3.9.3-eg-1

# 3.9.3-eg-1 - length 4 be denoted by Δ V = V(4 Δ x-V(4 for...

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Example The length of a side of a cube was measured to be 4 cm with a possible error of 1/10 cm. Use linear approximation to estimate the maximum error in the calculated volume of the cube. Solution: The volume V ( x ) of a cube with side x is given by V ( x ) = x 3 , and the linear approximation to V ( x ) at the measured side-length x = 4 is L ( x ) = V (4) + V 0 (4)( x - 4) , where V 0 ( x ) = 3 x 2 , so V 0 (4) = 3 · 4 2 = 48 and V (4) = 4 3 = 64 . Thus the linear approximation to V ( x ) at x = 4 is L ( x ) = 64 + 48( x - 4) , and it follows that V ( x ) L ( x ) = 64 + 48( x - 4) , for x near 4 . (1) Alternatively, if we let Δ x = x - 4 (so that x = 4 + Δ x ), then (1) becomes V (4 + Δ x ) L ( x + Δ x ) = 64 + 48 · Δ x, for Δ x near 0 . (2) Now let the increment of change of volume (relative to the volume of a cube with sides of
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Unformatted text preview: length 4) be denoted by Δ V = V (4 + Δ x )-V (4) , for Δ x given above. It then follows from (2) that Δ V = V (4 + Δ x )-V (4) ≈ L (4 + Δ x )-V (4) = (64 + 48Δ x )-64 = 48Δ x, for Δ x ≈ . So for an error in measurement of the magnitude | Δ x | ≤ 1 10 , the error made in calculating the volume in this case is ± ± ± ± V ² 4 + 1 10 ³-V (4) ± ± ± ± ≈ | 48Δ x | = 48 | Δ x | ≤ 48 · 1 10 = 24 5 cubic cm , so the approximate maximum calculated error in the volume is given by 24 5 cubic cm....
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## This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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