In class, you measured the isolength
stretchability and the isoforce stretchability of your systems in the
PDM. We found that for some systems these were very
different, while for others they were identical.
Show with algebra (NOT experiment) that the ratio of isolength stretchability to isoforce
stretchability is the same for both the left-hand side of the system and the right-hand side of the system.
i.e.:
\begin{align}
\frac{\left(\frac{\partial {x_L}}{\partial {F_L}}\right)_{x_R}}{\left(\frac{\partial {x_L}}{\partial {F_L}}\right)_{F_R}} &=
\frac{\left(\frac{\partial {x_R}}{\partial {F_R}}\right)_{x_L}}{\left(\frac{\partial {x_R}}{\partial {F_R}}\right)_{F_L}}
\label{eq:ratios}
\end{align}
Hint
You will need to make use of the
cyclic chain
rule:
\begin{align}
\left(\frac{\partial {A}}{\partial {B}}\right)_{C} = -\left(\frac{\partial {A}}{\partial {C}}\right)_{B}\left(\frac{\partial {C}}{\partial {B}}\right)_{A}
\end{align}
Hint
You will also need the
ordinary chain
rule:
\begin{align}
\left(\frac{\partial {A}}{\partial {B}}\right)_{D} = \left(\frac{\partial {A}}{\partial {C}}\right)_{D}\left(\frac{\partial {C}}{\partial {B}}\right)_{D}
\end{align}