Question

Show that the relative error in computing the volume of a sphere, due to an error in measuring the radius, is approximately equal to three times the relative error in the radius.

Answer 1

Let x be the radius of the sphere and y be its volume.
$$\begin{gathered} \mathrm{Let}∆x\mathrm{be}\mathrm{the}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{radius}\mathrm{and}∆V\mathrm{be}\mathrm{the}\mathrm{approximate}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{volume}. \\ y=\frac{4}{3}\pi x^3 \\ \Rightarrow \frac{dy}{dx}=4\pi x^2 \\ \Rightarrow ∆y=dy=\frac{dy}{dx}dx=4\pi x^2\times ∆x \\ \Rightarrow ∆y=3\times \frac{4}{3}\pi x^3\times \frac{∆x}{x} \\ \Rightarrow ∆y=3\times y\times \frac{∆x}{x} \\ \Rightarrow \frac{∆y}{y}=3\frac{∆x}{x} \end{gathered}$$
Hence proved.