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Linearization and Differentials

  1. Oct 15, 2015 #1
    I am currently taking Calculus 1 and we covered Linearization and Differentials. The title of the section in my textbook is called "Linear Approximations and Differentials," in the book by James Stewart. The book and lecture in this section made absolutely no sense to me. Like, I was COMPLETELY lost reading it prior to class, and after lecture I felt even more lost. I'm sure I am making it way more complicated than it needs to be, but typically when I have issues I use PatrickJMT to clarify things. He only has 1 video on Linearization. I was wondering if anyone has additional resources over these topics? Specifically differentials. I followed linearization to a degree, but I think I may have checked out once we hit differentials.

    Anyways, I am completely lost over these 2 topics. So, any resource would be fantastic. Thanks.

    Also, when referencing differentials, it separates dy/dx as it's own entity just dy and dx. Is this the same as delta y, delta x?
     
  2. jcsd
  3. Oct 15, 2015 #2

    mathman

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    It would be helpful if you gave specific examples of what is bothering you.
     
  4. Oct 15, 2015 #3
    It's the entire concept, not the examples. I didn't follow the concepts in deriving the formulas, nor their meaning. I understand it is simply a method of approximation with a point, a, that is near some x value. That's about all I understood, so I don't necessarily find it useful to present the examples, when the concept is lost on me. This is why I am asking for additional resources, rather than having someone solve some examples without me even understanding the underlying concept. I also don't wish it upon anyone to regurgitate a 2 hour lecture to me simply because I didn't understand it. I just didn't know if there were additional resources e.g. like PatrickJMT or KhanAcademy that people find useful in re-explaining the ideas of topics. If you'd like I can give you examples, I just really don't think it'll help me, so I don't want to waste anyone's time. Also, I hope I'm not coming off rude -- it's hard to read someones tone through text... but I'm not intending to be, so I'm sorry if I come off that way. I suppose I can sit a memorize the formulas, but without their underlying concept, I will inevitably forget them -- and I'm sure this topic will come up later and leave me in a similar spot.

    The formulas we were given are:

    L(x) = f(a) + f'(a) (x - a)
    dy = f'(x) dx (I understand this one symbolically, but not graphically)
    f(a + dx) = f(a) + dy (approx.)
    DeltaV / V = dV / V = [4(pi)r^3dr] / [(4/3)(pi)r^3] = (3dr) / r
     
  5. Oct 15, 2015 #4

    SteamKing

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    Let's take this formula first:

    L(x) = f(a) + f'(a) (x - a)

    For an approximate value of f'(a), we can use Δy / Δx, like in this pitcher:


    Introductory_Physics_fig_1.15.png
    f'(a) ≈ Δy / Δx, f(a) is point A on the red curve, and L(x) is the line going thru points A and B. L(x) is basically the formula for linear interpolation.

    The ratio, Δy / Δx, turns into dy/dx, only in the limit as Δx → 0. In the pitcher above, imagine point B moving closer and closer to point A along the red curve. The closer B gets to A, the closer the line segment AB comes to coinciding with the tangent to the curve at point A.

    The formula, dy = f'(x)dx, is just a way of saying that for each additional movement dx in the x-direction, there will be a change dy in the y-direction, which is what the ratio dy/dx tells you. It's approximately the slope of the tangent line in this portion of the curve.
     
  6. Oct 15, 2015 #5

    Mark44

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    L(x) represents the equation of the line that is tangent to the graph of f at the point (a, f(a)). If x is "close to" a, then L(x) will be "close to" f(x).

    To get a geometric feel for this, draw a curve (y = f(x)) and draw the tangent line at the point (a, f(a)).
    If y = f(x), the above is the definition of the differential of y.
    Assuming y = f(x),
    ##f(a + \Delta x) - f(a) = \Delta y \approx dy##. Provided that ##\Delta x## is suitably small, ##\Delta x## and dx are interchangeable, making this equation equivalent to the one above.
    Here V is the volume of a sphere. The actual relationship is
    ##\frac{\Delta V} V \approx \frac{dV} V = \frac{4 \pi r^3 dr}{(4/3) \pi r^3} = \frac{3 dr} r##
    The upshot here is that ##\frac{\Delta V} V \approx \frac{3 dr} r \approx \frac{3 \Delta r} r##
    IOW, the relative change in V is approximately (3/r) times the change in the radius.
     
  7. Oct 16, 2015 #6
    WOW. Thank you everyone. Between this and a book I have, I think I understand, but let me make sure.

    The first formula makes complete sense, now. But, I don't really see the sense in the second equation. It seems essentially the same, just showing that dy is dependent on dx, which seems obvious?

    Anyway, to be sure dy is the approximate change in y. The second equation is essentially using the change in x with the derivative of the function to approximate the change in y, so that when we add our f(a) to dy we get an pretty accurate answer as to what f(x) is. This seems like the same as the first equation, just in separate steps.

    The volume one still seems a little mysterious to me. But, is it saying the same thing as the above but with respect to r? That is 3dr/r gives me a pretty good estimate of the volume. Is this an equation that will typically be used continuously? I mean, I see how it's derived, but is it something I should commit to memory?
     
  8. Oct 16, 2015 #7

    Mark44

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    As I said before, that equation defined the differential dy.
    No. ##\frac{3dr}{r}## doesn't represent the change in volume -- it's an estimate of the relative change in volume, ##\frac{\Delta V}{V}##.
    You shouldn't commit that equation to memory, but you should understand how the equation is derived well enough so that you can apply the idea to any equation relating two variables.

    Here's an easy example with a surprising answer.
    Suppose a very long cable is laid along the earth's Equator (assume the earth is a perfect sphere, with no mountains or valleys or ocean depths where the cable runs). If the cable is lifted 1 meter all along its length, how much extra cable will be required?
     
  9. Oct 16, 2015 #8

    mathman

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    The basic idea is that if a function is "nice" it can be expanded in a Taylor series around the point of interest.

    f(x)=f(a)+f'(a)(x-a)+remainder, where the remainder is of the order [itex](x-a)^2[/itex]. For approximation, the remainder is small for x near a.
     
  10. Oct 17, 2015 #9
    I guess my confusion in dy=f'(x)dx is that in the homework set whenever we used differentials, we used them in the same way we used linearization. We would solve for dy then add f(a) to dy. Which is doing the same thing as linearization, just in two separate steps. We were only assigned one problem where we found dy, then solved for Δy to see the variance. It just seemed a little redundant based on the problem set, I guess. Is linearization typically used more, unless you need to find specifically a change in the y-values?

    Other than that, this helped a lot. I was able to finish my HW. Thanks everyone.
     
  11. Oct 17, 2015 #10
    2π?
    C = 2πr
    dC = 2πdr, where dr = 1 meter
    dC = 2π(1)
    dC = 2π meters

    Weird.
     
  12. Oct 17, 2015 #11

    Mark44

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    Yes, that's right. It's surprising, at least it was to me the first time I saw it, because you would think that lifting the cable up a meter all around the world would take a lot more than just a bit more than 6 meters.

    A nicer way to write what you did is this:
    ##\Delta C \approx dC = 2\pi dr \approx 2\pi \Delta r##
    If ##\Delta r## = 1 m., then ##\Delta C \approx 2\pi## meters.
     
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