J
Jonathan Kirwan
- Jan 1, 1970
- 0
Cool. Finally something I have the math for.![]()
Something to keep in mind about calculus is that it is really VERY
simple, in basic concept. The older books I used to learn from made
it lots more complex that it needed to be. I think there may be some
newer ones based on something that I discovered entirely for myself,
but then discovered that others had already discovered it -- namely
"non-standard analysis." (Abraham Robinson)
But the very simple way to view things in calculus is that it is just
like algebra, except that you get a new kind of special variable type
that adds to what's already in algebra. This special kind of
variable, unlike the usual ones, is ONLY ALLOWED to hold infinitely
small values (which aren't exactly 0, but are smaller than any finite
value.) In other words, in algebra, variables can only hold finite
values. In calculus, you get to keep that type of variable and add a
new one that can only hold infinitesimal values.
Take a look at a falling body under gravitation, assuming that at the
start of time (t=0) the velocity is zero (v=0.) Now you release the
object and it begins to fall. Under normal algebra, you often find
the simple formula that looks like:
D = v * t
So that the distance some object travels is the speed it is going
times the time it went at that speed. But this, of course, assumes
that the speed is the same for the entire time or that the speed given
is a useful average, at least.
In the case of a falling object, though, the velocity is constantly
changing. So what do we use for V? It's never the same twice.
Now the cheating answer is to simply say that under constant
acceleration, that the average velocity must be 1/2 of the final
velocity and we can compute that final velocity as g*t. So by that,
we can say:
D = [(1/2)*g*t]*t = g*t^2/2
Which is correct.
But if you want to think about things from a calculus perspective (a
better one, in general), then what you do instead is ask yourself:
"Well, since the velocity is constantly changing all the time, exactly
when is the velocity a precise value and not some average thing?"
In other words, looking at D=v*t, ask yourself for how long an exact
velocity is still exact. If you think about that, you have to answer,
"It is a particular finite value only for an instant of time."
Once you say this to yourself, now ask.. "Well, if the velocity is
exact only for an instant, how far will it travel in that instant?"
And to this, you answer yourself, "For a very small distance, namely
only an infinitely small distance." Kind of, well, an instant of
distance. So you write:
dD = v * dt
Which is to say that the object travels an infinitely small fraction
of the total distance, computed by multiplying the instantaneous and
finite velocity at some moment by the duration of that moment, and
that is itself infinitely small.
It makes sense that when you multiply a finite value by an infinitely
small one, that you should also get another infinitely small result.
Right?
Now remember, that dD isn't two things, but one thing. It's just a
variable. We could have called it Q, if you prefer. But we need some
way to keep track of the fact that it can ONLY HOLD infinitely small
values. So if we just remember the little-d in front as kind of
modifying the variable name so that it helps us remember this, then we
are fine. So dD is a variable and dt is a variable. They are
different variables, too.
In any case, keep in mind that these tiny, infinitesimally small
values are NOT necessarily the same as each other. Just as two
different variables in algebra are not necessarily the same as each
other, despite both representing some finite value. In like fashion,
the variables dD and dt, although each are infinitely small, they can
be different from each other and yet both infinitely small.
It's perfectly possible to relate one infinitely small variable to
another, like this:
dD/dt
In this case, all we are doing is dividing one infinitely small value
by another. And this often produces a finite value. For example, it
could be the case that:
dD/dt = 1
In this case, the two variables track each other in a 1:1
relationship. They are equally infinitely small. But we could have
written:
dD/dt = 2
In this case, dD is twice as big as dt. They are both infinitely
small, but it is still true that dD is twice as big as dt.
Now, as you already know about velocity, it is:
v = D/t
Well, an even better definition that works for objects that don't
always travel at uniform speed, is this:
V(t) = dD/dt
In this case, we are just saying that the speed at some moment (t) is
best computed as the instantaneous distance it traveled divided by the
instant of time it traveled. This is always true, even for things
that aren't always moving around at the same speed. So in that sense,
it is a more powerful and better way to see speed defined. It is a
more universal definition. The old one uses averages. The new one
uses instantaneous precision.
Of course, that's all just useless unless at some point you can get
back to finite values for real measurements.
How many infinitely small distances do you need to add up to make a
finite distance? Well, an infinite number of them, of course. So
calculus invents a new symbol, the integral sign, to indicate this
particular kind of infinite sum.
If you had an infinite number of dD pieces, what does it add up to?
Well, D of course! What else?
And like in algebra, what you do to one side of an equation you do to
another side. So if you infinite sum one side, you infinite sum the
other side, too. This is called "integration."
Going back to:
dD = v * dt
To clear it up and get rid of the infinitesimal variables, we just sum
both sides:
SUM(dD) = SUM(v * dt)
Since we also know that v=g*t, we can substitute to get:
SUM(dD) = SUM(g * t * dt)
Now comes the fun part. We already know what the SUM(dD) is, so:
D = SUM(g * t * dt)
That was easy. But what about the rest?
Well, since 'g' is a constant (as given), then we already know that
any common factor found in terms of a long sum can be extracted, just
as in (a*b+a*c+a*d) = a*(b+c+d). So our sum can be adjusted so that:
D = g*SUM(t * dt)
Okay, now we are stuck. 't' itself is NOT a constant, so we cannot
just pull it out. Similarly, dt, although a constant is infinitely
small and needs to be treated by the SUM(). So both stay in there.
Any time you see a product, think of areas. In this case, imagine
that you have an area that is 't' high and 'dt' wide. That inner part
could be seen that way, just fine. So we are summing up an infinite
number of tiny areas. What would that look like?
Well, imagine laying each tiny rectangle side by side, so that the
width is 'dt' and they are sorted according to their heights, 't'. As
in:
|
| |
| | |
.. | | | |
And so on. Each is very slim, dt wide. But when you stack them side
by side like that, they all add up to something that is 't' wide.
Similarly, how high are they? Well, as we go from 0 to t, they grow
from 0 high to 't' high. So in the end, we have a triangle that is
't' wide and 't' high. What is the total area of all that? Well, 1/2
of t^2, of course. So:
D = g*t^2/2
Just like before.
You can also cancel out various variables you see on different sides
of the equation.
Getting this back to electronics, which I suppose I should do, you are
told that the charge on a capacitor is:
Q = C * V
But if the V is changing, then the Q is changing. We could rewrite it
more precisely as:
dQ = C * dV
Just like we did earlier for speed, time, and distance relationships.
And if we want to, we could compare both sides at each moment of time,
so that:
dQ/dt = C * dV/dt
Dividing both sides by the same variable is just fine. It looks
funny, in a way, but it is just another variable and we can divide
both sides by this infinitesimal variable if we want to. Or we cancel
them back out by multiplying both sides by dt and get back what we
started with.
Just like in algebra.
By the way, dQ/dt is just I (current) so:
I = C * dV/dt
A useful thing to remember.
Anyway, calculus is just algebra with a new type of variable that can
only hold infinitely small values. That forces you, at some point
later on, to use an infinite sum if you want to see finite values.
But you can delay that while you think and cancel things just as in
regular algebra and, if you are lucky, there won't be that much left
on which to have to use an infinite sum.
Jon