Answer
(a) We use the linear approximation, Eq. (5) to estimate the change in $I$ and show that $\Delta I \approx 0$ if $\frac{{\Delta H}}{{\Delta W}} \approx \frac{H}{{2W}}$.
(b) An increase in $H$ by $0.022$ meters will leave $I$ (approximately) constant if $W$ is increased by $1$ kg.
Work Step by Step
(a) We are given $I = \frac{W}{{{H^2}}}$. The partial derivatives are
$\frac{{\partial I}}{{\partial W}} = \frac{1}{{{H^2}}}$, ${\ \ \ }$ $\frac{{\partial I}}{{\partial H}} = - \frac{{2W}}{{{H^3}}}$
Using the linear approximation, Eq. (5) the change in $I$ is estimated to be
$\Delta I \approx \frac{{\partial I}}{{\partial W}}\Delta W + \frac{{\partial I}}{{\partial H}}\Delta H$
$\Delta I \approx \frac{1}{{{H^2}}}\Delta W - \frac{{2W}}{{{H^3}}}\Delta H$
Let $\Delta I \approx 0$. Thus,
$0 \approx \frac{1}{{{H^2}}}\Delta W - \frac{{2W}}{{{H^3}}}\Delta H$
$\frac{1}{{{H^2}}}\Delta W \approx \frac{{2W}}{{{H^3}}}\Delta H$
Hence, $\Delta I \approx 0$ if $\frac{{\Delta H}}{{\Delta W}} \approx \frac{H}{{2W}}$.
(b) Recall from part (a):
$\Delta I \approx 0$ if $\frac{{\Delta H}}{{\Delta W}} \approx \frac{H}{{2W}}$.
If $I$ (approximately) stays constant and $W$ is increased by $1$ kg, then we have
$\Delta I \approx 0$ ${\ \ }$ and ${\ \ }$ $\Delta W = 1$
Since $\left( {W,H} \right) = \left( {25,1.1} \right)$, so
$\frac{{\Delta H}}{1} \approx \frac{{1.1}}{{2\cdot25}}$
$\Delta H \approx 0.022$
Thus, an increase in $H$ by $0.022$ meters will leave $I$ (approximately) constant if $W$ is increased by $1$ kg.
