Tuesday, October 18, 2011

∑ F = ma

… is a differential equation:

where acceleration a(t), velocity v(t), and displacement s(t) are all vectors and functions of time. This equation is second-order in position because the highest derivative is the second time derivative of position. Combined with the right boundary conditions, s(t) (also called the trajectory: path through space and time) can be determined.

This differential equation can be solved one component, or dimension, at a time. Let us focus on one of these, and call it the x component. The equations for y and z can be found exactly the same way.

Constant acceleration

If the graph of a(t) signifying acceleration in the x direction is constant

then the graph of v(t), the velocity in the x direction, is a straight line with slope a0

and the graph of x(t), the position along the x axis, is a parabola

It is also possible for the acceleration, or either of the initial velocity or initial position, to be negative. Thus the displacement/projectile motion formula is derived.

Friday, August 12, 2011
After being introduced to the concept of a limit and the derivative, the typical calculus student is asked to evaluate simple antiderivatives and apply a strange symbol, ∫, much like an elongated “S”, to their notation. He then completely shifts gears, and applies his knowledge of limits to summations of rectangular areas, of which there are an infinite amount, and all with vanishingly small width. After establishing the techniques used in finding exact areas bounded by curves, he is asked again to apply ∫ to the function whose area it is he must calculate…and is left to his own devices to interpret the Fundamental Theorem of Calculus (FTOC) — the theorem which relates the derivative to bounds of a definite integral, and the area bounded by a function to its antiderivative. Few introductory calculus courses take the time to prove the theorem, and simultaneously probe the intricate connections that definite integration has with differential calculus.
The first part of the FTOC is as follows: a function

is a general antiderivative of f such that A’(x) = f(x). The proof of this part is lengthy, but less conceptually rigorous than the second.
Second part of FTOC — Prove:

Proof: Consider F, a function which is a general antiderivative of f. Also consider an antiderivative of f, A — for simplicity, the same as that which was used in the first part of the FTOC. Because F and A are both antiderivatives of f which differ by a constant of integration, C, you may then make a relation such that

where C is a constant. Substituting x = a into this equation brings you to the following relation:

(by early properties of definite integrals, the region has a width of 0). Substituting x = b,

which, when rearranged, yields

(Note: there are many proofs of the FTOC and its corollaries, so it’d be wise to browse through as many as you can find!)
It helps to introduce a bit of generalization for the purpose of understanding the geometric significance of the FTOC (coupled with the picture above): the gradient of a small target region of a function is approximated by ∆y/∆x, whereas the area of that small region may be approximated by ∆y•∆x — inverse operations, correlating directly to the inverse relationship between differentiation and integration.

After being introduced to the concept of a limit and the derivative, the typical calculus student is asked to evaluate simple antiderivatives and apply a strange symbol, ∫, much like an elongated “S”, to their notation. He then completely shifts gears, and applies his knowledge of limits to summations of rectangular areas, of which there are an infinite amount, and all with vanishingly small width. After establishing the techniques used in finding exact areas bounded by curves, he is asked again to apply ∫ to the function whose area it is he must calculate…and is left to his own devices to interpret the Fundamental Theorem of Calculus (FTOC) — the theorem which relates the derivative to bounds of a definite integral, and the area bounded by a function to its antiderivative. Few introductory calculus courses take the time to prove the theorem, and simultaneously probe the intricate connections that definite integration has with differential calculus.

The first part of the FTOC is as follows: a function

is a general antiderivative of f such that A’(x) = f(x). The proof of this part is lengthy, but less conceptually rigorous than the second.

Second part of FTOC — Prove:

Proof: Consider F, a function which is a general antiderivative of f. Also consider an antiderivative of f, A — for simplicity, the same as that which was used in the first part of the FTOC. Because F and A are both antiderivatives of f which differ by a constant of integration, C, you may then make a relation such that

where C is a constant. Substituting x = a into this equation brings you to the following relation:

(by early properties of definite integrals, the region has a width of 0). Substituting x = b,

which, when rearranged, yields

(Note: there are many proofs of the FTOC and its corollaries, so it’d be wise to browse through as many as you can find!)

It helps to introduce a bit of generalization for the purpose of understanding the geometric significance of the FTOC (coupled with the picture above): the gradient of a small target region of a function is approximated by ∆y/∆x, whereas the area of that small region may be approximated by ∆y•∆x — inverse operations, correlating directly to the inverse relationship between differentiation and integration.