Fundamental Theorum of Calculus

[KtownChemist]

Correct me if i am wrong but isn't the fundamental Theorum of Calculus simply a formula to convert the energy of a system from x dimensions to either x+1 or x-1?

[Cap'n Refsmmat]

The fundamental theorem of calculus:

 

[math]f(x) = \int f'(x) dx[/math]

 

[math]\int_a^b f'(x) dx = f(b) - f(a)[/math]

 

That's not about the energy of systems, but integration and differentiation.

[KtownChemist]

isn't integration moving x dimensions to x+1 dimensions? and differentiation moving from x dimensions to x-1 dimensions?

 

4x^3 + C = 12x^2 this is simply restating the system or function y = 4x^3 + C as it appears in a one less dimensions

[Cap'n Refsmmat]

Not dimensions in the geometric sense, at least. For an introduction to what differentiation means I suggest you read my tutorial (link).

[KtownChemist]

Why does it not hold true for geometry as well? Just wondering. If you graph the velocity of a particle and want to know the instantaneous velocity of a particle you differentiate and take the Energy of the particle in four dimensions and convert it to only three dimensions. How does this not work? Thanks for enlightening me

Edited June 16, 2009 by KtownChemist

[ajb]

So, what you are saying is

 

[math]x(t)[/math] - graph (curve in some space)

[math]\dot{x}(t)= v(t)[/math] - instantaneous velocity

[math] \ddot{x}(t) = a(t)[/math] - instantaneous acceleration.

 

Where dot means take the derivative with respect to [math]t[/math]. (I have suppressed any further indices needed on [math]x[/math])

 

That is fine.

 

I don't understand your statements about the energy. In classical mechanics the energy is given by

 

[math]E = \frac{1}{2}v^{2}+ V(x)[/math]

 

(suppressed the dependence on [math]t[/math] )

[Bignose]

isn't integration moving x dimensions to x+1 dimensions? and differentiation moving from x dimensions to x-1 dimensions?

 

4x^3 + C = 12x^2 this is simply restating the system or function y = 4x^3 + C as it appears in a one less dimensions

 

I don't know what you mean by re-stating. Because while x^2 and x^3 can be close to each other in value for a small range... they aren't the same. You cannot use one to express the other. They are both different functions.

 

They are also both only 1 dimensional, because they have just the 1 input, x. To be 2 dimensional, it would have to be f(x,y). 3-D, f(x,y,z), and etc.

[KtownChemist]

I don't know what you mean by re-stating. Because while x^2 and x^3 can be close to each other in value for a small range... they aren't the same. You cannot use one to express the other. They are both different functions.

 

So your basically saying that F'(x) = f(x) is not actually equivalent?

 

They are also both only 1 dimensional, because they have just the 1 input, x. To be 2 dimensional, it would have to be f(x,y). 3-D, f(x,y,z), and etc.

 

So the area of a square is only one dimensional because its formula is A = x^2 ?

[Bignose]

So your basically saying that F'(x) = f(x) is not actually equivalent?

 

Sure, they are. But x^2 and x^3 aren't equivalent. You can't represent one with the other. Your equation: 4x^3 + C = 12x^2 is only true for certain values of x, no in general. That's the point. 4x^3 + C and 12xx^2 aren't equivalent.

 

 

So the area of a square is only one dimensional because its formula is A = x^2 ?

 

The function A(x)=x^2 is only one dimensional because it only has one input, x. The area of a rectangle with sides x and y would be 2 dimensional, A(x,y) = xy, because it has two inputs.

 

Just because the units of the terms in the equation have squared or cubed or some other power in them, does not necessarily imply that that function has the same dimension. A(x) =x^2 is 1 dimensional because if x=5, A=25 and only 25. In this case, the inverse is true also, in that if A=16, then x can only equal 4. Once you fix one of the values, the other is completely determined.

 

Now, look at A(x,y)=xy. Fix x=5. That doesn't tell you anything about the area of that rectangle or what y is. That's because it is a 2-D function and you've only fixed one of them. If you fix 2 of them, say x=5 and A=30, then y is fixed at 6.

 

And, it doesn't matter what the units are. B(x,y) = x/y would be dimensionless if both x and y had the same units, but it is still a function with 2 degrees of freedom.

 

Differentiation and integration don't necessarily change the degrees of freedom of a function. It can. Let [math]f(x)=x^4 + 7[/math] [math]\frac{df}{dy} = 0[/math] because f is not a function of y. In this case, a function with 1 degree of freedom when differentiated with respect to y becomes a function with no degrees of freedom because no matter what the input is, the output is 0.