Calculus Of One Real Variable By Pheng Kim Ving
Chapter 8: Applications Of The Derivative Part 2 Section 8.4: Approximations Of Errors In Measurement

 

8.4
Approximations Of Errors In Measurement

 

 

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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 = xa 1 then
xa = 1,001,
1st and 3rd axes: if 1,000 = xa + 1 then
xa = 999,
therefore xa is somewhere in [999, 1,001].

 

 

 

 

Example 1.1

 

 

Solution
Let s be the side and A the area of the square. Then A = s2. 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 m2.

EOS

 

Note that we calculate dA from the equation A = s2, 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 d1 yields d1 = 100 m with an error of 1 m. A measurement of distance d2 yields d2 = 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 d2 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 d1 is 1 m / 100 m = 1/100 = 0.01 and that for d2 is
1 m / 1,000 m = 1/1,000 = 0.001. As desired the relative error for d2 is smaller than that for d1.

 

The percentage error is the absolute error as a percentage of the quantity being expressed. For example, the percentage
error for d1 is (1 m / 100 m)(100/100) = (1/100)(100)% = (0.01)(100)% = 1% and that for d2 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:

 

 

 

 

Example 2.1

 

 

Solution

 

Thus the approximate percentage error of the calculated area is (0.006)(100/100) = 0.6%.

EOS

 

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Problems & Solutions

 

 

Solution

 

Let s be the side and A the area of the square. Then A = s2. 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 m2.

 

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Solution

 

 

So the approximate percentage error of the calculated volume of the sphere is (0.06)(100/100) = 6%.

 

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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?

 

Solution

 

Let a be the edge and V the volume of the cube. Then V = a3. 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%.

 

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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 half-life of the substance and determine an approximate
maximum size of the error in this half-life.

 

Solution

 

Let y0 be the initial quantity of the substance and y(t) the quantity remaining undecayed after t years. Then y(t) = y0ekt.
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) = y0ek(1) = y0ek.

 

But:

 

y(1) = py0.

 

Thus:

 

y0ek = py0,
ek = p,
k = ln p.

 

Let T be the half-life. Then:

 

 

The approximate half-life of the substance is 346.23 years and an approximate maximum size of the error in this half-life
is 17.33 years.

 

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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%.

 

Note

 

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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