since
v = 4/3 pi r^3
and c = 2pi r, r = c/2pi, and
v = 4/3 pi (c/2pi)^3
v = 1/(6pi^2) c^3
so, dv = 1/(2pi^2) c^2 dc
so, now you have dc and can evaluate dv
The circumference of a sphere was measured to be 89.000 cm with a possible error of 0.50000 cm.
A. Use linear approximation to estimate the maximum error in the calculated surface area.
B. Estimate the relative error in the calculated surface area.
1 answer