WEBVTT
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JOEL LEWIS: Hi.
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Welcome back to recitation.
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In lecture, you've been
learning about using gradients
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to compute tangent
planes to surfaces.
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So I have an example of a
practice problem here for you.
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So what I'd like
you to do in part
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a is to use gradients to
find the tangent plane
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to the surface z equals
x cubed plus 3x y squared
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at the point (1, 2, 13).
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And in part b, I'd like you
to do something similar, which
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is to use gradients to find
the tangent line to the curve x
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cubed plus 2xy plus y squared
equals 9 at the point (1, 2).
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So why don't you
pause the video,
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have a couple goes at those.
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Come back and we can
work on them together.
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So hopefully, you had some good
luck working on these problems.
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Let's just take a look at them.
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So for part a, you're
given a function
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in the sort of usual form
that we use to graph it,
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which is you're given z
equals a function of x and y.
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But in order to apply this
gradient method, what we really
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want is we want to
look at this surface
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as if it were a level surface
of some function of three
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variables.
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So in order to do that,
what we want to do always
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is to bring the x,
y and z all together
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on the same side with
just a zero or a constant
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on the other side.
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So let me do that.
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So I'm going to
rewrite the defining
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equation of this
surface as 0 equals
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x cubed plus 3x y
squared minus z,
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and I'm going to define
this right-hand side to be
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a new function w of x, y, z.
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All right?
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So if I call this thing w,
then our surface in question
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is just a level surface of w.
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It's the level
surface w equals 0.
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And so we know in that
situation that the gradient of w
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is perpendicular to
its level surfaces.
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It's orthogonal to
its level surfaces.
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So the normal to our surface
is exactly the gradient of w.
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All right?
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So gradient of w is the
normal to our surface,
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and a normal is what
we use to write down
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the equation for
a tangent line--
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oh, tangent plane, excuse me.
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So, OK, so let's compute
the gradient of w.
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Well, that's not hard to do.
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We just take the
partial derivatives
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with respect to x, y and z.
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So the partial derivative
of w with respect
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to x is 3 x squared
plus 3 y squared.
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The partial derivative
with respect to y is 6xy,
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and the partial derivative
with respect to z is minus 1.
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So one thing to notice is
that when you do this method,
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when you have the
function given at z,
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when you have the
surface given in the form
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z as a function
of x and y, you're
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going to bring the z over,
and you always have a minus 1
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there when you set the
problem up this way.
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Because you'll have
a minus z, and then
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you'll just take the
partial with respect to z,
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and the other terms will
only involve x and y,
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so they'll be killed by
the partial derivative.
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So in any case, this
is our gradient,
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so we want the normal vector.
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We were asked for the tangent
plane at a particular point,
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I believe.
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Yes, at the point (1, 2, 13).
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So we need to
compute the gradient
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at that particular point and
that will be our normal vector.
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So the gradient at this point
is-- well, we just plug in,
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so the gradient at (1, 2, 13).
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So x is 1.
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So this is 3 times
1 plus 3 times 4,
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so that's going to be 15,
and 6xy is 12, and minus 1
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is just minus 1.
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So this is the gradient vector
at our point (1, 2, 13).
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So now we have a point,
the point (1, 2, 13),
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and we have the normal
vector 15, 12, minus 1,
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so that gives us the equation
for the tangent plane right
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off.
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So the equation for
the tangent plane,
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I just dot the normal
vector with the vector
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connecting our point
to the point (x, y, z).
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so that gives us 15 times--
well, our point is (1, 2, 3)--
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so it's 15 times x minus 1
plus 12 times y minus 2 minus 1
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times z minus 13 equals 0.
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So in point-normal form, this
is the equation for that plane.
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Great.
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And if you wanted,
you could rewrite
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this a whole bunch
of different ways,
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but I'll just leave it there.
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So let's do part b.
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I guess I'll just start
it right below here.
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So for part b, we have a curve
x cubed plus 2xy plus y squared
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equals 9.
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So this is a curve
that is defined
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by this implicit
relationship between x and y.
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All right?
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And so what I want
to do is I can
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do exactly the same process.
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We're going to do
exactly the same thing.
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We're going to find
the normal-point form
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for the tangent
line, and so we're
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going to do that by defining
a function f of x, y.
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In this case, it's a function
of just two variables,
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because we're only working
with a curve in two dimensions.
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Before, we had a surface
in three dimensions,
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so we had a function
of three variables.
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So f of x, y, and
so then our curve
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is exactly a level curve
of the graph of f, right?
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It's the level curve f equals 9.
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So in order to find
the tangent line,
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I can do exactly the same thing.
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I can find the gradient.
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The gradient is normal
to the tangent line
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and then I can use
normal-point form.
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So the gradient of f is-- again,
f is just a polynomial function
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so its gradient is
easy to compute.
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It's 3 x squared plus
2y comma 2x plus 2y.
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And so we're interested
in this tangent line
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at a particular point.
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So we're interested
at the point (1, 2).
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So the gradient of f
at (1, 2), well, I just
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plug in again, so
I get 3 plus 4.
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That's 7.
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And 2 plus 4 is 6.
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And so again, the same analysis
as we used in the tangent plane
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case works in the
tangent line case.
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Let's come over here.
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So (x, y) is on the tangent
line if and only if we have that
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the gradient dot-- so that's
the gradient, [7, 6]--
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dot the vector x
minus 1, y minus 2--
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this is the vector
connecting the point (x,
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y) to our point (1,
2)-- is equal to 0,
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if and only if those two
things are orthogonal.
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So this is-- i.e.
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7 times x minus 1 plus 6
times y minus 2 is equal to 0.
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So this is the point-normal form
for the equation of that line.
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And again, you could,
you know, expand out
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and rewrite this
in whichever forms
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you happen to like to see
your equations of lines.
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So there you go.
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Using the gradient, we
can compute tangent planes
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to surfaces.
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Similarly, we can
use the same idea
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to compute tangent
lines to curves.
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The point is that the
gradient vector of a function
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is orthogonal to the level
curves of that function.
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And so we use that to
get the normal vectors
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to our curves or our surfaces,
and with the normal vector,
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we can then easily
compute the tangent plane
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or the tangent line.
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So I'll stop there.