Answer
The increase in $H$ is estimated to be $\Delta H \simeq 0.01912$ meters.
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
$\frac{{\partial I}}{{\partial W}}{|_{\left( {34,1.3} \right)}} = \frac{1}{{{{\left( {1.3} \right)}^2}}}$, ${\ \ \ }$ $\frac{{\partial I}}{{\partial H}}{|_{\left( {34,1.3} \right)}} = - \frac{{2\cdot34}}{{{{\left( {1.3} \right)}^3}}}$
Using the linear approximation, Eq. (5) the change in $I$ is estimated to be
$\Delta I \approx \frac{{\partial I}}{{\partial W}}{|_{\left( {34,1.3} \right)}}\Delta W + \frac{{\partial I}}{{\partial H}}{|_{\left( {34,1.3} \right)}}\Delta H$
$\Delta I \approx \frac{1}{{{{\left( {1.3} \right)}^2}}}\Delta W - \frac{{2\cdot34}}{{{{\left( {1.3} \right)}^3}}}\Delta H$
To keep $I$ constant if $W$ increases to $35$, we have
$\Delta I = 0$, ${\ \ \ }$ $\Delta W = 35 - 34 = 1$
Thus, $\Delta I$ becomes
$0 \simeq \frac{1}{{{{\left( {1.3} \right)}^2}}}\cdot1 - \frac{{2\cdot34}}{{{{\left( {1.3} \right)}^3}}}\cdot\Delta H$
$\frac{{2\cdot34}}{{{{\left( {1.3} \right)}^3}}}\cdot\Delta H \simeq \frac{1}{{{{\left( {1.3} \right)}^2}}}$
$\Delta H \simeq 0.01912$
Thus, the increase in $H$ is estimated to be $\Delta H \simeq 0.01912$ meters.
