Calculus Of One Real Variable –
By Pheng Kim Ving 
8.4 
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1. Approximations 
If a quantity x
(eg, side of a square) is obtained by measurement and a quantity y
(eg, area of the square) is calculated
as a function of x, say y
= f(x), then any error involved in the
measurement of x produces an error in the calculated
value of y as well.

Recall from Section 4.3 Part 2 that the 
Section 8.3 Part 1, we have:
That is, the error in x is dx and the corresponding approximate error in y is dy = f '(x) dx. 
Fig. 1.1

Fig. 1.2 – 1st and 2nd axes: if 1,000 = x_{a} – 1 then 

Solution
Let s be the
side and A the area
of the square. Then A
= s^{2}. The error of
the side is ds = 1 m.
The approximate error of
the calculated area is:
dA = 2s ds = 2(1,000)(1) = 2,000 m^{2}.
Note that we calculate dA from the equation A = s^{2}, since the values of s
and ds are
given. To find the differential of A
we must have an equation relating A
to s. So even
if the measured value of the side is given we still define the variable s
that takes on as a value the measured value.
In general, when the measured
value say V of a
quantity and the error say E
in the measurement are given, we define a
variable say x for the
quantity, so that x = V and dx = E,
which will be used later on in the solution. When using the
quantity, first use the variable x,
not the value V,
then use the value V
when a value is to be obtained.
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2. Types Of Errors 
A measurement of distance d_{1} yields d_{1} = 100 m with an error of 1 m. A measurement of distance d_{2} yields d_{2} = 1,000 m
with an error of 1 m. Both measurements have the same absolute error of 1 m.
However, intuitively we feel that
measurement of d_{2} has a smaller
error because it's 10 times larger and yet has the same absolute error. Clearly
the
effect of 1 m out of 1,000 m is smaller than that of 1 m out of 100 m. This
leads us to consider an error relative to the
size of the quantity being expressed. This relative error is accomplished by
representing the absolute error as a fraction
of the quantity being expressed. For example, the relative error for d_{1} is 1 m / 100 m = 1/100 = 0.01
and that for d_{2} is
1 m / 1,000 m = 1/1,000 = 0.001. As desired the relative error for d_{2} is smaller than that for d_{1}.
The percentage error is the
absolute error as a percentage of the quantity being expressed. For example,
the percentage
error for d_{1} is (1 m / 100
m)(100/100) = (1/100)(100)% = (0.01)(100)% = 1% and that for d_{2} is
(1 m / 1,000 m)(100/100) = (1/1,000)(100)% = (0.001)(100)% = 0.1%. We see that the
percentage error is the relative
error expressed as a percentage. If the relative error is r then the percentage error is p% = r . (100/100) = (r . 100)%.
So conversely if the percentage error is p%
then the relative error is r
= p/100.
In general:

Thus the approximate percentage error of the calculated area is (0.006)(100/100) = 0.6%.
Problems & Solutions 
Let s be the side and A the area of the square. Then A = s^{2}. The error of the side is ds = 60 cm = 0.6 m. The
approximate error of the calculated area is:
dA = 2s ds = 2(200)(0.6) = 240 m^{2}.
Solution
So the approximate percentage error of the calculated volume of the sphere is (0.06)(100/100) = 6%.
3. The edge of a cube is
measured to within 2% tolerance. Approximately what percentage error can result
in the
calculation of the volume of the
cube?
Let a be the edge and V the volume of the cube. Then V = a^{3}. The percentage error of the edge is 2% and so its
relative
error is da/a = 2/100 = 0.02. The
approximate relative error that can result in the calculation of the volume is:
Thus the approximate percentage error that can result in the calculation of the volume is (0.06)(100/100) = 6%.
4. The proportion of a
radioactive substance remaining undecayed after 1 year is measured to be 0.998
of the initial
quantity with an error of up to
0.0001. Find the approximate halflife of the substance and determine an
approximate
maximum size of the error in this
halflife.
Let y_{0} be the initial quantity of the substance and y(t) the quantity remaining undecayed after t years. Then y(t) = y_{0}e^{kt}.
Let p be the
proportion of the initial quantity remaining undecayed after 1 year, so that p = 0.998 and dp = 0.0001. After
1 year we have:
y(1) = y_{0}e^{k}^{(1)} = y_{0}e^{k}.
But:
y(1) = py_{0}.
Thus:
y_{0}e^{k}
= py_{0},
e^{k} = p,
k = ln p.
Let T be the halflife. Then:
The approximate halflife of the
substance is 346.23 years and an approximate maximum size of the error in this
halflife
is 17.33 years.
5. It is desired that the computed area of a
circle is with at most 2% error by measuring its radius. Approximate the
maximum allowable percentage error
that may be made in measuring the radius.
Solution
Thus the approximate maximum
allowable percentage error that may be made in measuring the radius is
(0.01)(100/100) = 1%.
Here the exact error of the function
(area) is given and the approximate error of the variable (radius) is to be
found. The
symbol:
represents the relative error,
not an approximate relative error, of the radius. It's the value 0.01 that's an
approximate
value of this relative error.
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