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All right. So let's do the linear things.
Linear curves.
So why linear?
Well this is because of the nature of Calculus again. Er ...
Calculus is a method of reducing nonlinear problems to linear problems.
So we are studying linear curves first.
Whatever we are going to study -- curves, surfaces ...
... all the objects we will study -- we will start with linear objects ...
... even equations, right?
And then we will progress to nonlinear using differentiation.
So what is ... what is going to be linear? What does it mean for a function to be linear?
Now I have a function of two variables.
I'm using algebraic language and ...
It is in the form a times x plus b times y plus c where ...
What? x and y are those variables, right? And a, b, c are constants.
All right. What should I be interested in the most from Calculus point of view?
What do you think?
Student: [...]. NB: The rate of change, yes.
So let's start with a change.
So how do I compute the change of that function?
Let me do some computations here.
Change of f.
Well in order to talk about the change I have to pick two points. Right?
So let me pick ...
... a point x zero and y zero.
Thinking geometrically that's one point on a plane.
And I want to compute the change from that value f of x zero, y zero to ...
... the value f at x zero plus delta x ...
... y zero plus delta y.
And [...] to change x by some value delta x, change y by delta y.
And what is that?
Well that's simply difference, right?
f of x zero plus delta x ...
... y zero plus delta y minus f of ...
... x zero, y zero.
How do I compute? Well I know the formula for f, so I substitute.
I substitute these to get a times x plus delta x ...
... plus b times y plus delta y plus c. And I subtract ...
... the whole value of f at x zero, y zero.
And those should be x zero, y zero of course.
And that will be a x zero plus y ...
... well b y zero ...
... plus c.
Now let me simplify. Of course c cancels.
So that difference is a times x zero plus a delta x ...
... plus b times y zero plus b delta y.
And I subtract a x zero plus b y zero.
Student: [...].
It is minus b y zero, yes. Thank you.
And is it better with minus?
It is better because something cancels, right?
b y zero cancels, a x zero cancels.
And the whole thing is a times ...
... delta x plus b times delta y.
Which looks like ...
Linear equation.
And?
Dot product, yes! It looks like a dot product.
And if you are to make this a dot product of two things ...
... what would you pick for those things?
Well a and b, right?
And delta x and delta y.
So the change of x ... well the change of f ...
... is equal to the dot product of those two things.
And now the advantage of that is that now we ...
... we can translate that into Algebra ... sorry, into Geometry.
Because dot product is in some sense geometric concept.
But unfortunately ...
... there is no way to visualize the dot product itself.
That's the fundamental problem with the dor product.
It tells you pretty much nothing visually.
So let me ...
... try to translate that into Geometry.
I'm looking at a plane, right? I have a point x zero, y zero.
I have another point on the plane.
x zero plus delta x comma y zero plus delta y.
And now [...] to call these points let's say P and Q.
And then what does it mean to change from P to Q?
Because this is the kind of change we are talking about, right?
That means somehow to consider a vector.
Right?
So this vector PQ ...
... which is about transition from P to Q.
What are the coordinates of that?
Now if you do simple subtraction it's delta x comma delta y.
So this is the vector we have there in the formula.
So abstractly if you think about a function that assigns a number to each point on the plane ...
... if you want to change the values ...
... change the points ...
... well that's the change involved in a plane.
And the change involved in the values of f ...
... as delta f ...
... is exactly the dot product ...
... a b times that vector PQ.
And as I said the problem is that ...
... we don't see that dot product.
So how to visualize that formula? How to interpret that geometrically?
Well the thing we can actually visualize is ...
... that product, divided by something.
Well let me ...
... introduce the vector here.
Well let's suppose that is the vector a, b.
OK? There is no way for us to see what the dot product is.
But if I divide that by magnitude of PQ ...
And of course I have to divide the left-hand side as well ...
... just to keep equality.
Well that makes sense geometrically.
What is that? Do you recognize ...
... that fraction?
Dot product divided by the magnitude of one of the vectors?
That's component, yes.
That is component ...
... of which vector?
Of the vector a, b ...
... with respect to direction PQ.
All right. So well we gain this geometric interpretation of the right-hand side. But ...
... we changed the left-hand side.
Can we still understand what left-hand side is?
Because it would be best to understand both sides.
Well think about that ...
Let us visualize this component of a, b with respect to PQ. Can we see that?
What do you do to see the component of this vector a, b with respect to that direction PQ?
Well you just project, right?
Well this is exactly that number.
That's how large it is.
This is component ...
... of the vector a, b with respect to the direction PQ.
Right? So the right-hand side is immediately geometrically visible.
What's the meaning of the left?
I moved from this point P to the point Q ...
... in the domain of the function.
I looked at the change of the function.
And then I divided it by ...
... the magnitude of the change of the ...
... variable P. Right?
What is that? What kind of ratio is that?
Change of a function divided by the change of the independent variable.
Isn't that the ratio you look at Calculus all the time?
Isn't that the rate of change ...
... of the function f?
That's exactly what it is.
So you see the [...] is that it changes itself.
It cannot be visualized. The rate of change can.
But the advantage of rate of change of the derivative over just change.
So this is the rate of change ...
... in case of linear function. Right?
So this is the special case when f is linear.
I don't need any limit here.
No matter how I change that ...
... the ratio will be equal to the rate of change.
In some sense. So ...
Er ... Rate of change of f ...
... at this point ...
... at the point x zero, y zero.
Is that rate different if I switch from P to some other point?
Well explain that.
What if I switch from P to a point there?
What would be the rate of change of a function ...
... in that direction?
Well the rate of change is going to be projection of this a, b vector ...
... onto that direction, right?
So the rate of change is quite different.
Now what if I want to change from that P ...
... to another point on the same line?
What if I want to change from P to this point?
What will be the rate of change?
Well it will be exactly the same.
So the rate of change is going to be the same along the whole line here.
No matter how far you go along this line the rate of change is the same.
Change is different, right?
The further you go the more change you have.
Isn't that what you expect?
The more you change x and y in the formula ...
... the more you change the value.
But ...
... if you go in the same direction the rate of change stays the same.
Well of course if you go in the different direction the rate of change changes.
It may be different.