So now, we know what is partial derivative.
So let us move on
towards the actual equation of tangent line.
So what we do know now,
we know two straight lines,
tangent lines in two cross sections with two streams,
x equal to a and y equal to b,
which are definitely into our tangent plane.
So is it enough
to actually write an equation of this plane?
Well, it turns out that it is.
Firstly, we need to
revisit some basic things about the equation of a plane.
As we start with a scalar equation of a plane,
which is as following.
It is scalar product or just as
a result of multiplication and
summation of two basic essentials.
So we assume that we are trying to characterize
arbitrary point of our plan x,
y, z, and so we are going to do so by two scenes.
One, given a point of our plane for example x_0,
y_0 and z_0, and
a vector which is orthogonal for all our plane.
In other words, it is orthogonal for
any vector that license a plane actually.
So what we're going to say is that scalar equation is
basically the idea that these two vectors are orthogonal,
or in other manner,
we are going to multiply our coordinates of
the orthogonal vector and
the vector which is definitely lying in the plane,
which is vector between point x, y,
z, our arbitrary point and point x_0, y_0 and z_0.
That's our equation of a plane.
It's not actually the only thing that we
can define our plane,
but it's quite convenient and it's
usually used in many many occasions.
So another thing I'm going
to remind you is the concept of scalar product,
which is a mapping from two vectors into
real numbers and it works pretty much as we stated above.
It's just a summation
of multiplied coordinates of two vectors.
It is pretty much what we have
done in our scalar equation plane,
and it is your responsibility to
know but let us reiterate is,
scalar project has its geometrical meaning,
which is basically a product of the lengths of
two vectors multiplied by
the cosine of an angle between them.
You also should consider that in two dimensional cases,
multidimensional cases and spaces of three,
four and other dimensions,
that's basically a definition of an angle.
So it is not quite an easy way
to roll with the scalar product here.
It's extremely complicated actually.
But it produces some other 3D scene
here because what we're going to say looking at
our multiplication by the cosine function
is that the orthogonal case
is actually characterized by the value of
a scalar product if two vectors are orthogonal,
which means that they have an angle of half of Pi.
So cosine of half of a Pi is zero,
so multiplication by the cosine function results
into zero as a scalar project result.
So basically, that's how one
can define that two vectors are orthogonal.
So how are we going to use it.
We are going to use it in the following manner.
First of all, we know the point that
definitely belongs to our tangent plane.
It is a point a,b,f from a, b.
That's what we are expecting,
that's the point of our approximation.
So we need to find
the orthogonal vector to all our tangent planes here,
but we do know we do know
two vectors which has directional for
our tangent lines which we have mentioned whilst
we were defining our partial derivatives.
They are directional for our tangent lines.
In other way, we do know that
one of the variables during
the calculation of partial derivatives were fixed.
For example, whilst we were defining
as the partial derivative y were fixed,
thus means that the y-coordinate of
the directional vector of
this tangent line equals to zero.
The same applies for the y case
and x-coordinate in this case.
So we have actually two vectors.
1,0 partial derivative towards x,
and 0,1 partial derivative derivative towards y.
Which results into 2x as
definitely lying in our tangent plane.
So yes, we do know it.
We need to construct another vector,
a third vector which is orthogonal to this two,
and it's quite easy.
Just let us revisit it.
It is a vector which is partial derivative towards x,
partial derivative towards y and minus one.
Let's say that whilst we can calculate
this scalar product with
the first vector here and the second vector here,
we are almost definitely going to get zero.
Well, it is easy because we multiply
1 by some partial derivative,
and then minus 1 by
very much partial derivative answers
in summation we get zero.
So now, we can write our scalar equation
of a plane and with just an easy transformation,
because it is nice to have that tangent
variable on the left side of the equation,
and to the right-hand,
they are going to just write this down.
So our equation of the tangent plane is
our tangent variable equals
the function at this very point,
and then partial derivatives was multiplied by the change
of arguments x and y respectively.
So let us for example
look at the case of previously discussed function,
square root from absolute value of product x and y.
We have actually already stated that that the point 0,0
both of our partial derivatives
are actually equals to zero.
Okay so I am going to just write it down.
Okay, that's fine.
Let us proceed with an actual equation.
We need to write z equals to.
First of all, the value of our function at point 0,0.
Well, the absolute value of zero is zero,
so it was quite easy.
Then we need to write
our partial derivative towards x multiplied
by x minus 0
plus partial derivative towards
y multiplied by y minus 0.
Well, that is quite an easy one.
So the thing is that we are
going to call our tangent plane,
these right points for function square root
from multiplication of x and y is horizontal plane.
The planes that coincides with 0 x,
y plane of the three-dimensional space.
Okay, and that's nice.
That's a fine way to define our tangent plane.