Integration/Differentiation of Formulas

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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.
 

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  • #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.
 
  • #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:
 
  • #4
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@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?
 
  • #5
Office_Shredder
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theoretical higher exponent type dimension?
I have no idea what these words in succession mean.

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?
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.
 
  • #6
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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?
 
  • #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.
 
  • #8
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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?
 
  • #9
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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)
 
  • #10
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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?
 
  • #11
pwsnafu
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Okay, now I understand that, thanks. But now what about using that same principal to solve for anything? What is circumference a function of?
C=2πr. So the only candidate is radius.
 
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