# Thread: linearization?

1. ## linearization?

Please help. Don't know how to do this...

The circumference of a sphere was measured to be 88 cm with a possible error of 0.5 cm.

(a) Use differentials to estimate the maximum and relative error in the calculated surface area. (Round maximum error to the nearest whole number and relative error to three decimal places.)
Maximum error=? cm^2
Relative error= ?
(b) Use differentials to estimate the maximum and relative error in the calculated volume. (Round maximum error to the nearest whole number and relative error to three decimal places.)
Maximum error= ? cm^3
Relative error=?

2. Hello, kwivo!

The circumference of a sphere was measured to be 88 cm with a possible error of 0.5 cm.
Since $\displaystyle C = 2\pi r$, we have: .$\displaystyle 2\pi r = 88\quad\Rightarrow\quad r \,=\,\frac{44}{\pi}$ cm

We are told that: .$\displaystyle \Delta r\:=\:dr\:=\:\pm0.5$ cm

(a) Use differentials to estimate the maximum and relative error
in the calculated surface area.
(Round maximum error to the nearest whole number
and relative error to three decimal places.)

Surface area: .$\displaystyle A\;=\;4\pi r^2$

The base sphere has area: .$\displaystyle A \:=\:4\pi\left(\frac{44}{\pi}\right)^2 \:=\:\frac{7744}{\pi}$ cm².

The approximate error in the area is: .$\displaystyle dA \;=\;8\pi r\,dr$

Hence: .$\displaystyle dA \;=\;8\pi\left(\frac{44}{\pi}\right)(\pm0.5)$

The maximum error is: .$\displaystyle dA \:=\:\pm176$ cm²

The relative error is: .$\displaystyle \frac{dA}{A} \;=\;\frac{\pm176}{\frac{7744}{\pi}} \;\approx\;\pm 0.071$

(b) Use differentials to estimate the maximum and relative error
in the calculated volume.
(Round maximum error to the nearest whole number
and relative error to three decimal places.)

Volume of a sphere: .$\displaystyle V \;=\;\frac{4}{3}\pi r^3$

The base sphere has volume: .$\displaystyle V \;=\;\frac{4}{3}\pi\left(\frac{44}{\pi}\right)^3 \;=\;\frac{340,736}{3\pi^2}$ cm³

The approximate error in the volume is: .$\displaystyle dV \;=\;4\pi r^2\,dr$

Hence: .$\displaystyle dV\;=\;4\pi\left(\frac{44}{\pi}\right)^2(\pm 0.5)$

The maximum error is: .$\displaystyle \pm\frac{3872}{\pi} \;\approx\;\pm1232$ cm³

The relvative error is: .$\displaystyle \frac{dV}{v}\;=\;\frac{\pm\frac{3872}{\pi}}{\frac{ 340,736}{3\pi^2}} \;\approx\;\pm0.107$