with equals 15.

00:02:16.060 --> 00:02:18.060
Now we can get rid
of this negative 10

00:02:18.060 --> 00:02:20.205
by adding 10 to both sides.

00:02:20.205 --> 00:02:21.580
You know, I like
that green color

00:02:21.580 --> 00:02:23.550
when I do stuff to both sides.

00:02:23.550 --> 00:02:26.680
So I can add 10 to both sides.

00:02:26.680 --> 00:02:30.860
And I'm left with
5x-- these negate

00:02:30.860 --> 00:02:34.460
each other-- is equal to 25.

00:02:34.460 --> 00:02:35.700
And this is the home stretch.

00:02:35.700 --> 00:02:37.510
You see where this is going.

00:02:37.510 --> 00:02:40.620
We can divide both sides by 5.

00:02:40.620 --> 00:02:44.860
And we are left with
x is equal to 5.

00:02:44.860 --> 00:02:48.170
Now let's verify that
this actually worked.

00:02:48.170 --> 00:02:50.030
So let's go back to
the original equation.

00:02:50.030 --> 00:02:54.430
We have 7 minus 10/5.

00:02:54.430 --> 00:02:56.160
This needs to be
equal to-- I'm just

00:02:56.160 --> 00:02:58.570
taking our 5 and
substituting it back here.

00:02:58.570 --> 00:03:01.020
This needs to be
equal to 2 plus 15/5.

00:03:03.790 --> 00:03:06.070
So this is 7 minus 10/5.

00:03:06.070 --> 00:03:07.980
This is just 2.

00:03:07.980 --> 00:03:12.780
It needs to be equal to 2
plus 15/5, which is just 3.

00:03:12.780 --> 00:03:18.580
So 2 plus 3, 7 minus
2 is 5, 2 plus 3 is 5,

00:03:18.580 --> 00:03:21.780
5 is indeed equal to 5.

00:03:21.780 --> 00:03:23.730
And we are done.


Ex 1: Multi step equation

WEBVTT
Kind: captions
Language: en

00:00:00.810 --> 00:00:06.660
So we have 14x plus 4
over negative 3x minus 2

00:00:06.660 --> 00:00:09.060
is equal to 8.

00:00:09.060 --> 00:00:10.680
And I'll give you a
few moments to see

00:00:10.680 --> 00:00:14.510
if you can tackle
it on your own.

00:00:14.510 --> 00:00:16.180
So this equation
right here, at first

00:00:16.180 --> 00:00:18.600
it doesn't look like a
straightforward linear

00:00:18.600 --> 00:00:19.220
equation.

00:00:19.220 --> 00:00:22.720
We have one expression on
top of another expression.

00:00:22.720 --> 00:00:24.720
But as we'll see,
we can simplify

00:00:24.720 --> 00:00:27.740
this to turn it into
a linear equation.

00:00:27.740 --> 00:00:29.250
So the first thing
that I want to do

00:00:29.250 --> 00:00:32.479
is, I don't like this negative
3x plus 2 sitting here

00:00:32.479 --> 00:00:35.320
in the denominator,
it makes me stressed.

00:00:35.320 --> 00:00:39.140
So I want to multiply both
sides of this equation

00:00:39.140 --> 00:00:42.760
times negative 3x minus 2.

00:00:46.870 --> 00:00:48.170
What does that do for us?

00:00:48.170 --> 00:00:51.140
Well, on the left-hand side, you
have this negative 3x minus 2,

00:00:51.140 --> 00:00:53.370
it's going to be over
negative 3x minus 2,

00:00:53.370 --> 00:00:57.600
they will cancel out.

00:00:57.600 --> 00:01:00.460
And so you're left with,
on the left-hand side,

00:01:00.460 --> 00:01:03.200
your 14x plus 4.

00:01:03.200 --> 00:01:05.190
And on your right-hand
side, you just

00:01:05.190 --> 00:01:08.090
have to multiply 8 times
negative 3x minus 2.

00:01:08.090 --> 00:01:12.560
So you are left with--
well, 8 times negative 3x

00:01:12.560 --> 00:01:18.460
is negative 24x, and
then 8 times negative 2

00:01:18.460 --> 00:01:21.154
is negative 16.

00:01:21.154 --> 00:01:22.070
And there you have it.

00:01:22.070 --> 00:01:25.590
We have simplified this to just
a traditional linear equation.

00:01:25.590 --> 00:01:27.330
We've got variables
on both sides,

00:01:27.330 --> 00:01:29.577
so we can just keep simplifying.

00:01:29.577 --> 00:01:31.160
So the first thing
I want to do, let's

00:01:31.160 --> 00:01:32.909
just say we want to
put all of our x terms

00:01:32.909 --> 00:01:34.330
on the left-hand side.

00:01:34.330 --> 00:01:39.469
So I want to get rid of this
negative 24x right over here.

00:01:39.469 --> 00:01:41.010
So the best way to
do that, I'm going

00:01:41.010 --> 00:01:43.630
to add 24x to the
right-hand side.

00:01:43.630 --> 00:01:45.380
I can't just do it to
the right-hand side,

00:01:45.380 --> 00:01:48.320
I have to also do it
to the left-hand side.

00:01:48.320 --> 00:01:52.130
And so I am left with,
on the left-hand side,

00:01:52.130 --> 00:01:56.220
14x plus 24x is 38x.

00:01:56.220 --> 00:02:00.130
And then I have the plus
4-- Is equal to, well,

00:02:00.130 --> 00:02:02.590
negative 24x plus 24x.

00:02:02.590 --> 00:02:05.335
Those cancel each
other out, and we

00:02:05.335 --> 00:02:07.330
are left with just
the negative 16.

00:02:09.860 --> 00:02:12.900
Now we just have to
get rid of this 4 here.

00:02:12.900 --> 00:02:16.880
Let's subtract that
4 from both sides.

00:02:16.880 --> 00:02:20.010
And we are left with-- and
this is the home stretch now--

00:02:20.010 --> 00:02:27.570
we are left with 38x is
equal to negative 16 minus 4

00:02:27.570 --> 00:02:29.860
is negative 20.

00:02:29.860 --> 00:02:33.373
And so we can divide both
sides of this equation by 38.

00:02:36.150 --> 00:02:40.420
And we are left with x
is equal to negative 20

00:02:40.420 --> 00:02:43.320
over 38, which can be
simplified further.

00:02:43.320 --> 00:02:46.220
Both the numerator the
denominator is divisible by 2.

00:02:46.220 --> 00:02:48.360
So let's divide the numerator
and denominator by 2,

00:02:48.360 --> 00:02:52.490
and we get negative 10 over 19.

00:02:52.490 --> 00:02:57.670
x is equal to negative 10
over 19, and we are done.

00:02:57.670 --> 00:03:00.200
I encourage you to
validate this for yourself.

00:03:00.200 --> 00:03:02.570
It's a little bit of a hairy
number right over here,

00:03:02.570 --> 00:03:04.570
but to take this number,
substitute it back

00:03:04.570 --> 00:03:07.510
into our original equation,
and validate that it actually

00:03:07.510 --> 00:03:10.390
does satisfy that equation.


Evaluating expressions with unknown variables 2

WEBVTT
Kind: captions
Language: en

00:00:00.620 --> 00:00:02.610
Let's do a few
more examples where

00:00:02.610 --> 00:00:06.120
we're evaluating expressions
with unknown variables.

00:00:06.120 --> 00:00:10.110
So this first one we're told 3x
plus 3y plus 3z is equal to 1,

00:00:10.110 --> 00:00:14.810
and then we're asked what's
12x plus 12y plus 12z equal to?

00:00:14.810 --> 00:00:18.060
And I'll give you a few
moments to think about that.

00:00:18.060 --> 00:00:21.560
Well let's rewrite
this second expression

00:00:21.560 --> 00:00:29.760
by factoring out the 12, so we
get 12 times x plus y plus z.

00:00:29.760 --> 00:00:31.260
That's this second
expression here,

00:00:31.260 --> 00:00:33.240
and you can verify that
by distributing the 12.

00:00:33.240 --> 00:00:35.820
You'll get exactly
this right up here.

00:00:35.820 --> 00:00:39.490
Now, what is 12 times
x plus y plus z?

00:00:39.490 --> 00:00:43.420
Well, we don't know yet
exactly what x plus y plus z

00:00:43.420 --> 00:00:46.400
is equal to, but this first
equation might help us.

00:00:46.400 --> 00:00:49.820
This first equation, we can
rewrite this left-hand side

00:00:49.820 --> 00:00:54.180
by factoring out the 3, so we
could rewrite this as 3 times

00:00:54.180 --> 00:00:58.640
x plus y plus z is equal to 1.

00:00:58.640 --> 00:01:02.070
All I did is I factored the
3 out on the left-hand side.

00:01:02.070 --> 00:01:04.310
And then if I want to
solve for x plus y plus z,

00:01:04.310 --> 00:01:09.300
I just divide both sides
of this equation by 3,

00:01:09.300 --> 00:01:15.520
and I'm left with x plus
y plus z is equal to 1/3,

00:01:15.520 --> 00:01:20.720
and so here, instead of x plus
y plus z, I can write 1/3.

00:01:20.720 --> 00:01:24.856
So this whole thing
simplified to 12 times 1/3.

00:01:24.856 --> 00:01:28.580
12 times 1/3 is the same
thing as 12 divided by 3,

00:01:28.580 --> 00:01:31.720
which is equal to 4.

00:01:31.720 --> 00:01:33.680
Let's try one more.

00:01:33.680 --> 00:01:39.800
So here we are told that
3a plus 5b is equal to 2,

00:01:39.800 --> 00:01:44.910
and then we're asked what's 15a
plus 15b going to be equal to?

00:01:44.910 --> 00:01:46.250
So we might-- let's see.

00:01:46.250 --> 00:01:50.792
I'll give you a few moments to
try to tackle this on your own.

00:01:50.792 --> 00:01:52.000
Let's see how we might do it.

00:01:52.000 --> 00:01:53.080
We could approach
it the way we've

00:01:53.080 --> 00:01:54.746
approached the last
few problems, trying

00:01:54.746 --> 00:01:56.210
to rewrite the
second expression.

00:01:56.210 --> 00:02:00.491
We could rewrite it as 15
times a plus b, and so we just

00:02:00.491 --> 00:02:01.990
have to figure out
what a plus b is,

00:02:01.990 --> 00:02:04.200
and we'll be able to
evaluate this expression.

00:02:04.200 --> 00:02:06.410
And so, it's tempting
to look up here,

00:02:06.410 --> 00:02:09.960
and say maybe we can solve
for a plus b somehow,

00:02:09.960 --> 00:02:11.840
but we really can't.

00:02:11.840 --> 00:02:15.370
If we divide-- if we
try to factor out a 3,

00:02:15.370 --> 00:02:21.050
we'll get 3 times a plus
5/3b, so this doesn't really

00:02:21.050 --> 00:02:23.810
simplify things in
terms of a plus b.

00:02:23.810 --> 00:02:33.450
If we try to factor out a 5,
we'd get 5 times 3/5a plus b

00:02:33.450 --> 00:02:35.720
is equal to 2, but
neither of these

00:02:35.720 --> 00:02:39.370
gets us in a form where we
can then solve for a plus b.

00:02:39.370 --> 00:02:41.120
So in this situation,
we actually

00:02:41.120 --> 00:02:45.760
do not have enough information
to solve this problem.

00:02:45.760 --> 00:02:47.360
So it's a little bit of a trick.

00:02:47.360 --> 00:02:53.280
Not enough info to solve.

00:02:56.170 --> 00:02:59.210
Anyway, hopefully
you enjoyed that.


Evaluating expressions where individual variable values are unknown

WEBVTT
Kind: captions
Language: en

00:00:00.770 --> 00:00:02.960
I'm now going to do a
few classic problems that

00:00:02.960 --> 00:00:04.940
show up a lot at
math competitions

00:00:04.940 --> 00:00:07.110
and sometimes on
standardized tests,

00:00:07.110 --> 00:00:09.697
and they seem like very daunting
and intimidating problems.

00:00:09.697 --> 00:00:11.530
But hopefully over the
course of this video,

00:00:11.530 --> 00:00:14.370
you'll realize that if you kind
of see what they're asking for,

00:00:14.370 --> 00:00:15.760
it's really not that bad.

00:00:15.760 --> 00:00:17.593
So let's try this first
one right over here.

00:00:17.593 --> 00:00:21.900
They tell us a plus b
plus c is equal to 7.

00:00:21.900 --> 00:00:27.380
And then they ask us, what is
5a plus 5b plus 5c equal to?

00:00:27.380 --> 00:00:29.040
And the first
reaction's like, well,

00:00:29.040 --> 00:00:31.040
they just gave me one
equation right over here.

00:00:31.040 --> 00:00:32.479
One equation with
three unknowns,

00:00:32.479 --> 00:00:34.010
how do I solve for a, b, or c?

00:00:34.010 --> 00:00:36.190
Don't I have to be able
to solve for a, b, or c

00:00:36.190 --> 00:00:40.830
in order to figure out
what 5a plus 5b plus 5c is?

00:00:40.830 --> 00:00:44.120
So I'll let you think
about that for a second.

00:00:44.120 --> 00:00:46.540
So the big idea
here is to realize

00:00:46.540 --> 00:00:50.200
that 5a plus 5b
plus 5c, that's just

00:00:50.200 --> 00:00:56.760
the same thing as 5
times a plus b plus c.

00:00:56.760 --> 00:01:01.610
If you distribute the 5 here, 5
times a, 5 times b, 5 times c,

00:01:01.610 --> 00:01:04.150
you get 5a plus 5b plus 5c.

00:01:04.150 --> 00:01:07.000
Another way you could think of
it is we are factoring a 5 out.

00:01:07.000 --> 00:01:09.030
If you factor a 5
out, you're left

00:01:09.030 --> 00:01:13.010
with 5 times a plus b plus c.

00:01:13.010 --> 00:01:16.060
Now, how do we evaluate this?

00:01:16.060 --> 00:01:19.540
Well, that first equation
gave us all the information

00:01:19.540 --> 00:01:20.720
that we needed.

00:01:20.720 --> 00:01:25.140
They told us that a plus
b plus c is equal to 7.

00:01:30.130 --> 00:01:34.474
So 5 times a plus b plus
c is the same exact thing.

00:01:34.474 --> 00:01:36.140
I think you see where
this is going now.

00:01:36.140 --> 00:01:43.120
It's 5 times 7, and now this
becomes pretty straightforward.

00:01:43.120 --> 00:01:46.280
That's going to be equal to 35.

00:01:46.280 --> 00:01:47.440
Let's do one more of these.

00:01:47.440 --> 00:01:49.220
And this is going to be a
little bit more involved,

00:01:49.220 --> 00:01:50.810
but hopefully you
see the same idea.

00:01:53.540 --> 00:01:57.660
So we're told a plus b plus
c is equal to negative 1.

00:01:57.660 --> 00:01:59.650
And we have two
more variables here.

00:01:59.650 --> 00:02:01.605
We're told that x
plus y is equal to 7.

00:02:01.605 --> 00:02:03.230
Then they ask us this
big, hairy thing.

00:02:03.230 --> 00:02:05.141
What is this equal to?

00:02:05.141 --> 00:02:06.890
And I'll give you a
few seconds to give it

00:02:06.890 --> 00:02:09.330
a little bit of thought.

00:02:09.330 --> 00:02:10.340
Well, you could imagine.

00:02:10.340 --> 00:02:12.610
They've given three
variables with one equation

00:02:12.610 --> 00:02:15.420
up here, another two variables
with another equation.

00:02:15.420 --> 00:02:17.610
We have five unknowns
with two equations.

00:02:17.610 --> 00:02:20.276
There's no way we're going to be
able to individually figure out

00:02:20.276 --> 00:02:22.620
what a, or b, or
c, or x, or y is.

00:02:22.620 --> 00:02:25.690
But maybe we can use what
we saw in the last example

00:02:25.690 --> 00:02:27.071
to solve this.

00:02:27.071 --> 00:02:29.570
And so what we might want do
is rearrange it so that we have

00:02:29.570 --> 00:02:32.616
the x's and the y's kind
of grouped together,

00:02:32.616 --> 00:02:34.740
and the a's, b'c, and c's
kind of grouped together.

00:02:34.740 --> 00:02:36.060
So let's try that out.

00:02:36.060 --> 00:02:38.350
So let's focus first on
the a's, b's, and c's.

00:02:38.350 --> 00:02:42.070
So we have negative
9a, negative 9b--

00:02:42.070 --> 00:02:45.010
I'll go in order, alphabetical
order-- negative 9b,

00:02:45.010 --> 00:02:49.530
and we have negative 9c, and I
think you see what's emerging.

00:02:49.530 --> 00:02:51.570
And then let's work on
the x's and the y's.

00:02:51.570 --> 00:02:55.770
And then we have
negative 7x, minus 7x,

00:02:55.770 --> 00:02:59.340
and then we have a minus 7y.

00:02:59.340 --> 00:03:03.530
So all I have done is
rearranged this expression here.

00:03:03.530 --> 00:03:07.140
But this makes it a
little bit clearer

00:03:07.140 --> 00:03:09.050
of what's going on here.

00:03:09.050 --> 00:03:14.440
These first three terms, I
can factor out a negative 9,

00:03:14.440 --> 00:03:19.950
and I get negative 9
times a plus b plus c.

00:03:19.950 --> 00:03:23.170
And these second two terms,
I can factor out a 7,

00:03:23.170 --> 00:03:26.551
minus 7 times x plus y.

00:03:26.551 --> 00:03:28.800
And just to verify, if you
wanted to go the other way,

00:03:28.800 --> 00:03:32.130
multiply this negative 7 times
x plus y, you'll get this.

00:03:32.130 --> 00:03:35.550
Multiply negative 9 times a
plus b plus c, distribute it,

00:03:35.550 --> 00:03:37.790
you'll get this right over here.

00:03:37.790 --> 00:03:39.760
And so this makes it
a little bit clearer.

00:03:39.760 --> 00:03:43.999
What is a plus b
plus c equal to?

00:03:43.999 --> 00:03:45.540
Well, they tell us
right over here. a

00:03:45.540 --> 00:03:48.885
plus b plus c is
equal to negative 1.

00:03:48.885 --> 00:03:51.740
This whole expression
is negative 1, at least

00:03:51.740 --> 00:03:52.990
in the parentheses.

00:03:52.990 --> 00:03:55.400
And what is x plus y equal to?

00:03:55.400 --> 00:03:56.670
Let me do this in a new color.

00:03:56.670 --> 00:03:58.020
What is x plus y equal to?

00:03:58.020 --> 00:04:03.400
Well, they tell us right over
here. x plus y is equal to 7.

00:04:03.400 --> 00:04:06.680
So this whole thing
simplifies to negative

00:04:06.680 --> 00:04:18.480
9 times negative
1 minus 7 times 7.

00:04:18.480 --> 00:04:21.200
And so this is equal to--
we're in the home stretch--

00:04:21.200 --> 00:04:22.970
negative 9 times negative 1.

00:04:22.970 --> 00:04:25.260
Well, that's positive 9.

00:04:25.260 --> 00:04:29.210
And then negative 7
times 7 is negative 49.

00:04:29.210 --> 00:04:35.840
So it's 9 minus 49, which
is equal to negative 40.

00:04:35.840 --> 00:04:38.068
And we are done.


Ex 3 age word problem

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:03.830
Let's do some more of these
classic age word problems.

00:00:03.830 --> 00:00:09.360
So we're told that William
is 4 times as old as Ben.

00:00:09.360 --> 00:00:14.980
12 years ago, William was
7 times as old as Ben.

00:00:14.980 --> 00:00:18.960
How old is Ben now?

00:00:18.960 --> 00:00:21.050
Once again, it's
a good idea to try

00:00:21.050 --> 00:00:25.750
to do this on your own first,
and then I'll work through it.

00:00:25.750 --> 00:00:27.440
What's the unknown here?

00:00:27.440 --> 00:00:31.120
Well, the unknown here
is how old is Ben now.

00:00:31.120 --> 00:00:34.610
So let's set a variable equal
to that, and we do x or y.

00:00:34.610 --> 00:00:38.180
But since Ben starts with
a b, let's use b for Ben.

00:00:38.180 --> 00:00:48.410
So let's let b equal Ben's
current age, Ben's age now.

00:00:48.410 --> 00:00:50.580
Let's see how all of
this other information

00:00:50.580 --> 00:00:54.300
relates to Ben's current
age, and then maybe we

00:00:54.300 --> 00:00:57.324
can set up some equation
and then solve for things.

00:00:57.324 --> 00:00:59.490
I'll do it a little bit
more structured in this one.

00:00:59.490 --> 00:01:01.406
You could have done many
of the problems we've

00:01:01.406 --> 00:01:02.780
been working on in this way.

00:01:02.780 --> 00:01:08.830
Let's think about Ben, and
then let's think about William.

00:01:08.830 --> 00:01:10.230
I'll do William in blue here.

00:01:10.230 --> 00:01:11.470
So let's think about William.

00:01:14.100 --> 00:01:17.180
And then there's two points
in time we're talking about.

00:01:17.180 --> 00:01:19.430
We're talking about
now, today, and we're

00:01:19.430 --> 00:01:23.580
going to talk
about 12 years ago.

00:01:23.580 --> 00:01:26.890
Over here, let's call that now.

00:01:26.890 --> 00:01:29.250
This will be our now
column, and then this

00:01:29.250 --> 00:01:30.580
will be our 12 years ago.

00:01:34.050 --> 00:01:36.160
Let's see what we
can fill in here.

00:01:36.160 --> 00:01:39.120
What is Ben's age now?

00:01:39.120 --> 00:01:41.060
Well, we just defined
that as the variable b.

00:01:41.060 --> 00:01:41.670
That's the unknown.

00:01:41.670 --> 00:01:43.086
That's what we
have to figure out.

00:01:43.086 --> 00:01:45.555
So let's just stick that there.

00:01:45.555 --> 00:01:47.520
That's just going to be b.

00:01:47.520 --> 00:01:50.850
Well, what's Ben's
age 12 years ago?

00:01:50.850 --> 00:01:54.680
So maybe we want to
express it in terms of b.

00:01:54.680 --> 00:01:59.390
If he's b now, 12 years
ago, he was just b minus 12.

00:01:59.390 --> 00:02:00.810
Fair enough.

00:02:00.810 --> 00:02:05.940
Now what is William's age today?

00:02:05.940 --> 00:02:09.250
Well, this first sentence
gave us the information.

00:02:09.250 --> 00:02:11.990
William is 4 times
as old as Ben.

00:02:11.990 --> 00:02:14.410
And we can assume that
they're talking about today,

00:02:14.410 --> 00:02:16.370
is 4 times as old as Ben.

00:02:16.370 --> 00:02:20.060
So if Ben is b, William
is going to be 4b.

00:02:22.930 --> 00:02:26.540
And so how old was
William 12 years ago?

00:02:26.540 --> 00:02:28.920
Well, if he's 4b right
now, 12 years ago,

00:02:28.920 --> 00:02:30.280
he'll just be 12 less than that.

00:02:30.280 --> 00:02:31.810
So he's 4b now.

00:02:31.810 --> 00:02:34.460
12 years ago, he
was 4b minus 12.

00:02:34.460 --> 00:02:36.250
So that's kind of interesting.

00:02:36.250 --> 00:02:40.095
But we haven't quite yet made
use of the second statement.

00:02:43.090 --> 00:02:45.480
This is William 12 years ago.

00:02:45.480 --> 00:02:49.480
12 years ago, William was
7 times as old as Ben.

00:02:49.480 --> 00:02:54.550
So 12 years ago, this number
is going to be 7 times

00:02:54.550 --> 00:02:55.680
this number.

00:02:55.680 --> 00:02:57.920
Or another way to think
about, take this number

00:02:57.920 --> 00:03:02.290
and multiply it by 7, and
you're going to get this number.

00:03:02.290 --> 00:03:07.925
12 years ago, Ben's age
is 1/7 of William's age,

00:03:07.925 --> 00:03:10.610
or William's age is
7 times Ben's age.

00:03:10.610 --> 00:03:14.530
So let's see if we can set
that up as an equation.

00:03:14.530 --> 00:03:22.620
We can have-- let me
write this down-- 7 times

00:03:22.620 --> 00:03:31.390
Ben's age 12 years
ago, b minus 12,

00:03:31.390 --> 00:03:34.030
is going to be equal
to William's age.

00:03:37.840 --> 00:03:40.220
And it seems like we've
done the hard part.

00:03:40.220 --> 00:03:41.370
We've set up the equation.

00:03:41.370 --> 00:03:43.710
Now we just use a little
bit of our algebraic tools

00:03:43.710 --> 00:03:45.260
to solve for b.

00:03:45.260 --> 00:03:46.990
So let's do that.

00:03:46.990 --> 00:03:49.040
The first thing we
might want to do,

00:03:49.040 --> 00:03:51.760
we could distribute
the 7, so 7 times

00:03:51.760 --> 00:03:55.040
b, 7 times, essentially,
a negative 12.

00:03:55.040 --> 00:03:59.400
We have 7b minus 7
times 12-- let's see.

00:03:59.400 --> 00:04:06.050
That's 84-- is going to
be equal to 4b minus 12.

00:04:09.800 --> 00:04:11.720
This whole expression
is literally 7 times

00:04:11.720 --> 00:04:13.770
Ben's age 12 years ago.

00:04:13.770 --> 00:04:16.470
Now what can we
do to solve this?

00:04:16.470 --> 00:04:21.160
Well, we can subtract 4b from
both sides, so let's do that.

00:04:23.970 --> 00:04:26.090
I could do two steps
at the same time.

00:04:26.090 --> 00:04:27.840
Well, actually, let's
just keep it simple.

00:04:27.840 --> 00:04:30.340
I'm going to subtract
4b from both sides.

00:04:30.340 --> 00:04:31.030
That goes away.

00:04:31.030 --> 00:04:33.500
On the right-hand side,
I have a negative 12.

00:04:33.500 --> 00:04:38.280
On the left-hand side, I am
left with 7b minus 4b is 3b.

00:04:38.280 --> 00:04:41.590
And then I still
have a minus 84.

00:04:41.590 --> 00:04:44.294
Well, I want to get rid of
this negative 84, this minus 84

00:04:44.294 --> 00:04:45.210
on the left-hand side.

00:04:45.210 --> 00:04:46.875
So let's add 84 to both sides.

00:04:50.360 --> 00:04:53.170
On the left-hand side,
I'm just left with 3b.

00:04:53.170 --> 00:04:57.390
And on the right-hand side,
I have negative 12 plus 84,

00:04:57.390 --> 00:05:00.635
or 84 minus 12, which is 72.

00:05:03.305 --> 00:05:05.450
Now if I want to
solve for b, I just

00:05:05.450 --> 00:05:08.700
have to divide both sides
of that equation by 3.

00:05:08.700 --> 00:05:11.860
And so I am left
with b is equal to--

00:05:11.860 --> 00:05:15.664
and now we have our
drum roll-- 72/3.

00:05:15.664 --> 00:05:17.580
And you might be able
to do that in your head.

00:05:17.580 --> 00:05:19.555
It would be 24, I believe.

00:05:19.555 --> 00:05:21.680
You could work it out on
paper if you have trouble.

00:05:21.680 --> 00:05:23.710
Let's just do it real quick.

00:05:23.710 --> 00:05:26.760
72, 3 goes into 7 two times.

00:05:26.760 --> 00:05:32.230
You get a 2 times 3 is 6,
subtract, you bring down the 2,

00:05:32.230 --> 00:05:33.640
3 goes into 12 four times.

00:05:33.640 --> 00:05:37.080
So b is equal to 24.

00:05:37.080 --> 00:05:40.070
Going back to our question,
what is Ben's age now?

00:05:40.070 --> 00:05:42.160
It is 24.

00:05:42.160 --> 00:05:45.610
And let's verify that
this is actually the case.

00:05:45.610 --> 00:05:48.250
They're telling us that William
is 4 times as old as Ben.

00:05:48.250 --> 00:05:50.620
So what is William's
current age?

00:05:50.620 --> 00:05:56.030
Well, 4 times 24 is 96,
so William is a senior.

00:05:56.030 --> 00:05:58.380
We should call him Mr. William.

00:05:58.380 --> 00:06:01.710
He is 96 years old.

00:06:01.710 --> 00:06:05.810
Maybe he's Ben's grandfather
or great-grandfather.

00:06:05.810 --> 00:06:08.500
Then they say 12 years ago,
William-- well, 12 years ago,

00:06:08.500 --> 00:06:11.690
William was 84 years old.

00:06:11.690 --> 00:06:12.980
So he was 84 years old.

00:06:12.980 --> 00:06:15.480
They say that's 7
times as old as Ben.

00:06:15.480 --> 00:06:20.790
Well, 12 years ago, if
he's 24 now, Ben was 12.

00:06:20.790 --> 00:06:26.738
And indeed, 84 is 7 times
12, so it all worked out.


Ex 2 age word problem

WEBVTT
Kind: captions
Language: en

00:00:00.670 --> 00:00:03.910
We're told that
in 40 years, Imran

00:00:03.910 --> 00:00:08.170
will be 11 times as
old as he is right now.

00:00:08.170 --> 00:00:12.650
And then we're asked,
how old is he right now?

00:00:12.650 --> 00:00:17.420
And so I encourage you
to try this on your own.

00:00:17.420 --> 00:00:21.210
Well, let's see if we can
set this up as an equation.

00:00:21.210 --> 00:00:24.120
So let's figure out what
our unknown is first.

00:00:24.120 --> 00:00:28.590
Well, our unknown is
how old he is right now.

00:00:28.590 --> 00:00:29.970
I'm just arbitrarily using x.

00:00:29.970 --> 00:00:31.515
We always like to use x.

00:00:31.515 --> 00:00:33.380
But I could've really
set it to be anything.

00:00:33.380 --> 00:00:36.460
But let's say x is equal
to how old he is right now.

00:00:36.460 --> 00:00:39.410
How old-- not how hold.

00:00:39.410 --> 00:00:43.800
How old he is now.

00:00:43.800 --> 00:00:48.560
Now, what do we know about how
old he will be in 40 years?

00:00:48.560 --> 00:00:51.970
Well, he's going to be
how old he is now plus 40.

00:00:51.970 --> 00:00:54.230
So let me write that down.

00:00:54.230 --> 00:01:05.570
So in 40 years Imran is
going to be x plus 40,

00:01:05.570 --> 00:01:08.534
plus this 40 right over here.

00:01:08.534 --> 00:01:10.450
But they give us another
piece of information.

00:01:10.450 --> 00:01:12.158
This by itself isn't
enough to figure out

00:01:12.158 --> 00:01:13.280
how old he is right now.

00:01:13.280 --> 00:01:18.120
But they tell us
in 40 years, Imran

00:01:18.120 --> 00:01:24.660
will be 11 times as
old as he is right now.

00:01:24.660 --> 00:01:27.100
So that's saying that this
quantity right over here,

00:01:27.100 --> 00:01:31.770
x plus 40, is going
to be 11 times x.

00:01:31.770 --> 00:01:34.470
In 40 years, he's
going to be 11 times as

00:01:34.470 --> 00:01:35.980
old as he is right now.

00:01:35.980 --> 00:01:38.990
So this is going to be times 11.

00:01:38.990 --> 00:01:41.440
You take x, multiply
it times 11,

00:01:41.440 --> 00:01:44.770
you're going to get how old
he's going to be in 40 years.

00:01:44.770 --> 00:01:48.100
So let's write that
down as an equation.

00:01:48.100 --> 00:01:55.640
You take x, multiply
it by 11, so 11 times

00:01:55.640 --> 00:01:59.920
as old as he is
right now is how old

00:01:59.920 --> 00:02:01.460
he is going to be in 40 years.

00:02:07.150 --> 00:02:10.057
And we have set up a nice
little, tidy linear equation

00:02:10.057 --> 00:02:10.650
now.

00:02:10.650 --> 00:02:14.462
So we just have to solve for x.

00:02:14.462 --> 00:02:16.420
So let's get all the x's
on the left-hand side.

00:02:16.420 --> 00:02:18.560
We have more x's here than
on the right-hand side.

00:02:18.560 --> 00:02:21.335
So we avoid negative numbers,
let's stick all the x's here.

00:02:21.335 --> 00:02:23.710
So if I want to get rid of
this x on the right hand side,

00:02:23.710 --> 00:02:24.852
I'd want to subtract an x.

00:02:24.852 --> 00:02:26.810
But obviously, I can't
just do it to the right.

00:02:26.810 --> 00:02:28.790
Otherwise, the equality
won't be true anymore.

00:02:28.790 --> 00:02:30.860
I need to do it on
the left as well.

00:02:30.860 --> 00:02:34.580
And so I am left with--
if I have 11 of something

00:02:34.580 --> 00:02:37.190
and I take away 1 of them, I'm
left with 10 of that something.

00:02:37.190 --> 00:02:44.950
So I'm left with 10 times x is
equal to-- well, these x's, x

00:02:44.950 --> 00:02:46.090
minus x is just 0.

00:02:46.090 --> 00:02:47.360
That was the whole point.

00:02:47.360 --> 00:02:49.099
It's going to be equal to 40.

00:02:49.099 --> 00:02:51.140
And you could do this in
your head at this point,

00:02:51.140 --> 00:02:52.880
but let's just
solve it formally.

00:02:52.880 --> 00:02:54.350
So if we want a 1
coefficient here,

00:02:54.350 --> 00:02:55.850
we'd want to divide
by 10, but we've

00:02:55.850 --> 00:02:58.060
got to do that to both sides.

00:02:58.060 --> 00:03:02.080
And so we are left with-- and
we could have our drum roll now.

00:03:02.080 --> 00:03:09.290
We are left with x is
equal to 4 years old.

00:03:09.290 --> 00:03:12.840
x is equal to 4.

00:03:12.840 --> 00:03:16.850
So our answer to the question,
how old is Imran right now?

00:03:16.850 --> 00:03:20.980
He is 4 years old.

00:03:20.980 --> 00:03:22.640
And let's verify this.

00:03:22.640 --> 00:03:27.250
If he's 4 years old
right now, in 40 years

00:03:27.250 --> 00:03:31.030
he's going to be 44 years old.

00:03:31.030 --> 00:03:37.590
And 44 years old is indeed 11
times older than 4 years old.

00:03:37.590 --> 00:03:42.270
This is a factor of 11
years, so it all worked out.


Ex 1 age word problem

WEBVTT
Kind: captions
Language: en

00:00:00.820 --> 00:00:05.120
Let's say that Arman
today is 18 years old.

00:00:07.990 --> 00:00:15.960
And let's say that Diya
today is 2 years old.

00:00:15.960 --> 00:00:19.210
And what I am curious
about in this video

00:00:19.210 --> 00:00:21.920
is how many years will it
take-- and let me write this

00:00:21.920 --> 00:00:31.390
down-- how many
years will it take

00:00:31.390 --> 00:00:43.370
for Arman to be three
times as old as Diya?

00:00:46.230 --> 00:00:48.330
So that's the
question right there,

00:00:48.330 --> 00:00:53.985
and I encourage you to try to
take a shot at this yourself.

00:00:53.985 --> 00:00:55.610
So let's think about
this a little bit.

00:00:55.610 --> 00:00:57.840
We're asking how many
years will it take.

00:00:57.840 --> 00:00:58.960
That's what we don't know.

00:00:58.960 --> 00:01:00.410
That's what we're curious about.

00:01:00.410 --> 00:01:03.480
How many years will
it take for Arman

00:01:03.480 --> 00:01:05.900
to be three times
as old as Diya?

00:01:05.900 --> 00:01:10.480
So let's set some variable--
let's say, y for years.

00:01:10.480 --> 00:01:17.520
Let's say y is equal
to years it will take.

00:01:17.520 --> 00:01:20.160
So given that, can we
now set up an equation,

00:01:20.160 --> 00:01:22.350
given this information,
to figure out

00:01:22.350 --> 00:01:26.360
how many years it will take
for Arman to be three times as

00:01:26.360 --> 00:01:28.510
old as Diya?

00:01:28.510 --> 00:01:33.140
Well, let's think about how
old Arman will be in y years.

00:01:33.140 --> 00:01:36.290
How old will he be?

00:01:36.290 --> 00:01:37.150
Let me write here.

00:01:37.150 --> 00:01:48.900
In y years, Arman is
going to be how old?

00:01:48.900 --> 00:01:52.450
Arman is going to be--
well, he's 18 right now--

00:01:52.450 --> 00:01:55.070
and in y years, he's
going to be y years older.

00:01:55.070 --> 00:01:59.130
So in y years, Arman is
going to be 18 plus y.

00:01:59.130 --> 00:02:00.720
And what about Diya?

00:02:00.720 --> 00:02:06.430
How old will she be in y years?

00:02:06.430 --> 00:02:08.490
Well, she's 2 now,
and in y years,

00:02:08.490 --> 00:02:12.160
she will just be 2 plus y.

00:02:12.160 --> 00:02:14.790
So what we're curious about,
now that we know this,

00:02:14.790 --> 00:02:18.960
is how many years will it
take for this quantity,

00:02:18.960 --> 00:02:23.170
for this expression, to be
three times this quantity?

00:02:23.170 --> 00:02:24.990
So we're really curious.

00:02:24.990 --> 00:02:28.590
We want to solve for
y such that 18 plus y

00:02:28.590 --> 00:02:34.922
is going to be equal
to 3 times 2 plus y.

00:02:38.080 --> 00:02:40.030
Notice, this is
Arman in y years.

00:02:40.030 --> 00:02:41.820
This is Diya in y years.

00:02:41.820 --> 00:02:44.720
And we're saying that what
Arman's going to be in y years

00:02:44.720 --> 00:02:49.990
is three times what Diya
is going to be in y years.

00:02:49.990 --> 00:02:51.780
So we've set up our equation.

00:02:51.780 --> 00:02:53.530
Now we can just solve it.

00:02:53.530 --> 00:02:56.067
So let's take this step by step.

00:02:56.067 --> 00:02:57.650
So the left hand
side-- and maybe I'll

00:02:57.650 --> 00:02:59.024
do this in a new
color, just so I

00:02:59.024 --> 00:03:01.410
don't have to keep switching--
so on the left hand side,

00:03:01.410 --> 00:03:03.450
I still have 18 plus y.

00:03:03.450 --> 00:03:05.830
And on the right hand side,
I can distribute this 3.

00:03:05.830 --> 00:03:08.040
So 3 times 2 is 6.

00:03:08.040 --> 00:03:10.920
3 times y is 3y.

00:03:10.920 --> 00:03:13.820
6 plus 3y.

00:03:13.820 --> 00:03:16.660
And then it's always nice
to get all of our constants

00:03:16.660 --> 00:03:18.750
on one side of the equation,
all of our variables

00:03:18.750 --> 00:03:20.460
on the other side
of the equation.

00:03:20.460 --> 00:03:21.974
So we have a 3y over here.

00:03:21.974 --> 00:03:24.390
We have more y's on the right
hand side than the left hand

00:03:24.390 --> 00:03:24.720
side.

00:03:24.720 --> 00:03:26.360
So let's get rid of the
y's on the left hand side.

00:03:26.360 --> 00:03:27.901
You could do it
either way, but you'd

00:03:27.901 --> 00:03:29.420
end up with negative numbers.

00:03:29.420 --> 00:03:32.800
So let's subtract
a y from each side.

00:03:32.800 --> 00:03:39.380
And we are left with, on
the left hand side, 18.

00:03:39.380 --> 00:03:43.410
And on the right hand side
you have 6 plus 3 y's.

00:03:43.410 --> 00:03:44.700
Take away one of those y's.

00:03:44.700 --> 00:03:47.210
You're going to be
left with 2 y's.

00:03:47.210 --> 00:03:50.270
Now we can get rid of
the constant term here.

00:03:50.270 --> 00:03:53.300
So we will subtract
6 from both sides.

00:03:56.320 --> 00:03:57.546
18 minus 6 is 12.

00:03:57.546 --> 00:03:59.670
The whole reason why we
subtracted 6 from the right

00:03:59.670 --> 00:04:02.630
was to get rid of this, 6
minus 6 is 0, so you have 12

00:04:02.630 --> 00:04:04.290
is equal to 2y.

00:04:04.290 --> 00:04:06.742
Two times the number of
years it will take is 12,

00:04:06.742 --> 00:04:08.700
and you could probably
solve this in your head.

00:04:08.700 --> 00:04:10.750
But if we just want a
one-coefficient year,

00:04:10.750 --> 00:04:12.290
we would divide
by 2 on the right.

00:04:12.290 --> 00:04:14.040
Whatever we do to one
side of an equation,

00:04:14.040 --> 00:04:15.590
we have to do it
on the other side.

00:04:15.590 --> 00:04:18.450
Otherwise, the equation will
not still be an equation.

00:04:18.450 --> 00:04:27.800
So we're left with y is equal
to 6, or y is equal to 6.

00:04:27.800 --> 00:04:30.880
So going back to the
question, how many years

00:04:30.880 --> 00:04:34.220
will it take for Arman to be
three times as old as Diya?

00:04:34.220 --> 00:04:37.540
Well, it's going
to take six years.

00:04:37.540 --> 00:04:39.450
Now, I want you to verify this.

00:04:39.450 --> 00:04:40.210
Think about it.

00:04:40.210 --> 00:04:43.480
Is this actually true?

00:04:43.480 --> 00:04:46.570
Well, in six years, how
old is Arman going to be?

00:04:46.570 --> 00:04:49.080
He's going to be 18 plus 6.

00:04:49.080 --> 00:04:51.040
We now know that
this thing is 6.

00:04:51.040 --> 00:04:55.530
So in six years,
Arman is going to be

00:04:55.530 --> 00:04:58.230
18 plus 6, which
is 24 years old.

00:04:58.230 --> 00:05:00.300
How old is Diya going to be?

00:05:00.300 --> 00:05:07.150
Well, she's going to be 2
plus 6, which is 8 years old.

00:05:07.150 --> 00:05:12.160
And lo and behold, 24 is,
indeed, three times as old

00:05:12.160 --> 00:05:13.000
as 8.

00:05:13.000 --> 00:05:16.170
In 6 years-- Arman
is 24, Diya is

00:05:16.170 --> 00:05:20.329
8-- Arman is three times as
old as Diya, and we are done.


Accreting mass due to gravity simulation

WEBVTT
Kind: captions
Language: en

00:00:00.790 --> 00:00:02.800
What we have here is
a simulation, created

00:00:02.800 --> 00:00:04.680
by Peter Collingridge
in response

00:00:04.680 --> 00:00:07.050
to one of our computer
science challenge

00:00:07.050 --> 00:00:10.980
to show how many particles
might interact gravitationally.

00:00:10.980 --> 00:00:12.610
And the whole point
of the simulation

00:00:12.610 --> 00:00:15.660
is to get an intuition
for how galaxies form,

00:00:15.660 --> 00:00:17.980
why they have the
structures they have,

00:00:17.980 --> 00:00:21.380
how solar systems form, how they
have the structures they have,

00:00:21.380 --> 00:00:24.165
and how gravity alone can
kind of define that structure.

00:00:24.165 --> 00:00:26.290
And what's really interesting
about the simulation,

00:00:26.290 --> 00:00:29.020
besides the fact that it's just
mesmerizing and extremely cool,

00:00:29.020 --> 00:00:30.520
is it shows the
particles collide.

00:00:30.520 --> 00:00:32.390
Once they get to a
certain critical mass,

00:00:32.390 --> 00:00:33.947
you see that they
get colored yellow,

00:00:33.947 --> 00:00:35.780
maybe to indicate that
there are now a star.

00:00:35.780 --> 00:00:37.470
Fusion can now occur.

00:00:37.470 --> 00:00:39.410
And you can zoom in
at different levels

00:00:39.410 --> 00:00:42.224
to really see how the
different particles

00:00:42.224 --> 00:00:43.890
or the different
masses are interacting.

00:00:43.890 --> 00:00:47.275
And then you can actually
rotate that to see

00:00:47.275 --> 00:00:48.150
a little bit clearer.

00:00:48.150 --> 00:00:50.520
This is if I'm looking
kind of right on top of it

00:00:50.520 --> 00:00:52.050
to see how they're interacting.

00:00:52.050 --> 00:00:54.040
And it's a three-dimensional
simulation,

00:00:54.040 --> 00:00:58.140
so it's a very rich way
of thinking about these.

00:00:58.140 --> 00:00:59.900
And what's exciting
for me is it's

00:00:59.900 --> 00:01:02.270
highly dependent on what
the initial conditions are.

00:01:02.270 --> 00:01:05.000
In an earlier version
of Peter's simulation,

00:01:05.000 --> 00:01:09.390
he did not give a net angular
momentum to the system.

00:01:09.390 --> 00:01:13.310
And so you did not have as
much of the planet satellite

00:01:13.310 --> 00:01:15.594
or as much of the disk
structures forming.

00:01:15.594 --> 00:01:17.010
Although right
here, we don't have

00:01:17.010 --> 00:01:18.240
too much of a disk structure.

00:01:18.240 --> 00:01:20.140
Although it does
seem to-- there does

00:01:20.140 --> 00:01:24.510
seem to be a dominant
plane in this scenario.

00:01:24.510 --> 00:01:27.650
And what's exciting is here
we have a binary system.

00:01:27.650 --> 00:01:29.590
Sometimes you restart it.

00:01:29.590 --> 00:01:31.120
You might not have
a binary system,

00:01:31.120 --> 00:01:32.270
depending on the
initial conditions.

00:01:32.270 --> 00:01:33.769
You might have
something that starts

00:01:33.769 --> 00:01:35.214
to look like the Milky Way.

00:01:35.214 --> 00:01:36.630
Sometimes you might
have something

00:01:36.630 --> 00:01:38.210
that looks very different
than the Milky Way.

00:01:38.210 --> 00:01:39.680
And it really gives
us clues of why

00:01:39.680 --> 00:01:42.760
we see such diversity,
especially when we're

00:01:42.760 --> 00:01:45.165
looking at galaxies, the
structure of galaxies,

00:01:45.165 --> 00:01:47.620
that it's highly dependent
on initial conditions.

00:01:47.620 --> 00:01:50.110
One can argue that
our own solar system

00:01:50.110 --> 00:01:52.400
did have some net
initial angular

00:01:52.400 --> 00:01:56.540
momentum because the current
theory, what really catalyzed

00:01:56.540 --> 00:01:59.450
it was a nearby supernova
that sent a shock wave

00:01:59.450 --> 00:02:02.920
and allowed the dust that would
form our solar system to reach

00:02:02.920 --> 00:02:04.700
a critical mass and
start to condense

00:02:04.700 --> 00:02:07.300
into the sun and the planets.

00:02:07.300 --> 00:02:09.169
And so this isn't,
at least in my mind,

00:02:09.169 --> 00:02:10.834
too unrealistic of a scenario.

00:02:10.834 --> 00:02:13.030
And it's really cool to
look at, and it really

00:02:13.030 --> 00:02:13.850
gives you a sense of things.

00:02:13.850 --> 00:02:15.474
You already see you
have a binary star.

00:02:15.474 --> 00:02:17.330
They're kind of orbiting
around each other,

00:02:17.330 --> 00:02:19.220
or orbiting around the center
of mass, which kind of looks

00:02:19.220 --> 00:02:20.364
like around each other.

00:02:20.364 --> 00:02:21.780
And then this star
right over here

00:02:21.780 --> 00:02:23.730
has its own kind
of captive planet

00:02:23.730 --> 00:02:25.702
that is just rotating around it.

00:02:25.702 --> 00:02:27.160
We can see it a
little bit clearer.

00:02:27.160 --> 00:02:30.400
If we had a very, at least from
this perspective, a very close

00:02:30.400 --> 00:02:32.550
range, we can zoom
in a little bit more

00:02:32.550 --> 00:02:34.267
to see it a little bit better.

00:02:34.267 --> 00:02:35.850
This has a satellite,
but then they're

00:02:35.850 --> 00:02:37.920
are also kind of dancing
around each other.

00:02:37.920 --> 00:02:41.179
So it's a really
fascinating simulation.

00:02:41.179 --> 00:02:43.470
I could really stare at this
and play with it for days.

00:02:43.470 --> 00:02:45.829
I encourage you to play
with it, restart it,

00:02:45.829 --> 00:02:47.620
see how the initial
conditions or what type

00:02:47.620 --> 00:02:50.760
of solar systems or galaxies
you might end up with.

00:02:50.760 --> 00:02:53.510
Whether they form disks, whether
you have binary systems or not,

00:02:53.510 --> 00:02:55.570
whether you have
planets with satellites.

00:02:55.570 --> 00:02:57.060
And then if you
are more advanced,

00:02:57.060 --> 00:02:59.010
actually play with
the code, and see

00:02:59.010 --> 00:03:01.150
if you can really change
the initial conditions,

00:03:01.150 --> 00:03:02.914
the starting
velocities of things,

00:03:02.914 --> 00:03:04.330
the number of
particles of things,

00:03:04.330 --> 00:03:07.830
the distribution of mass that
you start off with, the angular

00:03:07.830 --> 00:03:10.200
momentum that you
start off with.

00:03:10.200 --> 00:03:13.400
And see how that might change
the structure of the universes

00:03:13.400 --> 00:03:14.410
that you create.

00:03:14.410 --> 00:03:16.618
And I'm going to add an
annotation to this video that

00:03:16.618 --> 00:03:18.240
links directly to
this simulation,

00:03:18.240 --> 00:03:21.230
and I'll also put the link
inside of the description.

00:03:21.230 --> 00:03:21.834
So have fun.

00:03:21.834 --> 00:03:23.500
I could literally
spend hours with this.

00:03:23.500 --> 00:03:26.097
It's a fascinating,
fascinating module

00:03:26.097 --> 00:03:27.930
that he's created where
you zoom in and out.

00:03:27.930 --> 00:03:30.000
And I really thank
Peter Collingridge

00:03:30.000 --> 00:03:32.542
for this incredible
contribution.