teach you calculus. It simply provides an image of what calculus
methods can do, followed by some calculus results that can be applied to economic
problems.
Figure M.10-1 shows 2 functions:
y
= 2
x
on the left side and
y
=
x
2
on the right. We
want to answer two questions about each of these functions over the domain
x
[0,
∞
).
First, what is the
slope
of the function at any value of
x
? Second, what is the
area
between
the function and the horizontal axis and between 0 and the given value of
x
? For
y
= 2
x
,
the questions are fairly easy to answer. The slope is a
constant
+2, because
y
= 2
x
is a
lin-
ear
equation. The area required is just the area of a triangle, A = (b x h)/2, with base
b =
x
and height h =
y
= 2
x
. Hence A = (b x h)/2 = (
x y
)/2 = (
x
)(2
x
)/2 = 2
x
2
/2 =
x
2
.
For
x
= 1, A = 1 square unit; with
x
= 2, A = 2
2
= 4 square units; and with
x
= 3, A = 3
2
=
9 square units.
When we turn to Figure M.10-1(b), however, things are not so simple. The slope is
not
constant, but gets continuously steeper, and we cannot use our triangle formula to
calculate the area, because
y
=
x
2
is not a straight line. Calculus has been developed,
since its invention in the seventeenth century (virtually simultaneously) by Newton
and Liebniz, to deal with precisely these difficulties posed by nonlinear functions. We
shall assert (correctly, but without proof) that the slope of the function
y
=
x
2
at any
x
has the value 2
x
, and that the shaded area takes on the value A =
x
3
/3. [Any introduc-
tory calculus text will provide proofs.] Hence, when
x
= 1, the slope of the tangent to
the function (simply put, the slope of the function) = 2
x
=
2
and the area =
x
3
/3 = 1/3
square units; when
x
= 2, the slope = 2
x
= 4 and the area = 8/3 square units; and when
x
= 3, the slope = 6 and the area A = 9 square units.
Note that at any value of
x
, the value of the
slope
in (b) equals the value for
y
in (a).
This is as we would expect, since in (a), y = 2x, which
is
the value of the slope of the
function y = x
2
in (b). Note also that for
x
< 3, the shaded area in (a) exceeds that in (b),
at
x
= 3 the two areas are equal (both equal 9 square units), and for
x
> 3, the shaded
area in (b) exceeds that in (a).
What does all this mean? We have used the notation
m
= ∆
y/
∆
x
to denote the slope
m
of a function. Here both ∆
x
and ∆
y
represent
discrete differences
between the
x
values
and between the
y
values at two
separate
points. Suppose that we shorten the distance
between
x
1
and
x
2
(=
x
1
+ ∆
x
) until ∆
x
becomes infinitesimally small, and
x
2
therefore
virtually coincides with
x
1
,
and we do the same for
y
1
and y
2
. In the limit, we approach
the slope of the tangent to the functions, which we write
dy/dx
,
at x
1
. If
y
=
f
(
x
), then we
can write the slope (which we also call the
derivative
of the function) in several different
but equivalent ways:
dy/dx
,
y
’,
df
(
x
)
/dx
, or
f
’(
x
).
M10-2
MATH MODULE 10: CALCULUS RESULTS FOR THE NON-CALCULUS SPEAKER