the radius of a spherical ball is measured as 10 in., with a max error of 1/16 in. what is the max resulting error in its calculated surface area?
2007-04-22 14:47:01 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
A = 4*pi*r^2
You could say that r = R +/- e
where e is small compared to R
A = 4*pi*(R^2 +/- 2Re + e^2)
but the e^2 can be neglected because it is very small
So the error is +/- 4*pi*2*R*e
2007-04-22 14:53:03 · answer #1 · answered by Dr D 7 · 1⤊ 1⤋
Maximum error = 15.708 inches.
You take the exponent multiplied by the uncertainty over the measurement, then multiply that answer by the answer to the original equation (which is the surface area of a sphere) using the original measurement. Or:
(4 x pi x 10^2) x (2 x ((1/16) / 10)) = 15.708
2007-04-22 21:57:40 · answer #2 · answered by tech_geek 2 · 0⤊ 0⤋
A = 4pir^2
dA/dr = 8 pi r
dA = 2*pi*r*dr
dA = 2*3.14*10*(1/16) = 3.925 square inches <-- max error
2007-04-22 21:53:57 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
A = 4 * Ï * r^2
dA = 4 * Ï *2*r*dr = 8*Ï*r*dr
dr = 1/16, r = 10, find dA
dA = 15.71in
2007-04-22 21:55:42 · answer #4 · answered by gp4rts 7 · 0⤊ 0⤋