Integration/Differentiation of Formulas

  • Thread starter Periapsis
  • Start date
  • Tags
    Formulas
In summary, the key to understanding the relationship between a ball and a sphere is that if you slice up a ball into spheres of constant radius, 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 that can be sliced up into a bunch of three dimensional balls. Similarly, in calculus, the key is to understand that each derivative is a function of time and the integral is a function of the previous derivative, with the starting function being a function of time. This principle can be applied to solve for various quantities, such as finding the area from the circumference formula C = 2π
  • #1
26
0
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 doesn't get you anything.
 
Physics news on Phys.org
  • #2
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
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
@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
Periapsis said:
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
So let's 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
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
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
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
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
Periapsis said:
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.
 

1. What is integration and differentiation?

Integration and differentiation are two fundamental operations in calculus. Integration involves finding the area under a curve, while differentiation involves finding the rate of change or slope of a curve.

2. What are the basic rules for integration and differentiation?

The basic rules for integration include the power rule, constant multiple rule, and the sum and difference rule. The basic rules for differentiation include the power rule, product rule, quotient rule, and chain rule.

3. How are integration and differentiation related?

Integration and differentiation are inverse operations. This means that if you integrate a function, then differentiate the result, you will get back the original function. In other words, differentiation "undoes" integration.

4. What are the applications of integration and differentiation?

Integration and differentiation have many real-world applications, including calculating the area under a curve, finding the velocity and acceleration of an object, and solving optimization problems.

5. How can I improve my skills in integration and differentiation?

To improve your skills in integration and differentiation, it is important to practice regularly and work through a variety of problems. It can also be helpful to seek out additional resources, such as textbooks, online tutorials, or a tutor, to gain a deeper understanding of the concepts and techniques involved.

Suggested for: Integration/Differentiation of Formulas

Replies
1
Views
3K
Replies
4
Views
658
Replies
3
Views
1K
Replies
10
Views
3K
Replies
3
Views
2K
Replies
31
Views
628
Back
Top