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# Derivatives and Chain Rules

## Prerequisites

Students should be able to:

• Recognize the traditional calculus notation for partial derivatives (i.e., $\partial f/\partial x$).
• Calculate a partial derivative given a symbolic expression and the variable(s) to be held constant.
• Determine a total differential by zapping with d.
• Write the thermodynamic identity for both the Partial Derivative Machine and for a gas in a piston.
• Reason about the physical quantities related to the Partial Derivative Machine.

## In-class Content

• New Surfaces activity - “Squishability” of Water Vapor (SGA - 20 min)

## Homework for Energy and Entropy

1. (mbTreeDiagramMath) Use Tree Diagrams to find chain rules.

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.

1. Find $\left( \frac{dg}{dt} \right)$ for $g = (a + b)^2$, $a = \sin 2t$, and $b = t^{3/2}$.

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

3. 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!)

2. (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.

3. (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*}

1. Solve for the magnetic susceptibility, which is defined as: $$\chi_B=\left(\frac{\partial M}{\partial B}\right)_T$$

2. Also solve for almost the same derivative, now taken with the entropy $S$ held constant: $$\left(\frac{\partial M}{\partial B}\right)_S$$

3. Why does this second derivative turn out to be zero?

4. 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!)

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