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PxCxlsl_YwY | thing.
So, let's take the vector, OP. |
PxCxlsl_YwY | OK, so vector OP,
of course, has components x, |
PxCxlsl_YwY | y, z.
Now, we can think of this as |
PxCxlsl_YwY | actually a dot product between
OP and a mysterious vector that |
PxCxlsl_YwY | won't remain mysterious for very
long, |
PxCxlsl_YwY | namely, the vector one,
two, three. |
PxCxlsl_YwY | OK, so, this condition is the
same as OP.A equals zero, |
PxCxlsl_YwY | right?
If I take the dot product |
PxCxlsl_YwY | OPdotA I get x times one plus y
times two plus z times three. |
PxCxlsl_YwY | But now, what does it mean that
the dot product between OP and A |
PxCxlsl_YwY | is zero?
Well, it means that OP and A |
PxCxlsl_YwY | are perpendicular.
OK, so I have this vector, A. |
PxCxlsl_YwY | I'm not going to be able to
draw it realistically. |
PxCxlsl_YwY | Let's say it goes this way.
Then, a point, |
PxCxlsl_YwY | P, solves this equation exactly
when the vector from O to P is |
PxCxlsl_YwY | perpendicular to A.
And, I claim that defines a |
PxCxlsl_YwY | plane.
For example, |
PxCxlsl_YwY | if it helps you to see it,
take a vertical vector. |
PxCxlsl_YwY | What does it mean to be
perpendicular to the vertical |
PxCxlsl_YwY | vector?
It means you are horizontal. |
PxCxlsl_YwY | It's the horizontal plane.
Here, it's a plane that passes |
PxCxlsl_YwY | through the origin and is
perpendicular to this vector, |
PxCxlsl_YwY | A.
OK, so what we get is a plane |
PxCxlsl_YwY | through the origin perpendicular
to A. |
PxCxlsl_YwY | And, in general,
what you should remember is |
PxCxlsl_YwY | that two vectors have a dot
product equal to zero if and |
PxCxlsl_YwY | only if that's equivalent to the
cosine of the angle between them |
PxCxlsl_YwY | is zero.
That means the angle is 90°. |
PxCxlsl_YwY | That means A and B are
perpendicular. |
PxCxlsl_YwY | So, we have a very fast way of
checking whether two vectors are |
PxCxlsl_YwY | perpendicular.
So, one additional application |
PxCxlsl_YwY | I think we'll see actually
tomorrow is to find the |
PxCxlsl_YwY | components of a vector along a
certain direction. |
PxCxlsl_YwY | So, I claim we can use this
intuition I gave about dot |
PxCxlsl_YwY | product telling us how much to
vectors go in the same direction |
PxCxlsl_YwY | to actually give a precise
meaning to the notion of |
PxCxlsl_YwY | component for vector,
not just along the x, |
PxCxlsl_YwY | y, or z axis,
but along any direction in |
PxCxlsl_YwY | space.
So, I think I should probably |
PxCxlsl_YwY | stop here.
But, I will see you tomorrow at |
PxCxlsl_YwY | 2:00 here, and we'll learn more
about that and about cross |
PxCxlsl_YwY | products.
|
5DUQ3-Y_gX4 | [SQUEAKING] |
5DUQ3-Y_gX4 | [RUSTLING] |
5DUQ3-Y_gX4 | [CLICKING] |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: So I want
to revisit the things that Alan |
5DUQ3-Y_gX4 | talked about, but
just a little bit more |
5DUQ3-Y_gX4 | slowly and a bit more-- |
5DUQ3-Y_gX4 | just try and lay out the rules
for you as clearly as I can. |
5DUQ3-Y_gX4 | And what we're
going to try and do |
5DUQ3-Y_gX4 | is, again, just revisit
the notion of a derivative |
5DUQ3-Y_gX4 | to try and write it in a
way that we can generalize |
5DUQ3-Y_gX4 | to other kinds of objects. |
5DUQ3-Y_gX4 | And so I'm going
to start with 18.01 |
5DUQ3-Y_gX4 | and then go to
18.02 and so forth. |
5DUQ3-Y_gX4 | So as Alan said, the key
notion of a derivative, |
5DUQ3-Y_gX4 | just, I think, it's easy to get
so good at taking derivatives, |
5DUQ3-Y_gX4 | like knowing the rule
for the derivative |
5DUQ3-Y_gX4 | of sine or cosine or x squared. |
5DUQ3-Y_gX4 | You're so good at doing them
that you forget what they are, |
5DUQ3-Y_gX4 | right? |
5DUQ3-Y_gX4 | And so the very first thing
you learned about a derivative |
5DUQ3-Y_gX4 | is that it's the
slope of the tangent. |
5DUQ3-Y_gX4 | But what that really
is is linearization. |
5DUQ3-Y_gX4 | So you have some arbitrary
maybe nonlinear function f of x. |
5DUQ3-Y_gX4 | And you're at a point x. |
5DUQ3-Y_gX4 | And near that
point, you're going |
5DUQ3-Y_gX4 | to approximate the function
with a straight line. |
5DUQ3-Y_gX4 | That's the tangent. |
5DUQ3-Y_gX4 | So it's really the linear
approximation of f. |
5DUQ3-Y_gX4 | And then if you move a
little bit away from x-- so |
5DUQ3-Y_gX4 | let me call that
delta x, so not d. |
5DUQ3-Y_gX4 | Delta is going to
be a finite change. |
5DUQ3-Y_gX4 | d is going to be
infinitesimal pretty soon. |
5DUQ3-Y_gX4 | But if you move it just a finite
amount, a little finite amount |
5DUQ3-Y_gX4 | delta x away, of course,
the function value changes. |
5DUQ3-Y_gX4 | But in a linear approximation,
the new function value |
5DUQ3-Y_gX4 | is the red dot here. |
5DUQ3-Y_gX4 | So that linear
approximation is, if you're |
5DUQ3-Y_gX4 | taking the function f
of x at x plus delta x, |
5DUQ3-Y_gX4 | the new value is f of x. |
5DUQ3-Y_gX4 | And then this
linear thing is just |
5DUQ3-Y_gX4 | the slope, which we call f
prime of x times delta x. |
5DUQ3-Y_gX4 | That's just the
definition of the slope. |
5DUQ3-Y_gX4 | It's the little change in
y for a little change in x. |
5DUQ3-Y_gX4 | And, of course, these
two terms are not exact. |
5DUQ3-Y_gX4 | This red dot doesn't
exactly match where |
5DUQ3-Y_gX4 | you are in the real function. |
5DUQ3-Y_gX4 | So there are also corrections. |
5DUQ3-Y_gX4 | But the corrections
are higher order. |
5DUQ3-Y_gX4 | They're terms look like delta
x squared, delta x cubed maybe. |
5DUQ3-Y_gX4 | Maybe if the function
is not higher-- |
5DUQ3-Y_gX4 | it doesn't have
higher derivatives, |
5DUQ3-Y_gX4 | it might have square root of
delta x, so delta x to the 1.1. |
5DUQ3-Y_gX4 | But these are all
terms that are going |
5DUQ3-Y_gX4 | to be higher powers
of delta x terms |
5DUQ3-Y_gX4 | that, if delta x is
sufficiently small, |
5DUQ3-Y_gX4 | these terms will become
more and more negligible |
5DUQ3-Y_gX4 | compared to this linear term. |
5DUQ3-Y_gX4 | And a nice notation
for this that's |
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