Answer
(a) $\frac{{\partial \rho }}{{\partial T}} = - 0.1425$ and $\frac{{\partial \rho }}{{\partial S}} = 0.785$
(b) $\rho$ is concave down as a function of $T$ at a fixed $S=33$.
The quotients $\frac{{\Delta \rho }}{{\Delta T}}$ are decreasing.
The sign of $\frac{{{\partial ^2}\rho }}{{\partial {T^2}}}$ is negative.
Work Step by Step
(a)
1. Estimate $\frac{{\partial \rho }}{{\partial T}}$ by computing the average:
Consider the point $\left( {S,T} \right) = \left( {34,2} \right)$. From Table 1 we get
$\frac{{\partial \rho }}{{\partial T}} \approx \frac{{\Delta \rho }}{{\Delta T}} = \frac{{27 - 27.18}}{{4 - 2}} = \frac{{ - 0.18}}{2} = - 0.09$
Consider the point $\left( {S,T} \right) = \left( {35,10} \right)$. From Table 1 we get
$\frac{{\partial \rho }}{{\partial T}} \approx \frac{{\Delta \rho }}{{\Delta T}} = \frac{{26.6 - 26.99}}{{12 - 10}} = \frac{{ - 0.39}}{2} = - 0.195$
So, the average is
$\frac{{\partial \rho }}{{\partial T}} = \frac{{ - 0.09 - 0.195}}{2} = - 0.1425$
2. Estimate $\frac{{\partial \rho }}{{\partial S}}$ by computing the average:
Consider the point $\left( {S,T} \right) = \left( {34,2} \right)$. From Table 1 we get
$\frac{{\partial \rho }}{{\partial S}} \approx \frac{{\Delta \rho }}{{\Delta S}} = \frac{{28.01 - 27.18}}{{35 - 34}} = \frac{{0.83}}{1} = 0.83$
Consider the point $\left( {S,T} \right) = \left( {35,10} \right)$. From Table 1 we get
$\frac{{\partial \rho }}{{\partial S}} \approx \frac{{\Delta \rho }}{{\Delta S}} = \frac{{27.73 - 26.99}}{{36 - 35}} = \frac{{0.74}}{1} = 0.74$
So, the average is
$\frac{{\partial \rho }}{{\partial S}} = \frac{{0.83 + 0.74}}{2} = 0.785$
(b) To find if $\rho$ concave up or concave down as a function of $T$ at a fixed $S=33$, we determine whether the quotients $\frac{{\Delta \rho }}{{\Delta T}}$ are increasing or decreasing.
Moving up vertically along the $T$-coordinate we get from Table 1: $\Delta T > 0$ and $\Delta \rho < 0$. Therefore, $\frac{{\Delta \rho }}{{\Delta T}} < 0$. So, $\frac{{\Delta \rho }}{{\Delta T}}$ are decreasing. Since $\frac{{\Delta \rho }}{{\Delta T}}$ is the slope of the vertical trace parallel to the $\rho T$-plane, we conclude that $\rho$ is concave down. Therefore, the second derivative $\frac{{{\partial ^2}\rho }}{{\partial {T^2}}}$ is negative.
