A sphere is being manufactured with a radius of 60 cm and a maximum possible error of 0.029 cm for the radius. Find the approximate error and the approximate relative error in the volume of the sphere. Give the relative error as a percent.

The differential is $LaTeX: \displaystyle dV = 4 \pi r^{2}$ . Evaluating at $LaTeX: \displaystyle r = 60$ and $LaTeX: \displaystyle dr = 0.029$ gives $LaTeX: \displaystyle 417.6 \pi$ . The relative error is given by $LaTeX: \displaystyle \frac{dV}{V} = \frac{4 \pi r^{2}}{\frac{4 \pi r^{3}}{3}}\,dr=\frac{3}{r}\,dr$ . Evaluating gives $LaTeX: \displaystyle 0.00145 = 0.145$ %