The circumference of a sphere was measured to be 70 cm with a possible error of 0.9 cm. Use differentials to estimate the maximum error in the calculated surface area. _____?
Estimate the relative error in the calculated surface area. ______?
Hint: The circumference of a sphere of radius r is C = 2 \pi r, and its surface area is A = (4pi)(r^2). Eliminate r first!
2007-11-03 09:26:34 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
S=4pir^2 so dS = 8pir*dr but 2pi r= C and 2pidr =dC
so dS = 4 Cdr and dr dC/2pi so dS= 2/pi C*dC
So the maximum error is
dS=2/pi*70*0.9cm^2 =40.11 cm^2
Relative error = dS/S = 40.11/480 =8.36%
2007-11-03 09:43:04 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
the floor area S of a sphere is S = 4 pi R^2, the position R is the radius of the sector. The circumference C of a sphere is C = 2 pi R. Combining both supplies S = 4 pi (C / 2 pi)^2 = C^2 / pi. the total differential of it is dS = 2 C dC / pi. So, the blunders contained in the floor area is two ( 80 5 cm ) ( 0.5 cm ) / pi = 27.a million cm^2. The relative blunders contained in the floor area is dS / S = ( 2 C dC / pi ) / ( C^2 / pi ) = 2 dC / C = 2 ( 0.5 cm ) / ( 80 5 cm ) = 0.012.
2016-10-23 08:21:46 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Where did the 480 come from in the relative error? I've been able to figure everything else out, except for that.
2007-11-04 13:33:45 · answer #3 · answered by Mandi C 1 · 0⤊ 0⤋