Answer
Estimation of the change in $I$:
$\Delta I \simeq 0.5644$
Work Step by Step
We are given $I = \frac{W}{{{H^2}}}$. The partial derivatives are
$\frac{{\partial I}}{{\partial W}} = \frac{1}{{{H^2}}}$, ${\ \ \ }$ $\frac{{\partial I}}{{\partial H}} = - \frac{{2W}}{{{H^3}}}$
At $\left( {W,H} \right) = \left( {34,1.3} \right)$, we have
$\frac{{\partial I}}{{\partial W}}{|_{\left( {34,1.3} \right)}} = \frac{1}{{{{\left( {1.3} \right)}^2}}}$, ${\ \ \ }$ $\frac{{\partial I}}{{\partial H}}{|_{\left( {34,1.3} \right)}} = - \frac{{2\cdot34}}{{{{\left( {1.3} \right)}^3}}}$
If $\left( {W,H} \right)$ changes from $\left( {34,1.3} \right)$ to $\left( {36,1.32} \right)$, then
$\Delta W = 36 - 34 = 2$, ${\ \ \ }$ $\Delta H = 1.32 - 1.3 = 0.02$
Using the linear approximation, Eq. (5) to estimate the change in $I$ we get
$\Delta I \approx \frac{{\partial I}}{{\partial W}}{|_{\left( {34,1.3} \right)}}\Delta W + \frac{{\partial I}}{{\partial H}}{|_{\left( {34,1.3} \right)}}\Delta H$
$\Delta I \simeq \frac{1}{{{{\left( {1.3} \right)}^2}}}\cdot2 - \frac{{2\cdot34}}{{{{\left( {1.3} \right)}^3}}}\cdot0.02 \simeq 0.5644$
Estimation of the change in $I$:
$\Delta I \simeq 0.5644$
