Chapter 4 | Applications of Derivatives
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Calculating the Amount of Error Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study how differentials can be used to estimate the error. Consider a function f with an input that is a measured quantity. Suppose the exact value of the measured quantity is a , but the measured value is a + dx . We say the measurement error is dx (or Δ x ). As a result, an error occurs in the calculated quantity f ( x ). This type of error is known as a propagated error and is given by Δ y = f ( a + dx )− f ( a ). Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so we cannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can use differentials to approximate the propagated error Δ y . Specifically, if f is a differentiable function at a , the propagated error is Δ y ≈ dy = f ′( a ) dx . Unfortunately, we do not know the exact value a . However, we can use the measured value a + dx , and estimate Δ y ≈ dy ≈ f ′( a + dx ) dx . In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if we assume the measurement of the side length is made with a certain amount of accuracy. Example 4.10 Volume of a Cube Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm. a. Use differentials to estimate the error in the computed volume of the cube. b. Compute the volume of the cube if the side length is (i) 4.9 cm and (ii) 5.1 cm to compare the estimated error with the actual potential error.
Solution a. The measurement of the side length is accurate to within ±0.1 cm. Therefore, −0.1≤ dx ≤0.1. The volume of a cube is given by V = x 3 , which leads to dV =3 x 2 dx .
Using the measured side length of 5 cm, we can estimate that
−3(5) 2 (0.1) ≤ dV ≤3(5) 2 (0.1).
Therefore,
−7.5≤ dV ≤7.5. b. If the side length is actually 4.9 cm, then the volume of the cube is V (4.9) = (4.9) 3 = 117.649 cm 3 .
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