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TTU Math2450 Calculus3 Sec 11.6 and 11.7

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    what we discussed
    last time in 11.6
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    which was-- do you remember
    the topics we discussed?
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    We discussed the
    valuation of a derivative.
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    What else have we
    discussed about them?
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    I'm gonna split the fields,
    although they are related.
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    Gradient and the steepest
    ascent and descent.
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    So in which direction
    will z equals
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    f of x and y,
    differential cofunction,
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    and with a derivative
    that is continuous?
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    So will this have the
    maximum rate of change?
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    This is just review.
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    Why am I doing review?
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    Well, after talking to
    you on a personal basis,
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    like one-on-one basis by email
    and a little bit in person,
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    I realized that you like
    very much when I review.
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    When I briefly review
    some of the notions.
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    I will give you the essentials
    for the section, that
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    was 11.5 is embedded in this.
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    Embedded.
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    So 11.5 is embedded in
    11.6, and in one shot
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    we can talk of them.
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    In 11.7 you're gonna see
    some extrema of functions
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    of two variables,
    like max and min,
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    and then we have 11.8, which
    is Lagrange multipliers.
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    What have we done about
    this differentiable function
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    with continuous
    partial derivatives?
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    Let's assume that
    it's smooth, OK?
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    And in that case, we
    define the partial--
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    the directional derivative
    in the direction of u at u,
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    v with unit vector at the
    point x, y, but let's fix it
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    x0, y0, using a limit of
    a difference quotient,
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    just like any derivative
    should be introduced.
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    But I'm not gonna
    repeat that definition.
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    Why is that?
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    Because I want you to give
    me an alternative definition,
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    which is something
    that is simpler
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    to use in applications,
    which is what?
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    Who remembers?
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    The partial derivative
    of f with respect to x
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    measured at the point x0, y0.
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    I'm gonna use this color, and
    then I'll change the color.
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    And I'll say times u1.
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    Plus-- change the color again--
    f sub y at x0, y0, times u2.
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    And this is just a
    times-- multiplication
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    between real values.
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    Because u1 and u2 will be
    nobody but the components--
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    the real value components
    of the unit vector direction
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    that we have.
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    So, guys, remember, direction
    in this books means unit vector.
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    Every time I say
    direction, I should
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    say that's a unit vector.
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    Is there any way, any
    other way, to express this?
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    Maybe with a vector
    multiplication of some sort--
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    multiplication between vectors.
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    See, what if I had a pink
    vector with pink components,
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    and a blue vector
    with blue components?
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    I'm getting somewhere.
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    And I'm sneaky.
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    And you feel--
    you know where I'm
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    getting because you had
    chapter nine fresh in your mind
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    and a certain product
    between vectors.
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    So what is this?
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    It's a dot product, excellent.
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    Rachel, right?
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    So it's a dot product.
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    And I'm gonna run
    it, and it's going
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    to be just the gradient
    of f at the point p0.
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    But I'm going to write
    it again, x0, y0,
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    although it drives
    me crazy to have
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    to write that all the time.
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    Dot product or scalar
    product with what vector u?
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    I'm not gonna put a bar on
    that because the-- that would
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    be like an oxymoron.
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    The gradient is a bar thing.
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    It's always a vector, so
    I'm not gonna write a bar.
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    But I write a bar
    on u, reminding you
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    that u is a free vector.
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    All right, do you like this?
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    Yes.
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    It's a compactified form
    of the fluffy expression
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    we had before.
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    It's much easier to remember
    than the limit definition.
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    Of course, it's
    equivalent to it.
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    And in applications, I-- well,
    we ask about that all the time.
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    What was the gradient?
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    So see, in mathematics
    everything is related.
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    And talking about--
    speaking about the devil--
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    I mean, not you, but we
    weren't talking about you--
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    just this gradient.
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    Gradient f at the
    point p will simply
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    be the vector whose components
    in the direction of i and j
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    are the partials.
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    The partial derivative
    with respect
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    to x, plus the partial
    derivative with respect to y.
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    OK, last time we dealt even
    with gradients of-- functions
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    of more variables.
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    If I have n
    variables, so x1, x2,
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    x3, xn, then this vector will
    have n components-- f sub x1,
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    f sub x2, f sub x3, f sub x4.
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    Somebody stop me.
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    F sub xn.
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    So I have n of them, and
    it's an n [INAUDIBLE].
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    It's a sum ordered [INAUDIBLE].
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    OK.
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    A set of n elements
    in this order.
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    It's an order set,
    [INAUDIBLE] n.
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    All right.
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    So the order matters.
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    Now, steepest
    ascent and descent.
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    In which direction do I have
    the maximum rate of change?
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    And the answer is--
    this is Q and this
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    is A-- the maximum
    rate of change
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    happens in the direction of
    the gradient at every point--
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    every point of
    the domain where I
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    have smoothness or [INAUDIBLE]
    or [INAUDIBLE] whatever.
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    All right.
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    And what else?
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    I claimed last
    time, but I didn't
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    prove, that in that
    direction that I claimed c
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    from [INAUDIBLE].
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    The directional derivative
    is maximized exactly
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    in the direction
    of the gradient.
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    Can we prove that?
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    Now we can prove it.
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    Before I couldn't prove it
    because you couldn't see it,
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    because I didn't look
    at it as a dot product.
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    And we were all blind, like
    guiding each other in the dark.
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    Me blind, you blind.
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    We couldn't see.
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    Now we can see.
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    So now we can see
    how to prove my claim
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    that the directional
    derivative, which
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    is measuring the
    maximum-- measuring
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    the instantaneous rate of
    change in a direction-- compass
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    direction, like,
    what is that, east?
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    East, northeast,
    southwest, whatever.
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    Those are the compass
    directions like i, j, i
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    plus j over square
    root 2 and so on,
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    those are called
    compass directions
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    because you hold the compass
    in your hand as horizontal
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    as you can, and you
    refer to the floor.
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    Even if you are on a slope.
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    Maybe you imagine me on a
    slope, hiking or whatever.
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    Going down, going up.
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    But the compass should
    always be kept horizontal.
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    Do you hike, Alex?
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    STUDENT: Yeah.
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    PROFESSOR: I'm sorry,
    I've put you on the spot.
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    So whatever you do when you
    think of a path up or down
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    is measured in the direction
    of the horizontal plane,
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    the compass direction.
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    And that is the gradient
    I was talking about.
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    You see, it's a function.
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    It's a vector in plane.
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    That means a geographic
    compass direction.
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    Prove it.
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    Let's prove the claim.
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    Let's prove the claim,
    because this is Tuesday,
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    and it's almost weekend, and
    we have to prove something
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    this week, right?
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    Now, do you like what you see?
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    Well, I have no idea.
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    What if I want to measure this.
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    This could be a negative
    number, but it doesn't matter.
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    Assume that I take
    the dot product,
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    and I think, well, it's a scalar
    product, it must be positive.
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    What do I get?
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    This guy-- if I'm a physicist,
    I'm going to say this guy
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    is the length of the
    first vector at p0.
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    Gradient at p0.
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    That's the length of u.
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    Duh, that is 1.
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    Last time I checked, u was
    unitary, so that's silly of me,
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    but I'll write it anyway.
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    Times the cosine of the
    angle between-- oh, cosine
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    of the parenthesis angle
    between nabla f and u.
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    OK.
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    When is this maximized?
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    STUDENT: When the angle
    between nabla f and u is 0.
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    PROFESSOR: Exactly.
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    When this is pi, so this
    quantity is maximized-- gosh,
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    I hate writing a lot.
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    I had to submit homework
    in the past two days,
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    and one was about biological
    research and the other one
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    about stress management.
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    The stress management class
    stresses me out the most.
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    I shouldn't make it public.
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    Really, because we have to write
    these essays of seven or eight
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    pages every Tuesday,
    every end of the week.
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    Twice a week.
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    , So I realized how much
    I hate writing down a lot,
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    and what a blessing it
    is to be a mathematician.
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    You abbreviate everything.
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    You compress everything.
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    I love formulas.
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    So what we have here is
    maximized when the cosine is 1.
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    And if you have
    become 0 to 2 pi open,
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    then theta 0 is
    your only option.
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    Well, if you take
    an absolute value,
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    you could also have it
    in the other direction,
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    cosine pi, but in that
    case you change the sign.
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    So what you get--
    you get a maximum.
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    Let's say you hike, right?
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    I'm just hiking in my brain.
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    The maximum rate of
    change in this direction,
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    climbing towards the peak.
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    And then the steepest descent
    is the exactly minus gradient
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    direction.
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    So I could have 0
    or pi for the angle.
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    That's the philosophical
    meaning of that.
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    All right.
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    So the directional
    derivative, which is this guy,
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    when does it become maximum?
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    When the angle is 0.
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    So I'm done.
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    QED.
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    What does it mean?
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    That I know this happens
    when-- when direction u is
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    the direction of the gradient.
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    Can I write u equals
    gradient of f?
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    Not quite.
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    I should say divided
    by norm or magnitude.
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    Why is that?
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    And you say, Magdalena,
    didn't you say like 10 times
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    that u is a unit vector?
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    You want u to be a
    unit vector direction.
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    So the direction should be
    the direction of the gradient
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    in order to maximize this
    directional derivative.
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    But then you have
    to take the gradient
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    and divide it by its magnitude.
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    Let's compute it.
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    Let's see what we get.
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    Let's see what we get.
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    Now, I'm sorry about my
    beautiful handwriting,
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    but I'm-- well, I'm gonna
    have to-- I have room here.
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    And actually I can
    use this formula.
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    So in the direction
    of the gradient,
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    when u is the gradient.
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    Let's take u to be-- what
    did I say over there?
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    Gradient of f over the
    magnitude of gradient of f.
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    I'll take this guy
    and drag him here.
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    [INAUDIBLE]
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    What will the value of the
    directional derivative be?
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    We've done last time that--
    the same thing on an example.
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    We've done it on a function,
    beautiful f of x, y
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    equals x squared plus y squared.
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    This is the type
    of function I like,
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    because they are the
    fastest to deal with.
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    But anyway, we'll have all
    sorts of other functions.
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    What am I going to write here?
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    I have to write.
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    Well, by definition,
    now, by my new way
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    to look at the
    definition, I'm gonna
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    have a gradient at
    the point, the vector,
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    dot-- who is u again?
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    The gradient this time,
    that special value,
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    divided by absolute-- by
    magnitude of the gradient.
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    But what in the world is that?
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    This animal is--
    what if you take
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    a vector multiplied by itself?
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    Dot product.
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    No, not dot product.
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    What you get is the
    magnitude squared.
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    All right.
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    So although the pinkie
    guy and the pinkie guy
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    are magnitude of f squared
    divided by magnitude of f.
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    And Alex said, but wait,
    that's just-- I know.
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    I didn't want to just
    jump ahead too fast.
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    We get gradient
    of f in magnitude.
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    So, beautiful.
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    So we know who that maximum is.
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    The maximum of
    the rate of change
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    will be for-- this
    equals f of x, y.
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    The max of the rate of change
    is-- what is that again?
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    Magnitude of lambda f, in
    the direction nabla-- nabla f
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    divided by its magnitude.
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    This is what we discovered.
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    And now I'm going
    to ask you, what
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    is the minimum rate of
    change at the same point?
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    STUDENT: [INAUDIBLE].
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    PROFESSOR: Just
    parallel opposite.
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    So it's gonna-- I'm gonna
    have the so-called highest--
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    steepest, not highest,
    highest means maximum.
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    The lowest value.
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    So I'm going to have
    the lowest value, which
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    indicates the steepest descent.
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    Me going down on-- in the snow,
    I'm dreaming, on a sleigh,
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    or on a plastic bag.
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    That would give me
    the steepest descent,
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    and the steepest descent
    will correspond to what?
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    I'm going to make an NB.
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    Nota bene.
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    In Latin.
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    Note the minimum will
    be minus magnitude
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    of f, [INAUDIBLE] nabla f,
    in the opposite direction.
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    Shall I write it in words?
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    Let me write it in as
    O-P-P from opposite-- no,
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    from opposite--
    opposite direction.
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    What do I mean
    opposite direction?
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    Opposite direction
    to the gradient.
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    Which is the same direction,
    if you think about,
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    because it's the same line.
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    So it's going to be minus nabla
    f over magnitude of nabla f.
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    It's like when we were
    in [INAUDIBLE], which
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    was 1 minus x squared minus
    y squared, whatever it was--
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    we had i plus j for the
    descent, and minus i minus
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    j after the ascent--
    the steepest descent
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    and the steepest ascent.
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    We started with examples
    because it's easier
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    to understand
    mathematics-- actually
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    it's easier to understand
    anything on an example.
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    And then-- if the
    example is good.
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    If the example is
    bad, it's confusing.
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    But if the example is
    good, you understand just
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    about any concept, and then
    you move on to the theory,
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    and this is the theory.
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    And it looks very abstract.
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    When somebody steps
    in this classroom
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    and they haven't taken more than
    calc 2 they will get scared,
  • 19:06 - 19:09
    and they will never
    want to take calc 3.
  • 19:09 - 19:13
    Well, that's why-- I didn't
    want to scare you off yet.
  • 19:13 - 19:20
    OK, so this is what you have
    to remember from section 11.6
  • 19:20 - 19:25
    with 11.5 embedded in it.
  • 19:25 - 19:32
    Now, one thing that
    I would like to see
  • 19:32 - 19:38
    would be more examples and
    connections to other topics.
  • 19:38 - 19:42
    So one example that I picked--
    and I think it's a nice one.
  • 19:42 - 19:44
    I just copied from the book.
  • 19:44 - 19:47
    I usually don't
    bring cheat sheets.
  • 19:47 - 19:52
    I don't like professors who
    bring books to the class
  • 19:52 - 19:55
    and start reading
    out of the book.
  • 19:55 - 19:56
    I think that's ridiculous.
  • 19:56 - 20:00
    I mean, as if you guys couldn't
    read your own book at home.
  • 20:00 - 20:06
    And I try to make up examples
    that are easier than the ones
  • 20:06 - 20:08
    from the book to start with.
  • 20:08 - 20:13
    But here's example 4, which
    is not so easy to deal with,
  • 20:13 - 20:15
    but it's not hard either.
  • 20:15 - 20:18
    And I picked it because
    I saw this browsing
  • 20:18 - 20:22
    through the previous
    finals, I saw it
  • 20:22 - 20:25
    as a pattern coming
    every now and then.
  • 20:25 - 20:31
    Find the direction in which
    f increases or decreases
  • 20:31 - 20:32
    most rapidly.
  • 20:32 - 20:34
    I have to write
    down beautifully.
  • 20:34 - 20:53
    Find the directions in which
    f increases or decreases most
  • 20:53 - 21:01
    rapidly at p0 coordinates 2, 1.
  • 21:01 - 21:05
  • 21:05 - 21:11
    And, what is the
    maximum-- that's
  • 21:11 - 21:21
    a question-- what is the maximum
    rate of change or of increase?
  • 21:21 - 21:24
    This type of problem is also
    covered in the Khan Academy
  • 21:24 - 21:28
    videos and also the MIT
    library, but I don't
  • 21:28 - 21:30
    feel they do a very good job.
  • 21:30 - 21:32
    They cover it just
    lightly, as if they
  • 21:32 - 21:35
    were afraid to speak too much
    about-- give too many examples
  • 21:35 - 21:38
    and talk too much
    about the subject.
  • 21:38 - 21:43
    So what do you guys want
    to do with this one?
  • 21:43 - 21:44
    Help me solve the problem.
  • 21:44 - 21:46
    That was kind of the idea.
  • 21:46 - 21:49
    So we start with
    computing the what?
  • 21:49 - 21:53
    The animal called--
    what's the animal?
  • 21:53 - 21:54
    STUDENT: Gradient.
  • 21:54 - 21:55
    PROFESSOR: Gradient.
  • 21:55 - 21:56
    Thank you very much.
  • 21:56 - 22:02
    So we do that and we
    start differentiating.
  • 22:02 - 22:07
    With respect to x, we get
    a product [INAUDIBLE].
  • 22:07 - 22:13
    So your product [INAUDIBLE] 1--
    it's differentiated-- times e
  • 22:13 - 22:19
    to the 2y minus x [INAUDIBLE]
    plus x undifferentiated.
  • 22:19 - 22:22
    The second guy, prime.
  • 22:22 - 22:24
    Copy and paste, Magdalena.
  • 22:24 - 22:26
    Times-- don't
    forget the minus 1,
  • 22:26 - 22:29
    because-- I'm talking to myself.
  • 22:29 - 22:34
    Because if you do, you get
    a 0 on this in the final.
  • 22:34 - 22:35
    I'm talking to myself.
  • 22:35 - 22:36
    OK?
  • 22:36 - 22:36
    All right.
  • 22:36 - 22:43
    Plus, parenthesis-- the same
    procedure with respect to y.
  • 22:43 - 22:45
    When I do it with
    respect to y, this
  • 22:45 - 22:47
    is good review of
    the whole chapter.
  • 22:47 - 22:48
    Yes, sir.
  • 22:48 - 22:49
    STUDENT: [INAUDIBLE].
  • 22:49 - 22:58
  • 22:58 - 23:01
    PROFESSOR: This primed
    with respect to x.
  • 23:01 - 23:02
    Am I doing something wrong?
  • 23:02 - 23:04
    No.
  • 23:04 - 23:05
    Are you with me?
  • 23:05 - 23:08
    So this guy primed
    with respect to x, I'm
  • 23:08 - 23:13
    going to write it es copied.
  • 23:13 - 23:18
    And take that out, you maintain
    and differentiate with respect
  • 23:18 - 23:19
    to x.
  • 23:19 - 23:22
    So I did it right.
  • 23:22 - 23:29
    OK, the second one, x is a
    constant for me right now.
  • 23:29 - 23:31
    Who is the variable y?
  • 23:31 - 23:37
    So I copy and paste e
    to the whole argument
  • 23:37 - 23:40
    times-- I cover everything
    else with my hand,
  • 23:40 - 23:43
    and I differentiate the
    argument with respect to y.
  • 23:43 - 23:47
    And I get prime sub 2
    and a j, and I say, thank
  • 23:47 - 23:49
    god, this was a little bit long.
  • 23:49 - 23:53
    You realize that if you make
    the slightest algebra mistake,
  • 23:53 - 23:55
    it's all over for you.
  • 23:55 - 23:58
    In that case, I
    ask my colleagues,
  • 23:58 - 24:03
    what do you guys do when guy
    missed that or missed this?
  • 24:03 - 24:09
    0, 0, no points-- OK,
    maybe a little bit,
  • 24:09 - 24:12
    maybe a tiny bit
    of extra credit.
  • 24:12 - 24:14
    But pay attention to your math.
  • 24:14 - 24:18
    So you know what you need to do.
  • 24:18 - 24:24
    Now I'm going to go on and
    say, but I am at the point 0.
  • 24:24 - 24:28
    By the way, I really don't
    like what we did in the book.
  • 24:28 - 24:32
    OK, I should not say that
    out loud, but it's too late.
  • 24:32 - 24:37
    The book denotes that
    sometimes, well, we
  • 24:37 - 24:41
    try not to do that too
    often, but not by F sub 0.
  • 24:41 - 24:44
    Because some other
    books use that.
  • 24:44 - 24:45
    I don't like that.
  • 24:45 - 24:49
    So every time-- you
    should never do that.
  • 24:49 - 24:51
    Because it gives
    the feeling that
  • 24:51 - 24:54
    you're differentiating
    a constant or something.
  • 24:54 - 24:59
    OK, so I always try to
    say, gradient [? up ?]
  • 24:59 - 25:03
    is-- which means, I
    have a fixed value.
  • 25:03 - 25:06
    But I don't fix the value
    before I took the gradient.
  • 25:06 - 25:08
    This is too confusing
    as a notation.
  • 25:08 - 25:10
    Don't do it.
  • 25:10 - 25:11
    Close your eyes
    when you get to it
  • 25:11 - 25:12
    when you're reading the book.
  • 25:12 - 25:21
    OK, now I have to plug
    in instead of x sub 2--
  • 25:21 - 25:25
    I tried to remember
    that-- y equals 1.
  • 25:25 - 25:29
    So I go 1 times e to
    the 2 times 1 minus 2
  • 25:29 - 25:33
    plus 2 times e to
    the 2 times 1 minus--
  • 25:33 - 25:39
    I wish I had Data with me, I
    mean the guy from Star Trek.
  • 25:39 - 25:42
    Because he could do this in
    just a fraction of a second
  • 25:42 - 25:46
    without me having to bother
    with this whole thing.
  • 25:46 - 25:51
  • 25:51 - 25:53
    STUDENT: But then
    if Data existed,
  • 25:53 - 25:54
    why would we be math majors?
  • 25:54 - 25:57
    PROFESSOR: Exactly,
    so we do this
  • 25:57 - 26:02
    so that we can program
    people, I mean androids,
  • 26:02 - 26:06
    and eventually learn
    how to clone ourselves.
  • 26:06 - 26:09
    So let's see what we have.
  • 26:09 - 26:12
    e to the 0 is 1.
  • 26:12 - 26:16
    e to the 0 is 1, 1 minus
    2-- are you guys with me?
  • 26:16 - 26:18
    I'm going too fast?
  • 26:18 - 26:19
    STUDENT: No, the
    [? board ?] says
  • 26:19 - 26:24
    the y component is equal to
    1, x component is equal to 2.
  • 26:24 - 26:28
    PROFESSOR: x
    component equals to 2.
  • 26:28 - 26:31
    x0 is 2, and y0 is 1.
  • 26:31 - 26:39
    And I plug in x0
    equals [INAUDIBLE].
  • 26:39 - 26:45
    So 2 times e to the 2 times 1
    minus 2-- you think I like it?
  • 26:45 - 26:46
    I don't like it.
  • 26:46 - 26:47
    But anyway, it's my life.
  • 26:47 - 26:49
    I have to go on.
  • 26:49 - 26:54
    So we have 1 minus 2
    plus minus 1, right?
  • 26:54 - 27:00
    Minus 1i-- minus
    1i sounds scary.
  • 27:00 - 27:09
    OK, plus 4j-- minus i plus 4j.
  • 27:09 - 27:10
    Is it lovely?
  • 27:10 - 27:12
    No, I hate it.
  • 27:12 - 27:16
    The magnitude is going
    to be square root of 17.
  • 27:16 - 27:17
    But that's life.
  • 27:17 - 27:21
    I mean, you as
    engineering majors
  • 27:21 - 27:23
    see that all the time,
    and even worse than that.
  • 27:23 - 27:30
    So I'm going to say the
    gradient of F at P0 in magnitude
  • 27:30 - 27:34
    will be square root of 17.
  • 27:34 - 27:36
    What is that?
  • 27:36 - 27:41
    That is the maximum rate
    of change, right guys?
  • 27:41 - 27:44
    But in which direction
    does that happen?
  • 27:44 - 27:48
  • 27:48 - 27:54
    My beloved book says, in the
    direction of minus i plus 4j.
  • 27:54 - 27:58
    That's our answer, so
    in the direction of--
  • 27:58 - 28:00
    STUDENT: Over root 17.
  • 28:00 - 28:02
    PROFESSOR: No, let
    me tell you what.
  • 28:02 - 28:05
    We fought about that as the
    authors when we wrote the book.
  • 28:05 - 28:09
    So I said, if you're
    saying, in the direction of,
  • 28:09 - 28:11
    then you have to
    say, over root 17.
  • 28:11 - 28:14
    And my coauthor said,
    no, actually, Magdalena,
  • 28:14 - 28:17
    as a matter of English--
    since you're not a native,
  • 28:17 - 28:19
    you don't understand.
  • 28:19 - 28:22
    In the direction
    of a certain vector
  • 28:22 - 28:25
    means the direction
    of a certain vector,
  • 28:25 - 28:28
    the direction could be that.
  • 28:28 - 28:36
    But I say equivalently, in
    the direction double dot
  • 28:36 - 28:39
    minus i plus 4j over
    square root of 17,
  • 28:39 - 28:43
    meaning that this
    is the direction.
  • 28:43 - 28:45
    It's a matter of interpretation.
  • 28:45 - 28:47
    I don't understand it,
    but it's your language.
  • 28:47 - 28:47
    Yes, sir.
  • 28:47 - 28:48
    STUDENT: [INAUDIBLE].
  • 28:48 - 28:53
  • 28:53 - 28:55
    PROFESSOR: I'm saying--
  • 28:55 - 28:56
    STUDENT: Which one do you like?
  • 28:56 - 28:57
    PROFESSOR: Which one do I like?
  • 28:57 - 28:58
    This one.
  • 28:58 - 28:59
    STUDENT: [INAUDIBLE].
  • 28:59 - 29:01
    PROFESSOR: Thank you.
  • 29:01 - 29:04
    So if we define direction
    to be a unit vector,
  • 29:04 - 29:09
    let's be consistent and not say,
    of me, of you, of my cousin,
  • 29:09 - 29:11
    of whatever.
  • 29:11 - 29:15
    All right, was this hard?
  • 29:15 - 29:17
    No.
  • 29:17 - 29:22
    Do I have a caveat about
    this kind of problem?
  • 29:22 - 29:24
    Yeah, that was kind of the idea.
  • 29:24 - 29:26
    If I put that in
    the midterm, guys,
  • 29:26 - 29:29
    please do it two or
    three times, make
  • 29:29 - 29:31
    sure you didn't make
    any algebra mistakes.
  • 29:31 - 29:35
    Because if you know the
    theory, I will still
  • 29:35 - 29:37
    give you like 30% or something.
  • 29:37 - 29:39
    But if you mess up
    with the algebra,
  • 29:39 - 29:45
    I have no choice but giving
    up the 70%, whatever that is.
  • 29:45 - 29:49
    I try to be fair and give
    you something for everything
  • 29:49 - 29:51
    you do and know.
  • 29:51 - 29:53
    But try not to mess
    it up too badly,
  • 29:53 - 29:55
    because it's very
    easy to mess it up.
  • 29:55 - 29:56
    Yes, sir.
  • 29:56 - 29:58
    STUDENT: So which
    way is increasing?
  • 29:58 - 30:00
    PROFESSOR: OK,
    find the direction
  • 30:00 - 30:04
    in which r increases or
    decreases most rapidly.
  • 30:04 - 30:08
    OK, the direction in
    which I could draw it,
  • 30:08 - 30:12
    this is increasing in
    the direction of that.
  • 30:12 - 30:13
    At what rate?
  • 30:13 - 30:15
    At the rate of
    square root of 17.
  • 30:15 - 30:18
    Good question.
  • 30:18 - 30:20
    How about the other one?
  • 30:20 - 30:24
    In the direction of plus i
    minus 4j over square root of 17,
  • 30:24 - 30:29
    I get the rate of change
    minus square root of 17.
  • 30:29 - 30:30
    And that's it.
  • 30:30 - 30:32
    Do we have to say that?
  • 30:32 - 30:36
    Ehh, I give you extra
    credit if you say that.
  • 30:36 - 30:39
    But at this point, I'm saying
    I'm happy with what I just
  • 30:39 - 30:41
    wrote on the board.
  • 30:41 - 30:42
    STUDENT: [INAUDIBLE].
  • 30:42 - 30:46
  • 30:46 - 30:49
    PROFESSOR: Yeah, so
    it's like finding
  • 30:49 - 30:53
    on which path are you
    going to get to the top
  • 30:53 - 30:56
    the fastest when you
    climb a mountain.
  • 30:56 - 30:59
    It's the same kind of question.
  • 30:59 - 31:01
    Because it's all about
    z being an altitude.
  • 31:01 - 31:03
    z equals F of x.
  • 31:03 - 31:07
    I will go ahead and
    erase the whole thing.
  • 31:07 - 31:16
    And I wish you good luck now.
  • 31:16 - 31:22
    I wish you good luck because
    I was asked by [INAUDIBLE]
  • 31:22 - 31:29
    to solve a problem like the one
    you gave me in the web work,
  • 31:29 - 31:30
    but I don't remember it.
  • 31:30 - 31:32
    But it's OK.
  • 31:32 - 31:36
    it was a review of
    what we did last time.
  • 31:36 - 31:44
    And we said, instead of that,
    where the tangent plane--
  • 31:44 - 31:45
    we know what that is.
  • 31:45 - 31:50
    At P0 was-- guys, by the
    final I want this memorized.
  • 31:50 - 31:51
    There is no question.
  • 31:51 - 31:56
    So z minus z0 is
    like Taylor's formula
  • 31:56 - 31:59
    in the linear approximation.
  • 31:59 - 32:00
    You truncate.
  • 32:00 - 32:02
    You throw away the
    second order, and so on.
  • 32:02 - 32:04
    So what is that?
  • 32:04 - 32:06
    f sub x at delta x.
  • 32:06 - 32:08
    Oh, x equals 0.
  • 32:08 - 32:09
    I'm lazy.
  • 32:09 - 32:13
    Times x minus x0--
    this is the delta x.
  • 32:13 - 32:17
    This is the delta z.
  • 32:17 - 32:21
    And this is not the round
    surface, curvy and everything.
  • 32:21 - 32:29
    This is the plane approximation,
    the planar approximation--
  • 32:29 - 32:39
    A-P-P-R. I'm not done--
    plus f sub y y minus y0.
  • 32:39 - 32:50
    So this is the equation of the
    plane that's tangent pi at 0.
  • 32:50 - 32:53
    And let me draw
    the surface pink.
  • 32:53 - 33:01
    Because I'm a girl, and because
    I want to draw this in pink.
  • 33:01 - 33:07
    Let's call it some
    S, for Surface.
  • 33:07 - 33:10
    Can we paint an S here?
  • 33:10 - 33:18
    OK, but if I'm giving
    the same picture
  • 33:18 - 33:24
    for a different equation--
    so I have F of x, y, z
  • 33:24 - 33:27
    equals constant, that's
    the implicit form.
  • 33:27 - 33:31
  • 33:31 - 33:32
    I make this face.
  • 33:32 - 33:33
    Why do I make this face?
  • 33:33 - 33:36
    Because I've got
    three confessions.
  • 33:36 - 33:40
    I'm more like a
    priestess in mathematics.
  • 33:40 - 33:44
    People don't like
    implicit differentiation.
  • 33:44 - 33:46
    We will do a little
    bit of that today,
  • 33:46 - 33:50
    because you told me your stories
    from implicit differentiation,
  • 33:50 - 33:52
    and I got scared.
  • 33:52 - 33:54
    Those were horror stories.
  • 33:54 - 33:56
    And I don't want those to repeat
    in the final or the midterm.
  • 33:56 - 33:59
    So I'm going to do something
    with implicit differentiation
  • 33:59 - 34:01
    as well.
  • 34:01 - 34:04
    What did I want to say?
  • 34:04 - 34:07
    If we apply this, we'd be wrong.
  • 34:07 - 34:11
    But we have to remember
    what the normal would be.
  • 34:11 - 34:17
    And the normal was
    the gradient of big F.
  • 34:17 - 34:24
    So those will be F sub xi plus
    F sub yj plus big F sub zk.
  • 34:24 - 34:29
    Then force, such a
    surface, even implicitly,
  • 34:29 - 34:32
    the tangent plane looks
    a little bit differently.
  • 34:32 - 34:34
    But it's the same story.
  • 34:34 - 34:35
    And I prove that.
  • 34:35 - 34:38
    You may not believe it,
    or may not remember.
  • 34:38 - 34:43
    But I proved that, that
    it's one and the same thing.
  • 34:43 - 34:44
    So I get F sub x.
  • 34:44 - 34:47
    I get 0i0.
  • 34:47 - 34:52
    This was the A dot,
    times x minus x plus.
  • 34:52 - 34:53
    What's next?
  • 34:53 - 34:57
    From the coefficients
    coming from the gradient--
  • 34:57 - 35:00
    that was the gradient.
  • 35:00 - 35:02
  • 35:02 - 35:06
    So this is F of
    x, y, z equals C.
  • 35:06 - 35:13
    And the gradient was-- this is
    the normal, F sub x, F sub y,
  • 35:13 - 35:21
    F sub z angular bracket
    equals gradient of F, which
  • 35:21 - 35:23
    is more or less than normal.
  • 35:23 - 35:31
    The unit normal will be just
    gradient of F over length of F.
  • 35:31 - 35:40
    So let's continue-- F sub
    y at x0y0 times delta y.
  • 35:40 - 35:48
    This is B. I had the problem
    with memorizing these,
  • 35:48 - 35:52
    especially since when I was
    18 I did not understand them
  • 35:52 - 35:56
    whatsoever first
    year of college.
  • 35:56 - 36:01
    I had to use markers,
    put them in markers
  • 36:01 - 36:03
    and glue them to my closet.
  • 36:03 - 36:04
    Because when I
    was 18, of course,
  • 36:04 - 36:07
    I was looking in the
    mirror all the time.
  • 36:07 - 36:08
    So whenever I got
    into the closet,
  • 36:08 - 36:12
    opened the door to the
    mirror, next to the mirror
  • 36:12 - 36:14
    there was this formula.
  • 36:14 - 36:20
    So whether I liked it or not--
    I didn't-- I memorized it just
  • 36:20 - 36:23
    by seeing it every day
    when I opened the door.
  • 36:23 - 36:24
    Yes, sir.
  • 36:24 - 36:26
    STUDENT: That green one, should
    that be x minus 0 or z minus 0?
  • 36:26 - 36:27
    PROFESSOR: z minus 0.
  • 36:27 - 36:28
    That is my mistake.
  • 36:28 - 36:30
    Thank you.
  • 36:30 - 36:35
    So again, you have delta
    x, delta y, delta z.
  • 36:35 - 36:37
    Thank you.
  • 36:37 - 36:39
    Is it hard to memorize?
  • 36:39 - 36:41
    No, not if you
    put it in markers.
  • 36:41 - 36:45
    I think you will do just fine.
  • 36:45 - 36:48
    What have we done as time?
  • 36:48 - 36:50
    Let me review it really quickly.
  • 36:50 - 36:51
    We said, wait a minute.
  • 36:51 - 36:54
    How come they are
    one and the same?
  • 36:54 - 36:56
    You said, oh I'm
    getting a headache.
  • 36:56 - 37:00
    I don't understand why
    they are one and the same.
  • 37:00 - 37:07
    And we said, yes, but you
    see, this guy is nothing but z
  • 37:07 - 37:10
    minus F of x, y.
  • 37:10 - 37:15
    So from this form, you can make
    it implicit it by pulling out F
  • 37:15 - 37:25
    to the left and creating this
    big F of x, y, and z equals 0.
  • 37:25 - 37:26
    And what does it mean?
  • 37:26 - 37:31
    It means that F sub x will be
    oh my god, this has nothing
  • 37:31 - 37:36
    to do with us minus F sub x.
  • 37:36 - 37:40
    F sub y will be minus F sub y.
  • 37:40 - 37:41
    Am I right?
  • 37:41 - 37:48
    And F sub z, big F sub z, is
    simply-- there's no z here-- 1.
  • 37:48 - 37:56
    So coming back to the guide's
    A, B, C, forget about A,
  • 37:56 - 38:00
    B, C. I'll take the A, and
    I'll replace it with minus F
  • 38:00 - 38:04
    sub x, which doesn't write.
  • 38:04 - 38:09
    And it's time for him to
    go away-- minus F sub x.
  • 38:09 - 38:14
    And this is F sub
    y with the minus.
  • 38:14 - 38:18
  • 38:18 - 38:22
    And finally, Mr. C,
    who is happy, he is 1.
  • 38:22 - 38:24
    So he says, I'm happy.
  • 38:24 - 38:27
    You're going to separate me.
  • 38:27 - 38:29
    We are going to separate him.
  • 38:29 - 38:34
    So this equation is
    nothing but what?
  • 38:34 - 38:37
    Let's write it from
    the left to the right.
  • 38:37 - 38:40
    Let's keep the green guy
    in the left hand side.
  • 38:40 - 38:43
  • 38:43 - 38:46
    And everybody else
    goes for a moving sale.
  • 38:46 - 38:49
    The blue and the pink go away.
  • 38:49 - 38:51
    And when they go
    to the other side,
  • 38:51 - 38:55
    they have a minus,
    pick up a minus sign.
  • 38:55 - 38:58
    But with the minus, the
    minus here is a plus sign.
  • 38:58 - 39:00
    Are you guys with me?
  • 39:00 - 39:03
  • 39:03 - 39:07
    So the blue guy has
    moved and changed sign.
  • 39:07 - 39:09
    Where's the pink?
  • 39:09 - 39:12
    The pink guy will also move.
  • 39:12 - 39:17
    And he picked up this.
  • 39:17 - 39:21
    So practically, this and that
    formula are one and the same.
  • 39:21 - 39:25
    They're both used for
    the same tangent plane.
  • 39:25 - 39:28
    It depends how you
    introduce the tangent plane.
  • 39:28 - 39:33
    And [INAUDIBLE], before class
    started, 20 or 25 minutes
  • 39:33 - 39:38
    before class, when I was in a
    hurry, I answered you briefly.
  • 39:38 - 39:42
    You made some algebra
    mistake in the-- you got it?
  • 39:42 - 39:46
    I would like to make
    up one like those.
  • 39:46 - 39:48
    But I forgot what
    your surface was.
  • 39:48 - 39:49
    Was it an ellipsoid?
  • 39:49 - 39:51
    STUDENT: Ellipsoid.
  • 39:51 - 39:53
    PROFESSOR: OK, I'll
    make up an ellipsoid.
  • 39:53 - 39:54
    STUDENT: [INAUDIBLE].
  • 39:54 - 39:58
    PROFESSOR: Yeah, mhmm,
    if you have it with you.
  • 39:58 - 40:00
    If you don't have it
    with you, that's fine.
  • 40:00 - 40:02
    So I'm going to
    go ahead and keep
  • 40:02 - 40:04
    just the implicit equation.
  • 40:04 - 40:07
    Because the ellipsoid is given
    by the implicit equation.
  • 40:07 - 40:13
    And everything
    else I will erase.
  • 40:13 - 40:16
    And that was problem 24.
  • 40:16 - 40:17
    How many problems?
  • 40:17 - 40:21
    Well, you still have
    a lot, up to 49.
  • 40:21 - 40:22
    STUDENT: 42.
  • 40:22 - 40:24
    PROFESSOR: 42, OK, I reduced it.
  • 40:24 - 40:29
    Now, when I sent you
    an email on Sunday,
  • 40:29 - 40:33
    I said I was giving
    you an extension.
  • 40:33 - 40:35
    STUDENT: Till March.
  • 40:35 - 40:37
    PROFESSOR: Till a lot of March.
  • 40:37 - 40:41
    Because I thought March the
    2nd, and I gave a few more days.
  • 40:41 - 40:43
    So you have one more week.
  • 40:43 - 40:44
    STUDENT: When's spring break?
  • 40:44 - 40:46
    PROFESSOR: Spring
    break is the 14th,
  • 40:46 - 40:49
    but this is due
    on the 9th right?
  • 40:49 - 40:51
    STUDENT: The 10th.
  • 40:51 - 40:54
    PROFESSOR: The 10th--
    maybe I don't remember.
  • 40:54 - 40:56
    So what was your problem?
  • 40:56 - 40:58
    What's your problem?
  • 40:58 - 40:59
    What?
  • 40:59 - 41:00
    Can I take it?
  • 41:00 - 41:06
  • 41:06 - 41:10
    So you have an
    ellipsoid, which comes
  • 41:10 - 41:16
    from the ellipse 3x
    squared [INAUDIBLE].
  • 41:16 - 41:21
    Then you have a 2z
    squared equals 9.
  • 41:21 - 41:25
    Then it says, you
    look at the point P
  • 41:25 - 41:27
    of coordinates minus
    1-- I haven't even
  • 41:27 - 41:30
    checked if it's correct,
    but it should be.
  • 41:30 - 41:34
    So 3 plus-- I did not
    program the problem.
  • 41:34 - 41:39
    3 plus 4, 7, plus 2, 9,
    so he did a good job.
  • 41:39 - 41:42
    We want the tangent plane.
  • 41:42 - 41:45
    I'll put it here.
  • 41:45 - 41:48
    We want the tangent plane.
  • 41:48 - 41:50
    How do we compute
    the tangent plane?
  • 41:50 - 41:54
    You say, this is F
    of x, y, z, right?
  • 41:54 - 41:56
    So F sub x equals 6x.
  • 41:56 - 41:59
    F sub y equals 2y.
  • 41:59 - 42:02
    F sub z equals 4z.
  • 42:02 - 42:07
    Computing it at P0,
    what do we have?
  • 42:07 - 42:11
    x is minus 1, y is
    minus 2, z is minus 1.
  • 42:11 - 42:18
    I should get negative 6,
    negative 4, and minus 4.
  • 42:18 - 42:21
  • 42:21 - 42:25
    And then I should plug
    in and get minus 6 times
  • 42:25 - 42:28
    x minus minus 1.
  • 42:28 - 42:30
    I have to pay attention myself.
  • 42:30 - 42:36
    It's not easy to get the
    algebra right-- minus 4 times y
  • 42:36 - 42:45
    plus 2 minus 4 times
    z plus 1 equals 0.
  • 42:45 - 42:51
    And I hope I get what you got--
    minus 6x minus 4y minus 4z,
  • 42:51 - 42:52
    so many of those.
  • 42:52 - 42:53
    You've got to divide by 2.
  • 42:53 - 42:55
    I'm not getting that.
  • 42:55 - 42:59
    And then minus 6-- I'm
    going to write it down.
  • 42:59 - 43:03
  • 43:03 - 43:05
    So it's even, right?
  • 43:05 - 43:08
  • 43:08 - 43:16
    The whole thing minus 18, divide
    by 2 should be-- divide by 2,
  • 43:16 - 43:19
    and did you change the
    signs, [INAUDIBLE]?
  • 43:19 - 43:21
    What's your password?
  • 43:21 - 43:22
    No.
  • 43:22 - 43:27
    [LAUGHING] Check if I'm
    getting the same thing you got.
  • 43:27 - 43:33
    So I get 3x plus 2y plus 2z.
  • 43:33 - 43:37
    I divide by minus
    2, right, plus 9?
  • 43:37 - 43:39
    Did you both get the same thing?
  • 43:39 - 43:41
    STUDENT: [INAUDIBLE].
  • 43:41 - 43:43
    PROFESSOR: You didn't?
  • 43:43 - 43:47
    Well, I'm trying to
    simplify all my answers
  • 43:47 - 43:50
    with this simplification.
  • 43:50 - 43:53
    So I guess of course
    if you enter it
  • 43:53 - 43:55
    like that, it's going to work.
  • 43:55 - 44:04
    Now, I need you guys
    to help me on this one.
  • 44:04 - 44:06
    Find the parametric
    form-- it's so easy,
  • 44:06 - 44:11
    but this is a problem
    session-- of the line passing
  • 44:11 - 44:13
    through the same point
    that's perpendicular
  • 44:13 - 44:15
    to the tangent plane.
  • 44:15 - 44:19
    Express your answer in the
    parametric form of the type
  • 44:19 - 44:21
    the one that you know that
    I don't like very much,
  • 44:21 - 44:30
    but I will write it down-- a2t
    [INAUDIBLE] b2 a3t plus b3.
  • 44:30 - 44:35
    This is how he wants me to
    write it, which I don't like.
  • 44:35 - 44:38
    I would've even preferred
    it to be in symmetric form.
  • 44:38 - 44:39
    It's the same.
  • 44:39 - 44:41
    I'll put the t, and I'm fine.
  • 44:41 - 44:47
    So x minus x0 over
    l, y minus y0 over m,
  • 44:47 - 44:52
    z minus z0 over n, that
    was the symmetric form.
  • 44:52 - 44:55
    I make it parametric
    by saying equal to t.
  • 44:55 - 44:59
    So what were the
    parametric equations?
  • 44:59 - 45:02
    x equals lt plus x0.
  • 45:02 - 45:03
    That's the normal.
  • 45:03 - 45:06
    y equals mt plus y0.
  • 45:06 - 45:09
    z equals nt plus z0.
  • 45:09 - 45:14
    So finally, my answer-- I'll
    check with [INAUDIBLE] answer
  • 45:14 - 45:20
    in a second-- should be take the
    normal from the tangent plane,
  • 45:20 - 45:25
    3, 2, 2, right?
  • 45:25 - 45:32
    2t plus whatever,
    this is 2t plus z
  • 45:32 - 45:46
    equals-- first it's 3, 3, 2,
    2, 3, 2, 2, plus x0 y0 z0.
  • 45:46 - 45:48
    So erase the pluses and minuses.
  • 45:48 - 45:51
  • 45:51 - 45:52
    And those should
    be the equations.
  • 45:52 - 45:58
    And I should write them down.
  • 45:58 - 46:00
    It's OK, you have the same.
  • 46:00 - 46:05
    Now you just have to take
    them, this, this, and that,
  • 46:05 - 46:09
    and put it in this form.
  • 46:09 - 46:11
    This is a combination problem.
  • 46:11 - 46:13
    Why do I say
    combination problem?
  • 46:13 - 46:17
    It's combining Chapter
    11 with Chapter 9.
  • 46:17 - 46:19
    This was the review
    from Chapter 9.
  • 46:19 - 46:22
  • 46:22 - 46:23
    Did you have trouble
    understanding
  • 46:23 - 46:27
    the radiant or
    the tangent planes
  • 46:27 - 46:30
    or anything like that,
    implicit form, explicit form?
  • 46:30 - 46:37
    Let me do an application,
    since I'm doing review anyway.
  • 46:37 - 46:40
    I'm done with the Section 11.6.
  • 46:40 - 46:42
    But before I want
    to go further, I
  • 46:42 - 46:44
    want to do some
    review of Chapter
  • 46:44 - 47:01
    11 sections 11.1 through 11.6.
  • 47:01 - 47:04
    I said something about
    implicit differentiation
  • 47:04 - 47:07
    being a headache
    for many of you.
  • 47:07 - 47:19
    One person asked me, how do
    you compute z sub x and/or z
  • 47:19 - 47:32
    sub y based on the equation x
    squared plus y squared plus z
  • 47:32 - 47:33
    squared equals 5?
  • 47:33 - 47:36
  • 47:36 - 47:38
    And of course this is
    implicit differentiation.
  • 47:38 - 47:46
  • 47:46 - 47:48
    Why implicit?
  • 47:48 - 47:56
    OK, because this is an implicit
    equation of the type F of x, y,
  • 47:56 - 48:00
    z equals constant.
  • 48:00 - 48:02
    When do we call it explicit?
  • 48:02 - 48:06
    When one of the
    variables, x or y or z,
  • 48:06 - 48:10
    is given explicitly in
    terms of the other two.
  • 48:10 - 48:15
    So if this would be-- well,
    here it's hard to pull it out.
  • 48:15 - 48:18
    But whether it be upper
    part then lower hemisphere,
  • 48:18 - 48:20
    z would be plus or minus.
  • 48:20 - 48:24
    So you have two caps,
    two hemispheres,
  • 48:24 - 48:28
    plus/minus square root 5 minus
    x squared minus y squared.
  • 48:28 - 48:30
    Well, that's two functions.
  • 48:30 - 48:32
    We don't like that.
  • 48:32 - 48:34
    We want to be able to do
    everything in one shot
  • 48:34 - 48:38
    without splitting it into
    two different graphs.
  • 48:38 - 48:43
    So how do we view z
    to be a function of x,
  • 48:43 - 48:45
    you're going to ask yourself.
  • 48:45 - 48:49
    You imagine inside
    this thing that x and y
  • 48:49 - 48:54
    are independent variables--
    independent variables.
  • 48:54 - 48:56
    They can take
    whatever they want.
  • 48:56 - 48:58
    One is temperature.
  • 48:58 - 48:58
    One is time.
  • 48:58 - 49:01
    They run like crazies.
  • 49:01 - 49:05
    But z depends on both
    temperature and time, like us
  • 49:05 - 49:06
    unfortunately.
  • 49:06 - 49:07
    It's so cold outside.
  • 49:07 - 49:09
    I hate it.
  • 49:09 - 49:12
    OK, you promised me,
    and it came true.
  • 49:12 - 49:14
    Who promised me?
  • 49:14 - 49:15
    Matthew, I give you a
    brownie point for that.
  • 49:15 - 49:18
    Because you said last week
    it's going to be 80 degrees.
  • 49:18 - 49:19
    And it was.
  • 49:19 - 49:21
    So the prophecy came true.
  • 49:21 - 49:25
    On the other hand,
    it came back too bad.
  • 49:25 - 49:27
    And of course it's
    not Matthew's fault.
  • 49:27 - 49:31
    He didn't say what's
    going to happen this week.
  • 49:31 - 49:36
    All right, in this case,
    implicit differentiation
  • 49:36 - 49:39
    is just a philosophical thing.
  • 49:39 - 49:42
    It's a very important
    philosophical step
  • 49:42 - 49:44
    that you're taking-- think.
  • 49:44 - 49:48
  • 49:48 - 49:56
    Think of z being a
    function of x and y.
  • 49:56 - 50:05
    And two, differentiate
    z with respect to x.
  • 50:05 - 50:10
  • 50:10 - 50:14
    So what do you
    mean, differentiate
  • 50:14 - 50:15
    with respect to x?
  • 50:15 - 50:21
    By differentiating
    the entire equation,
  • 50:21 - 50:31
    both sides of an equation
    with respect to x.
  • 50:31 - 50:35
    So for you, x is
    the wanted variable.
  • 50:35 - 50:37
    y is like a constant.
  • 50:37 - 50:42
    z is a function of x
    is not hard at all.
  • 50:42 - 50:48
    So what is going to happen
    actually if you were to do it?
  • 50:48 - 50:50
    Theoretically, you
    would go like that.
  • 50:50 - 50:54
    If I'm going to differentiate
    this guy with respect to x,
  • 50:54 - 50:56
    what is the philosophy?
  • 50:56 - 51:00
    The chain rule tells
    me, differentiate F
  • 51:00 - 51:06
    with respect to the first
    variable, and then times dx/dx.
  • 51:06 - 51:10
    And you say, god, now
    that was silly, right?
  • 51:10 - 51:12
    Differentiate with respect to x.
  • 51:12 - 51:14
    That's the chain rule.
  • 51:14 - 51:17
    Plus differentiate
    F with respect
  • 51:17 - 51:22
    to the second place,
    second variable, and then
  • 51:22 - 51:28
    say, dy with respect to x.
  • 51:28 - 51:30
    But are dx and dy married?
  • 51:30 - 51:32
    Do they depend on one another?
  • 51:32 - 51:35
    Do they file an income
    tax return together?
  • 51:35 - 51:38
    They don't want to have
    anything to do with one another.
  • 51:38 - 51:42
    Thank god, so x and y are
    independent variables.
  • 51:42 - 51:46
    If you're taking
    statistics or researching
  • 51:46 - 51:48
    any other kind of
    physics, chemistry,
  • 51:48 - 51:51
    you know that these are
    called independent variables,
  • 51:51 - 51:54
    and this is called the
    dependent variable.
  • 51:54 - 51:57
    And then you have what's
    called the constraint.
  • 51:57 - 52:02
    In physics and engineering and
    mechanics, F of some variable
  • 52:02 - 52:05
    equals C. It's called
    constraint usually.
  • 52:05 - 52:09
    OK, so this guy is all silly.
  • 52:09 - 52:13
    These guys don't want to have
    to do anything with one another.
  • 52:13 - 52:15
    And then you get plus.
  • 52:15 - 52:21
    Finally, dF with respect
    to the third place,
  • 52:21 - 52:25
    and then that third place, z,
    is occupied by a function that's
  • 52:25 - 52:26
    a function of x.
  • 52:26 - 52:28
    So you go, dz/dx.
  • 52:28 - 52:31
    Why del and not d?
  • 52:31 - 52:36
    Because poor z is a function
    of two variables, x and y.
  • 52:36 - 52:38
    So you cannot say, dz/dx.
  • 52:38 - 52:41
    You have to say
    del z dx, equals 0.
  • 52:41 - 52:46
    Thank god, I got to the
    end where I wanted to get.
  • 52:46 - 52:52
    Now, if I want to see what's
    going on, it's a piece of cake.
  • 52:52 - 52:54
    That's 1.
  • 52:54 - 52:59
    And I get that Mr. z sub x,
    which other people write dz/dx,
  • 52:59 - 53:06
    but I don't, because I don't
    like it-- I keep mixing x.
  • 53:06 - 53:12
    Equals-- how do I
    pull this guy out?
  • 53:12 - 53:14
    How do I substitute for that?
  • 53:14 - 53:22
    I get Mr. First Fellow
    here to the other side.
  • 53:22 - 53:28
    He's going to pick up
    a minus at whatever d
  • 53:28 - 53:35
    I have divided by-- so this
    guy divided by this guy.
  • 53:35 - 53:46
  • 53:46 - 53:51
    STUDENT: What happened
    to dF/dy, dy/dx?
  • 53:51 - 53:53
    PROFESSOR: So again, that
    is a very good thing.
  • 53:53 - 53:55
    So dF/dy was behaving.
  • 53:55 - 53:56
    He was nice.
  • 53:56 - 54:00
    But when we got to dy with
    respect to dx, y said,
  • 54:00 - 54:03
    I'm not married to dx.
  • 54:03 - 54:05
    I have nothing to do with dx.
  • 54:05 - 54:07
    I'm independent from this.
  • 54:07 - 54:10
    So dy/dx is 0.
  • 54:10 - 54:15
    And so this guy
    disappears. dx/dx is 1.
  • 54:15 - 54:17
    Duh, that's a piece of cake.
  • 54:17 - 54:18
    So I'm done.
  • 54:18 - 54:23
    This is actually a formula
    that looks sort of easy.
  • 54:23 - 54:26
    But there is a lot
    hidden behind it.
  • 54:26 - 54:29
    This is the implicit
    function theorem.
  • 54:29 - 54:31
  • 54:31 - 54:35
    where you of course assume
    that these partial derivatives
  • 54:35 - 54:38
    exist, are continuous,
    everything is nice.
  • 54:38 - 54:41
    It's a beautiful result.
    People actually get
  • 54:41 - 54:46
    to learn it only when they are
    big, I mean big mathematically,
  • 54:46 - 54:49
    mature, in graduate school,
    first or second year
  • 54:49 - 54:50
    of graduate school.
  • 54:50 - 54:52
    We call that
    intermediate analysis
  • 54:52 - 54:55
    or advanced-- very
    advanced-- calculus.
  • 54:55 - 54:57
    Because this calculus
    is advanced enough,
  • 54:57 - 55:00
    but I'm talking about
    graduate level calculus.
  • 55:00 - 55:03
    And this is the so-called
    implicit function theorem.
  • 55:03 - 55:08
    So if you will ever be even not
    necessarily a graduate student
  • 55:08 - 55:12
    in mathematics, but a graduate
    student in physics or something
  • 55:12 - 55:16
    related to pure science,
    remember this result. So let me
  • 55:16 - 55:18
    see what's going to
    happen in practice.
  • 55:18 - 55:23
    In practice, do we
    have to learn this?
  • 55:23 - 55:27
    No, in practice we can build
    everything from scratch, again,
  • 55:27 - 55:32
    just the way we did
    it with the formula.
  • 55:32 - 55:34
    So for the example
    I gave you, it
  • 55:34 - 55:37
    should be a piece of cake
    to do the differentiation.
  • 55:37 - 55:39
    But I'm going to step by step.
  • 55:39 - 55:42
    Step one, think.
  • 55:42 - 55:46
  • 55:46 - 55:47
    You have to think.
  • 55:47 - 55:50
    If you don't think,
    you cannot do math.
  • 55:50 - 55:53
    So you have x squared
    plus y squared,
  • 55:53 - 55:57
    the independent guys, and z,
    who is married to both of them.
  • 55:57 - 55:59
    Or maybe z is the baby.
  • 55:59 - 56:02
    These are two spouses that are
    independent from one another.
  • 56:02 - 56:04
    And z is their baby.
  • 56:04 - 56:06
    Because he depends
    on both of them.
  • 56:06 - 56:10
  • 56:10 - 56:12
    So you thought you had
    a different approach
  • 56:12 - 56:17
    to the problem, different
    vision of what's going on.
  • 56:17 - 56:23
    Now finally, step two,
    differentiate with respect
  • 56:23 - 56:25
    to one only, x only.
  • 56:25 - 56:29
  • 56:29 - 56:32
    You could of course do the
    same process with respect to y.
  • 56:32 - 56:35
    And in some of the
    final exam problems,
  • 56:35 - 56:38
    we are asking, compute
    z sub x and z sub y.
  • 56:38 - 56:40
    The secret is that--
    maybe I shouldn't
  • 56:40 - 56:45
    talk too much again-- when
    I grade those finals, if you
  • 56:45 - 56:48
    do z sub x, I give you 100%.
  • 56:48 - 56:51
    Because z sub y is the same.
  • 56:51 - 56:54
    So I really don't care.
  • 56:54 - 56:57
    Sometimes there
    are so many things
  • 56:57 - 57:01
    to do that all I care
    is, did he or she cover
  • 57:01 - 57:03
    the essential work?
  • 57:03 - 57:09
    So with respect to x, x squared
    differentiated with respect
  • 57:09 - 57:11
    to x-- 2x.
  • 57:11 - 57:16
    Good first step, now, y squared
    differentiated with respect
  • 57:16 - 57:18
    to x.
  • 57:18 - 57:19
    0-- am I going to write 0?
  • 57:19 - 57:21
    Yes, because I'm silly.
  • 57:21 - 57:25
    But I don't have to.
  • 57:25 - 57:28
    2 times z of x, y.
  • 57:28 - 57:32
  • 57:32 - 57:36
    2 jumps down, z of x, y-- I'm
    not done with the chain rule.
  • 57:36 - 57:40
  • 57:40 - 57:41
    STUDENT: z sub x.
  • 57:41 - 57:43
    PROFESSOR: It's z
    sub x, very good.
  • 57:43 - 57:44
    This is dz/dx.
  • 57:44 - 57:48
    I'm not going to hide
    it completely like that.
  • 57:48 - 57:50
    That is the same thing.
  • 57:50 - 57:52
    y prime is 0, thank god.
  • 57:52 - 57:56
  • 57:56 - 58:00
    So you say, if I were to
    keep in mind that that's
  • 58:00 - 58:06
    the derivative of big
    F with respect to x,
  • 58:06 - 58:09
    I could plug in
    everything in here.
  • 58:09 - 58:10
    I could plug in the formula.
  • 58:10 - 58:12
    But why memorize the
    formula and plug it
  • 58:12 - 58:16
    in when you can do everything
    from scratch all over again?
  • 58:16 - 58:19
    Math is not about memorization.
  • 58:19 - 58:23
    If you are good, for example,
    some people here-- I'm
  • 58:23 - 58:28
    not going to name them--
    are in sciences that involve
  • 58:28 - 58:30
    a lot of memorization.
  • 58:30 - 58:32
    More power to them.
  • 58:32 - 58:34
    I was not very good at that.
  • 58:34 - 58:38
    So I'm going to go ahead and
    write z sub x pulled down
  • 58:38 - 58:43
    minus 2x divided by 2z.
  • 58:43 - 58:47
    I'm too lazy to remind
    you that z is the baby,
  • 58:47 - 58:50
    and he depends on
    his parents x and y.
  • 58:50 - 58:52
    I'm not going to write that.
  • 58:52 - 58:54
    And that's the answer.
  • 58:54 - 58:58
    So you have minus x/z.
  • 58:58 - 59:04
    So for example, if somebody
    says, compute z sub x
  • 59:04 - 59:17
    at the point on the sphere,
    that is 0, root 5, and 0,
  • 59:17 - 59:19
    what do you have to do?
  • 59:19 - 59:24
    You have to say,
    z sub x equals--
  • 59:24 - 59:31
    and now I'm asking you
    something that is minus 0/0.
  • 59:31 - 59:36
  • 59:36 - 59:49
    Assuming that the expressions,
    the derivatives, are defined
  • 59:49 - 59:57
    and the denominator one
    is different from 0-- so
  • 59:57 - 60:00
    whenever you do the
    implicit function theorem,
  • 60:00 - 60:05
    you can apply with the
    condition that you are away
  • 60:05 - 60:10
    from points where derivative
    of F with respect to z are 0.
  • 60:10 - 60:13
    So this is a problem
    that's not well posed.
  • 60:13 - 60:16
    So to give you a
    well-posed problem, what
  • 60:16 - 60:18
    do I need to do on the final?
  • 60:18 - 60:24
    I have to say the
    same-- 2, 1, and 0.
  • 60:24 - 60:27
  • 60:27 - 60:28
    STUDENT: z can't be 0.
  • 60:28 - 60:30
    PROFESSOR: No, I know.
  • 60:30 - 60:35
    So I go, z is 0 is too easy.
  • 60:35 - 60:36
    Let's have y to be 0.
  • 60:36 - 60:37
    STUDENT: 2, 0, 1.
  • 60:37 - 60:40
    PROFESSOR: Very good, x equals
    2, z equals 1, excellent.
  • 60:40 - 60:48
    So z sub x at the
    point 2, 0, 1 will
  • 60:48 - 60:55
    be by the implicit
    function theorem minus 2/1
  • 60:55 - 60:56
    equals negative.
  • 60:56 - 61:00
    You see, that's a slope
    in a certain direction
  • 61:00 - 61:06
    if you were to look at z with
    respect to x in the plane x, z.
  • 61:06 - 61:09
    OK, what else?
  • 61:09 - 61:13
    Nothing-- that was review
    of chain rule and stuff.
  • 61:13 - 61:15
    And you have to
    review chain rule.
  • 61:15 - 61:18
  • 61:18 - 61:20
    Make yourself a note.
  • 61:20 - 61:23
    Before the midterm, I have
    to memorize the chain rule.
  • 61:23 - 61:24
    Yes, sir.
  • 61:24 - 61:25
    STUDENT: [INAUDIBLE].
  • 61:25 - 61:28
  • 61:28 - 61:32
    PROFESSOR: I will do that either
    in the review session today
  • 61:32 - 61:37
    or in the review
    for the midterm, OK?
  • 61:37 - 61:40
    And I'm thinking about that.
  • 61:40 - 61:44
    In March, I want to
    dedicate at least 10 days
  • 61:44 - 61:46
    for the review for the midterm.
  • 61:46 - 61:46
    Yes, sir.
  • 61:46 - 61:48
    STUDENT: When is the midterm?
  • 61:48 - 61:50
    PROFESSOR: The midterm
    is on the 2nd of April.
  • 61:50 - 61:55
  • 61:55 - 61:58
    Several people asked me--
    OK, I forgot about that.
  • 61:58 - 61:59
    I have to tell you guys.
  • 61:59 - 62:01
    Several people
    asked me questions
  • 62:01 - 62:04
    by email about the midterm.
  • 62:04 - 62:07
    So the midterm-- write
    down for yourselves-- will
  • 62:07 - 62:09
    be over the following chapters.
  • 62:09 - 62:11
  • 62:11 - 62:21
    Chapter 10, no Chapter 9.
  • 62:21 - 62:23
    Chapter 9 is [INAUDIBLE].
  • 62:23 - 62:31
    Chapter 11 all,
    Chapter 10 only what
  • 62:31 - 62:37
    we have required-- 10.1, 10.2,
    and 10.3-- and Chapter 12,
  • 62:37 - 62:44
    all but Section 12.6.
  • 62:44 - 62:47
    Because I see that some of
    you study ahead of time.
  • 62:47 - 62:48
    More power to you.
  • 62:48 - 62:50
    You know what to read.
  • 62:50 - 62:51
    Skip Section 12.6.
  • 62:51 - 62:57
    And I'm planning to not give
    you anything after Chapter 13
  • 62:57 - 62:58
    on the midterm.
  • 62:58 - 63:02
    But of course, Chapter 13 will
    be on the final emphasized
  • 63:02 - 63:07
    in at least six problems
    out of the 15 problems
  • 63:07 - 63:11
    you'll have on the
    final, all right?
  • 63:11 - 63:15
    We still have plenty of time.
  • 63:15 - 63:18
    Chapter 9, guys, you
    were concerned about it.
  • 63:18 - 63:22
    It's some sort of
    embedded, you see?
  • 63:22 - 63:23
    Wherever you go,
    wherever you turn,
  • 63:23 - 63:27
    you bump into some parametric
    equations of a line
  • 63:27 - 63:28
    or bump into a tangent line.
  • 63:28 - 63:34
    That's the dot product that
    you dealt with, delta F dot N.
  • 63:34 - 63:38
    So it's like an obsession,
    repetitive review of Chapter 9
  • 63:38 - 63:40
    at ever step.
  • 63:40 - 63:41
    Vector spaces are
    very important.
  • 63:41 - 63:45
    Vectors in general
    are very important.
  • 63:45 - 63:48
    I'm going to move
    onto 11.7 right now.
  • 63:48 - 63:49
    We'll take a break.
  • 63:49 - 63:53
    Why don't we take a short
    break now, five minutes.
  • 63:53 - 63:59
    And then we have to
    go on until 2:50.
  • 63:59 - 64:02
    So practically we
    have one more hour.
  • 64:02 - 64:04
    Take a break, eat,
    drink something.
  • 64:04 - 64:05
    I don't want a big break.
  • 64:05 - 64:08
    Because then a big break
    we'll just fall asleep.
  • 64:08 - 64:11
    I'm tired as well.
  • 64:11 - 64:13
    So we have to keep going.
  • 64:13 - 64:16
  • 64:16 - 64:20
    [BACKGROUND CHATTER]
  • 64:20 - 68:40
  • 68:40 - 68:41
    PROFESSOR: All right.
  • 68:41 - 68:44
    I will start with a
    little bit of a review
  • 68:44 - 68:46
    of some friend of yours.
  • 68:46 - 68:49
    And since we are
    in Texas, of course
  • 68:49 - 68:51
    this counts as an obsession.
  • 68:51 - 68:54
  • 68:54 - 69:00
    This is going to be extrema of
    functions of several variables.
  • 69:00 - 69:15
  • 69:15 - 69:17
    Do I draw better lately?
  • 69:17 - 69:20
    I think I do.
  • 69:20 - 69:22
    That's why I stopped
    drinking coffee.
  • 69:22 - 69:24
    I'm drinking white tea.
  • 69:24 - 69:26
    It's good for you.
  • 69:26 - 69:28
    All right.
  • 69:28 - 69:28
    White tea.
  • 69:28 - 69:31
    For some reason, the
    black tea was giving me
  • 69:31 - 69:33
    the shaking and all that.
  • 69:33 - 69:35
    Too much black tea.
  • 69:35 - 69:38
    I don't know, maybe
    it has less caffeine.
  • 69:38 - 69:40
    Jasmine is good,
    green, or white.
  • 69:40 - 69:42
    STUDENT: I think green
    has less [INAUDIBLE].
  • 69:42 - 69:44
    PROFESSOR: OK.
  • 69:44 - 69:47
    So above this
    saddle is a function
  • 69:47 - 69:52
    of two variables-- you
    know a lot already,
  • 69:52 - 69:57
    but I'm asking you to compute
    the partial derivatives
  • 69:57 - 70:00
    and the gradient.
  • 70:00 - 70:01
    And you're going to
    jump on it and say
  • 70:01 - 70:04
    I'm doing [INAUDIBLE] anyway.
  • 70:04 - 70:09
    So I've got 2x, and
    this is minus 2y.
  • 70:09 - 70:11
    If I want to ask
    you the differential
  • 70:11 - 70:14
    on the final or
    midterm, you will say
  • 70:14 - 70:19
    that f sub xdx plus x of ygy.
  • 70:19 - 70:22
    Everybody knows that.
  • 70:22 - 70:24
    Don't break my heart.
  • 70:24 - 70:28
    Don't say 2x minus y,
    because I'll never recover.
  • 70:28 - 70:31
    Every time I see that,
    I die 100 deaths.
  • 70:31 - 70:35
    So don't forget about
    the x and the y,
  • 70:35 - 70:40
    which are the important guys
    of infinitesimal elements.
  • 70:40 - 70:42
    This is a 1 form.
  • 70:42 - 70:48
    In mathematics, any
    combination of a dx and dy
  • 70:48 - 70:51
    in a linear combination
    in the 1 form.
  • 70:51 - 70:55
    It's a consecrated terminology.
  • 70:55 - 70:56
    But I'm not asking you
    about the differential.
  • 70:56 - 71:00
    I'm asking you
    about the gradient.
  • 71:00 - 71:07
    All righty, and that is a f
    sub xi plus f sub yk, which
  • 71:07 - 71:11
    is exactly 2xi minus 2yj.
  • 71:11 - 71:15
  • 71:15 - 71:18
    And you say all right,
    but I want to take a look,
  • 71:18 - 71:20
    I always have started
    with examples.
  • 71:20 - 71:23
    Hopefully they are good.
  • 71:23 - 71:27
    Let's look at the tangent
    vectors to the surface.
  • 71:27 - 71:30
    We discussed about the
    notion of tangent vector
  • 71:30 - 71:34
    before, remember, when
    we had r sub u and r sub
  • 71:34 - 71:36
    v form the parametrization.
  • 71:36 - 71:39
    Now look at the tangent
    vectors for this graph
  • 71:39 - 71:43
    along the x direction
    going this way,
  • 71:43 - 71:47
    and along the y
    direction going this way.
  • 71:47 - 71:52
    We see that both of them are
    horizontal at the origin.
  • 71:52 - 71:55
    And that's a beautiful thing.
  • 71:55 - 72:01
    And so this origin is a
    so-called critical point.
  • 72:01 - 72:05
    Critical point for a
    differentiable function.
  • 72:05 - 72:13
  • 72:13 - 72:24
    Z equals f of xy is
    a point in plane x0i0
  • 72:24 - 72:31
    where the partial
    derivatives vanish.
  • 72:31 - 72:36
  • 72:36 - 72:41
    And according to the book, and
    many books, all don't exist.
  • 72:41 - 72:45
    Well I don't like that.
  • 72:45 - 72:53
    Even our book says if you
    have a function in calc 1--
  • 72:53 - 73:02
    let's say b equal g
    of u, critical point.
  • 73:02 - 73:05
    Do you remember what
    a critical point was?
  • 73:05 - 73:12
    U0, in calc 1 we said either a
    point where g prime of u was 0,
  • 73:12 - 73:17
    or g prime of u 0 doesn't exist.
  • 73:17 - 73:19
  • 73:19 - 73:23
    Although u is 0
    is in the domain.
  • 73:23 - 73:24
    I don't like that.
  • 73:24 - 73:27
    You say wait a minute,
    why don't you like that?
  • 73:27 - 73:30
    I don't like that for
    many reasons practically.
  • 73:30 - 73:38
    If you have the
    absolute value function,
  • 73:38 - 73:42
    you'll say yeah, yeah, but
    look, I considered the corner
  • 73:42 - 73:44
    to be a point of
    non-differentiability,
  • 73:44 - 73:50
    but it's still an extreme
    value, a critical point.
  • 73:50 - 73:54
    According to our book
    in Calculus 1, yeah.
  • 73:54 - 73:57
    We extended this
    definition to ugly points,
  • 73:57 - 74:00
    points where you don't have
    a [? pick ?] or a value
  • 74:00 - 74:05
    or an inflection, but you
    have something ugly like
  • 74:05 - 74:09
    a [? cusp, ?] a
    corner, the ugliness.
  • 74:09 - 74:10
    I don't like that
    kind of ugliness,
  • 74:10 - 74:15
    because I want to have
    more information there.
  • 74:15 - 74:20
    I maybe even have a point with
    a bigger problem than that.
  • 74:20 - 74:22
    First of all, when I
    say critical point,
  • 74:22 - 74:25
    I have to assume the point is
    in the domain of the function.
  • 74:25 - 74:28
    But then what kind of
    ugliness I can have there?
  • 74:28 - 74:30
    I don't even want
    to think about it.
  • 74:30 - 74:35
    So in the context of
    my class-- in context
  • 74:35 - 74:47
    of my class-- calc 3 honors, I
    will denote a critical point.
  • 74:47 - 74:51
  • 74:51 - 75:01
    Is the x0y0 such that
    f sub x at x0y0 is 0.
  • 75:01 - 75:05
    One slope is 0, the
    other slope is 0.
  • 75:05 - 75:08
    f sub y is x0y0, of course.
  • 75:08 - 75:14
    And no other are the points.
  • 75:14 - 75:18
    What am I going to call the
    [INAUDIBLE] points where
  • 75:18 - 75:24
    derivatives don't exist?
  • 75:24 - 75:31
    I simply say I
    have a singularity.
  • 75:31 - 75:34
    I have a singularity.
  • 75:34 - 75:38
    What type of singularity we can
    discuss in an advanced calculus
  • 75:38 - 75:39
    setting.
  • 75:39 - 75:42
    If you're math majors, you're
    going to have the chance
  • 75:42 - 75:44
    to discuss that later on.
  • 75:44 - 75:49
    So remember that I would
    prefer both in the context
  • 75:49 - 75:55
    of calculus 1 and calculus
    3 to say critical value
  • 75:55 - 75:59
    is where the derivative
    becomes zero.
  • 75:59 - 76:03
    Not undefined, plus, minus,
    infinity, or something
  • 76:03 - 76:06
    really crazy, one on the
    left, one on the right.
  • 76:06 - 76:10
    So I don't want to have
    any kind of complications.
  • 76:10 - 76:15
    Now you may say, but I
    thought that since you
  • 76:15 - 76:17
    have those slopes
    both zero, that
  • 76:17 - 76:21
    means that the tangent plane
    at the point is horizontal.
  • 76:21 - 76:23
    And that's exactly what it is.
  • 76:23 - 76:24
    I agree with you.
  • 76:24 - 76:27
    If somebody would draw the
    tangent plane to the surface,
  • 76:27 - 76:30
    S-- S is for surface,
    but it's funny,
  • 76:30 - 76:32
    S is also coming from saddles.
  • 76:32 - 76:37
    So that's a saddle
    point, saddle surface.
  • 76:37 - 76:39
    Origin is so-called
    saddle point.
  • 76:39 - 76:40
    We don't know yet why.
  • 76:40 - 76:43
  • 76:43 - 76:49
    The tangent plane
    at 0, at the origin,
  • 76:49 - 76:53
    will be 0.0 in this case.
  • 76:53 - 76:54
    Why?
  • 76:54 - 76:56
    Well, it's easy to see.
  • 76:56 - 77:01
    z minus 0 equals f sub x,
    x minus x0 plus f sub y,
  • 77:01 - 77:03
    y minus y0.
  • 77:03 - 77:06
    But this is 0 and that's
    0, so z equals zero.
  • 77:06 - 77:08
    So thank you very much.
  • 77:08 - 77:09
    Poor horse.
  • 77:09 - 77:15
    I can take a horizontal
    plane, imaginary plane
  • 77:15 - 77:22
    and make it be tangent to
    the saddle in all directions
  • 77:22 - 77:24
    at the point in the middle.
  • 77:24 - 77:30
  • 77:30 - 77:31
    All right.
  • 77:31 - 77:33
    STUDENT: So you're saying
    [? the critical ?] point
  • 77:33 - 77:33
    is where both--
  • 77:33 - 77:36
    PROFESSOR: Where both
    partial derivatives vanish.
  • 77:36 - 77:38
    They have to both vanish.
  • 77:38 - 77:40
    In case of calculus
    1, of course there
  • 77:40 - 77:45
    is only one derivative that
    vanishes at that point.
  • 77:45 - 77:47
    What if I were
    in-- now, you see,
  • 77:47 - 77:50
    the more you ask me
    questions, the more I think
  • 77:50 - 77:52
    And it's a dangerous thing.
  • 77:52 - 77:56
    What if I had z equals
    f of x1, x2, x3, xn?
  • 77:56 - 77:59
    Critical point would be where
    all the partial derivatives
  • 77:59 - 78:00
    will be zero.
  • 78:00 - 78:03
    And then the situation
    becomes more complicated,
  • 78:03 - 78:05
    but it's doable.
  • 78:05 - 78:11
    The other is the classification
    of special points.
  • 78:11 - 78:25
    Classification of
    critical points
  • 78:25 - 78:31
    based on second
    partial derivatives.
  • 78:31 - 78:40
  • 78:40 - 78:45
    The objects you want to study
    in this case are several.
  • 78:45 - 78:48
  • 78:48 - 78:53
    One of the most important ones
    is the so-called discriminant.
  • 78:53 - 78:55
    What is the discriminant?
  • 78:55 - 78:59
    You haven't talked about
    discriminants since a long time
  • 78:59 - 78:59
    ago.
  • 78:59 - 79:02
    And there is a relationship
    between discriminant
  • 79:02 - 79:09
    in high school algebra and
    discriminant in calculus 3.
  • 79:09 - 79:11
    The discriminant
    the way we define
  • 79:11 - 79:14
    it is D, or delta--
    some people denote it
  • 79:14 - 79:17
    like this, some
    people by delta--
  • 79:17 - 79:19
    and that is the following.
  • 79:19 - 79:22
    This is the determinant.
  • 79:22 - 79:30
    f sub xx, f sub xy,
    f sub yx, f sub yy,
  • 79:30 - 79:32
    computed at the point
    p0, which is critical.
  • 79:32 - 79:36
  • 79:36 - 79:40
    So p0 first has to satisfy
    those two equations,
  • 79:40 - 79:43
    and then I'm going to have
    to compute the [INAUDIBLE]
  • 79:43 - 79:45
    at that point.
  • 79:45 - 79:46
    But you say wait a
    minute, Magdalena,
  • 79:46 - 79:48
    what the heck is this?
  • 79:48 - 79:51
    Well this is the second
    partial with respect
  • 79:51 - 79:55
    of x, one after the other,
    second partial with respect
  • 79:55 - 79:57
    to y, one after the other.
  • 79:57 - 79:58
    These guys are equal.
  • 79:58 - 80:01
    Remember that there was
    a German mathematician
  • 80:01 - 80:07
    whose name was Schwartz, the
    black cavalier, the black man.
  • 80:07 - 80:09
    Schwartz means black in German.
  • 80:09 - 80:10
    And he came up with
    this theorem that it
  • 80:10 - 80:15
    doesn't matter in which order
    you differentiate, f sub xy
  • 80:15 - 80:19
    or f sub yx is the same thing as
    long as the function is smooth.
  • 80:19 - 80:22
    So I'm very happy about that.
  • 80:22 - 80:26
    Now there are these
    other guys, A, B,
  • 80:26 - 80:31
    C. It's very easy to
    remember, it's from the song
  • 80:31 - 80:33
    that you all learned
    in kindergarten.
  • 80:33 - 80:37
    Once you know your ABC, you
    come back to the discriminant.
  • 80:37 - 80:45
    So f sub xx at the point p0,
    f sub xy at the point p0,
  • 80:45 - 80:50
    and f sub yy at the point p0.
  • 80:50 - 80:53
    Second partial with respect to
    x, second partial with respect
  • 80:53 - 80:57
    to x and y, mixed one,
    mixed derivative, and second
  • 80:57 - 80:58
    partial with respect to y.
  • 80:58 - 81:01
  • 81:01 - 81:07
    You have to plug in the values
    for the p0 will be x0, y0.
  • 81:07 - 81:10
    The critical point
    you got from what?
  • 81:10 - 81:13
    From solving this system.
  • 81:13 - 81:16
    So you got x0y0 by
    solving that system.
  • 81:16 - 81:22
    Come back, plug in, compute
    those, get ABC as numbers.
  • 81:22 - 81:26
    And who is D going
    to be based on ABC?
  • 81:26 - 81:30
    According to the diagram that
    I drew, it's easy for you guys
  • 81:30 - 81:36
    to see that A and
    B and C are what?
  • 81:36 - 81:46
    Related to D. So D will
    simply be A, B, B, and C,
  • 81:46 - 81:48
    computed at the point p0.
  • 81:48 - 81:51
  • 81:51 - 81:54
    So it's going to
    be now-- now that
  • 81:54 - 81:57
    will remind you of something.
  • 81:57 - 81:58
    AC minus B-squared.
  • 81:58 - 82:03
  • 82:03 - 82:04
    OK?
  • 82:04 - 82:07
    When we had the quadratic
    formula in school--
  • 82:07 - 82:09
    I'm not going to write it.
  • 82:09 - 82:10
    I'm going to write it here.
  • 82:10 - 82:12
    So what was the
    quadratic formula?
  • 82:12 - 82:15
    ax-squared plus bx
    plus c equals 0.
  • 82:15 - 82:17
    That was algebra.
  • 82:17 - 82:18
    Baby algebra.
  • 82:18 - 82:19
    What do we call that?
  • 82:19 - 82:21
    High school algebra?
  • 82:21 - 82:28
    x12 plus minus b plus minus
    square root of b-squared minus
  • 82:28 - 82:34
    4ac divided by 2a.
  • 82:34 - 82:37
    Now don't don;t know what
    kind of professors you had.
  • 82:37 - 82:42
    But I had a teacher when
    I was in high school.
  • 82:42 - 82:45
    Every time she taught me
    something and I did not
  • 82:45 - 82:47
    absorb it, she was all over me.
  • 82:47 - 82:49
    She was preparing me for
    some math competitions,
  • 82:49 - 82:51
    and she taught me a trick.
  • 82:51 - 82:54
    She said look,
    Magdalena, pay attention.
  • 82:54 - 83:01
    If b would be an even number--
    take b to be 2b prime,
  • 83:01 - 83:04
    2-- give me another letter.
  • 83:04 - 83:08
    2 big B. Right?
  • 83:08 - 83:12
    Then, the quadratic formula
    would be easier to use.
  • 83:12 - 83:18
    Because in that case, you get
    x1 2 equals minus-- b is 2b.
  • 83:18 - 83:21
    So you have just 2b like that.
  • 83:21 - 83:25
    Plus minus square
    root 4b squared
  • 83:25 - 83:29
    minus 4ac divided by 2a.
  • 83:29 - 83:31
    She explained this
    to me once and then
  • 83:31 - 83:35
    she expected me to remember
    it for the rest of my life.
  • 83:35 - 83:42
    And then she said minus big B
    plus minus square root of bb
  • 83:42 - 83:44
    squared minus ac.
  • 83:44 - 83:45
    Do you see why?
  • 83:45 - 83:50
    It's because you pull out the
    factor of 4, square root of 4
  • 83:50 - 83:51
    is 2.
  • 83:51 - 83:54
    2, 2, and 2 simplify.
  • 83:54 - 83:56
    And then she gave me
    to solve problems.
  • 83:56 - 83:57
    STUDENT: What about the a?
  • 83:57 - 83:59
    STUDENT: How about the a?
  • 83:59 - 84:00
    STUDENT: Because you
    divide it by [INAUDIBLE].
  • 84:00 - 84:01
    PROFESSOR: Divided by.
  • 84:01 - 84:04
    I forgot to write it down.
  • 84:04 - 84:06
    Because I didn't have space.
  • 84:06 - 84:09
    I said, I'm not going
    to bend and doodle.
  • 84:09 - 84:18
    So when you have x-squared
    plus 2x-- let's say minus 3.
  • 84:18 - 84:20
    And she gave me that.
  • 84:20 - 84:22
    And I said OK, let me do it.
  • 84:22 - 84:22
    Let me do it.
  • 84:22 - 84:31
    x1 2 minus 2 plus minus square
    root b-squared minus 4ac,
  • 84:31 - 84:34
    which is 12, divided by 2.
  • 84:34 - 84:36
    And she started screaming.
  • 84:36 - 84:38
    And she started
    screaming big time.
  • 84:38 - 84:39
    Do you know why?
  • 84:39 - 84:44
    She said, I just told
    you the half formula.
  • 84:44 - 84:48
    By half formula, I mean
    she meant this one.
  • 84:48 - 84:51
    So when-- and I said OK,
    OK, the half formula.
  • 84:51 - 84:54
    But then for maybe
    another seven years,
  • 84:54 - 84:57
    I did this with the
    formula-- with the formula
  • 84:57 - 85:00
    that everybody knows.
  • 85:00 - 85:02
    And at the end, I
    would remember I
  • 85:02 - 85:03
    could have done
    the half formula,
  • 85:03 - 85:08
    but I didn't do it
    because I'm in a routine.
  • 85:08 - 85:12
    So the way she wanted
    me to do this was what?
  • 85:12 - 85:14
    Who is the half of 2?
  • 85:14 - 85:15
    1.
  • 85:15 - 85:21
    So put minus 1 plus minus
    square root of big B
  • 85:21 - 85:27
    is 1-squared minus
    a times c, which
  • 85:27 - 85:31
    is plus 3, divided by
    1 divided by nobody.
  • 85:31 - 85:33
    This way you don't have
    to simplify it further,
  • 85:33 - 85:36
    and you do it faster.
  • 85:36 - 85:46
    So you get minus 1 plus minus
    root 4, which is minus 5 and 3.
  • 85:46 - 85:49
    But of course, you could
    have done this by factoring.
  • 85:49 - 85:52
    So you could have
    said wait a minute.
  • 85:52 - 85:55
    Two numbers that multiply-- um--
  • 85:55 - 85:56
    STUDENT: [INAUDIBLE]
    square root.
  • 85:56 - 85:58
    PROFESSOR: I didn't do right.
  • 85:58 - 85:58
    So it's 4--
  • 85:58 - 86:01
    STUDENT: [INAUDIBLE].
  • 86:01 - 86:02
    PROFESSOR: Yeah.
  • 86:02 - 86:07
    So you get x plus 5 times--
  • 86:07 - 86:09
    STUDENT: It's x minus 1--
  • 86:09 - 86:11
    [INTERPOSING VOICES]
  • 86:11 - 86:12
    PROFESSOR: Oh, I think I--
  • 86:12 - 86:13
    STUDENT: Square root.
  • 86:13 - 86:14
    PROFESSOR: Square root.
  • 86:14 - 86:16
    I'm sorry, guys.
  • 86:16 - 86:16
    OK.
  • 86:16 - 86:18
    Thank you for that.
  • 86:18 - 86:19
    1 and minus 3.
  • 86:19 - 86:25
    So x plus 3 times
    x minus 1, which
  • 86:25 - 86:29
    is the same-- the exact same
    as x-squared plus 2x minus 3
  • 86:29 - 86:30
    equals.
  • 86:30 - 86:31
    All right?
  • 86:31 - 86:37
    So just the way she insisted
    that I learn the half formula.
  • 86:37 - 86:42
    I'm not insisting that you learn
    the half formula, god forbid.
  • 86:42 - 86:46
    But see here there is
    some more symmetry.
  • 86:46 - 86:49
    The four doesn't appear anymore.
  • 86:49 - 86:52
    b-squared minus 4ac appeared
    here, but here it doesn't.
  • 86:52 - 86:55
    Here you're going to
    have b-squared minus ac.
  • 86:55 - 86:56
    There is a reason.
  • 86:56 - 87:00
    This comes from a
    discriminant just like that.
  • 87:00 - 87:02
    And this is why I told
    you the whole secret
  • 87:02 - 87:06
    about the half
    quadratic formula.
  • 87:06 - 87:08
    Not because I wanted
    you to know about it,
  • 87:08 - 87:14
    but because I wanted you to see
    that there is a pattern here.
  • 87:14 - 87:16
    You have-- for the
    half formula, you
  • 87:16 - 87:21
    have plus minus square root
    of a new type of discriminant.
  • 87:21 - 87:24
    People even call this
    discriminant b-squared
  • 87:24 - 87:25
    minus 4ac.
  • 87:25 - 87:27
    b-squared minus ac.
  • 87:27 - 87:31
    So for us, it is
    ac minus b-squared.
  • 87:31 - 87:35
    It's just the opposite of
    that discriminant you have.
  • 87:35 - 87:40
    Now depending on the sign
    of this discriminant,
  • 87:40 - 87:48
    you can go ahead and classify
    the critical values you have.
  • 87:48 - 87:52
    So classification
    is the following.
  • 87:52 - 87:57
    Classification of
    special critical points.
  • 87:57 - 88:00
  • 88:00 - 88:07
    If delta at p0 is negative,
    then p0 is a saddle point.
  • 88:07 - 88:12
  • 88:12 - 88:21
    If delta at p0 is 0,
    nothing can be said yet
  • 88:21 - 88:24
    about the nature of the point.
  • 88:24 - 88:27
    So I make a face, a sad face.
  • 88:27 - 88:41
    If delta at p0 is greater than
    0, then I have to ramify again.
  • 88:41 - 88:47
    And I get if a is positive,
    it's going to look like a smile.
  • 88:47 - 88:49
    Forget about this side.
  • 88:49 - 88:51
    It's going to look like a smile.
  • 88:51 - 88:58
    So it's going to be a valley
    point, what do we call that?
  • 88:58 - 89:03
    Relative minimum,
    or valley point.
  • 89:03 - 89:05
    Don't say valley
    point on the exam, OK?
  • 89:05 - 89:06
    Relative minimum.
  • 89:06 - 89:10
    If a is less than
    0 at the point,
  • 89:10 - 89:14
    then locally the
    surface will look
  • 89:14 - 89:21
    like I have a peak-- a relative
    maximum Peaks and valleys.
  • 89:21 - 89:24
    Just the way you
    remember them in Calc 1.
  • 89:24 - 89:26
    Now it's a little
    bit more complicated
  • 89:26 - 89:29
    because the functions
    have two variables.
  • 89:29 - 89:33
    But some of the patterns
    can be recognized.
  • 89:33 - 89:36
  • 89:36 - 89:43
    Let's go back to
    our original example
  • 89:43 - 89:45
    and say wait a
    minute, Magdalena.
  • 89:45 - 89:48
    You just gave us a
    saddle, but we didn't
  • 89:48 - 89:50
    do the whole classification.
  • 89:50 - 89:55
    Yes, we didn't, because I
    didn't go over the next steps.
  • 89:55 - 89:57
    z equals x-squared
    minus y-squared.
  • 89:57 - 89:59
    Again, we computed the gradient.
  • 89:59 - 90:02
    We computed the
    partial derivatives.
  • 90:02 - 90:08
    And then what was that one in
    finding the critical points?
  • 90:08 - 90:12
    So f sub x equals
    0, f sub y equals 0.
  • 90:12 - 90:14
    Solve for x and y.
  • 90:14 - 90:17
  • 90:17 - 90:19
    And that's good,
    because that's going
  • 90:19 - 90:23
    to give me a lot of
    information, a lot that
  • 90:23 - 90:26
    will give me exactly where
    the critical points may be.
  • 90:26 - 90:32
    So that is if and only if I need
    to solve 2x equals 0 minus 2y
  • 90:32 - 90:34
    equals 0.
  • 90:34 - 90:36
    Is this system hard to solve?
  • 90:36 - 90:37
    No.
  • 90:37 - 90:39
    That's exactly why I picked it.
  • 90:39 - 90:41
    Because it's easy to solve.
  • 90:41 - 90:47
    The only solution is
    x0 equals y0 equals 0.
  • 90:47 - 90:52
    So the origin--
    that's exactly where
  • 90:52 - 91:00
    you put your butt on the
    saddle when you ride the horse.
  • 91:00 - 91:04
    That is the only
    critical point you have.
  • 91:04 - 91:06
    The only one.
  • 91:06 - 91:12
    Now if we want to classify that,
    what kind of-- is it a valley?
  • 91:12 - 91:13
    No.
  • 91:13 - 91:15
    It looks like a valley
    in the direction
  • 91:15 - 91:21
    of the axis of the horse,
    Because the saddle's
  • 91:21 - 91:23
    going to look like that.
  • 91:23 - 91:24
    This is the horse.
  • 91:24 - 91:26
    That's the head of
    the horse I'm petting.
  • 91:26 - 91:29
    And this is the
    tail of the horse.
  • 91:29 - 91:35
    So in this direction, the saddle
    will be shaped like a parabola,
  • 91:35 - 91:36
    like a valley.
  • 91:36 - 91:41
    But in the
    perpendicular direction,
  • 91:41 - 91:43
    it's going to be
    shaped going down,
  • 91:43 - 91:45
    like a parabola going down.
  • 91:45 - 91:49
    So it's neither a
    valley nor a peak.
  • 91:49 - 91:51
    It's a valley in one
    direction, and a peak
  • 91:51 - 91:52
    in another direction.
  • 91:52 - 91:55
    And that's the saddle point.
  • 91:55 - 91:56
    So say it again.
  • 91:56 - 91:57
    What is that?
  • 91:57 - 92:01
    It looks like a valley in
    one principle direction
  • 92:01 - 92:05
    and the peak in the other
    principle direction.
  • 92:05 - 92:09
    And then that's going
    to be a saddle point.
  • 92:09 - 92:14
    Indeed, how do we figure this
    out by the method I provided?
  • 92:14 - 92:17
    Well, who is A?
  • 92:17 - 92:22
    A is f sub xx at the point.
  • 92:22 - 92:25
    2x goes primed one time.
  • 92:25 - 92:28
    f sub x was 2x.
  • 92:28 - 92:30
    f sub y was 2y.
  • 92:30 - 92:32
    f sub xx is 2.
  • 92:32 - 92:35
  • 92:35 - 92:40
    f sub xB is f sub xy.
  • 92:40 - 92:42
    What is that?
  • 92:42 - 92:44
    0.
  • 92:44 - 92:46
    Good, that makes my life easier.
  • 92:46 - 92:50
    C equals f sub yy.
  • 92:50 - 92:52
    What is that?
  • 92:52 - 92:53
    2.
  • 92:53 - 92:56
    OK, this is looking beautiful.
  • 92:56 - 92:59
    Because I don't have
    to plug in any values.
  • 92:59 - 93:00
    The D is there for me to see it.
  • 93:00 - 93:05
    And it's going to consist of the
    determinant having the elements
  • 93:05 - 93:11
    2, 0, 0, 2-- minus 2, minus 2.
  • 93:11 - 93:14
    I'm sorry, guys, I
    missed here the minus.
  • 93:14 - 93:19
    And it cost me my life-- 2x
    and minus 2y, and here minus 2.
  • 93:19 - 93:21
    STUDENT: It didn't
    cost you your life,
  • 93:21 - 93:22
    because you caught it before
    you were done with the problem.
  • 93:22 - 93:23
    PROFESSOR: I caught it up there.
  • 93:23 - 93:24
    I'm taking the final exam.
  • 93:24 - 93:28
    I still get 100%,
    because I caught it up
  • 93:28 - 93:30
    at the last minute.
  • 93:30 - 93:34
    So 2, 0, 0, minus
    2-- I knew that I
  • 93:34 - 93:36
    had to get something negative.
  • 93:36 - 93:40
    So I said, for god's sake, I
    need to get a saddle point.
  • 93:40 - 93:42
    That's why it's the
    horse in the saddle.
  • 93:42 - 93:47
    So I knew I should
    get minus 4, negative.
  • 93:47 - 93:51
    All right, so the
    only thing I have
  • 93:51 - 93:53
    to say as a final
    answer is the only
  • 93:53 - 93:57
    critical point of this surface
    that I'm too lazy to write
  • 93:57 - 93:59
    about-- don't write that.
  • 93:59 - 94:02
    So the only critical
    point on the surface z
  • 94:02 - 94:05
    equals x squared
    minus y squared will
  • 94:05 - 94:10
    be at the origin O
    of corner 0, 0, 0
  • 94:10 - 94:14
    where the discriminant
    being negative
  • 94:14 - 94:17
    indicates it's going
    to be a saddle point.
  • 94:17 - 94:20
    And that's it-- nothing else.
  • 94:20 - 94:22
    You don't need more.
  • 94:22 - 94:25
    But there are more examples.
  • 94:25 - 94:27
    Because life is hard.
  • 94:27 - 94:28
    And I'm going to give
    you another example.
  • 94:28 - 94:32
  • 94:32 - 94:35
    Well, OK, this one.
  • 94:35 - 94:38
  • 94:38 - 94:48
    Suppose we have the
    surface-- that's
  • 94:48 - 94:50
    still going to be very easy.
  • 94:50 - 94:52
    But I want to make the
    first examples easy.
  • 94:52 - 94:55
  • 94:55 - 94:56
    I have a reason why.
  • 94:56 - 95:05
  • 95:05 - 95:07
    This is a function of
    two variables, right?
  • 95:07 - 95:12
    It's still a polynomial in
    two variables of order 2.
  • 95:12 - 95:18
    And how do I solve for the
    classification of the extrema?
  • 95:18 - 95:24
    I'm looking for local extrema,
    not absolute-- local extrema.
  • 95:24 - 95:25
    I'm not constrained.
  • 95:25 - 95:28
    I'm saying, what do you
    mean, no constraint?
  • 95:28 - 95:31
    Constrained would have
    been, let's say that x and y
  • 95:31 - 95:34
    are in the unit disc.
  • 95:34 - 95:39
    Or let's say x and y are
    on the circle x squared
  • 95:39 - 95:40
    plus y squared equals 1.
  • 95:40 - 95:42
    That would be a constraint.
  • 95:42 - 95:45
    But they're not
    constrained about anything.
  • 95:45 - 95:47
    x and y are real numbers.
  • 95:47 - 95:51
    They can take the whole
    plane as a domain.
  • 95:51 - 95:59
    So I get f sub x equals 0, f sub
    y equals 0, solve for x and y,
  • 95:59 - 96:01
    get the critical values.
  • 96:01 - 96:05
    I get very nice 2x.
  • 96:05 - 96:06
    I have to pay attention.
  • 96:06 - 96:10
    Because now this is
    not so easy anymore--
  • 96:10 - 96:17
    plus prime with respect to x,
    2y, prime with respect to x, 0,
  • 96:17 - 96:26
    prime with respect to x, plus
    3, prime of this, OK, equals 0.
  • 96:26 - 96:32
    f sub y-- 0 plus
    prime with respect
  • 96:32 - 96:41
    to y, 2x, plus prime
    with respect to y, 2y,
  • 96:41 - 96:47
    plus nothing, prime with
    respect to y equals 0.
  • 96:47 - 96:53
  • 96:53 - 96:58
    And now you have
    to be very smart.
  • 96:58 - 97:03
    Well, you have to be perceptive
    and tell me what I got.
  • 97:03 - 97:05
    What is this that we mean?
  • 97:05 - 97:09
  • 97:09 - 97:11
    Look at this system.
  • 97:11 - 97:13
    It looks like crazy.
  • 97:13 - 97:23
  • 97:23 - 97:26
    STUDENT: [INAUDIBLE] the
    origin or-- because can't you
  • 97:26 - 97:28
    just subtract it down?
  • 97:28 - 97:30
    PROFESSOR: Is this possible?
  • 97:30 - 97:32
    And what does this mean?
  • 97:32 - 97:33
    What do we call such a system?
  • 97:33 - 97:38
  • 97:38 - 97:42
    Inconsistent system--
    we call it inconsistent.
  • 97:42 - 97:45
    How can I make this
    problem to be possible,
  • 97:45 - 97:48
    to have some critical points?
  • 97:48 - 97:50
    STUDENT: If you add 3x.
  • 97:50 - 97:52
    PROFESSOR: How about
    that, just remove the 3x
  • 97:52 - 97:56
    and see what's going to happen.
  • 97:56 - 97:59
    Oh, in that case,
    I have something
  • 97:59 - 98:06
    that's over-determined, right?
  • 98:06 - 98:12
    I have something that
    tells me the same thing.
  • 98:12 - 98:14
    So I'm priming
    with respect to x.
  • 98:14 - 98:15
    I get that.
  • 98:15 - 98:16
    I'm priming with respect to y.
  • 98:16 - 98:18
    I get this.
  • 98:18 - 98:19
    I get 0.
  • 98:19 - 98:22
    So I don't even need
    the second equation.
  • 98:22 - 98:28
    And that means
    the critical point
  • 98:28 - 98:41
    is any point of
    the form-- shall I
  • 98:41 - 98:44
    put a Greek letter alpha minus
    alpha or lambda minus lambda?
  • 98:44 - 98:46
    What shall I do?
  • 98:46 - 98:52
    So any point that is situated
    on the second bisector,
  • 98:52 - 98:54
    I mean the x, y plane.
  • 98:54 - 98:57
    And this is the x,
    and this is the y.
  • 98:57 - 99:01
    And I say, what does it
    mean, x plus y equals 0?
  • 99:01 - 99:03
    Not this line-- don't draw it.
  • 99:03 - 99:05
    That is x equals y.
  • 99:05 - 99:09
    The other one, called
    the second bisector-- y
  • 99:09 - 99:13
    equals negative x, so not
    this one, the diagonal,
  • 99:13 - 99:22
    but the diagonal that's
    on the corridor, this one.
  • 99:22 - 99:26
    All right, so any point of
    the form alpha minus alpha,
  • 99:26 - 99:28
    here's the critical point.
  • 99:28 - 99:33
    The question is, how am I going
    to get to the classification
  • 99:33 - 99:35
    for such points?
  • 99:35 - 99:37
    Can anybody help me?
  • 99:37 - 99:38
    So step two--
  • 99:38 - 99:40
    STUDENT: Solve the equation.
  • 99:40 - 99:43
    STUDENT: Solve alpha for one
    of the two variables first.
  • 99:43 - 99:46
    PROFESSOR: Take alpha minus
    alpha-- could be anything.
  • 99:46 - 99:51
    And then I'll say, f
    sub-- this is f sub x.
  • 99:51 - 99:54
    And this is f sub y.
  • 99:54 - 99:55
    What is f sub x?
  • 99:55 - 99:57
    f sub xx, I'm sorry.
  • 99:57 - 100:00
  • 100:00 - 100:02
    STUDENT: [INAUDIBLE].
  • 100:02 - 100:02
    PROFESSOR: Huh?
  • 100:02 - 100:03
    2.
  • 100:03 - 100:05
    OK, are you with me?
  • 100:05 - 100:08
    So you know what it is.
  • 100:08 - 100:12
    f sub xy equals?
  • 100:12 - 100:13
    STUDENT: 2.
  • 100:13 - 100:15
    PROFESSOR: 2.
  • 100:15 - 100:18
    f sub yy equals?
  • 100:18 - 100:19
    STUDENT: 2
  • 100:19 - 100:22
    PROFESSOR: 2-- that's
    the mystery man.
  • 100:22 - 100:23
    The book doesn't
    give this example,
  • 100:23 - 100:25
    and it drives me crazy.
  • 100:25 - 100:29
    And I wanted to give you
    some bad example where
  • 100:29 - 100:31
    the classification doesn't work.
  • 100:31 - 100:35
    Because we always cook
    up nice examples for you
  • 100:35 - 100:39
    and claim everything
    is beautiful.
  • 100:39 - 100:41
    Life is not always beautiful.
  • 100:41 - 100:44
    So you get 0.
  • 100:44 - 100:48
    In that case, nothing can be
    said with this classification.
  • 100:48 - 100:50
    I make a face, sad face.
  • 100:50 - 100:52
    So what do I hope?
  • 100:52 - 100:59
    To get to Maple or MATLAB and
    be able to draw that, or a TI-92
  • 100:59 - 101:03
    if my mother would
    give me $200 and some.
  • 101:03 - 101:04
    I told her.
  • 101:04 - 101:08
    She asked me what to
    buy for my birthday.
  • 101:08 - 101:11
    I have a TI-83 or something.
  • 101:11 - 101:13
    And it was cheap.
  • 101:13 - 101:17
    I bought it on eBay, and
    then I stopped using it.
  • 101:17 - 101:21
    And then I saw this TI-92
    that can draw surfaces
  • 101:21 - 101:23
    in three dimensions.
  • 101:23 - 101:24
    And I said, this is like MATLAB.
  • 101:24 - 101:26
    You just carry it
    in your pocket.
  • 101:26 - 101:29
    It's only a little
    bit too expensive.
  • 101:29 - 101:35
    All right, how
    about another kind?
  • 101:35 - 101:39
  • 101:39 - 101:40
    Look at this one.
  • 101:40 - 101:55
  • 101:55 - 101:59
    You cannot tell
    with the naked eye.
  • 101:59 - 102:02
    But you can go ahead
    and do this step one
  • 102:02 - 102:05
    looking for critical values.
  • 102:05 - 102:18
    So the system, f sub x will
    be 6x plus 2y equals 0.
  • 102:18 - 102:24
    f sub y will be--
    who's going to tell me?
  • 102:24 - 102:30
    2x plus 2y equals 0.
  • 102:30 - 102:34
    Now, by elimination or by
    substitution or by anything
  • 102:34 - 102:37
    I want, I subtract the
    second from the first.
  • 102:37 - 102:38
    What do I get?
  • 102:38 - 102:41
    I get 4x equals 0.
  • 102:41 - 102:46
    And that gives me the only
    possibility is x0 equals 0.
  • 102:46 - 102:53
    And then I say, OK, if my
    only one is 0, then y is 0.
  • 102:53 - 102:54
    0 is 0.
  • 102:54 - 103:01
    So I only have one critical
    point, which is the origin.
  • 103:01 - 103:04
    Now, do I know, what
    am I going to get?
  • 103:04 - 103:06
    Not unless I'm a
    genius and I can
  • 103:06 - 103:08
    see two steps ahead of time.
  • 103:08 - 103:11
    I would need to do ABC
    quickly in my head.
  • 103:11 - 103:13
    Some of you are able, thank god.
  • 103:13 - 103:15
    But some of you,
    like me, are not.
  • 103:15 - 103:19
    So I have to take a few
    seconds to see what's going on.
  • 103:19 - 103:24
    A-- f sub xx at the point is 0.
  • 103:24 - 103:28
    B-- f sub xy.
  • 103:28 - 103:30
    C-- f sub yy.
  • 103:30 - 103:37
  • 103:37 - 103:39
    What do we do?
  • 103:39 - 103:40
    We get 6.
  • 103:40 - 103:41
    Are we happy about it?
  • 103:41 - 103:45
    We don't know yet,
    to be happy or not.
  • 103:45 - 103:49
    f sub xy or f sub yx,
    you see, Mr. Schwarz
  • 103:49 - 103:52
    is now happy that
    he proved to you
  • 103:52 - 103:55
    that it doesn't matter
    which order you're
  • 103:55 - 103:59
    taking for a polynomial
    that's a smooth function.
  • 103:59 - 104:04
    You always have the same.
  • 104:04 - 104:08
    And finally, C is 2.
  • 104:08 - 104:13
    And you are ready to do the
    D. And I could smell that D,
  • 104:13 - 104:15
    but I didn't want
    to say anything.
  • 104:15 - 104:24
    6, 2, 2, and 2-- is
    that a nice thing?
  • 104:24 - 104:28
    Yeah, we haven't encountered
    this example yet.
  • 104:28 - 104:31
    Because according to
    the classification,
  • 104:31 - 104:32
    this is greater than 0.
  • 104:32 - 104:34
    Does it really matter
    what value it is?
  • 104:34 - 104:38
    No, it only matters
    that it is positive.
  • 104:38 - 104:42
    And if it's positive, that
    means I can move on with my life
  • 104:42 - 104:45
    and look at the classification.
  • 104:45 - 104:50
    From this point where
    delta or v is positive,
  • 104:50 - 104:57
    I'm going to get a ramification
    into separate cases.
  • 104:57 - 105:01
    And who is going to
    tell me next what to do?
  • 105:01 - 105:04
    Look at A. Oh, by
    the way, talking
  • 105:04 - 105:08
    about the quadratic
    formula from school,
  • 105:08 - 105:13
    from kindergarten,
    when we computed
  • 105:13 - 105:19
    the-- I'll use the general one,
    minus b plus minus square root
  • 105:19 - 105:21
    of b squared minus 4ac over 2a.
  • 105:21 - 105:27
  • 105:27 - 105:34
    We were afraid of
    some special cases
  • 105:34 - 105:36
    when we were looking at that.
  • 105:36 - 105:38
    Especially when
    delta was negative,
  • 105:38 - 105:40
    that was really
    imaginary and so on.
  • 105:40 - 105:43
    But one thing we remember
    from ninth grade--
  • 105:43 - 105:47
    was this ninth grade
    or eighth grade?
  • 105:47 - 105:52
    The parabola opens up
    when a is positive.
  • 105:52 - 105:58
    Just the same way, something
    opens up when A, big A,
  • 105:58 - 105:59
    is positive here.
  • 105:59 - 106:02
    Then you have opening up.
  • 106:02 - 106:06
    When big A is negative,
    then you have opening down.
  • 106:06 - 106:10
    So remember-- I'm going to make
    smile here so you remember.
  • 106:10 - 106:12
    So I have it like that.
  • 106:12 - 106:16
    So I suspect that
    it's going to look
  • 106:16 - 106:23
    like a surface of some
    sort that maybe is not
  • 106:23 - 106:26
    surface of revolution.
  • 106:26 - 106:27
    You should tell me what it is.
  • 106:27 - 106:32
    You should think about this and
    do the cross sections with z
  • 106:32 - 106:35
    constant and tell me
    what surface that is.
  • 106:35 - 106:37
    But in any case, what do I care?
  • 106:37 - 106:42
    I care that I'm
    looking at the origin.
  • 106:42 - 106:44
    And this is where
    my special point is.
  • 106:44 - 106:47
    That's going to be
    the value point.
  • 106:47 - 106:48
    How do I know?
  • 106:48 - 106:53
    Because A, which
    is 6, is positive.
  • 106:53 - 106:56
    At this point, I know
    what I'm left with.
  • 106:56 - 107:00
    I know that my surface is
    going to look like a valley.
  • 107:00 - 107:02
    So how do I know again?
  • 107:02 - 107:04
    I'm not going to draw it.
  • 107:04 - 107:08
    But it's going to look
    something like that.
  • 107:08 - 107:10
    At the origin, this
    is going to be 7.
  • 107:10 - 107:12
    Are you guys with me?
  • 107:12 - 107:15
    And it's going to open up.
  • 107:15 - 107:19
    And so you should not attempt
    intersecting with z equals 5
  • 107:19 - 107:20
    or z equals 1.
  • 107:20 - 107:22
    Because you're not
    going to get anything.
  • 107:22 - 107:25
    But if you intersect,
    for example, at z
  • 107:25 - 107:27
    equals 9, what are
    you going to get?
  • 107:27 - 107:31
    If you intersect
    at z equals 9, you
  • 107:31 - 107:37
    get 3x plus 2xy plus
    y squared equals 2.
  • 107:37 - 107:39
    And what is that?
  • 107:39 - 107:44
    It's a rotated
    form of an ellipse.
  • 107:44 - 107:48
    It's hard to see, because
    it's missing [INAUDIBLE].
  • 107:48 - 107:51
    But this is exactly what
    discriminant is saying.
  • 107:51 - 107:56
    So this is going to be an x.
  • 107:56 - 107:59
    Good, so I know what
    I'm going to get.
  • 107:59 - 108:01
    What do you have to
    say on the midterm
  • 108:01 - 108:04
    or on the final
    about this problem?
  • 108:04 - 108:05
    STUDENT: The point is--
  • 108:05 - 108:08
    PROFESSOR: The
    point is the origin.
  • 108:08 - 108:09
    I classified it.
  • 108:09 - 108:10
    I got delta positive.
  • 108:10 - 108:12
    I got A positive.
  • 108:12 - 108:13
    So it's a valley.
  • 108:13 - 108:15
    It's a relative minimum.
  • 108:15 - 108:16
    And that's it.
  • 108:16 - 108:22
    I have a relative min at the
    point P of coordinates 0, 0,
  • 108:22 - 108:23
    and 7.
  • 108:23 - 108:26
  • 108:26 - 108:28
    And that's the valley.
  • 108:28 - 108:29
    Yes, sir.
  • 108:29 - 108:31
    STUDENT: Why is it
    A that determines
  • 108:31 - 108:34
    whether it's a relative
    min or a relative max?
  • 108:34 - 108:35
    PROFESSOR: It's a whole story.
  • 108:35 - 108:36
    You can prove it.
  • 108:36 - 108:40
    I don't remember if we proved
    this in the book or not.
  • 108:40 - 108:43
    But it can be
    proved, so the fact
  • 108:43 - 108:49
    that it has to do with
    concavity and convexity.
  • 108:49 - 108:54
    When you had a second
    derivative, let's say,
  • 108:54 - 108:57
    what's the equivalent
    of the Calculus I
  • 108:57 - 108:59
    notion that you know about?
  • 108:59 - 109:02
    In Calculus I, you had
    functions of one variable,
  • 109:02 - 109:04
    and life was so easy like that.
  • 109:04 - 109:08
    And f prime positive meant
    that the function increased.
  • 109:08 - 109:12
    And f prime negative meant
    that the function decreased.
  • 109:12 - 109:18
    And f double prime was
    just like your-- you sense
  • 109:18 - 109:22
    that the second partials
    must have something
  • 109:22 - 109:26
    to do with it, especially the
    first one with respect to x.
  • 109:26 - 109:29
    If you were in
    plane, and you have
  • 109:29 - 109:34
    f double prime with respect
    to x, when was this a valley?
  • 109:34 - 109:35
    When you had the smile.
  • 109:35 - 109:38
    When did you have a smile?
  • 109:38 - 109:43
    When f double prime was
    positive, you have concave up.
  • 109:43 - 109:45
    When f double
    prime was negative,
  • 109:45 - 109:48
    you have concave down.
  • 109:48 - 109:49
    Remember, guys?
  • 109:49 - 109:54
    So you have a smile or a frown.
  • 109:54 - 109:54
    This is how we know.
  • 109:54 - 109:58
    For the same reason that
    would take about two pages
  • 109:58 - 110:04
    to write down the proof, you
    have a smile for A positive.
  • 110:04 - 110:08
    And the smile means
    actually in all directions
  • 110:08 - 110:13
    you have a smile locally
    around the origin.
  • 110:13 - 110:18
    OK, look in the book.
  • 110:18 - 110:20
    I'm not sure how
    much should we do.
  • 110:20 - 110:24
    Do we give a sketch of a proof,
    or we give the entire proof?
  • 110:24 - 110:26
    But more likely, a sketch.
  • 110:26 - 110:29
  • 110:29 - 110:31
    Yes.
  • 110:31 - 110:36
    STUDENT: I asked the
    slightly wrong question,
  • 110:36 - 110:37
    but I answered it myself.
  • 110:37 - 110:42
    I wanted to ask, why is it
    dependent on A and not on C?
  • 110:42 - 110:43
    PROFESSOR: Not on C.
  • 110:43 - 110:45
    STUDENT: But then I realized
    that it is dependent on C
  • 110:45 - 110:46
    as well, because
    if A is positive,
  • 110:46 - 110:47
    then C must be positive.
  • 110:47 - 110:51
    PROFESSOR: Yes, yes, it
    is dependent on both.
  • 110:51 - 110:52
    STUDENT: OK, there we go.
  • 110:52 - 110:55
    That was my question.
  • 110:55 - 110:57
    PROFESSOR: So guys, remember.
  • 110:57 - 111:03
    Imagine what happens when
    you had no B, B was 0.
  • 111:03 - 111:06
    Then the matrix
    is diagonalizable.
  • 111:06 - 111:12
    And here you have
    A and C. And Alex
  • 111:12 - 111:16
    says, why would A be
    more important than C?
  • 111:16 - 111:17
    It's not.
  • 111:17 - 111:21
    But practically, if A is
    positive and C is negative,
  • 111:21 - 111:24
    that means these are the
    principal directions in which
  • 111:24 - 111:29
    one bends like a valley up and
    one bends like a peak down.
  • 111:29 - 111:36
    So this is what happens in the
    direction of x, f double prime
  • 111:36 - 111:39
    in the direction of x, kind of.
  • 111:39 - 111:43
    And this is in the
    direction of y.
  • 111:43 - 111:45
    So this is f double prime
    in the direction of y,
  • 111:45 - 111:47
    which we don't denote like that.
  • 111:47 - 111:52
    We call it f sub xx and f
    sub yy, which is A and C.
  • 111:52 - 112:00
    So A positive, A being 1
    and C being negative 2,
  • 112:00 - 112:05
    means a valley here, means
    the valley meets the horse.
  • 112:05 - 112:11
    Look, I'm drawing the
    tail of the horse.
  • 112:11 - 112:14
    He's a little bit
    fat, this horse.
  • 112:14 - 112:19
    And that's his mane, his eye.
  • 112:19 - 112:22
    I'm just taking a break.
  • 112:22 - 112:24
    STUDENT: That's a
    pretty good drawing.
  • 112:24 - 112:31
    PROFESSOR: It looks more
    like a dog or a plush horse
  • 112:31 - 112:32
    or something.
  • 112:32 - 112:37
    So A equals 1, and
    C equals minus 2.
  • 112:37 - 112:42
    But if it were diagonalizable,
    and A would be 1
  • 112:42 - 112:47
    and C would be 7, both of
    them positive in any case,
  • 112:47 - 112:52
    then you'll have valley and
    valley, an x direction valley
  • 112:52 - 112:54
    and y direction valley.
  • 112:54 - 112:57
    So it has to be a
    valley everywhere.
  • 112:57 - 113:01
    These are the principal
    directions that I have 1 and 2.
  • 113:01 - 113:06
    But then the ultimate
    case, what happens
  • 113:06 - 113:15
    when A is negative
    and-- hmm, OK,
  • 113:15 - 113:22
    then either you have them both
    one positive, one negative,
  • 113:22 - 113:27
    or you have plus,
    plus and minus, minus.
  • 113:27 - 113:33
    And then you have this
    as your surface, right?
  • 113:33 - 113:34
    Which one is the x direction?
  • 113:34 - 113:35
    That's the y direction.
  • 113:35 - 113:37
    That's the second one.
  • 113:37 - 113:40
    The x direction is that.
  • 113:40 - 113:43
    In the x direction,
    you have a frown.
  • 113:43 - 113:47
    So f sub yy is negative.
  • 113:47 - 113:51
    In the y direction,
    you also have a frown.
  • 113:51 - 113:53
    So both of them are negative.
  • 113:53 - 113:55
    So you have a relative max.
  • 113:55 - 113:59
    Yes, sir, Matthew, tell me.
  • 113:59 - 114:02
    STUDENT: So isn't it possible
    to have both A and C positive,
  • 114:02 - 114:07
    but then yet still not be
    more positive than B squared?
  • 114:07 - 114:11
    PROFESSOR: No, because
    there's a theorem that--
  • 114:11 - 114:13
    STUDENT: I was just
    wondering like numbers-wise.
  • 114:13 - 114:17
    PROFESSOR: You have this matrix.
  • 114:17 - 114:20
    And there is a theorem that
    shows you that you can actually
  • 114:20 - 114:22
    diagonalize this matrix.
  • 114:22 - 114:25
    You'll learn your linear
    algebra [INAUDIBLE].
  • 114:25 - 114:27
    STUDENT: It makes
    sense, because when
  • 114:27 - 114:30
    you were saying A is this way,
    and that way there's no way
  • 114:30 - 114:34
    you could have 2 come
    up, and then yet,
  • 114:34 - 114:37
    not be a-- you know
    what I'm saying?
  • 114:37 - 114:39
    Because then they'd
    be less than 0.
  • 114:39 - 114:43
    PROFESSOR: You can if
    you don't have or the 2.
  • 114:43 - 114:45
    That's an excellent question.
  • 114:45 - 114:48
    If I would have x
    to the 4 y to the 4
  • 114:48 - 114:52
    added together, like Ax to
    the 4 plus By to the 4 plus
  • 114:52 - 114:56
    something, then I have the
    so-called monkey saddle.
  • 114:56 - 114:58
    That's so funny.
  • 114:58 - 115:06
    You can have something
    that looks like that.
  • 115:06 - 115:08
    So in your direction,
    you can have this.
  • 115:08 - 115:10
    Then I've reached
    two equal peaks
  • 115:10 - 115:11
    in the x and the y direction.
  • 115:11 - 115:14
    But in the between,
    I also went down.
  • 115:14 - 115:18
    So depending on a higher
    degree symmetric polynomial,
  • 115:18 - 115:21
    you can have a monkey saddle.
  • 115:21 - 115:25
    And then it's not just
    like you can predict what's
  • 115:25 - 115:27
    going to happen in between.
  • 115:27 - 115:31
    In between, if I go
    up, if I go valley
  • 115:31 - 115:34
    in the x direction and
    valley in the y direction,
  • 115:34 - 115:37
    I know that's going to be
    a valley everywhere-- no.
  • 115:37 - 115:39
    If a polynomial
    is high in order,
  • 115:39 - 115:42
    it can go down,
    valley, and up again,
  • 115:42 - 115:46
    and monkey saddle it looks like.
  • 115:46 - 115:49
    Guys, you have dealt
    with it when you
  • 115:49 - 115:53
    went to Luna Park or Joyland.
  • 115:53 - 116:00
    It's one of those things
    that look like-- I'm trying.
  • 116:00 - 116:01
    I cannot draw.
  • 116:01 - 116:03
    STUDENT: It sounds
    more like an octopus.
  • 116:03 - 116:04
    PROFESSOR: Like an octopus.
  • 116:04 - 116:07
    And one of those
    things-- exactly--
  • 116:07 - 116:11
    that are shaped so that
    they are undulated,
  • 116:11 - 116:14
    in some directions are
    going up, in some directions
  • 116:14 - 116:15
    are going down.
  • 116:15 - 116:17
    STUDENT: Like an
    egg carton, almost?
  • 116:17 - 116:20
    PROFESSOR: Yeah,
    really undulated.
  • 116:20 - 116:23
    Imagine even a surface made
    of metal that's undulated
  • 116:23 - 116:25
    and rotating at the same time.
  • 116:25 - 116:29
    They have some of
    those in Disney World.
  • 116:29 - 116:32
    Have you been to Orlando?
  • 116:32 - 116:35
    STUDENT: I was
    there last semester.
  • 116:35 - 116:38
    PROFESSOR: But you didn't take
    me with you, which is bad.
  • 116:38 - 116:42
    Because that's one of
    my favorite places.
  • 116:42 - 116:45
    STUDENT: I was just trying to
    think of what you were talking
  • 116:45 - 116:45
    about so I could visualize it.
  • 116:45 - 116:48
    PROFESSOR: Maybe we
    could make a proposal
  • 116:48 - 116:51
    to teach Calculus
    III at Disney World
  • 116:51 - 116:55
    so that we could have examples
    of motion and surfaces
  • 116:55 - 117:00
    all around and study the
    motion of all sorts of gadgets,
  • 117:00 - 117:01
    velocity and trajectory.
  • 117:01 - 117:06
  • 117:06 - 117:10
    Last night I couldn't sleep
    until 1:00, and I was thinking,
  • 117:10 - 117:14
    I gave examples of
    the winter sports
  • 117:14 - 117:18
    like bobsled and all
    sorts of skiing and so on.
  • 117:18 - 117:22
    But I never thought
    about a screw curve
  • 117:22 - 117:28
    with curvature and torsion that
    is based on the roller coaster.
  • 117:28 - 117:30
    And the roller
    coaster is actually
  • 117:30 - 117:35
    the best place to study the
    [INAUDIBLE], the velocity,
  • 117:35 - 117:38
    the tangent unit, the
    normal, the bi-normal.
  • 117:38 - 117:43
    And when you have in a
    plane the roller coaster
  • 117:43 - 117:47
    goes like that, like this and
    like that, like in a plane,
  • 117:47 - 117:50
    you have nothing but bending,
    which means curvature.
  • 117:50 - 117:53
    But then when the roller coaster
    goes away from the plane,
  • 117:53 - 117:55
    you have the torsion.
  • 117:55 - 117:58
    And that makes you sick
    really to the stomach.
  • 117:58 - 118:01
    So we would have to
    experience that to understand
  • 118:01 - 118:03
    Calculus III better.
  • 118:03 - 118:06
    So our next proposal is
    we ask the administration
  • 118:06 - 118:11
    instead of study abroad
    courses, the domestic study
  • 118:11 - 118:16
    at Disney World for Calc III.
  • 118:16 - 118:17
    It's Applied Calculus III.
  • 118:17 - 118:20
  • 118:20 - 118:24
    OK, something else that
    I want you to do-- I
  • 118:24 - 118:27
    had prepared an example.
  • 118:27 - 118:31
  • 118:31 - 118:33
    This is an absolute extrema.
  • 118:33 - 118:41
  • 118:41 - 118:44
    And you say, what the heck
    are the absolute extrema?
  • 118:44 - 118:48
    Because she only talked to
    us about relative maximum
  • 118:48 - 118:50
    and relative minimum.
  • 118:50 - 118:54
    And she never said anything
    about absolute extrema.
  • 118:54 - 118:56
  • 118:56 - 118:59
    And that will be the table.
  • 118:59 - 119:02
    And these will be the extrema.
  • 119:02 - 119:06
    I want to refresh your memory
    first just a little bit.
  • 119:06 - 119:07
    This will be the last example.
  • 119:07 - 119:10
    Because it's actually
    two examples in one.
  • 119:10 - 119:13
  • 119:13 - 119:26
    And what if you have, let's say,
    f of x equals e to the minus
  • 119:26 - 119:38
    x squared over the
    interval minus 1, 1?
  • 119:38 - 119:41
    You are in Calc I.
    You will build a time
  • 119:41 - 119:44
    machine from Disney World.
  • 119:44 - 119:50
    And we went back in time when
    you actually took Calc I.
  • 119:50 - 119:52
    And you struggled
    with this at first.
  • 119:52 - 119:54
    But then you loved
    it so much that you
  • 119:54 - 119:57
    said, oh, that's my favorite
    problem on the final.
  • 119:57 - 120:05
    They asked us for two things--
    relative extrema, min or max,
  • 120:05 - 120:09
    min/max theory, and
    they say absolute.
  • 120:09 - 120:11
    But for the absolute, your
    teacher said, attention,
  • 120:11 - 120:15
    you have to know how
    to get to the absolute.
  • 120:15 - 120:20
    You are constrained to be
    on the segment minus 1, 1.
  • 120:20 - 120:22
    You see, the fact
    that they introduced
  • 120:22 - 120:25
    this extra constraint
    and they don't
  • 120:25 - 120:31
    let you move with x on the whole
    real line is a big headache.
  • 120:31 - 120:32
    Why is that a big headache?
  • 120:32 - 120:35
    Your life would be much
    easier if it were just e
  • 120:35 - 120:37
    to the negative x squared.
  • 120:37 - 120:40
    Because in that
    case, you say, OK,
  • 120:40 - 120:47
    f prime of x equals minus
    2xe to the minus x squared.
  • 120:47 - 120:48
    Piece of cake.
  • 120:48 - 120:50
    x0 is 0.
  • 120:50 - 120:55
    That's the only critical point.
  • 120:55 - 121:00
    And I want to study what kind
    of critical point that is.
  • 121:00 - 121:04
    So I have to do f
    double prime of x.
  • 121:04 - 121:08
    And if I don't know the
    product rule, I'm in trouble.
  • 121:08 - 121:12
    And I go, let's
    say, minus 2 times
  • 121:12 - 121:16
    e to the negative x
    squared from prime of this
  • 121:16 - 121:20
    and this non-prime, plus
    minus 2x un-prime times e
  • 121:20 - 121:25
    to the minus x squared
    times minus 2x again.
  • 121:25 - 121:28
    So it's a headache.
  • 121:28 - 121:31
    I pull out an e to
    the minus x squared.
  • 121:31 - 121:38
    And I have 4x squared--
    4x squared-- minus 2.
  • 121:38 - 121:41
  • 121:41 - 121:43
    But you say, but wait
    a minute, Magdalena,
  • 121:43 - 121:46
    I'm not going to compute
    the inflection points.
  • 121:46 - 121:49
    The inflection points
    will be x equals
  • 121:49 - 121:53
    plus/minus 1 over root 2.
  • 121:53 - 121:56
    I only care about
    the critical point.
  • 121:56 - 121:59
    And the only critical
    point I have is at 0.
  • 121:59 - 122:05
    Compute f double prime of 0 to
    see if it's a smile or a frown.
  • 122:05 - 122:08
  • 122:08 - 122:10
    And you do it.
  • 122:10 - 122:15
    And you plug in, and you say,
    e to the 0 is 1, 0 minus 2.
  • 122:15 - 122:17
    So you get a negative.
  • 122:17 - 122:19
    Do you care what it is?
  • 122:19 - 122:20
    No, but you care it's negative.
  • 122:20 - 122:23
  • 122:23 - 122:31
    So at 0, so you
    draw, and you know
  • 122:31 - 122:37
    at 0 you're going to
    have some sort of a what?
  • 122:37 - 122:38
    Relative max.
  • 122:38 - 122:40
  • 122:40 - 122:41
    Where?
  • 122:41 - 122:44
    At 0, and when you
    plug 0 again, 1.
  • 122:44 - 122:46
    So you draw a table.
  • 122:46 - 122:52
    And you say,
    relative max at 0, 1.
  • 122:52 - 122:56
    And then you're not done.
  • 122:56 - 122:58
    Because you say,
    wait a minute, I
  • 122:58 - 123:04
    am to study my function in
    Calc I at minus 1 and 1.
  • 123:04 - 123:08
    It's like you have a
    continuous picture,
  • 123:08 - 123:13
    and you chop, take scissors,
    and cut and cut at the extrema.
  • 123:13 - 123:16
    And there you can
    get additional points
  • 123:16 - 123:20
    where you can get a relative
    max or relative min.
  • 123:20 - 123:25
    Absolute max or min will be
    the lowest of all the values
  • 123:25 - 123:28
    and the highest
    of all the values.
  • 123:28 - 123:35
    OK, so I get at the point
    minus 1-- how shall I put here?
  • 123:35 - 123:37
    x equals minus 1.
  • 123:37 - 123:39
    What do I get for y?
  • 123:39 - 123:41
    And for plus 1,
    what do I get for y?
  • 123:41 - 123:43
    This is the question.
  • 123:43 - 123:48
    I plug it in, and I
    get minus minus 1/e.
  • 123:48 - 123:52
  • 123:52 - 123:56
    And then when I have
    1, what do I get?
  • 123:56 - 124:00
    1/e again.
  • 124:00 - 124:02
    So do I have a
    relative min here?
  • 124:02 - 124:07
    No, but I have an
    absolute something.
  • 124:07 - 124:10
  • 124:10 - 124:12
    And what do I have here?
  • 124:12 - 124:16
    Here I have an absolute max.
  • 124:16 - 124:22
    So how do we check the absolute
    maxima and absolute minima?
  • 124:22 - 124:24
    We look for critical points.
  • 124:24 - 124:27
    We get many of them,
    finitely many of them.
  • 124:27 - 124:30
    We compute all the
    values of z for them,
  • 124:30 - 124:33
    all the function values.
  • 124:33 - 124:36
    And then we look
    at the end points,
  • 124:36 - 124:38
    and we compare all three
    of them, all the three
  • 124:38 - 124:39
    values in the end.
  • 124:39 - 124:43
    So in the end, you
    compare 1/e to 1/e to 1.
  • 124:43 - 124:46
    And that's all you can get.
  • 124:46 - 124:49
    So the lowest in one will
    be the highest in one.
  • 124:49 - 124:51
    Good.
  • 124:51 - 124:55
    In Calculus III, it's
    more complicated.
  • 124:55 - 124:58
    But it's not much
    more complicated.
  • 124:58 - 125:02
    Let's see what's
    going to happen.
  • 125:02 - 125:04
    You can have a
    critical point inside.
  • 125:04 - 125:07
    We are just praying we
    don't have too many.
  • 125:07 - 125:10
    So how do I get to step one?
  • 125:10 - 125:15
    Critical point means f sub
    x equals e to the x squared
  • 125:15 - 125:19
    minus y squared times 2x.
  • 125:19 - 125:21
    All righty, it looks good.
  • 125:21 - 125:25
    f sub y equals e to the x
    squared minus y squared times
  • 125:25 - 125:26
    minus y.
  • 125:26 - 125:27
    I'm full of hope.
  • 125:27 - 125:30
    Because I only have one
    critical point, thank god.
  • 125:30 - 125:35
    Origin is my only
    critical point.
  • 125:35 - 125:38
    I don't know what that
    is going to give me.
  • 125:38 - 125:40
    But it can give
    me a relative max
  • 125:40 - 125:42
    or relative min or a saddle.
  • 125:42 - 125:46
    I don't know what
    it's going to be.
  • 125:46 - 125:50
    Who tells me what
    that is going to be?
  • 125:50 - 125:52
    Well, did I do this further?
  • 125:52 - 125:55
  • 125:55 - 126:01
    I did it further and
    a little bit lazy.
  • 126:01 - 126:05
    But I'm not asking the
    nature of the point.
  • 126:05 - 126:09
    So for the time being, I only
    want to see what happens at 0,
  • 126:09 - 126:10
    0.
  • 126:10 - 126:12
    So I have 1.
  • 126:12 - 126:18
    So in my table I will put, OK,
    this is x, y, and this is z.
  • 126:18 - 126:21
    For 0, 0, I'm interested.
  • 126:21 - 126:24
    Because that's the critical
    point inside the domain.
  • 126:24 - 126:26
    The domain will be the unities.
  • 126:26 - 126:30
    And inside the origin,
    something interesting happens.
  • 126:30 - 126:31
    I get a 1.
  • 126:31 - 126:35
    And I hope that's going to
    be my absolute something.
  • 126:35 - 126:37
    But I cannot be sure.
  • 126:37 - 126:38
    Why?
  • 126:38 - 126:43
    There may be other values
    coming from the boundary.
  • 126:43 - 126:46
    And just like in
    Calculus I, the only guys
  • 126:46 - 126:49
    that can give you other
    absolute max or min,
  • 126:49 - 126:52
    they can come from the
    boundary, nothing else.
  • 126:52 - 126:55
    Nowhere else in the interior
    of the disc am I going to look.
  • 126:55 - 126:57
    I'm not interested.
  • 126:57 - 127:00
    I'm only interested in x
    squared plus y squared equals 1.
  • 127:00 - 127:03
    This is where
    something can happen,
  • 127:03 - 127:07
    nothing else interesting in the
    inside, just like in Calc I.
  • 127:07 - 127:12
    So to take x squared plus y
    squared equals 1 into account,
  • 127:12 - 127:16
    I pull y squared, who is
    married to x-- the poor guy.
  • 127:16 - 127:18
    He's married to x.
  • 127:18 - 127:22
    He's dependent on x
    completely, y squared
  • 127:22 - 127:24
    equals 1 minus x squared.
  • 127:24 - 127:28
    And I have to push him
    back into the function.
  • 127:28 - 127:37
    So at the boundary, f becomes
    a function of one variable.
  • 127:37 - 127:43
    He becomes f of x
    only equals e to the x
  • 127:43 - 127:47
    squared minus 1 plus x squared.
  • 127:47 - 127:50
    Are you guys with me?
  • 127:50 - 127:57
    So f of x will become e
    to the 2x squared minus 1
  • 127:57 - 128:03
    along the boundary, along
    the circle, only here.
  • 128:03 - 128:08
  • 128:08 - 128:11
    Now what else do I need to do?
  • 128:11 - 128:14
    I need to compute the critical
    values for this function of one
  • 128:14 - 128:18
    variable, just the way
    I did it in Calc I.
  • 128:18 - 128:23
    So f prime of x will give
    me e to the 2x squared
  • 128:23 - 128:28
    minus 1 times-- what comes
    down from the chain rule?
  • 128:28 - 128:28
    STUDENT: 4x.
  • 128:28 - 128:31
    PROFESSOR: 4x, so life is
    hard but not that hard.
  • 128:31 - 128:35
    Because I can get what?
  • 128:35 - 128:40
    I can get only x
    at 0 equals 0 here.
  • 128:40 - 128:45
    OK, so that's a critical point
    that comes from the boundary.
  • 128:45 - 128:48
    But guys, you have
    to pay attention.
  • 128:48 - 128:53
    When x is 0, how many
    y's can I have for that 0
  • 128:53 - 128:55
    on the boundary?
  • 128:55 - 128:59
    This is on the boundary--
    on the boundary.
  • 128:59 - 129:00
    STUDENT: Two.
  • 129:00 - 129:03
    PROFESSOR: Two of
    them-- I can have 1,
  • 129:03 - 129:04
    or I can have negative 1.
  • 129:04 - 129:07
  • 129:07 - 129:10
    There is one more tricky thing.
  • 129:10 - 129:12
    This is a function
    of one variable only.
  • 129:12 - 129:16
    But this stinking
    function is not
  • 129:16 - 129:19
    defined for arbitrary x real.
  • 129:19 - 129:21
    So I make a face again.
  • 129:21 - 129:24
    So I go, oh, headache.
  • 129:24 - 129:24
    Why?
  • 129:24 - 129:26
    x is constrained.
  • 129:26 - 129:28
    x is constrained, you see?
  • 129:28 - 129:36
    If you were inside the disc, x
    must be between minus 1 and 1.
  • 129:36 - 129:40
    So I have to take into account
    that x is not any real number,
  • 129:40 - 129:44
    but x is between minus 1 and 1.
  • 129:44 - 129:46
    Those are endpoints
    for this function.
  • 129:46 - 129:48
    And in Calc I, I
    learned, OK, I have
  • 129:48 - 129:52
    to also evaluate what
    happens at those endpoints.
  • 129:52 - 129:56
    But thank god that will exhaust
    my list, so I have a list.
  • 129:56 - 130:00
    Minus 1 for x and 1
    for x-- thank god.
  • 130:00 - 130:04
    That will give you
    what y on the boundary?
  • 130:04 - 130:07
    When x is 1 and x
    is minus 1, you're
  • 130:07 - 130:10
    interested in what happens,
    maximization or minimization,
  • 130:10 - 130:13
    for this function
    at the endpoints.
  • 130:13 - 130:21
    But fortunately, since you are
    on the boundary, y must be 0.
  • 130:21 - 130:25
    Because that's
    how you got y out.
  • 130:25 - 130:29
    If x is plus/minus 1 on
    the boundary, y must be 0.
  • 130:29 - 130:33
    So my list contains how
    many interesting points?
  • 130:33 - 130:36
    One, two, three, four,
    five-- for all of them,
  • 130:36 - 130:38
    we need to compute,
    and we are done.
  • 130:38 - 130:43
    Of all of them, the lowest z
    is called absolute minimum.
  • 130:43 - 130:46
    And the highest z is
    the absolute maximum.
  • 130:46 - 130:46
    And we are done.
  • 130:46 - 130:50
    You guys need to help me,
    because I'm running out of gas.
  • 130:50 - 130:53
    So x is 0.
  • 130:53 - 130:54
    Y is 1.
  • 130:54 - 130:55
    What is z?
  • 130:55 - 130:57
    STUDENT: [INAUDIBLE].
  • 130:57 - 131:00
    PROFESSOR: e to the
    minus 1, you were
  • 131:00 - 131:02
    fast, 1/e, thank you, guys.
  • 131:02 - 131:06
    So when x is 0 and y is minus 1?
  • 131:06 - 131:07
    STUDENT: [INAUDIBLE].
  • 131:07 - 131:08
    PROFESSOR: Huh?
  • 131:08 - 131:09
    STUDENT: [INAUDIBLE].
  • 131:09 - 131:11
    PROFESSOR: No, no,
    no, it's the same.
  • 131:11 - 131:13
    Because x is 0. y is minus 1.
  • 131:13 - 131:17
    I get e to the minus
    1, which is 1/e.
  • 131:17 - 131:19
    So far, so good-- I'm
    circling all the guys
  • 131:19 - 131:22
    that I want to compare after.
  • 131:22 - 131:29
    So for the final four points
    there, what do I have?
  • 131:29 - 131:34
    My final candidates could
    be x equals plus/minus 1
  • 131:34 - 131:39
    and y equals 0-- e and e.
  • 131:39 - 131:40
    Who's the biggest?
  • 131:40 - 131:41
    Who's the smallest?
  • 131:41 - 131:43
    STUDENT: e is the biggest.
  • 131:43 - 131:46
    PROFESSOR: e is the biggest,
    and 1/e is the smallest.
  • 131:46 - 131:49
    So how do I write conclusion?
  • 131:49 - 132:01
    Conclusion-- we have
    two absolute maxima
  • 132:01 - 132:08
    at minus 1, 0 and 1, 0.
  • 132:08 - 132:25
    And we have two absolute
    minima at 0, minus 1 and 0, 1.
  • 132:25 - 132:32
  • 132:32 - 132:42
    OK, now I have to-- now
    that's like a saddle.
  • 132:42 - 132:44
    Can you see it with the
    eyes of your imagination?
  • 132:44 - 132:46
    It's hard to see it.
  • 132:46 - 132:49
  • 132:49 - 132:50
    This is the disc.
  • 132:50 - 132:53
    And the four
    points, the cardinal
  • 132:53 - 132:58
    points-- OK, this is the disc.
  • 132:58 - 133:02
    We are looking at this
    disc from perspective.
  • 133:02 - 133:07
    And the five points,
    one is in the middle.
  • 133:07 - 133:09
  • 133:09 - 133:12
    One is here.
  • 133:12 - 133:13
    Minus 1 is 0.
  • 133:13 - 133:15
    One is here, 1, 0.
  • 133:15 - 133:20
    One is here, 0, minus
    1, and one here, 0, 1.
  • 133:20 - 133:26
    At minus 1, 0 and 1,
    0 I get the maximum.
  • 133:26 - 133:29
    So the way it's going to be
    shaped would be like that.
  • 133:29 - 133:31
    In this direction,
    it will be like that.
  • 133:31 - 133:33
    And cut the cake here.
  • 133:33 - 133:35
    You see it's like that.
  • 133:35 - 133:37
    It's going to be like this.
  • 133:37 - 133:41
    OK, passing through the
    origin, with my hands
  • 133:41 - 133:46
    I'm molding the surface made of
    Play-Doh or something for you.
  • 133:46 - 133:49
    So I'm starting here,
    and I'm going up.
  • 133:49 - 133:52
    And at this points, I'm here.
  • 133:52 - 133:53
    Are you with me?
  • 133:53 - 133:55
    The same height.
  • 133:55 - 134:01
    In the other direction,
    I'm going from 0.
  • 134:01 - 134:04
    But I'm not so high.
  • 134:04 - 134:08
    I'm going only up
    to-- what is 1/e?
  • 134:08 - 134:12
    About 1/3, meh,
    something like that.
  • 134:12 - 134:18
    So I'm going to get here.
  • 134:18 - 134:23
    So the problem is that
    one will be in between.
  • 134:23 - 134:27
    So if you really want to
    see what it looks like,
  • 134:27 - 134:29
    we are here at 1.
  • 134:29 - 134:32
    We grow from 1,
    altitude 1, you see?
  • 134:32 - 134:40
    We grow from 1 to about
    2.71718283 for both of these.
  • 134:40 - 134:48
    And from 1, in this direction
    I have to go down to 1/e.
  • 134:48 - 134:51
    So it looks like that.
  • 134:51 - 134:53
    I'll try to draw, OK?
  • 134:53 - 135:03
  • 135:03 - 135:08
    Do you see the patch
    around the origin?
  • 135:08 - 135:11
    So here's e.
  • 135:11 - 135:16
    And here's 1/e
    above the sea level.
  • 135:16 - 135:19
    And this is 1.
  • 135:19 - 135:27
    And you have one just like that
    in the back that is the-- it
  • 135:27 - 135:28
    still looks like a saddle.
  • 135:28 - 135:29
    It is a saddle.
  • 135:29 - 135:30
    It's symmetric.
  • 135:30 - 135:34
    But it's another kind of saddle.
  • 135:34 - 135:35
    There are all sorts
    of saddles made
  • 135:35 - 135:39
    in Texas, different
    ranches, different saddles.
  • 135:39 - 135:45
    So that was the harder one.
  • 135:45 - 135:48
    The ones that I actually
    saw on the finals,
  • 135:48 - 135:50
    some of the last
    three or four finals,
  • 135:50 - 135:55
    were much easier in the sense
    that the table you had to draw
  • 135:55 - 136:01
    was much shorter than
    this one-- in principle,
  • 136:01 - 136:05
    one critical value and
    one max and one min point.
  • 136:05 - 136:10
    But you have to be prepared
    more, rehearse more,
  • 136:10 - 136:12
    so when you see the
    problem in the midterm,
  • 136:12 - 136:16
    you say, oh, well that is
    easier than I'm used to.
  • 136:16 - 136:18
    That's the idea.
  • 136:18 - 136:20
    OK, go home.
  • 136:20 - 136:22
    Send me emails by WeBWorK.
  • 136:22 - 136:25
    We still have time to talk about
    the homework if you get stuck.
  • 136:25 - 136:28
  • 136:28 - 136:32
    And I'll see you Thursday.
  • 136:32 - 136:35
    [BACKGROUND CHATTER]
  • 136:35 - 136:59
Title:
TTU Math2450 Calculus3 Sec 11.6 and 11.7
Description:

Gradient, Tangent plane and Extrema of a functions of several variables

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Video Language:
English

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