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Why is rate of change related to the area under a function

  1. Feb 16, 2013 #1
    I dont understand why integrals and derivatives work, and i dont understand why theyre so closely related.

    Let's take a function y= x^2 + 2x + 9

    y' = 2x + 2

    Why do the rules for taking derivatives work? Why does reducing the power of a term by 1 and adding that as a coefficient work to find the rate of change? This might be like asking why gravity exists, but i'm curious.

    Now in regards to my main question, take that derivative, y' = 2x + 2 and graph it.

    If you want to calculate the area of the curve under the derivative, you can just do so using the original term it's derived from (by dropping any constants that would get lost in the derivation).

    So i dont understand why a function can also serve to express the area under the function of it's derivative (when you drop constant terms.

    I know mathematically it makes sense because the process of derivation and antiderivation are opposites, so i mathematically understand the connection, but i dont get the relationship between RoC and area.
     
  2. jcsd
  3. Feb 16, 2013 #2

    micromass

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    It works because we can prove that it works. Have you ever proven the derivative laws? Or were you just required to memorize them?

    Consider the following graph:

    500px-FTC_geometric.svg.png

    Let [itex]A(x)[/itex] be the area from 0 to x under the graph. The value [itex]A(x+h)-A(x)[/itex] is given by the red piece in the graph. If h is small, then we can approximate it by a small rectangle. The rectangle has sides h and f(x). So A(x+h) is the area up to x+h is given by the area up to x plus the rectangle. So [itex]A(x+h)\sim A(x) + f(x)h[/itex]. This gives us that the rate of change of A is given by [itex]\frac{A(x+h)-A(x)}{h}\sim f(x)[/itex]. So if h is small, then the rate of change of A is f(x). This is an (informal) explanation of why the connection works.
     
  4. Feb 16, 2013 #3
    First question: it is just a trick which prevents you from doing the whole boring computation. Formally you would write:
    [tex]
    y'(x) = \lim_{t\rightarrow0}\frac{y(x+t)-y(x)}{t}
    [/tex]
    Apllying this to whatever polynomial you'll find the trick you're used to.

    Second question: how would one compute the area under y'(x)? Well, one way is to divide the x-axis is small intervals of measure t, ([itex]\left[0,t\right] [/itex] etc. ) and compute the area of the little rectangle made by each interval and sum all rectangles: (for simplicity consider an interval (0,a) )

    [tex]
    A=\sum_{i=0}^{N}t\cdot y'(it)
    [/tex]
    With a=tN.
    This is just
    [tex]
    \sum_{i=0}^{N}t\cdot\lim_{t\rightarrow0}(y(it+t)-y(it))/t =\lim_{t\rightarrow0}\sum_{i=0}^{N}y(it+t)-y(it)
    [/tex]
    The only terms that do not vanish (they are not summed and subtracted in consequent terms) are the first and the last one, so that:

    [tex]
    A=\lim_{t\rightarrow0}y(a)-y(0)=y(a)-y(0)
    [/tex]
     
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