The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated volume.
Hello, drahcirnaw!
I had to snake my way through this one.
Hope I got it right . . .
The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm.
Use differentials to estimate the maximum error in the calculated volume.
We have: .$\displaystyle C \:=\:2\pi r$
We are given: .$\displaystyle C = 90 \quad\Rightarrow\quad 90 \:=\:2\pi r \quad\Rightarrow\quad r \:=\:\frac{45}{\pi}$ .[1]
We have: .$\displaystyle dC \:=\:2\pi dr$
We are given: .$\displaystyle dC \,=\,\pm0.5 \quad\Rightarrow\quad \pm 0.5 \:=\:2\pi dr \quad\Rightarrow\quad dr \:=\:\frac{\pm0.5}{2\pi}$ .[2]
The volume of a sphere is: .$\displaystyle V \:=\:\frac{4}{3}\pi r^3 $
. . Then: .$\displaystyle dV \;=\;4\pi r^2dr $
Substitute [1] and [2]: .$\displaystyle dV \;=\;4\pi\left(\frac{45}{\pi}\right)^2\left(\frac{ \pm0.5}{2\pi}\right) \;=\;\pm\frac{2025}{\pi^2}$
Therefore, the maximum error in volume is about: .$\displaystyle \pm205.2\text{ cm}^3$