Answer
(a) $I$ at $\left( {T,H} \right) = \left( {95,50} \right)$ is
$I\left( {95,50} \right) = 73.1913$
(b) The partial derivative $\frac{{\partial I}}{{\partial T}}$ tells us the increase in $I$ per degree increase in $T$ when $\left( {T,H} \right) = \left( {95,50} \right)$.
$\frac{{\partial I}}{{\partial T}}{|_{\left( {T,H} \right) = \left( {95,50} \right)}} \simeq 1.66$
Work Step by Step
(a) We have an approximate formula for the heat index:
$I\left( {T,H} \right) = 45.33 + 0.6845T + 5.758H - 0.00365{T^2}$
$ - 0.1565HT + 0.001H{T^2}$
Calculate $I$ at $\left( {T,H} \right) = \left( {95,50} \right)$:
$I\left( {95,50} \right) = 45.33 + 0.6845\cdot95 + 5.758\cdot50 - 0.00365\cdot{\left( {95} \right)^2}$
$ - 0.1565\cdot50\cdot95 + 0.001\cdot50\cdot{\left( {95} \right)^2}$
$ = 73.1913$
(b) The partial derivative $\frac{{\partial I}}{{\partial T}}$ tells us the increase in $I$ per degree increase in $T$ when $\left( {T,H} \right) = \left( {95,50} \right)$.
So,
$\frac{{\partial I}}{{\partial T}} = 0.6845 - 2 \times 0.00365T - 0.1565H + 2 \times 0.001HT$
$\frac{{\partial I}}{{\partial T}} = 0.6845 - 0.0073T - 0.1565H + 0.002HT$
$\frac{{\partial I}}{{\partial T}}{|_{\left( {T,H} \right) = \left( {95,50} \right)}} \simeq 1.66$
