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Well, let's do algebra.
Now as we remember what a linear object looks like ...
And we remember the meanings of [following] things.
We begin with an arbitrary function.
x is some function. y is some function.
And ... Well, let me take an example.
So let's take x equals t squared.
And y equals t cubed.
Are those good functions?
All right. Those together give me ...
... a curve.
And probably you don't even know how it looks like, but it doesn't matter.
And that is the strength of algebra.
You don't have to know how the curve looks like geometrically.
You [...] to study something about it.
Right? As long as you have some algebraic methods to study something you don't need pictures.
Well, it's good to support your study by pictures and understand what you are doing, but ...
... not totally necessary.
Er ... So what I would look at is ...
... I would look at this curve ...
... at the time t equals one.
Or maybe minus one.
One hour back in the past what happened to the particle?
And that's what I am interested in: what happened back there?
Well, of course, I can substitute.
And I can figure out that ...
... x of minus one is one.
And one of minus one is ...
... minus one.
[...] asking those simple questions just to wake you up.
Er ... what else can I say?
Is this ...
So out of this I can make a presentation.
x equals one.
y equals minus one.
Is that a linear ...
... parametric presentation of a curve?
What kind of curve is that?
Described by this parametric system.
Student: A point. NB: It's a point.
A particle is staying at that point forever.
Right?
OK.
Well, that's a linear, but not quite interesting linear.
So eventually we can add something.
What is it that we should add?
To get a better understanding of ...
... of that curve at this point.
Because now we know at that time t equals minus one the particle was at the position ...
... x equals one, y equals minus one.
So the particle was there.
And the simplest approximation of the whole motion described here ...
... would be to say ' Well, the particle was there at this point forever.'
Well, that's not the best approximation, right?
So let's give it more ...
Let's get some more information about it.
What is it that I should add?
Well, we expect to add some linear terms. Right?
We add something times t.
We add something times t.
What should those coefficients be?
Student: [...].
NB: Um? Student: [...].
Well, you see at this point ...
... what we always going to do ...
... is we are going to look back at ...
... linear expression of a linear object.
And what we are going to do is ...
... we are going to recall the meaning ...
... of that coefficient ...
... in that linear object.
So what was the meaning of this?
The Calculus meaning.
Student: Rate of change.
Rate of change.
Rate of change of x with respect to t.
Right?
Now, if that is ...
If that is the ...
... rate of change ...
... of x with respect to t ...
... can you find it?
Student: [...].
Exactly. So you know what the rate of change of x with respect to t is if the function is given.
Right? So it is ...
... x prime which is two t.
So is that ...
What the number should it be?
Could I put two t right there?
Sure?
And what do I get?
x equals one plus two t times t.
I get one plus two t squared.
Is that linear?
I cannot put two t right there.
Because that should be a constant.
It cannot be any function.
And what I get is derivative as a function.
I'm stuck. I don't know.
Student: [...].
I can substitute that value. Well should I substitute that value?
Does it make sense to substitute this value into that derivative?
Well, what's the meaning of that?
That's the rate of change of y with respect to t.
Right? y prime equals three t squared.
Student: [...].
All right. So, would it make sense to study the rate of change of x at that time?
Well, probably would, right?
Because I'm interested now in studying what happen to the point at that particular time.
When the point is at this position.
And isn't it an interesting information what the rate of change of x-position was at that particular time?
Ad that is something I can compute.
Right? If I substitute t plus minus one I get x prime equals minus two.
So instead of substituting two t ...
... I substitute two times that t.
So, minus two.
[...] substitute t plus minus one for y prime.
And put what?
Three.
Right there.
And what do these two numbers together make?
Minus two and three together.
Remember the meaning of linear presentation.
When I put these two they make a vector, right?
And the useful meaning of that vector was -- it's a velocity vector.
So what do you think the meaning of this particular vector minus two, three is?
Isn't it ... Isn't it velocity?
... of my particle?
At that point?
Well, that's exactly what it is.
So minus two, three will make er ...
... minus two, three, a vector like ...
... like this.
So this is velocity.
Velocity vector at that time, at the time minus one.
And of course, as the particle moves ...
... it will probably change the velocity.
Why would it?
Well, do you think it will change the velocity?
Well, of course, it will because ...
... if it doesn't ...
... it would be the constant velocity motion.
And therefore the formula should be linear.
And the formulas are not.
So that means the velocity will change.
Any time after this minus one and any time before that the velocity was different.
But at this time t plus minus one ...
... position was right here, velocity is right there.
As how fast the particle was moving.
Are we getting better understanding of this motion now?
Although I still don't know how that curve looks like.
I know first that the curve passes through this point.
And now I'm extracting geometric information from this algebra analysis, right?
It not only passes through this point.
It also what?
If I zoom in, right? That's what I was doing here.
If I zoom in to that curve that I don't even know what it looks like.
If I zoom in then around that point the curve will look like ...
Student: A line. NB: Like a lime. And that line is ... I know what that line is.
It will look like this.
Well, probably short time.
But still that's useful information.
I know the position. I know how the particle was moving around that time.
And then I can keep analyzing.
That time, a different time, a different time.
And then ...
... once I ...
... sketch a little bit about this curve ...
Well, I can try to join all those together.
And get a complete picture of the curve.
And that is the next step.
Well, the first step is called differentiation.
When you localize, linearize and get this piece of information.
The next step is going to be integration.
When you are putting together all those pieces.
And get exactly ...
... well, information about where the particle is exactly at any time.
So we will come to integration later. It's much more complicated procedure than differentiation.
And we will practice differentiation since now for many different settings.
Er ... Studying curves is just one setting when we can practice differentiation.
Well it, so er ...
So that's what ...
... one can do algebraically.
Start with any curve, parametric curve.
And replace that description by a linear description at any given point.