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Math Help - Need clarification regarding definite integration!

  1. #1
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    Need clarification regarding definite integration!

    Ok, we all know that definite integration of velocity functions, dy/dx, or f' gives you displacement/distance of the function (depending on the situation).

    However, I have encountered situations where I used definite integration of differentiated functions to get the "quantity" of something. For example, suppose a graph shows the rate of leaves being produced over a season; you would use definite integration of the rate function in a given interval to calculate the sum of leaves produced. Nothing to do with displacement/distance!

    This example brings my point. Why can definite integration be used in two completely different ways? I'm just worried that I may be missing something. Maybe standard velocity/dy&dx/rate/f' equations are derived in terms of the x-axis and the other rate equation forms (designed for quantity problems) are actually in terms of "t" (time)? I just don't know. I'd appreciate it so much if someone could help me out. Thanks!!!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Recall f ' is the rate of change of a quantity --is there not an infinity of meanings of f ' depending on what f represents the quantity of ?

    Turning it around f ' is the rate of change of a quantity and the def integral is the total change in that quantity as the independent variable varies over an interval . Again an infinity of possibilities.

    displacement and velocity is just one application of a general situation.

    Remember displacement is just the change in position s(t2) - s(t1) and velocity is the derivative of position
    v(t) = s ' (t)
    Last edited by Calculus26; April 14th 2009 at 06:05 PM.
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    I'm sorry but I've no idea what you just said.

    Please, answer this simply - when you definite integrate a differentiated function (therefore a rate function), how would you know if you found distance/displacement or the total quantity of something? Or are they the same thing?


    Just simple answers please!
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Whenever you integrate f ' from a to b you are calculating the change in
    f as x vaies from a to b. Period.

    If f ' is the velocity you are calculating the change in position which is what the displacement is.

    I f ' is the rate at which leaves are produced you are calculating the change in the number of leaves produced.

    etc etc
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    I see. Allright, here's one question. Let's say that, after definite integrating a leaf rate function from a to b, you get 100 leaves produced. Does that also mean that the original undifferentiated function experiences a displacement of 100 units from a to b (if you're talking about in terms of graph position)?

    Thanks for trying to clear this up, I'm getting there.
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    MHF Contributor Calculus26's Avatar
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    A matter of terminology

    displacement is a physics term which refers to the change in position of a particle with velocity f '.

    It is not a math term which somehow refers to the displacement of a graph in some way.

    Consider the FTC

    f(b)-f(a) is the change in f period-- The change in f is called the displacement only when f ' is the velocity.

    In normal conversation you wouldn't say the the number of leaves was displaced by 100 you would say the number of leaves changed by 100--same in math.
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    Deleted.
    Last edited by Grandad; April 15th 2009 at 07:44 AM. Reason: Delete, please. Corrected version posted.
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    Hello Kaitosan
    Quote Originally Posted by Kaitosan View Post
    I see. Allright, here's one question. Let's say that, after definite integrating a leaf rate function from a to b, you get 100 leaves produced. Does that also mean that the original undifferentiated function experiences a displacement of 100 units from a to b (if you're talking about in terms of graph position)?

    Thanks for trying to clear this up, I'm getting there.
    Bear in mind that a displacement is just a measure of a quantity - just as volume, mass, number of leaves, population of a city, the number of inches in someone's waist measurement, etc, etc, are. Displacement happens to measure distance, that's all. Apart from that, there's nothing special about it.

    We very often need to talk about the rate at which displacement is changing over time. In fact, it's so important that we have a special word for it: velocity. But velocity is just the rate of change of displacement with respect to time, that's all. No more, no less.

    Suppose we know how the population of a city, P, varies over time - other words we have P = f(t) ( P is a function, f, of t) - and we want to know how fast the population is changing at a particular moment in time. Then we shall need to find the value of \frac{dP}{dt}. But because this isn't as common a problem as those involving displacement, we don't have a special word for this quantity - it's just the rate of change of the population with respect to time.

    If the population is measured in 1000's of people and the time in years, then \frac{dP}{dt} will be measured in 1000's of people per year. So if at a given moment, \frac{dP}{dt} = 5, it means that the population of the city is growing at the rate of 5000 people per year, at that moment.

    OK, now suppose we are given the problem the other way around: we know the rate of change of the population as a function of t - in other words,we know that \frac{dP}{dt} = f'(t), and need to know how much the population has increased between t = a and t = b.

    Then we simply say:

    \frac{dP}{dt} = f'(t)

    \Rightarrow P = \int_a^bf'(t)\,dt = f(a) - f(b)

    This quantity f(b) - f(a), then, is the population of the city at time t = b minus the population at time t = a. In other words, it's the increase in the population (measured in 1000's of people, or whatever) over the period t = a to t = b.

    Does that help to make it clearer?

    Grandad
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    Thanks guys, I got it.
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    MHF Contributor Calculus26's Avatar
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    Grandad --- not to put too fine of a point on it but dispalcement is not a measure of distance generally.

    displacement is the integral of v(t) whereas distance is the integral of
    |v(t)|

    for example if a projectile is fired from ground level to a height of 100ft and then falls to earth the disatance is 200 ft but the displacement is 0.
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  11. #11
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    Hello Calculus26
    Quote Originally Posted by Calculus26 View Post
    Grandad --- not to put too fine of a point on it but dispalcement is not a measure of distance generally.

    displacement is the integral of v(t) whereas distance is the integral of
    |v(t)|

    for example if a projectile is fired from ground level to a height of 100ft and then falls to earth the disatance is 200 ft but the displacement is 0.
    Of course, you are quite correct. However, I tend to tailor my replies in a way that I think will suit the needs of the one who is asking the question. So forgive me if, in trying to keep things simple, I have used the term sloppily.

    Grandad
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