Now, if he can't come up with
a definition of directional derivative,
this is almost certainly nice,
but it's wake, yes,
you don't want to calculate it all
over again when you
need to find the directional derivative.
So we need to find an alternative way, an easier way.
Maybe more computable way
to find the directional derivative in a given point.
In order to do it, and then I go into first of all
remind ourselves some concepts that we already know.
First of all, let's start with gradient,
which is a vector of first derivatives written out here.
The scalar product of two vectors which is a rule in
our two-dimensional case is
multiplication of coordinates summed.
Also, it has a geometrical meaning which is basically
multiplication of extra stance towards
the cosine of the vector in between them,
which is schematically drawn
in the picture on the slide here.
So first of all,
let us revisit some basic scalar product properties
here is symmetrical with,
well, no surprise here
because obviously is geometrical census symmetrical.
You do not care on
the order of multiplication of two real numbers.
For the others, it defines the vector lengths.
It is bilinear,
and it defines the angle
between vectors x, we are rewriting new.
It's the easiest part of all our course.
But what we also then know,
we also do know what is differential function.
The differentiable function is a function which is
approximated by its tangent plane thoroughly.
In our case it,
is infinitesimally towards distance
between a given point and the point of approximation.
So why do we building all
of these stuff on top of each other?
Just to define directional derivative more easily,
let us just try to look at our definition of
differentiation in terms of directional derivative.
If we're computing the actual derivative,
then the change of argument we've it earlier.
It's t multiplied by x coordinate of the direction,
and the nature of y-coordinate
is t multiplied by y-coordinate of direction.
More importantly, the distance
between the function and the approximation function at
given point and the approximation point
is simply t since our direction is normalized.
So let us try to write the definition of
our directional derivative down. What do we get?
We get a limit when t approaches zero,
change of function divided by
t. Assume that our function
is differentiable at the point a, b.
Then the change of function can be rewritten in terms of
first derivative and infinite testable function
towards the distance.
So let us write this down and we are going to
substitute the change of argument with tl_x,
and the change of y argument as tl_y.
Well, the actual distance
between two dots with a t as we computed total there.
So what we do have here?
We do have here sum in
the denominator proportional to the denominator,
to the t. We can
divide both nominator and denominator by t,
and thus, we get,
well, the first term which is independent
from the value of the limit.
As a second term,
well, the same thing similarly.
The last thing we have here is small or from one,
which is simply infinitesimal function
as you do remember.
Thus, its limit is just equals to zero.
As a result, we get a pretty nice notation.
We get partial derivative multiplied by
the x coordinate of
the direction plus partial derivative targets y,
multiplied by the y coordinate of
the direction or simply scalar product
of gradient and our direction.
That's extraordinarily nice formula.
So we established an easy relation
between the directional derivative,
hence, the gradient, and
derivative itself with a scalar product.
So let us consider some example here.
Just let us take a look at
function x squared plus 2y squared.
For example, at point 1, 2.
Well, we are going to ask ourselves a question,
what is the direction with
the maximal value of directional derivative in each one?
So what we need to do firstly,
we do not argue even that,
because the function f is differentiable at
the point m. Because it's polynomial,
thus, it has a decent partial derivatives.
We do remember our sufficient condition
of differentiability.
So we need just to find
here the gradient of the function at the point 1, 2.
In order to do so, I am going just
to write a partial derivative towards x which is 2x,
and partial derivative towards y which is 4y.
Then I'm going to substitute y
and x with one and two respectively.
Thus, we get two,
eight as a result, that's our gradient.
So what we are going to think about here.
We have some arbitrary
direction cosine Alpha, sine Alpha.
We are going to speak about
what its maximum value of this very vivid expression.
So firstly, we need to compute
our directional derivative,
which is simple scalar product
of the written and we've just found,
and our arbitrary direction,
which is two cosine,
functions plus eight sine out of Alpha,
and which is nice and pretty obvious.
But it's tricky to
answer what is the maximum value and why.
So in order to do it,
we are going to draw some pretty approximate picture.
Assume that that's our point 1, 2,
doesn't look like it, but never mind.
We have here our gradient.
So we're talking about is computing
our directional derivative with arbitrary direction
l. Let us revisit the idea what scalar products.
What we do have here?
We do have here also
not only simple formula for coordinates,
but also eight geometrical mean and [inaudible].
We can state that it's the length
of our gradient multiplied by the length of l,
multiply it by some angle.
We are going to call it, for example, Gamma.
Well, the length of the gradient.
Well, it's quite easy.
It is two squared plus a squared.
That's not square. It's the square root of 68.
Let it be. The length of the direction is always one,
and this thing that is left is, well,
cosine of some arbitrary angle Gamma.
Thus, in order to come up with the maximum value,
you just need to understand what's
the maximum value of cosine function which is one.
So the maximum here is square root of 68.
By doing so, we've actually come
to very interesting idea here.
Because not only we spoke about whether or not
in geometrical meaning is more important
than the actual computation
of rule for the scalar product rule,
we also found that
the maximum value of directional derivative somehow
equals to the length of the gradient.
We're going to use
that rule for sine in order to establish
the direction of maximum gross and the speed of
maximal gross of the function at a given point.