a) V=x^3 ... dV=3x^2dx = 3(30)^2*(+/-.20)= +/- 540 cc
b)+/-540/30^3=+/- .02
c) +/- 2 %
Landon B.
asked • 10/02/23Calculus Question
The edge of a cube was found to be 30 cm with a possible error of 0.2 cm. Use differentials to estimate:
(a) the maximum possible error in the volume of the cube?
(b) the relative error in the volume of the cube ?
(c) the percentage error in the volume of the cube ?
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