The measurement of the edge of a cube is found to be 14 inches, with a possible error of 0.04 inch. Use differentials to approximate the maximum possible propagated error in computing the volume of the cube and its surface area?
Estimate the maximum allowable percent error in measuring the side if the error in computing the volume cannot exceed 2%. ds/s% = ?
2007-11-05 18:57:41 · 3 answers · asked by Daniel B 1 in Science & Mathematics ➔ Mathematics
V = s^3, so dV = 3s^2 ds = 23.52 in^3 (= 24 in^3 to 2 s.f.)
A = 6s^2, so dA = 12s ds = 6.72 in^2 (= 6.7 in^2 to 2 s.f.)
dV/V = 3s^2 ds / s^3 = 3 ds / s ≤ 0.02.
So ds/s ≤ 0.02 / 3 = 0.0067 (2 s.f.), i.e. at most a 0.67% error in s can be allowed.
2007-11-05 19:03:45 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
I'll give you the formulas but I will not do the work for you
V = s^3
S.A. = 6s^2
To find the minimum volume and surface area, use s=14-0.04 in
Max use s=14+0.04
Middle use s=14
2007-11-06 03:05:31 · answer #2 · answered by Anonymous · 0⤊ 0⤋
volume is s^3
surface area is 6(s^2)
2007-11-06 03:14:03 · answer #3 · answered by Anonymous · 0⤊ 0⤋