# use of differentials

The length, width and height of a rectangular box are measured to be 3cm, 4cm and 5cm respectively, with a maximum error of 0.05cm in each measurement. Use differentials to approximate the maximum error in the calculated volume.

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Let $V=xyz$ be the actual volume of the box, where $x,y,z$ are respectively its actual length, width and height, and let $V_{0}=x_{0}y_{0}z_{0}=3\cdot 4\cdot 5=60$ cm$^{3}$ be the measured volume.

The maximum error of 0.05 cm in each measurement means that $\left\vert x-3\right\vert \leq 0.05$, $\left\vert y-4\right\vert \leq 0.05$, $% \left\vert z-5\right\vert \leq 0.05$ cm.

Since these three measurement errors are small, respectively, $\pm \frac{5}{3}\%$, $\pm \frac{5}{4}\%$ and $\pm \frac{5}{5}\%$, of the measured length, width and height, we can approximate the error in the computed volume by the differential $dV$ evaluated at $(x_{0},y_{0},z_{0})=(3,4,5)$, and taking $dx=dy=dz=\max \left\{ \left\vert x-3\right\vert ,\left\vert y-4\right\vert ,\left\vert z-5\right\vert \right\} =0.05$ cm:

$$\begin{eqnarray*} dV &=&\left. \frac{\partial }{\partial x}\left( xyz\right) \right\vert _{(3,4,5)}dx+\left. \frac{\partial }{\partial y}\left( xyz\right) \right\vert _{(3,4,5)}dy+\left. \frac{\partial }{\partial z}\left( xyz\right) \right\vert _{(3,4,5)}dz \\ &=&4\cdot 5\cdot 0.05+3\cdot 5\cdot 0.05+3\cdot 4\cdot 0.05 \\ &=&2.35\text{ cm}^{3}. \end{eqnarray*}$$

Of course, for $dx=dy=dz=-0.05$ cm the value of the differential $dV$ would be symmetric: $dV=-2.35\text{ cm}^{3}.$

Indeed these values compare very well with the maximum error $\varepsilon _{\max }$ in the computed volume:

$$-2.3201=2.95\cdot 3.95\cdot 4.95-V_{0}\leq \varepsilon\leq 3.05\cdot 4.05\cdot 5.05-V_{0}=2.3801.$$

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Hint: express V as a function of L, W, H. Take the partial derivative with respect to each variable to see how V changes with each.

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