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Integration/Differentiation of Formulas

  1. Oct 15, 2013 #1
    So I am new to calculus, and have been playing around with integration and differentiation. I have learned that integrating and deriving formulas, you get other formulas that can be used, like integrating surface area of a sphere to get volume. I was wondering why this is, and what limits you from continuing to integrate on and on. For example, you can integrate the surface area of a sphere, but integrating the volume of a sphere doesnt get you anything.
     
  2. jcsd
  3. Oct 15, 2013 #2

    Office_Shredder

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    For the sphere - ball relationship, the key is that if you slice up a ball into spheres of constant radius, then each spherical slice has volume (surface area)*(delta r) and adding all of those up gives the volume of the ball. To integrate the ball's volume and get something useful you need to have a four dimensional shape which can be sliced up into a bunch of three dimensional balls.
     
  4. Oct 15, 2013 #3

    LCKurtz

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    I was slicing up a 4 dimensional orange with my high-tech 4 dimensional knife and accidentally cut 4 fingers at once. 4D is tricky. :smile:
     
  5. Oct 15, 2013 #4
    @Office_Shredder
    So to integrate anything, you need a theoretical higher exponent type dimension? Then why can you take a two dimensional surface area and integrate it to find the volume? V = ∫ f(r) 4πr2 dr right? Integrating that would result in a cubed radius?
     
  6. Oct 15, 2013 #5

    Office_Shredder

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    I have no idea what these words in succession mean.

    You can take a two dimensional surface area and integrate it to find the volume, in the sense that you can integrate the surface area of a sphere to get the volume of a ball. I don't know what your function f(r) is supposed to be here - if it's completely arbitrary, then you're just integrating an arbitrary function.
     
  7. Oct 15, 2013 #6
    So lets take position for example, a function of time. s(t). Why is it that when you take the derivative of position, you get velocity, and from there, acceleration, jerk, and jounce? Why does it start with a function of time?
     
  8. Oct 15, 2013 #7

    Office_Shredder

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    What are your definitions of velocity and acceleration and jerk and jounce? Velocity is typically defined as the instantaneous change in position as a function of time, which is the definition of the derivative. Acceleration is the instantaneous change in velocity as a function of time, which is again a derivative by definition.

    If you have a different definition then we can work through how one is a derivative of another.
     
  9. Oct 15, 2013 #8
    I suppose what i'm really trying to ask, why is position, a function of time. and why is velocity, a function of position? why is acceleration a function of velocity? and why is jerk a function of acceleration?
     
  10. Oct 15, 2013 #9

    Office_Shredder

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    Acceleration is not a function of velocity, it's also a function of time (in the typical physics setup). Jerk is a function of time as well. If s(t) is position, v(t) velocity, a(t) acceleration and j(t) jerk, we have
    s'(t) = v(t)
    s''(t) = v'(t) = a(t)
    s'''(t) = v''(t) = a'(t) = j(t)
     
  11. Oct 15, 2013 #10
    Okay, now I understand that, thanks. But now what about using that same principal to solve for anything? What is circumference a function of? I learned the integral of the circumference, is the area. But what is the beginning formula?
     
  12. Oct 16, 2013 #11

    pwsnafu

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    C=2πr. So the only candidate is radius.
     
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