Measuring Error in a Manufacturing Process

A company manufactures spherical ball bearings of radius \(3{\,{\rm{cm}}}\). The customer accepts a tolerance of 1% in the radius. Use differentials to approximate the relative error for the surface area of the acceptable ball bearings.

Solution The tolerance of 1% in the radius \(R\) means that the relative error in the radius \(R\) must be within \(0.01.\) That is, \(\dfrac{\vert \Delta R\vert }{R}≤ 0.01\). The surface area \(S\) of a sphere of radius \(R\) is given by the formula \(S=4\pi R^{2}.\) We seek the relative error in \(S, \dfrac{\vert \Delta S\vert }{S},\) which can be approximated by \(\dfrac{\vert dS\vert}{S}\). \[ \begin{eqnarray*} && \frac{\vert \Delta S\vert }{S}\approx \frac{\vert dS\vert }{S} \underset{\underset{\color{#0066A7}{\hbox{\(dS=S^{\,\prime} dR\)}}}{\color{#0066A7}{\uparrow}}}{=} \frac{( 8\pi R) \vert dR \vert }{4\pi R^{2}}=\frac{2\vert dR\vert }{R}= 2\cdot \frac{\vert \Delta R\vert }{R} \underset{\underset{\color{#0066A7}{\hbox{\(\tfrac{\vert \Delta R\vert }{R}≤ 0.01\)}}}{\color{#0066A7}{ \uparrow}}}{\le} 2(0.01)= 0.02 \end{eqnarray*} \]

The relative error in the surface area will be less than or equal to \(0.02\).