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Chapter 4 | Applications of Derivatives
If the side length is actually 5.1 cm, then the volume of the cube is V (5.1) = (5.1) 3 = 132.651 cm 3 .
Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is measured to be 5 cm, the computed volume is V (5) =5 3 =125. Therefore, the error in the computed volume is 117.649 − 125 ≤ Δ V ≤ 132.651 − 125.
That is,
−7.351≤Δ V ≤7.651.
We see the estimated error dV is relatively close to the actual potential error in the computed volume.
4.10 Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an accuracy of 0.2 cm.
The measurement error dx (=Δ x ) and the propagated error Δ y are absolute errors. We are typically interested in the size of an error relative to the size of the quantity being measured or calculated. Given an absolute error Δ q for a particular quantity, we define the relative error as Δ q q , where q is the actual value of the quantity. The percentage error is the relative error expressed as a percentage. For example, if we measure the height of a ladder to be 63 in. when the actual height is 62 in., the absolute error is 1 in. but the relative error is 1 62 =0.016, or 1.6%. By comparison, if we measure the width of a piece of cardboard to be 8.25 in. when the actual width is 8 in., our absolute error is 1 4 in., whereas the relative error is 0.25 8 = 1 32 , or 3.1%. Therefore, the percentage error in the measurement of the cardboard is larger, even though 0.25 in. is less than 1 in. Example 4.11 Relative and Percentage Error An astronaut using a camera measures the radius of Earth as 4000 mi with an error of ±80 mi. Let’s use differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume of Earth, assuming the planet is a perfect sphere.
Solution If the measurement of the radius is accurate to within ±80, we have −80≤ dr ≤80. Since the volume of a sphere is given by V = ⎛ ⎝ 4 3 ⎞ ⎠ πr 3 , we have
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