\section*{Problem 3}
\subsection*{Part A}

The moderator temperature coefficient must be controlling power. With all other
factors constant, the moderator temperature coefficient is the only thing adding
negative reactivity to the system.

\subsection*{Part B}

\[\rho_{net} = \frac{\partial \rho_{net}}{\partial T}dT 
+ \frac{\partial \rho_{net}}{\partial H} dH
+ \frac{\partial \rho_{net}}{\partial Poison} dPoison
+ \frac{\partial \rho_{net}}{\partial Power} dPower
\]

But with ignoring fuel temperature feedback and no boron effects, 

\[\rho_{net} = -10 [\frac{\text{pcm}}{^\circ F}]dT 
+ \frac{\partial \rho_{net}}{\partial H} dH
\]


Given that there is no poison or fuel temperature feedback, and steam demand
does not change, reactor power will stay the same after the control rod drops
into the core. Only moderator temperature can change reactivity in this problem.

\subsection*{Part C}

\[0 = -10 [\frac{\text{pcm}}{^\circ F}]dT 
 - 100[pcm]
\]

\[dT = \frac{100[pcm]}{-10[\frac{\text{pcm}}{^\circ F}]}\]

\[dT = -10^\circ F\]
\[\boxed{T_{final} = 577^\circ F}\]

\subsection*{Part D}

Power will remain the same, and therefore steam pressure should remain the same
as well.