## Why Calculus?

```Why do we use calculus to determine the exact expressions for
derivatives?  For example, to determine the derivative of
y = ax^2 + bx + c  we typically consider two points on this
curve, (x,y) and (x+dx,y+dy), where dx and dy are initially
regarded as small finite quantities.  So we have the two
equations
y =       ax^2   +   bx   +  c

(y+dy) = a(x+dx)^2 + b(x+dx)  +  c

Subtracting the first from the second gives

dy  =  2ax dx + a(dx)^2 + b dx

So far, so good.  Then, since dx is considered to be non-zero,
it is legitimate to divide by dx, giving the result

dy/dx   =  2ax  + a dx  + b

Now for the interesting part: "Let dx go to zero".  Of course if
dx actually EQUALS zero then we weren't justified in dividing the
previous equation by zero.  To avoid this little inconsistency
the epsilon-delta limit device is used to establish that the
LIMIT of the ratio dy/dx is  2ax + b  as dx APPROACHES zero.

It could be argued that the "limit" concept has been introduced
here because we had reached the "limit" of our ability to
algebraically solve equations.  To illustrate, consider again
the case
y = ax^2 + bx + c

We want the slope q of the line  y = qx + p tangent to this curve.
Equating y values, we have

ax^2 + (b-q)x + (c-p) = 0

Solving this quadratic for x gives the point(s) of intersection

(q-b) +- sqrt[ (b-q)^2 - 4a(c-p) ]
x = -----------------------------------
2a

There is a single point of intersection (i.e., the line is tangent
to the curve) iff the sqrt term vanishes, which leaves q = 2ax + b.
There are no divisions by "something approaching zero", no "limits",
"differentials", or "infintesimals" involved in this derivation.

Of course, this purely algebraic approach to finding derivatives
becomes unwieldy when dealing with more complicated functions, and
the "x+dx" trick is a far more efficient way of determining the
derivative of a function F(x) without having to solve explicitly
for the roots of F(x)-L(x).  On the other hand, if our objective
was an explicit algebraic solution of F(x)-L(x), then calculus
doesn't help. Hence it could be said that we resort to calculus when
algebra becomes too difficult (somewhat as we resort to algebraic
numbers when integer arithmetic becomes too difficult).

What about integration?  Suppose we want to know the average
height of all the point in the triangle shown below:
______
|                   /\
|                 /   \
|               /      \
/         \
H           /            \
/               \
|       /                  \
___|__   /_____________________\

| ------- B -----------|

The straight-forward calculus approach (without much thinking)
would probably be to treat the width of the triangle w(h) at
height h above the base as the "weight" at h, and evaluate the
mean as the weighted average given by the ratio of integrals

H
INT  h w(h)  dh
h=0
h_mean  =  ----------------
H
INT  w(h) dh
h=0

The width varies linearly from a value of B at h=0 to a value
of 0 at h=H.  Therefore, we have w(h) = B(1-h/H).  Inserting
this expression into the above formula and evaluating the
integrals gives

/ 1           1 B    \  |H
(  - B h^2  -  - - h^3 ) |
\ 2           3 H    /  |0
h_mean  =  ---------------------------
/         1 B    \  |H
(  B h  -  - - h^2 ) |
\         2 H    /  |0

B H^2 (1/2 - 1/3)
=  -----------------  =  H/3
B H (1 - 1/2)

Of course, with a bit of foresight we could have made the
substutition of variables h = H - h' (so h' is the vertical
distance from the TOP of the triangle), which would give the
simpler expression w(h') = (B/H)h', leading to the formula

H
INT  (B/H) (h')^2 dh'
h'=0                      (H^3)/3
h'_mean  =  ---------------------  =  -------  = 2H/3
H                       (H^2)/2
INT  (B/H) (h') dh'
h'=0

Thus we again have h_mean = H - h'_mean = H/3.  This isn't too bad,
but do we really need calculus to solve this simple problem?  The
ancient Greeks would probably have noted that any triangle can be
split into four precisely similar triangles as shown below
______
|                   /\
|                 /   \
|               /      \
/_________\
H           / \         /\
/    \      /   \
|       /       \   /      \
___|__   /__________\/_________\

Suppose the average height of a triangle of this shape is k
times its height.  It follows that the average height of the
overall triangle is kH, and the average heights of the four
sub-triangles (above the original base) are kH/2, H/2-kH/2,
kH/2, and H/2+kH/2, so the average height of the combined set
is (H+kH)/4, and this must equal kH for the original triangle.
Thus we have H + kH = 4kH and so k = 1/3.

This illustrates how it's often possible, by exploiting some
symmetry in a problem, to solve a "calculus problem" without
resorting to calculus.  The most impressive example of this is
Isaac Newton's Principia, in which hundreds of complex problems
of kinematics and dynamics are solved by Newton's own "synthetic"
methods, modelled on the geometrical presentations of the
ancient Greeks.  Regardless of whether this was the original form
in which he conceived of his results, or (as he later suggested in
the midst of the calculus priority dispute) simply reformulations
of results he had derived by means of his fluxional calculus,
it's tremendously impressive to behold how he was always able to
see a "synthetic" form of demonstration.

Of course, the very brilliance and vaiety of the insights needed
to produce the demonstrations in the Principia reveals the reason
that calculus prevailed over synthetic methods.  As has been said,
Newton was a giant, but the instruments he used are too heavy for
anyone else to lift.  We can't all dream up an inspired synthetic
approach each time we're faced with a new problem.  So we use
calculus.  It may not always be elegant, but it's easy and can
be applied to a wide variety of problems without much thought.
```