Gradient of a scalar field

Intuition of the gradient of a scalar field (temperature in a room) in 3 dimensions.

In the last video we had a
three-dimensional surface,

where the height z was
a function of x and y.

And it gave us surface in
three-dimensional space.

Now let's try to get our heads
around what the gradient

of a function of three
variables looks like.

So the easiest one for me to
imagine is a scalar field.

So what's a scalar field?

One that I find fairly
intuitive is temperature in

a three-dimensional room.

So let's say the temperature
in a room is a function of

where I am in the room.

So let's say it's a function of
my x, y, and z coordinates.

And I don't know, I have never
actually modeled temperature.

But let's say I have, I don't
know, a 20 kelvin-- actually,

let me make it so that our
vector field works out right.

Let's say we have a 10
kelvin heat force in

the center of our room.

I can imagine as you go further
and further away from that

heat source it's going to
get colder and colder.

So let's say that the
temperature function.

And let's say that center of
the room is at the coordinates

x, y, and z is equal to 0.

So let's say our temperature
function-- I'm just making this

up, I don't know if this is an
accurate model of temperature--

it's equal to 10 times e
to the minus r squared.

Now why did I say r?

I said it's a function
of x, y, and z.

Well I'm just saying that it
exponentially decays as you

get further and further
away from that source.

Kind of radially further and
further away from that source.

So what's the radial
distance away?

And this actually isn't that
relevant to learning gradients,

but let's get a little
intuition about what that

actual temperature function--
how it actually changes as

you go through the room.

So the radius away from the
center, that's just going to be

r squared is just x squared
plus y squared plus z squared.

That's just the Pythagorean
theorem in three dimensions.

So let's write our
temperature function.

So let's write temperature as
a function of x, y, and z is

equal to 10 e to the minus x
squared plus y squared plus z

squared-- which is exactly
what I wrote up here.

Instead of x squared plus y
squared plus z squared, I wrote

r squared, just to kind of give
you the intuition that this

expression is just saying the
square of the distance as we

get away from the center of
our room, or from the

coordinate 0, 0, 0.

But that's not what
we're learning here.

But I want you to understand,
at least conceptualize this,

it's hard to draw
a scalar field.

All a scalar field means is
that in any point in this

base-- and in this case we're
dealing with three-dimensional

space-- at any point in that
space we can associate a value.

And that makes sense.

If you were to take a
thermometer and measure any

point in space in the room
that you're in right now,

you would get a temperature.

You wouldn't get a temperature
and a direction, so it's

not a vector field.

You would just get
a temperature.

And that's why it's
called a scalar field.

Associated with every
coordinate is just

a temperature.

So how would we view the
gradient of this function?

Well the gradient of this
function is going to tell us in

which direction-- and actually,
the gradient of this function

is going to generate a vector
field, because it's going to

tell us in which direction
do we have the largest

increase in temperature.

And also, the magnitude of
those vectors in that vector

field will tell us how large
of an increase in temperature

we are looking at.

Or you can kind of
view it as almost a

three-dimensional slope.

Hope that doesn't confuse you.

So let's compute the gradient,
and then I'll show you a

diagram that might make things
a little bit more intuitive.

Let me erase this
thing down here.

And I'm going to switch from
this blue color, because

it's a little nauseating.

So the gradient of T is going
to be equal to the partial

derivative T with respect to x
times the unit vector in the x

direction, plus the partial
derivative of the temperature

function with respect to y
times the unit vector in the y

direction, plus the partial
derivative of the temperature

function with respect to z
times the unit vector

in the z direction.

And now we just plug and
chug and figure out the

partial derivatives.

So the gradient of T is equal
to-- now you might be daunted.

Oh, I have an e to this three
variable function, how do I

take the partial derivative?

Remember, if you're taking the
partial derivative with respect

to x you just pretend like the
y's and the z's are constants.

So let's do that.

So let's take the derivative
of the inside function.

That's the way I view it.

So minus x squared plus y
squared plus z squared,

with respect to x.

So you could distribute
this minus if you like.

So it'd be minus x
squared minus y squared

minus z squared.

So the derivative of that with
respect to x is just going to

be-- these are just constants,
so the derivative with

respect to x is just 0.

So the derivative is minus 2x.

Right?

Minus 2x is the derivative
of minus x squared.

Minus 2x times the
derivative of the outside.

Well, what's the
derivative of e to the x?

The derivative of e to
the x is e to the x.

That's why e is such
an amazing number.

And this 10 here, this is just
a constant that when you take

the derivative of a constant
times something the

constant carries over.

So the derivative of the
outside expression, the way I

imagine it, is equal to 10 e
to the minus x squared plus

y squared plus z squared.

And then all of that times the
unit vector in the i direction.

Right?

And now we can do the same
thing for the y direction.

So plus-- what's the
partial derivative of

this with respect to y?

Well it's going to
look very similar.

The partial derivative of this
inner function with respect

to y, it's minus y squared.

So it's minus 2y.

And then the derivative
of the whole thing is

just itself again.

So times 10 e to the
minus x squared plus y

squared plus z squared.

And then all of that times
the unit vector in the

y direction times j.

And then finally, the partial
derivative of the temperature

function with respect to z.

And that's just minus 2z times
10 e to the minus x squared

plus y squared plus z squared.

This is just the chain rule.

And I'm treating the other two
variables that I'm not taking

the partial derivative with
respect to, as constants.

And then all of that times the
unit vector in the k direction.

And we could simplify
this a little bit.

You could have
minus 2x times 10.

That's minus 20x.

Let me write it up here.

So the gradient of the
temperature function is equal

to minus 20 e to the minus x
squared plus y squared-- you

probably can't read this-- plus
z squared, times i minus 20y.

And actually, I'm not going to
go into that, because I realize

I'm running out of time.

I think you can simplify
this algebraically.

But anyway, the more important
thing is I always find with

gradients it's easy to
calculate them, but the

intuition-- oh sorry.

This is also included.

This is a k right here.

The harder part is
the intuition.

So let's get an intuition of
what this gradient function

will actually look like.

So what would happen.

If you wanted to know the
gradient at any point in space,

you would substitute an
x, y, and z in here.

So you could write it as
the gradient function is a

function of x, y, and z.

Remember, T, the temperature at
any point, was a scalar field.

At any point in three
dimensions it just

gave you a number.

Now when you have the gradient,
at any point in three

dimensions it gives
you a vector.

Right?

Because it has i, j,
and k components.

Where the magnitude are the
partial derivatives, and

then the direction is
given by i, j, and k.

So we've gone from having a
scalar field to a vector field.

And let's see what
it looks like.

And let me make it bigger so we
can explore it a little bit.

I think that's pretty good.

So this is the vector field.

This is actually the gradient
of the function that

we just solved for.

And as you can see, at any
point-- and when this graphing

program that did it, it just
picked different points and it

calculated the gradients at
that point, and then it

graphed them as vectors.

So the length of the vectors
are just the magnitudes of

the x, y, and z components.

And then you add them together
like you would add any vectors.

And then the direction is given
by the relative weighting of

the i, j, and k components.

And as you can see, the
intuition is pretty

interesting.

As you get closer and closer to
our heat source, the rate at

which the temperature
increases, increases!

Right?

The vectors as you get closer,
get bigger and bigger.

And let me zoom in.

Let's actually fly in
to the vector field.

So we're now within
the vector field.

And you can see as we get
closer and closer to the center

of our heat source, the
vectors, the rate at which the

temperature increases, gets
bigger and bigger and bigger.

Anyway, I hope I
didn't confuse you.

When I first learned gradients,
I think the computation is

relatively straightforward.

It's just partial derivatives.

But the intuition is always
the interesting thing.

And hopefully this temperature
analogy-- and not even

analogy-- this temperature
model will make a

little sense to you.

But it applies to pretty
much any scalar field.

Anyway, I'll see you
in the next video.