calculus?

can anyone explain to me what calculus is and the sort of mathematics it is orientated towards? thnx

 

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Heres a great place to start, it should answer your basic questions.

http://hyperphysics.phy-astr.gsu.edu/hbase/calc.html#calch

 

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Calculus is actually a very intuitive branch of mathematics; the problem is that converting from intuition to rigorous formulas can involve complicated details.

Differential calculus interests itself primarily with the problem of determining rates of change. For example, velocity is the rate at which your position is changing, and acceleration is the rate at which your velocity is changing.

Suppose I throw a ball into the air, and the formula for its height at time t happens to be:
x(t) = -4.9 t2 + 5t
Now, suppose I want to compute the average velocity for the ball. I compute the position at some time t, and I compute its position again after h seconds have passed (time t + h). To find velocity, we compute the difference in position by the duration.
Computing position at these two times gives:
x(t) = -4.9 t2 + 5 t
and
x(t + h) = -4.9 (t + h)2 + 5 (t + h)
= -4.9 t2 - 9.8 t h - 4.9 h2 + 5 t + 5 h
So the difference in position is
x(t + h) - x(t) = (-4.9 t2 - 9.8 t h - 4.9 h2 + 5 t + 5 h) - (-4.9 t2 + 5 t)
And if we simplify this, we get:
x(t + h) - x(t) = -9.8 t h - 4.9 h2 + 5 h
So now if we divide by the duration to get velocity:
v = (x(t + h) - x(t)) / h = -9.8 t - 4.9 h + 5
And we have the formula for the average velocity over the interval of time [t, t + h].
However, if you look at the equation, you'll notice that the equation for velocity works when we plug in zero for duration! Doing so gives:
v = -9.8 t + 5 over the interval [t, t]
So one is compelled to say that -9.8 t + 5 is the velocity at time t.

However, there is some hand waving here; notice that if we plug in 0 for h at the beginning of the problem instead of the end, the equation for velocity would be:
v = 0 / 0
Which is bad, since 0/0 is undefined! (Or if you use a special number system for this sort of thing, we call it an indeterminate form)

The theoretical foundation of differential calculus involves using the idea of limits to rigorously prove our results instead of having to rely on "hand waving". The proper way to compute the velocity is:
v = limh -> 0 (x(t + h) - x(t)) / h
Which says that velocity is given by the limit of that formula as we let h approach 0. Sometimes this is written as:
(x(t + h) - x(t)) / h -> v as h -> 0
Where the -> stands for "approaches".

Since we're interested in finding rates of change quite often, it is handy to have a special notation for it, so the term derivative was invented and is defined as (although I'm being slightly sloppy):
dx / dt = limh -> 0 (x(t + h) - x(t)) / h
so v is the derivative of x with respect to t (dx / dt), and we can write:
v = dx / dt

The major breakthrough in the development analysis was coming up with a completely rigorous definition of precisely what a limit is... it's elementary but complicated so I'll skip that for now.

Once we have this rigorous algebraic definition and this powerful notation, we can now come up with some formulae for greatly simplifying the work needed to compute derivatives. I'll need to introduce some new notation first though:
(d/dx) f(x)
This is essentially the same thing as df/dx, but this notation is sometimes more convenient. It could be read as "The derivative with respect to x of f(x)".
Some examples of the formulae are (in these formulas, c is some constant):
(d/dt) t = 1
(d/dt) c = 0
(d/dt) tn = n tn - 1
(d/dx) (f(t) + g(t)) = (d/dx) f(t) + (d/dx) g(t)
(d/dt) (c * f(t)) = c * (d/dt) f(t)
(d/dt) (f(t) * g(t)) = f(t) * (d/dt) g(t) + g(t) * (d/dt) f(t)
(d/dt) f(g(t)) = (d/dt) g(t) * (d/dg(t)) f(g(t))
Where d/dg(t) f(g(t)) means that when we compute velocities, instead of dividing by the duration, we divide by the difference in g(t). This can be written more simply as:
d/dg(t) f(g(t)) = df/dg
And the last equation becomes in the compact notation:
df/dt = df/dg * dg/dt

These equations can shorten the amount of work needed to find derivatives. For example, going back to the original problem of:
x(t) = -4.9 t2 + 5t
if we use the formulae:
v = dx/dt = (d/dt) (-4.9 t2 + 5t)
= (d/dt) (-4.9 t2) + (d/dt) (5t)
= -4.9 (d/dt) t2 + 5 (d/dt) t
= -4.9 * 2 t + 5 * 1
= -9.8 t + 5
Which is exactly what we got before!
Differential calculus is essentially two things:
(1) Learning how to compute derivatives
(2) Learning how to use derivatives to solve problems
Integral calculus is the opposite thing. Differential calculus essentially deals with infinitesimal differences, but integral calculus involves adding up a lot of infinitesimal things!
The simplest and clearest use for integral calculus is finding areas. You can imagine any shape as being filled up with tiny rectangles, and you can imagine that the area of the shape is the sum of the areas of these rectangles!
This conceptual process has been known for millennia. The Greeks used it to compute the area of a circle by the following reasoning:
Ok, from the center of the circle draw evenly spaced radii, and connect the endpoints to form triangles.
Clearly, since all these triangles are inside the circle, if we add up their area it must be less than that of the circle.
The formula for the areas of a triangle is (1/2) * base * height. Now, all the triangles have the same height, so if we add up all the bases and multiply by (1/2) * height, that gives us the total area.
Now, the sum of all the bases must be less than the perimeter of the circle, and the height of each triangle must be less than the radius of the circle, so the total area of the triangles must be less than:
Atriangles < (1/2) * P * r = (1/2) * (2 * pi * r) * r = pi * r2
And we can make the error as small as we like by inscribing more triangles.
Similarly, if we repeat the process, but instead of forming triangles inside the circle we make them outside of the triangle and let the radii be the heights of the triangle, we use the same type of logic to get:
Atriangles > (1/2) * P * r = pi * r2
And again we can get the error as small as we like.
Now, since the area of the circle has to be between these two values, the only possible value for the area of the circle is:
Acircle = pi * r2

Integral calculus is essentially a more rigorous version of this concept. The standard "Riemann Integral" uses a standard system of figures: that of rectangles (though reasoning like that for the circle can be adapted into this standard format). Theoretically, one uses a set of rectangles that are bigger than the area in question and smaller than the area in question, and use the same logic above, but it is done rigorously using the notion of a limit.
Just like differential calculus, a set of rules for manipulating integrals allows us to compute areas more efficiently.

And since integrals are concerned with adding up lots of small things while derivatives are concerned with small differences, it should come as no surprise that integrals and derivatives act somewhat as opposites, and the proof was fairly profound and it's called the Fundamental Theorem of Calculus. For example:
0ab v(t) dt = x(b) - x(a)
The left hand side reads: The integral of velocity with respect to time between time a and time b is... this can be thought of with the aforementioned "infinitesimal" sum idea. This formula says that if we start at time a, and we keep adding the velocity times an "infinitesimal" duration, and end at time b, we should get the difference in position.
Compare this to the ordinary fact about average velocity; average velocity times duration = distance! So (v dt) part in the integral is essentially saying multiply velocity times a small amount time, and that gives us a small amount of distance (dx). If we add up all of these (v dt) quantities from time a to time b, the result should be the distance we traveled between time a and time b.

Hope this helps.

 

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Velocity is not just the rate at which your position is changing, its a vector which means it also has direction as well as a magnitude.

 

luckily, position is also a vector, with magnitude and direction, so this statement is perfectly acceptable.

 

god-of-course: perhaps you want something a simpler explanation of calculus.

Suppose a particle is traveling on a curved path. A person might want to know its direction and speed at a particular point on its path.

Intuitively, its direction is along a tangent to the curve. If you have some equations describing the curve and the speed along the curve, calculus can be used to determing the precise direction and speed of the object at any point on the curve.

There are some cute problems like the one about a man being chased by a bull. Suppose a man is in a field near a high stone wall and is seen by a bull who starts to chase him. He runs along the wall toward a gate. The bull is too dumb to run toward the gate and head him off. Instead the bull always runs runs directly toward the man, correcting his course as the man runs along the wall.

Calculus can be used to determine the path the bull follows and whether or not he will catch the man before he gets to the gate.

 

Hmm - I don't find calculus to be intuitive - if it was it might have been developed a lot earlier. In fact there are still aspects of limits and differentials that I'm not convinced I understand fully.

Nor do I think it's intuitive that velocity is a tangent vector. A common question in 1st year mechanics is:

A ball attached to a string is being whirled round in a circle. What is the path of the ball if the string breaks? Does it:

a) carry on in a circle?
b) fly out radially?
c) carry on tangentially?
d) something in between these?

You'd be surprised how many wrong answers you get.

Cheers,

Ron.

 

C!

 

Here is, I hope, a more intuitive way to think about calculus.
Imagine that you are a device that can read data and write data.
Be this program to differentiate.
Diff
Old I = 0
Loop:
read I
O = I - old I ;
write O
goto Loop:

To integrate
Integrate
OldO = 0
Loop:
Read I
O = OldO + I
OldO = O
Write O
goto Loop:

Differentials provide instructions for making a curve.
Integrals read those instructions and make the curve.
(This is called numerical calculus or difference calculus, but this is the real calculus. The limits calculusl is for two valued coordinate systems.)

 

The numerical way is simply to let x start as a base 0 number and write the numerical sequence of results when the base is incremented.
f(x) = 3x^2 + x - 4
f(x)0 = -4 d(fx)) d2(f(x))
f(x)1 = 0 4
f(x)2 = 10 10 6
f(x)3 = 26 16 6
f(x)4 = 48 22 6
... from this you know the equation was a power of 2.
Numerical differentiation is usefull when you don't have an equation but do have some numbers. This occurs when you have a sensor sending data and you need to differentiate it.
It also works on sequences.

Of course the symbolic differential is:
d(f(x)) = 6x + 1
d2(f(x)) = 6

I think this is how one should teach calculus. Start with dt = 1, which is always true with a stream of data.
Then learn on x,y coordinates where limits are required to sneak up on a point.

Also, I think all math operations should be taught as algorithms so you can see into them.
I hope I did not mess with the poster's head.

Harold