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Differtential Calculus

  1. Apr 25, 2005 #1
    Can anyone explain to me about Differnetial calculus and Integral calculus.
     
  2. jcsd
  3. Apr 25, 2005 #2

    James R

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    That's a very broad topic. What do you want to know?
     
  4. Apr 25, 2005 #3
    About using differential coeffecient or derivative to define velocity and accelleration and finding area of irregular objects.
     
  5. Apr 25, 2005 #4
    Given a curve f(x) on an x-y plane, the area between the curve and the x-axis can be found by integrating the function.
    The slope at any point can be found by taking the derivative of the function.
    Graphs of position, velocity, and acceleration all tend to the above principles given that acceleration is the derivative of velocity, and velocity is the derivative of position.

    The integral of a derivative of a function yields the original funciton, aka they are inverse processes, similar to arcsin and sin.
     
  6. Apr 25, 2005 #5
    Here are a few of the fundamental definitions of calculus:

    Calculus can be defined with the study of limits, so I first define limits of various types:

    If [itex]f(x)[/itex] is a real function of a real variable [itex]x[/itex], and [itex]a \in \mathbb{R}[/itex], we define

    [tex]\lim_{x \rightarrow a} f(x) = L \in \mathbb{R}[/tex]

    to mean

    [tex]\exits \delta \in \mathbb{R} \ \forall \varepsilon > 0 \ \mbox{s.t} \ | x - a | < \delta \Longrightarrow | f(x) - L | < \varepsilon.[/tex]

    If this definition is satisfied, we say that [itex]L[/itex] is the limit of [itex]f(x)[/itex] as [itex]x[/itex] goes to [itex]a[/itex].

    If there is no [itex]L \in \mathbb{R}[/itex] satisfying the definition, we say that the limit of [itex]f(x)[/itex] as [itex]x[/itex] goes to [itex]a[/itex] does not exist.

    In a similar fashion, we define (with the same assumptions on f(x))

    [tex]\lim_{x \rightarrow \infty} f(x)= L[/tex]

    to mean that

    [tex]\exists D \in \mathbb{R} \ \forall \varepsilon > 0 \ \mbox{s.t.} \ x > D \Longrightarrow |f(x) - L | < \varepsilon.[/tex]

    In this case, we say that [itex]L[/itex] is the limit of [itex]f(x)[/itex] as [itex]x[/itex] goes to infinity. If there is no such [itex]L \in \mathbb{R}[/itex], we again say that the limit does not exist.

    There are other analogous definitions to generalize the concept of a limit as well, but I don't need them for the rest of this post.

    Now, we define the derivative of a function. Let [itex]f(x)[/itex] be as before. Then the derivative of [itex]f(x)[/itex] at the point [itex]x=a[/itex] with respect to [itex]x[/itex] is defined to be

    [tex]\frac{d}{dx} f(x) = \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}[/tex]

    if this limit exists. If the limit does not exist, we say the derivative of [itex]f(x)[/itex] is undefined at the point [itex]x=a[/itex]. If the derivative of [itex]f(x)[/itex] (with respect to [itex]x[/itex] exists at every point in some interval [itex]I[/itex], then we say that [itex]f(x)[/itex] is differentiable on [itex]I[/itex].

    This definition is motivated geometrically. If the derivative of a function exists at some point, then the derivative is the slope of the tangent to the function at that point.

    Finally I'll define the most common type of integral that you'll run into (at least for a while), the Riemann integral, in a limited context.

    Consider a real function [itex]g(x)[/itex] of a real variable [itex]x[/itex], defined, bounded, and differentiable (I'm too lazy to define continuity seperately!) on an interval [itex]I = \left[ a, b\right][/itex]. Let

    [tex]\Delta = \frac{b-a}{n}[/tex]

    and let [itex]I_n = \left[ a + (n-1)\Delta, \ a + n\Delta\right][/itex], and let [itex]x^*_n \in I_n[/itex]. We define the definite integral of [itex]f(x)[/itex] from [itex]a[/itex] to [itex]b[/itex] with respect to [itex]x[/itex] to be

    [tex]\int_a^b f(x) \ dx = \lim_{n \rightarrow \infty} \sum_{i=0}^n f(x^*_n)\Delta.[/tex]

    This definition is also motivated geometrically. If you have a positive function [itex]f(x)[/itex], the integral

    [tex]\int_a^b f(x) \ dx[/tex]

    is the area bounded by [itex]x=a, \ x=b, \ y=0,[/itex] and [itex]y = f(x)[/itex]. Essentially, you are adding up the areas of infinitely many infinitessimally thick rectangles of height [itex]f(x)[/itex], all along the interval [itex]\left[ a , \ b \right][/itex] to get the area of the whole shape.


    I will conclude by stating an important theorem, commonly known as the Fundamental Theorem of Calculus. It is as follows (again in a limited context):

    Suppose that [itex]F(x)[/itex] is a real function of a real variable, differentiable on [itex]I = \left[ a, \ b \right] [/itex], and that

    [tex]\frac{d}{dx}F(x) = f(x)[/itex]

    is differentiable on [itex]I[/itex]. Then

    [tex] F(b) - F(a) = \int_a^b f(x) \ dx.[/tex]

    In addition, for a bounded, real function [itex]f(x)[/itex] of a real variable, differentiable on [itex]\left[a, \ t \right][/itex], we find

    [tex]f(t) = \frac{d}{dt}\int_a^t f(x) \ dx.[/tex]

    This theorem essentially gives technical meaning to what whozum said, that differentiation and integration are in some sense inverse operations.



    In terms of physics, if the function [itex]x(t)[/itex] describes the position of a particle as a function of time, then

    [itex]\frac{d}{dt}x(t)[/itex]

    is the velocity of the particle as a function of time, and

    [itex] \frac{d}{dt}\left(\frac{d}{dt} x(t)\right),[/itex]

    also denoted

    [tex] \frac{d^2}{dt^2} x(t)[/tex]

    is the acceleration of the particle as a function of time.
     
  7. Apr 26, 2005 #6
    I want to know this:
    if x(t) = 5m t=5s then what is the instantaneous velocity at t=2s
    Is it: dx(t)/dt = 2/2 = 1m/s
     
  8. Apr 26, 2005 #7
    I don't really understand what you're asking. Do you mean

    [tex]x(t) = \left( 5\frac{\mbox{m}}{\mbox{s}}\right) t[/tex]

    ? If so, then

    [tex]\frac{d}{dt}x(t) = 5\frac{\mbox{m}}{\mbox{s}}[/tex]

    and the speed is 5 m/s no matter what t is.

    On the other hand, let's say

    [tex]x(t) = 5t^2.[/tex]

    In this case, it's a little more exciting. You get

    [tex]\frac{d}{dt} x(t) = 10t[/tex]

    so at [itex]t=2[/itex], the speed is 20, but at [itex]t=5[/itex], the speed is 50.
     
  9. Apr 27, 2005 #8

    Gza

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    lol, calculus has just a few more applications than that. :smile:
     
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