Retourner vers la vidéo

Implicit Differentiation and Linear Approximation | MIT 18.01SC Single Variable Calculus, Fall 2010

  • 0:07 - 0:09
    Hi! Welcome back to recitation.
  • 0:09 - 0:15
    In lecture, you've been talking about implicitly defined functions and implicit differentiation.
  • 0:15 - 0:20
    One of the reasons that these are important, or that implicit differentiation
  • 0:20 - 0:23
    is important, is that sometimes you have a function defined implicitly
  • 0:23 - 0:28
    and you can't solve for it. You don't have any algebraic method for computing
  • 0:28 - 0:31
    the function values as a formula.
  • 0:31 - 0:36
    For example, this function that I've written on the board
  • 0:36 - 0:40
    that I've called w of x is defined implicitly by the equation that
  • 0:40 - 0:46
    w of x plus one, quantity times e to the w of x, is equal to x for all x.
  • 0:46 - 0:51
    So this function, some of its values you can guess.
  • 0:51 - 0:57
    Like at x equals zero, the function value is going to be negative one.
  • 0:57 - 1:01
    The reason is that this can't ever be zero, so the only way to get this side to be zero
  • 1:01 - 1:04
    is if w is negative one, if this term is zero.
  • 1:04 - 1:08
    So, some of its values are easy to compute, but some of its values aren't.
  • 1:08 - 1:13
    So for example, if I asked you what w of three halves is,
  • 1:13 - 1:18
    it's very hard, there's no algebraic way you can manipulate this equation
  • 1:18 - 1:22
    that will let you figure that out. So in that situation you might still care
  • 1:22 - 1:26
    about what the function value is. So what can you do?
  • 1:26 - 1:29
    Well, you can try and find a numerical approximation.
  • 1:29 - 1:33
    So in this problem I'd like you to try and estimate the value w of three halves
  • 1:33 - 1:48
    by using a linear approximation of the function w of x in order to compute this value.
  • 1:48 - 1:59
    As a hint, I've given you that w of one is zero. Right? If you put in
  • 1:59 - 2:09
    x equals one and w of one equals zero on the left hand side, you do indeed get one, as you should.
  • 2:09 - 2:19
    So, good! So that will give you a hint about where you could base your linear approximation.
  • 2:19 - 2:23
    Why don't you pause the video, and take a few minutes to work this out.
  • 2:23 - 2:32
    Come back and we can work it out together.
  • 2:32 - 2:37
    All right, welcome back! So, hopefully, you've had a chance to work on this question a little bit.
  • 2:37 - 2:43
    In order to do this linear approximation that we want, we need to know a base point
  • 2:43 - 2:46
    and we need to know the derivative of the function at that base point.
  • 2:46 - 2:51
    Those are the two pieces of data you need in order to construct a linear approximation.
  • 2:51 - 2:56
    We have a good candidate for a base point here, which is the point one, zero.
  • 2:56 - 3:00
    So this curve, whatever it looks like, it passes through the point one, zero.
  • 3:00 - 3:03
    And that's the point we're going to use for our approximation.
  • 3:03 - 3:16
    So we're going to use the linear approximation w of x is approximately equal to
  • 3:16 - 3:33
    w prime of one, times x minus one, plus w of one when x is approximately equal to one.
  • 3:33 - 3:38
    So this is the linear approximation we're going to use. And we have that w of one here is zero.
  • 3:38 - 3:48
    So this is equal to w prime of one, times x minus one.
  • 3:48 - 3:51
    (So the w one of zero just goes away.)
  • 3:51 - 3:58
    In order to estimate w of x, in particular w of three halves, what we need to know is
  • 3:58 - 4:03
    we need to know the derivative of w. And to get the derivative of w
  • 4:03 - 4:08
    we need to use, well, we have only one piece of information about w,
  • 4:08 - 4:16
    that it's defined by this implicit equation. So in order to get the derivative of w
  • 4:16 - 4:19
    we have to use implicit differentiation. OK?
  • 4:19 - 4:26
    So let's do that. So if we implicitly differentiate this equation --
  • 4:26 - 4:28
    so let's start with -- the right hand side's going to be really easy --
  • 4:28 - 4:32
    we're going to differentiate with respect to x. The right hand side's going to be one.
  • 4:32 - 4:35
    On the left hand side, it's going to be a little more complicated.
  • 4:35 - 4:39
    We have a product, and then this piece, we're going to have a chain rule situation.
  • 4:39 - 4:54
    We have e to the w of x. So, [implicit diff 'n], we're going to take an implicit derivative.
  • 4:54 - 4:59
    On the left, so, the product rule first. We take the derivative of the first part.
  • 4:59 - 5:06
    So that's just w prime of x, times the second part. That's e to the w of x.
  • 5:06 - 5:15
    Plus the first part. That's w of x plus one, times the derivative of the second part.
  • 5:15 - 5:24
    So the second part is e to the w of x. So that gives me an e to the w of x, times w prime of x.
  • 5:24 - 5:29
    That's the chain rule. So that's what happens when I differentiate the left hand side.
  • 5:29 - 5:33
    And on the right hand side, I take the derivative of x and I get one.
  • 5:33 - 5:40
    OK? Good. So now I've got this equation and I need to solve this equation for w prime.
  • 5:40 - 5:45
    Because if you look up here, that's what I want. I want a particular value of w prime.
  • 5:45 - 5:51
    And as always happens in implicit differentiation, from the point of view of this w prime,
  • 5:51 - 5:54
    it's only involved in the equation in a very simple way.
  • 5:54 - 6:01
    So there's it multiplied by functions of x and w of x. But not... it's just, you know,
  • 6:01 - 6:06
    it's just multipled by something that doesn't involve w prime at all.
  • 6:06 - 6:09
    And here it's multiplied by something that doesn't involve w prime at all.
  • 6:09 - 6:14
    So you can just collect your w primes and divide through. It's just like solving a linear equation.
  • 6:14 - 6:29
    So here if we collect our w primes, this is w prime of x, times, it looks like, w of x, plus two
  • 6:29 - 6:36
    times e to the w of x. Did I get that right?
  • 6:36 - 6:41
    Looks good. OK, so that's still equal to one.
  • 6:41 - 6:48
    That means w prime of x is just, well not just, it's equal to
  • 6:48 - 6:58
    one over w of x plus two, times e to the w of x.
  • 6:58 - 7:04
    OK, so this is true for every x. But I don't need this equation for every x.
  • 7:04 - 7:08
    I just need the particular value of w prime at one.
  • 7:08 - 7:12
    So I'm going to take this equation then and I'm just going to put in
  • 7:12 - 7:17
    x equals one. So I put in x equals one. I'll do it over here.
  • 7:17 - 7:22
    So I get w prime of one. And everywhere I had an x, I put in a one.
  • 7:22 - 7:27
    So actually, in this equation, the only place x appears is in the argument of w.
  • 7:27 - 7:34
    So this is w of one, plus two, times e to the w of one.
  • 7:34 - 7:39
    OK, so in order to get w prime of one, I need to know what w of one is.
  • 7:39 - 7:44
    But I had that. I had it -- it was right back here. There was the - that was my hint to you.
  • 7:44 - 7:51
    This is why we are using this point as a base point. Which is, we know the value of w for this value of x.
  • 7:51 - 8:04
    So we take that value. So w of one at zero, so this is just one over two.
  • 8:10 - 8:14
    OK? So I take that back upstairs to this equation that I had here.
  • 8:14 - 8:21
    And I have that w of x is approximately equal to w prime of one, times x minus one.
  • 8:21 - 8:31
    So w of x is approximately equal to (w prime of one we saw is one half)
  • 8:31 - 8:41
    times x minus one. And that approximation was good near our base point,
  • 8:41 - 8:47
    so that's good when x is near one.
  • 8:47 - 8:51
    All right. So this is the linear approximation. I asked for the linear approximation
  • 8:51 - 8:55
    -- its value at the particular point x equals three halves.
  • 8:55 - 9:04
    So w of three halves is approximately one half times (well, three halves
  • 9:04 - 9:13
    minus one is also a half) so this is a quarter.
  • 9:13 - 9:16
    OK. So this is our estimate for w of three halves.
  • 9:16 - 9:19
    W of three halves is approximately one fourth.
  • 9:19 - 9:22
    If you wanted a better estimate, you could try iterating this process,
  • 9:22 - 9:28
    or choosing some base point even closer if you could figure out
  • 9:28 - 9:35
    the value of w and x near this point that you're interested in -- three halves.
  • 9:35 - 9:41
    Just to sum up what we did was we had this implicitly defined function w.
  • 9:41 - 9:45
    We wanted to estimate its value at a point where we couldn't compute it explicitly.
  • 9:45 - 9:48
    So what we did, was we did our normal linear approximation method.
  • 9:48 - 9:54
    We wrote down our normal linear approximation formula.
  • 9:54 - 9:59
    The only thing that was slightly unusual is that we had to use implicit differentiation.
  • 9:59 - 10:04
    In order to compute the derivative that appears in the linear approximation, we implicitly differentiated.
  • 10:04 - 10:08
    OK? So that happened just like normal. And then at the end, we plugged in the values
  • 10:08 - 10:15
    that we were interested in to actually compute the particular value of that approximation.
  • 10:15 -
    So I'll end there.
Titre:
Implicit Differentiation and Linear Approximation | MIT 18.01SC Single Variable Calculus, Fall 2010
Description:

Implicit Differentiation and Linear Approximation

Instructor: Joel Lewis

View the complete course: http://ocw.mit.edu/18-01SCF10

License: Creative Commons BY-NC-SA
More information at http://ocw.mit.edu/terms
More courses at http://ocw.mit.edu

plus » « moins
Langue de la vidéo:
English
Équipe:
MIT OpenCourseware
Durée:
10:17

sous-titres en English

Révisions