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 
             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 = -----------------------------------

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

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

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

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

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.

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