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