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Section 8.5 A Physical Application: Equilibrium and Stability

Consider a marble that rolls on the surface \(z = f(x,y)\) under the influence of gravity. If we measure heights from the \(xy\)-plane, the potential energy of the marble at the point \((x,y)\) of the surface is
\begin{equation} U(x,y) = mgz = mg\,f(x,y),\tag{8.16} \end{equation}
where \(m\) is the mass of the marble and \(g\) is the acceleration due to gravity. The marble is in equilibrium at a point where the force along the surface vanishes, which happens exactly where both partial derivatives of the potential energy are zero:
\begin{equation} U_x = mg\,f_x = 0 \quad \text{and} \quad U_y = mg\,f_y = 0.\tag{8.17} \end{equation}
In other words, the equilibrium points of the marble are precisely the critical points of \(f\text{.}\) The second derivative test then tells us whether the equilibrium is stable (a displaced marble rolls back) or unstable (a displaced marble rolls away).

Example 8.22. Equilibrium and Stability of a Marble.

A marble rests at the origin on each of the following three surfaces:
\begin{equation} \text{(i)}\; z = x^2 + y^2, \qquad \text{(ii)}\; z = -x^2 - y^2, \qquad \text{(iii)}\; z = \frac{y^2}{3} - \frac{x^2}{3}.\tag{8.18} \end{equation}
Show that \((0,0)\) is an equilibrium point in each case, and determine whether the equilibrium is stable or unstable.

Solution.

In each case the potential energy is \(U(x,y) = mg\,f(x,y)\text{,}\) so by (8.17) the equilibrium points are the critical points of \(f\text{.}\) For all three surfaces we have \(f_x(0,0) = f_y(0,0) = 0\text{,}\) so the origin is an equilibrium point in each case. We classify each equilibrium with the second derivative test, as shown in FigureΒ 8.23.
Case (i): For \(f(x,y) = x^2 + y^2\) we have \(f_{xx} = f_{yy} = 2\) and \(f_{xy} = 0\text{,}\) so \(H(0,0) = 4 \gt 0\) and \(f_{xx} \gt 0\text{,}\) and the origin is a local minimum of the potential energy. A marble displaced slightly in any direction rolls back toward the bottom: the equilibrium is stable.
Case (ii): For \(f(x,y) = -x^2 - y^2\) we have \(f_{xx} = f_{yy} = -2\) and \(f_{xy} = 0\text{,}\) so \(H(0,0) = 4 \gt 0\) and \(f_{xx} \lt 0\text{,}\) and the origin is a local maximum of the potential energy. A marble displaced slightly in any direction rolls away from the top: the equilibrium is unstable.
Case (iii): For \(f(x,y) = y^2/3 - x^2/3\) we have \(f_{xx} = -2/3\text{,}\) \(f_{yy} = 2/3\text{,}\) and \(f_{xy} = 0\text{,}\) so \(H(0,0) = -4/9 \lt 0\text{,}\) and the origin is a saddle point. Along the \(y\)-axis the potential energy rises, so a marble displaced in the \(y\)-direction rolls back; but along the \(x\)-axis the potential energy falls, so a marble displaced in the \(x\)-direction rolls away. Since some displacements grow, the equilibrium is unstable.
The general principle illustrated by this example, shown in FigureΒ 8.24, is that an equilibrium is stable exactly when the potential energy has a local minimum there.
Figure 8.23. A marble at an equilibrium point on each of the three surfaces. On the bowl \(z = x^2 + y^2\) a displaced marble rolls back (stable); on the dome \(z = -x^2 - y^2\) it rolls away (unstable); on the saddle \(z = y^2/3 - x^2/3\) it rolls back along the \(y\)-direction but away along the \(x\)-direction (unstable).
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Figure 8.24. Cross-sections of the potential energy near an equilibrium point. Along a direction where \(U\) has a minimum (blue), the force is restoring and pushes the marble back: stable. Along a direction where \(U\) has a maximum (red), the force pushes the marble away: unstable. For the bowl, every cross-section is the blue type; for the dome, every cross-section is the red type; a saddle has one of each.
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