The circumference of a sphere was measured to be 74.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area

Answer 1

Duplicate post. See "CALC"

Answer 2

C = 2 pi r
dr/dC = 1/(2pi)

A = 4 pi r^2
dA/dr = 8 pi r

dA/dC = dA/dr * dr/dC
dA/dC = 8 pi r *1/(2pi)
= 4 r
but r = C/(2 pi)
so
dA/dC = 4 C/(2 pi)
= 2 C/pi
dA = 2 * 74/pi * .5 = 74/pi