WEBVTT

00:00:00.000 --> 00:00:07.860
JOEL LEWIS: Hi.

00:00:07.860 --> 00:00:09.410
Welcome back to recitation.

00:00:09.410 --> 00:00:11.890
In lecture, you've been
learning about using gradients

00:00:11.890 --> 00:00:14.290
to compute tangent
planes to surfaces.

00:00:14.290 --> 00:00:19.700
So I have an example of a
practice problem here for you.

00:00:19.700 --> 00:00:21.930
So what I'd like
you to do in part

00:00:21.930 --> 00:00:25.860
a is to use gradients to
find the tangent plane

00:00:25.860 --> 00:00:30.570
to the surface z equals
x cubed plus 3x y squared

00:00:30.570 --> 00:00:32.940
at the point (1, 2, 13).

00:00:32.940 --> 00:00:36.570
And in part b, I'd like you
to do something similar, which

00:00:36.570 --> 00:00:42.015
is to use gradients to find
the tangent line to the curve x

00:00:42.015 --> 00:00:47.010
cubed plus 2xy plus y squared
equals 9 at the point (1, 2).

00:00:47.010 --> 00:00:48.610
So why don't you
pause the video,

00:00:48.610 --> 00:00:51.519
have a couple goes at those.

00:00:51.519 --> 00:00:53.310
Come back and we can
work on them together.

00:01:01.520 --> 00:01:04.450
So hopefully, you had some good
luck working on these problems.

00:01:04.450 --> 00:01:06.830
Let's just take a look at them.

00:01:06.830 --> 00:01:10.130
So for part a, you're
given a function

00:01:10.130 --> 00:01:13.390
in the sort of usual form
that we use to graph it,

00:01:13.390 --> 00:01:17.690
which is you're given z
equals a function of x and y.

00:01:17.690 --> 00:01:20.120
But in order to apply this
gradient method, what we really

00:01:20.120 --> 00:01:23.390
want is we want to
look at this surface

00:01:23.390 --> 00:01:27.070
as if it were a level surface
of some function of three

00:01:27.070 --> 00:01:27.830
variables.

00:01:27.830 --> 00:01:30.160
So in order to do that,
what we want to do always

00:01:30.160 --> 00:01:33.210
is to bring the x,
y and z all together

00:01:33.210 --> 00:01:36.770
on the same side with
just a zero or a constant

00:01:36.770 --> 00:01:37.720
on the other side.

00:01:37.720 --> 00:01:39.511
So let me do that.

00:01:39.511 --> 00:01:41.010
So I'm going to
rewrite the defining

00:01:41.010 --> 00:01:46.750
equation of this
surface as 0 equals

00:01:46.750 --> 00:01:54.930
x cubed plus 3x y
squared minus z,

00:01:54.930 --> 00:01:58.780
and I'm going to define
this right-hand side to be

00:01:58.780 --> 00:02:02.376
a new function w of x, y, z.

00:02:02.376 --> 00:02:03.830
All right?

00:02:03.830 --> 00:02:07.370
So if I call this thing w,
then our surface in question

00:02:07.370 --> 00:02:11.680
is just a level surface of w.

00:02:11.680 --> 00:02:14.260
It's the level
surface w equals 0.

00:02:14.260 --> 00:02:20.030
And so we know in that
situation that the gradient of w

00:02:20.030 --> 00:02:22.820
is perpendicular to
its level surfaces.

00:02:22.820 --> 00:02:24.920
It's orthogonal to
its level surfaces.

00:02:24.920 --> 00:02:30.066
So the normal to our surface
is exactly the gradient of w.

00:02:30.066 --> 00:02:32.040
All right?

00:02:32.040 --> 00:02:45.120
So gradient of w is the
normal to our surface,

00:02:45.120 --> 00:02:46.960
and a normal is what
we use to write down

00:02:46.960 --> 00:02:48.790
the equation for
a tangent line--

00:02:48.790 --> 00:02:51.380
oh, tangent plane, excuse me.

00:02:51.380 --> 00:02:53.940
So, OK, so let's compute
the gradient of w.

00:02:53.940 --> 00:02:56.380
Well, that's not hard to do.

00:02:56.380 --> 00:02:58.210
We just take the
partial derivatives

00:02:58.210 --> 00:02:59.520
with respect to x, y and z.

00:02:59.520 --> 00:03:01.910
So the partial derivative
of w with respect

00:03:01.910 --> 00:03:07.200
to x is 3 x squared
plus 3 y squared.

00:03:07.200 --> 00:03:13.290
The partial derivative
with respect to y is 6xy,

00:03:13.290 --> 00:03:17.100
and the partial derivative
with respect to z is minus 1.

00:03:17.100 --> 00:03:19.910
So one thing to notice is
that when you do this method,

00:03:19.910 --> 00:03:22.770
when you have the
function given at z,

00:03:22.770 --> 00:03:25.020
when you have the
surface given in the form

00:03:25.020 --> 00:03:27.450
z as a function
of x and y, you're

00:03:27.450 --> 00:03:29.900
going to bring the z over,
and you always have a minus 1

00:03:29.900 --> 00:03:32.585
there when you set the
problem up this way.

00:03:32.585 --> 00:03:34.210
Because you'll have
a minus z, and then

00:03:34.210 --> 00:03:36.200
you'll just take the
partial with respect to z,

00:03:36.200 --> 00:03:38.880
and the other terms will
only involve x and y,

00:03:38.880 --> 00:03:41.180
so they'll be killed by
the partial derivative.

00:03:41.180 --> 00:03:43.740
So in any case, this
is our gradient,

00:03:43.740 --> 00:03:45.150
so we want the normal vector.

00:03:45.150 --> 00:03:48.620
We were asked for the tangent
plane at a particular point,

00:03:48.620 --> 00:03:49.480
I believe.

00:03:49.480 --> 00:03:51.774
Yes, at the point (1, 2, 13).

00:03:51.774 --> 00:03:53.190
So we need to
compute the gradient

00:03:53.190 --> 00:03:56.360
at that particular point and
that will be our normal vector.

00:03:56.360 --> 00:04:02.510
So the gradient at this point
is-- well, we just plug in,

00:04:02.510 --> 00:04:07.570
so the gradient at (1, 2, 13).

00:04:07.570 --> 00:04:09.720
So x is 1.

00:04:09.720 --> 00:04:14.300
So this is 3 times
1 plus 3 times 4,

00:04:14.300 --> 00:04:21.320
so that's going to be 15,
and 6xy is 12, and minus 1

00:04:21.320 --> 00:04:23.170
is just minus 1.

00:04:23.170 --> 00:04:27.460
So this is the gradient vector
at our point (1, 2, 13).

00:04:27.460 --> 00:04:30.890
So now we have a point,
the point (1, 2, 13),

00:04:30.890 --> 00:04:34.080
and we have the normal
vector 15, 12, minus 1,

00:04:34.080 --> 00:04:36.930
so that gives us the equation
for the tangent plane right

00:04:36.930 --> 00:04:37.740
off.

00:04:37.740 --> 00:04:47.860
So the equation for
the tangent plane,

00:04:47.860 --> 00:04:51.250
I just dot the normal
vector with the vector

00:04:51.250 --> 00:04:53.700
connecting our point
to the point (x, y, z).

00:04:53.700 --> 00:04:57.465
so that gives us 15 times--
well, our point is (1, 2, 3)--

00:04:57.465 --> 00:05:09.430
so it's 15 times x minus 1
plus 12 times y minus 2 minus 1

00:05:09.430 --> 00:05:15.460
times z minus 13 equals 0.

00:05:15.460 --> 00:05:20.180
So in point-normal form, this
is the equation for that plane.

00:05:20.180 --> 00:05:20.680
Great.

00:05:20.680 --> 00:05:22.356
And if you wanted,
you could rewrite

00:05:22.356 --> 00:05:24.310
this a whole bunch
of different ways,

00:05:24.310 --> 00:05:26.110
but I'll just leave it there.

00:05:26.110 --> 00:05:27.540
So let's do part b.

00:05:27.540 --> 00:05:29.420
I guess I'll just start
it right below here.

00:05:32.440 --> 00:05:44.200
So for part b, we have a curve
x cubed plus 2xy plus y squared

00:05:44.200 --> 00:05:44.960
equals 9.

00:05:44.960 --> 00:05:47.300
So this is a curve
that is defined

00:05:47.300 --> 00:05:50.036
by this implicit
relationship between x and y.

00:05:50.036 --> 00:05:51.700
All right?

00:05:51.700 --> 00:05:53.290
And so what I want
to do is I can

00:05:53.290 --> 00:05:55.002
do exactly the same process.

00:05:55.002 --> 00:05:56.710
We're going to do
exactly the same thing.

00:05:56.710 --> 00:05:59.220
We're going to find
the normal-point form

00:05:59.220 --> 00:06:01.120
for the tangent
line, and so we're

00:06:01.120 --> 00:06:05.420
going to do that by defining
a function f of x, y.

00:06:05.420 --> 00:06:08.030
In this case, it's a function
of just two variables,

00:06:08.030 --> 00:06:11.500
because we're only working
with a curve in two dimensions.

00:06:11.500 --> 00:06:14.380
Before, we had a surface
in three dimensions,

00:06:14.380 --> 00:06:16.570
so we had a function
of three variables.

00:06:16.570 --> 00:06:19.240
So f of x, y, and
so then our curve

00:06:19.240 --> 00:06:24.080
is exactly a level curve
of the graph of f, right?

00:06:24.080 --> 00:06:26.590
It's the level curve f equals 9.

00:06:29.220 --> 00:06:32.650
So in order to find
the tangent line,

00:06:32.650 --> 00:06:34.380
I can do exactly the same thing.

00:06:34.380 --> 00:06:35.760
I can find the gradient.

00:06:35.760 --> 00:06:37.850
The gradient is normal
to the tangent line

00:06:37.850 --> 00:06:40.390
and then I can use
normal-point form.

00:06:40.390 --> 00:06:49.170
So the gradient of f is-- again,
f is just a polynomial function

00:06:49.170 --> 00:06:51.190
so its gradient is
easy to compute.

00:06:51.190 --> 00:07:02.230
It's 3 x squared plus
2y comma 2x plus 2y.

00:07:02.230 --> 00:07:06.220
And so we're interested
in this tangent line

00:07:06.220 --> 00:07:07.260
at a particular point.

00:07:07.260 --> 00:07:10.640
So we're interested
at the point (1, 2).

00:07:10.640 --> 00:07:15.660
So the gradient of f
at (1, 2), well, I just

00:07:15.660 --> 00:07:18.690
plug in again, so
I get 3 plus 4.

00:07:18.690 --> 00:07:21.120
That's 7.

00:07:21.120 --> 00:07:25.890
And 2 plus 4 is 6.

00:07:25.890 --> 00:07:30.140
And so again, the same analysis
as we used in the tangent plane

00:07:30.140 --> 00:07:32.130
case works in the
tangent line case.

00:07:32.130 --> 00:07:35.290
Let's come over here.

00:07:35.290 --> 00:07:51.520
So (x, y) is on the tangent
line if and only if we have that

00:07:51.520 --> 00:07:56.600
the gradient dot-- so that's
the gradient, [7, 6]--

00:07:56.600 --> 00:08:01.700
dot the vector x
minus 1, y minus 2--

00:08:01.700 --> 00:08:04.170
this is the vector
connecting the point (x,

00:08:04.170 --> 00:08:07.310
y) to our point (1,
2)-- is equal to 0,

00:08:07.310 --> 00:08:09.730
if and only if those two
things are orthogonal.

00:08:09.730 --> 00:08:13.340
So this is-- i.e.

00:08:13.340 --> 00:08:19.630
7 times x minus 1 plus 6
times y minus 2 is equal to 0.

00:08:19.630 --> 00:08:23.060
So this is the point-normal form
for the equation of that line.

00:08:23.060 --> 00:08:24.940
And again, you could,
you know, expand out

00:08:24.940 --> 00:08:27.130
and rewrite this
in whichever forms

00:08:27.130 --> 00:08:30.970
you happen to like to see
your equations of lines.

00:08:30.970 --> 00:08:31.980
So there you go.

00:08:31.980 --> 00:08:34.520
Using the gradient, we
can compute tangent planes

00:08:34.520 --> 00:08:35.460
to surfaces.

00:08:35.460 --> 00:08:38.000
Similarly, we can
use the same idea

00:08:38.000 --> 00:08:40.470
to compute tangent
lines to curves.

00:08:40.470 --> 00:08:45.570
The point is that the
gradient vector of a function

00:08:45.570 --> 00:08:48.310
is orthogonal to the level
curves of that function.

00:08:48.310 --> 00:08:50.800
And so we use that to
get the normal vectors

00:08:50.800 --> 00:08:54.810
to our curves or our surfaces,
and with the normal vector,

00:08:54.810 --> 00:08:57.460
we can then easily
compute the tangent plane

00:08:57.460 --> 00:08:58.850
or the tangent line.

00:08:58.850 --> 00:09:00.374
So I'll stop there.