For each of the sets of functions below, draw a chain rule diagram for the indicated derivative and use it to write a chain rule, then evaluate your chain rule to find the derivative.

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## Homework for Energy and Entropy

- (mbTreeDiagramMath)
*Use Tree Diagrams to find chain rules.*Find $\left( \frac{dg}{dt} \right)$ for $g = (a + b)^2$, $a = \sin 2t$, and $b = t^{3/2}$.

Find $\left( \frac{\partial h}{\partial v} \right)_u$ for $h = \sqrt{a - b}$, $a = uv^2 - 1/v$, and $b = \frac{uv}{u + v}$.

Find $\left( \frac{\partial A}{\partial B} \right)_F$ for $A(B,C)$ and $F(B,C)$. (Of course, you don't have to evaluate this derivative!)

- (mbCyclic)
*Check cyclic chain rule for realistic equation of state*A possible equation of state for a gas takes the form $$pV=N k_B T \exp\left(-\frac{\alpha V}{N k_B T}\right)$$ in which $\alpha$, $N$, and $k_B$ are constants. Calculate expressions for: $$\left(\frac{\partial p}{\partial V}\right)_T\qquad\qquad \left(\frac{\partial V}{\partial T}\right)_p\qquad\qquad \left(\frac{\partial T}{\partial p}\right)_V$$ and show that these derivatives satisfy the cyclic chain rule.

- (mbParamagnet)
*Use chain rules to solve physics problem***Paramagnetism**\hfill\break We have the following equations of state for the*total magnetization*$M$, and the entropy $S$ of a paramagnetic system: \begin{eqnarray*} M&=&N\mu\, \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}}\\ \noalign{\smallskip} S&=&Nk_B\left\{\ln 2 + \ln \left(e^{\frac{\mu B}{k_B T}}+e^{-\frac{\mu B}{k_B T}}\right) +\frac{\mu B}{k_B T} \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}} \right\}\\ \end{eqnarray*}Solve for the

*magnetic susceptibility*, which is defined as: $$\chi_B=\left(\frac{\partial M}{\partial B}\right)_T $$Also solve for almost the same derivative, now taken with the entropy $S$ held constant: $$\left(\frac{\partial M}{\partial B}\right)_S $$

Why does this second derivative turn out to be zero?

Sense-making: solve explicitly for the chain rule that allows you to evaluate $$\left(\frac{\partial M}{\partial B}\right)_S $$ using both total differentials (zapping with d) and a chain rule diagram. (Your chain rule should be the same!)