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Why are differentials used?

  1. Apr 5, 2004 #1

    ShawnD

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    I've posted something like this before, but this one questions usage rather than reliability.

    If f(x) = x^2, and I want to find the change between x at 10 and 11. I could just fill in values
    11^2 - 10^2 = 21

    Or I could make a differential
    dy = 2x dx
    dy = 2(10)(1)
    dy = 20

    Just filling in values took about 3 seconds, but making a differential actually required thinking and time.
    If a differential is less accurate and takes more time, why are they even used?
     
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  3. Apr 5, 2004 #2

    matt grime

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    Suppose you were asked to estimate the change in going from 10 ro 10.000000043214

    do it by hand without differentials if you like but I can say that to within, what, 12 sigfigs the answer is 0.00000086428

    edit (I didnt' count the zeroes exactly but you get the idea)
     
  4. Apr 5, 2004 #3

    Zurtex

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    Those are 2 different things you have worked out. The 1st one is 11^2 - 10^2, the 2nd one is the gradient of the tangent of x^2 at x = 10.

    You could approximate gradients by simply working out [tex]\frac{\Delta y}{\Delta x}[/tex] but it is far less accurate.

    So if you meant that as the gradient you have done the equivalent of drawing a straight line through the points x=10 and x=11 on the line and worked out the gradient of that line.
     
    Last edited: Apr 5, 2004
  5. Apr 5, 2004 #4

    ShawnD

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    Are there any cases where working out the gradient is easier?

    In math classes, I've only seen equations where it would simply be easier to fill in values. Here is a question I had on a quiz once:

    The equivalent resistance R when a resistance of 20 ohm and a variable resistance Rv are connected in parallel is

    [tex]R = \frac{20R_v}{20 + R_v}[/tex]

    If the variable resistance is changed from 10.0 to 10.1 ohms, what is the change in the equivalent resistance R?

    To make a differential and solve for it, it took me 5 lines of writing. If I just filled in values, it would take 2 lines.
     
  6. Apr 5, 2004 #5

    Kurdt

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    A differential measures the RATE of change of something whereas filling in values measures the actual change. For example your first example states that the change from x=10 to x=11 is 21 by filling in values but your integral is calculated incorrectly. It should read.

    [tex]\int_{10}^{11} 2x dx = \left[ x^2 \right]_{10}^{11}=11^2-10^2=21[/tex]

    which is your original fitting in of values and is perfectly accurate. Differentials only really come into play when modelling things that do not have a linear relationship with some reference frame or other quantity, such as the density of the sun which varies with radius or heat capacities which vary with temperature.
     
    Last edited: Apr 5, 2004
  7. Apr 5, 2004 #6

    ShawnD

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    What you just did was take the derivative then integrate it. I'm talking about just taking the derivative.

    [tex]dy = 2x dx[/tex]


    here is the question
    Notice the way it was done.
     
    Last edited: Apr 5, 2004
  8. Apr 5, 2004 #7

    Kurdt

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    Ok. Well again in your previous example the use of the derivative on [tex] x^2 [/tex] would tell you the rate of changeof the function. Say [tex] x^2 [/tex] represented a change of position with time, then the derivative would tell us what speed a particle was travelling at at a certain point on the graph. so if you changed position from 10 meters to 11 meters the corresponding difference in the derivative would tell you how much your speed changed in that interval as opposed to the rate of change at a point. so a value of 2 just happens to be the acceleration if you took the second derivative which is constant throughout but thats because you took a unit change.
     
  9. Apr 5, 2004 #8

    matt grime

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    But writing a couple of lines is not necessarily less computationally intensive than writing 5 lines if the question is significantly different. Once you've computed the differential stuff it becomes very easy to work it out for many different values of dx, and if they are sufficently small this is all you require. JUst imagine cases where the derivative is easier than the original function to calculate! Say I want you to work out an estimate for changing 4x^5+x^4 at x=1 f dx=0.1, 0.01, 0.0023. Wouldn't it be easier touse the linearizations that actually compute the damn function at all those points? Suppose I want to charge you a million dollars per floating point operation, which would you rather use then? Remember, the SIMPLIFIED situations you are using things for aren't all there is. If you weren't using linearization all over the place you wouldn't be able to to calculate anything worth a damn.
     
  10. Apr 5, 2004 #9

    ShawnD

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    When you put it in computer terms it makes sense.

    Thanks Grime.
     
  11. Apr 7, 2004 #10

    pig

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    it is also useful when it is hard to calculate the value of the function itself, for example, for some reason you need to calculate the square root of 45 by hand..

    knowing that 7^2 is 49, you approximate it to, lets say 6.8. 6.8^2 is 46.24 which is off by 1.24.

    but then the root of 45 ~ 6.8 - 1.24/13.6 = 6.8 - 31/340.
    this gives 6,7088.. while 45^(1/2) = 6,7082.. which is quite close for the amount of work involved :)
     
    Last edited: Apr 7, 2004
  12. Jul 23, 2004 #11

    mathwonk

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    I agree that soemtimes I feel the only reason we teach the use of differentials for approximating differences is to introduce the concept of differentials. I.e. that approximation technique is not really the best use of them.

    More important is the method of substitution, where you substitute x = x(u) and dx = (dx/du) du. I.e. differentials are a convenient calculating tool in many situations, such as path integrals where the general integrand looks like A(x,y)dx + B(x,y)dy, and the differentials are visible in there.

    Also differential equations are called that for a reason. For instance solving them by separation of variables is another important use of differentials, where you solve

    dy/dx = y by changing it to dy/y = dx, so ln(y) = x+c so y = a e^x.
     
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