= Solution
\b[The power coefficient is approximately $-2.28\times10^{-11}\ \mathrm{W^{-1}}$, or $-2.28\times10^{-5}\ \mathrm{MW^{-1}}$.] From part (c),
$$
R_1=R_2=\frac{75}{236\times10^6}
=3.17797\times10^{-7}\ \mathrm{K\,W^{-1}},
\qquad
\frac1{2W_Cc_C}=2.11864\times10^{-7}\ \mathrm{K\,W^{-1}}.
$$
Substituting in the <power coefficient of reactivity> gives
$$
\begin{aligned}
\alpha_P={}&(-2.0\times10^{-5})(3.17797\times10^{-7})\\
&+(-1.5\times10^{-5})(3.17797\times10^{-7})\\
&+(-5.5\times10^{-5})(2.11864\times10^{-7})\\
={}&-2.27754\times10^{-11}\ \mathrm{W^{-1}}.
\end{aligned}
$$
The negative sign means that a slow increase in thermal <power> produces a decrease in <reactor reactivity> along this family. This is a restoring thermal <feedback control> effect despite the positive moderator coefficient. \b[A negative quasistatic slope alone does not prove stability of every transient:] the separate <heat capacities>, transport delays and neutron-population dynamics matter for a time-dependent stability analysis.
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