Answer
(a) The maximum error is $26.74~cm^2$
The relative error is $0.0119$
(b) The maximum error is $178.7~cm^3$
The relative error is $0.01786$
Work Step by Step
(a) $C = 2\pi ~r$
$r = \frac{C}{2\pi}$
Also:
$dC = 2\pi~dr$
$dr = \frac{dC}{2\pi}$
We can find the maximum error in the surface area:
$A = 4\pi ~r^2$
$dA = 8\pi~r~dr$
$dA = 8\pi~(\frac{C}{2\pi})~(\frac{dC}{2\pi})$
$dA = \frac{2C~dC}{\pi}$
$dA = \frac{(2)(84~cm)(0.5~cm)}{\pi}$
$dA = \frac{84}{\pi}~cm^2$
$dA = 26.74~cm^2$
We can find the relative error:
$\frac{dA}{A} = \frac{8\pi~r~dr}{4\pi~r^2}$
$\frac{dA}{A} = \frac{2~dr}{r}$
$\frac{dA}{A} = \frac{2~(\frac{dC}{2\pi})}{C/2\pi}$
$\frac{dA}{A} = \frac{2~dC}{C}$
$\frac{dA}{A} = \frac{(2)~(0.5~cm)}{84~cm}$
$\frac{dA}{A} = \frac{1}{84}$
$\frac{dA}{A} = 0.0119$
(b) We can find the maximum error in the volume:
$V = \frac{4}{3}\pi ~r^3$
$dV = 4\pi~r^2~dr$
$dV = (4\pi)~(\frac{C}{2\pi})^2~(\frac{dC}{2\pi})$
$dV = \frac{C^2~dC}{2\pi^2}$
$dV = \frac{(84~cm)^2~(0.5~cm)}{2\pi^2}$
$dV = \frac{1764}{\pi^2}~cm^3$
$dV = 178.7~cm^3$
We can find the relative error:
$\frac{dV}{V} = \frac{4\pi~r^2~dr}{(4/3)\pi~r^3}$
$\frac{dV}{V} = \frac{3~dr}{r}$
$\frac{dV}{V} = 3\frac{(\frac{dC}{2\pi})}{(\frac{C}{2\pi})}$
$\frac{dV}{V} = 3\frac{dC}{C}$
$\frac{dV}{V} = 3(\frac{0.5~cm}{84~cm})$
$\frac{dV}{V} = \frac{1}{56}$
$\frac{dV}{V} = 0.01786$
