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Exercises 7.8 Exercises: Scalar Fields and Potentials

Each exercise below repeats the three questions of Example 7.23 in a different physical setting: find the direction of most rapid decrease, the rate along a prescribed direction, and the directions of no change.

1. Electric Potential on a Coated Plate.

The electric potential on a thin coated plate, in volts, is
\begin{equation*} V(x,y) = 100 - 2x^2 - 6y^2, \end{equation*}
with \(x\) and \(y\) in centimeters. Consider the point \(P_0(2,1)\text{.}\)
  1. Find \(\nabla V\) at \(P_0\) and the electric field \(\mathbf E = -\nabla V\) there. Along which unit vector does the potential drop most rapidly, and at what rate?
  2. A test charge is nudged along \(\mathbf u = \tfrac{4}{5}\mathbf i - \tfrac{3}{5}\mathbf j\text{.}\) Compute \(D_{\mathbf u}V\) at \(P_0\text{.}\)
  3. Find the two unit vectors along which the potential does not change, and name the curve they are tangent to.
Answer.
  1. \(\nabla V = -4x\,\mathbf i - 12y\,\mathbf j\text{,}\) so \((\nabla V)_{(2,1)} = -8\,\mathbf i - 12\,\mathbf j\) and \(\mathbf E = 8\,\mathbf i + 12\,\mathbf j\) V/cm. The potential drops fastest along \(\mathbf u = \tfrac{1}{\sqrt{13}}(2\,\mathbf i + 3\,\mathbf j)\text{,}\) at the rate \(-|\nabla V| = -4\sqrt{13} \approx -14.4\) V/cm.
  2. \begin{equation*} D_{\mathbf u}V = (-8)\left(\tfrac{4}{5}\right) + (-12)\left(-\tfrac{3}{5}\right) = \tfrac{4}{5} = 0.8 \end{equation*}
    in V/cm. The potential rises slightly along this heading.
  3. \(\mathbf u = \pm\tfrac{1}{\sqrt{13}}(3\,\mathbf i - 2\,\mathbf j)\text{.}\) These are tangent to the equipotential ellipse \(2x^2 + 6y^2 = 14\) through \(P_0\text{,}\) and they are orthogonal to \(\mathbf E\text{,}\) as they must be.
Figure 7.25. The potential \(V(x,y) = 100 - 2x^2 - 6y^2\) on the plate. One route holds \(V = 86\) V along the equipotential \(2x^2 + 6y^2 = 14\text{;}\) the other follows a field line, on which \(D_{\mathbf u}V = -|\nabla V|\) at every point.

2. The Fall Line on a Ski Slope.

A mountainside has elevation, in feet,
\begin{equation*} h(x,y) = 3000 - 0.0004x^2 - 0.001y^2, \end{equation*}
with \(x\) and \(y\) in feet. A skier stands at \(P_0(200,100)\text{.}\) Her gravitational potential energy is \(U = mgh\text{,}\) so the downhill force she feels is a positive multiple of \(-\nabla h\text{.}\)
  1. Find the fall line, the unit vector of steepest descent, and the grade of the slope there, expressed as a percent (feet dropped per hundred feet travelled).
  2. A traverse is a route on which she neither climbs nor descends. Find the two unit vectors that accomplish this.
  3. She wants a gentler line of exactly \(12\%\) grade. Using \(D_{\mathbf u}h = |\nabla h|\cos\theta\text{,}\) find the angle \(\theta\) between her heading and the fall line.
Answer.
  1. \(\nabla h = -0.0008x\,\mathbf i - 0.002y\,\mathbf j\text{,}\) so \((\nabla h)_{(200,100)} = -0.16\,\mathbf i - 0.20\,\mathbf j\text{.}\) Steepest descent is along \(\mathbf u = \tfrac{1}{\sqrt{41}}(4\,\mathbf i + 5\,\mathbf j)\text{,}\) and \(|\nabla h| = \sqrt{0.0656} \approx 0.256\text{,}\) a grade of about \(25.6\%\text{.}\)
  2. \(\mathbf u = \pm\tfrac{1}{\sqrt{41}}(5\,\mathbf i - 4\,\mathbf j)\text{,}\) tangent to the contour line of constant elevation through \(P_0\text{.}\)
  3. \(\cos\theta = 0.12/0.256 \approx 0.469\text{,}\) so \(\theta \approx 62^\circ\) off the fall line, on either side.
Figure 7.26. Part (a) says the grade is \(|\nabla h| = 0.2561\text{.}\) Here is what that number is, on the hill itself. The vertical plane through \(P_0\) in the direction \(\mathbf u\) cuts the surface in the gold curve; the red line is that curve’s tangent at \(P_0\text{;}\) and \(\alpha\) is the angle it makes with the horizontal. Since \(\tan\alpha = D_{\mathbf u}h\text{,}\) the grade is the tangent of that angle. Swing \(\mathbf u\) with the slider: \(\alpha\) is largest along the fall line, where \(\tan\alpha = 0.2561\) and \(\alpha = 14.4^\circ\text{,}\) and it closes to nothing along the contour, where the tangent line lies flat.
Figure 7.27. The mountainside \(h(x,y) = 3000 - 0.0004x^2 - 0.001y^2\text{.}\) Compare the fall line, the level traverse, and the \(12\%\) line. Watch the grade readout drift away from \(12\%\) as the skier leaves \(P_0\text{,}\) since the angle in part (c) is computed at \(P_0\) only.

3. Sailing the Isobars of a Low.

Near the center of a low-pressure system the sea-level pressure, in millibars, is modeled by
\begin{equation*} P(x,y) = 988 + 0.004\left(x^2 + y^2\right), \end{equation*}
with \(x\) and \(y\) in miles from the center of the low. A ship sits at \(P_0(20,10)\text{.}\)
  1. Find the pressure at the ship and \(\nabla P\) there. The pressure-gradient force per unit volume of air is a positive multiple of \(-\nabla P\text{;}\) in which unit direction does it push, and what is \(|\nabla P|\) in millibars per mile?
  2. A second ship 40 miles from the center reports a stronger wind. Using this model, explain why \(|\nabla P|\) supports that report.
  3. The captain decides to hold a constant barometer reading. Give the two possible unit headings at \(P_0\text{,}\) and describe the shape of the full route.
Answer.
  1. \(P(20,10) = 988 + 0.004(500) = 990\) mb, and \(\nabla P = 0.008x\,\mathbf i + 0.008y\,\mathbf j\) gives \((\nabla P)_{(20,10)} = 0.16\,\mathbf i + 0.08\,\mathbf j\text{.}\) The force points along \(-\mathbf u = -\tfrac{1}{\sqrt{5}}(2\,\mathbf i + \mathbf j)\text{,}\) that is, inward toward the low, and \(|\nabla P| = 0.08\sqrt{5} \approx 0.18\) mb/mi.
  2. Here \(|\nabla P| = 0.008\sqrt{x^2+y^2}\) grows linearly with distance from the center, so at 40 miles it is \(0.32\) mb/mi, nearly double the value at the first ship. (Real storms reverse this trend close to the eye; the model is only reasonable over a limited annulus.)
  3. \(\mathbf u = \pm\tfrac{1}{\sqrt{5}}(\mathbf i - 2\,\mathbf j)\text{.}\) Since the level curves of \(P\) are the circles \(x^2+y^2 = \text{constant}\text{,}\) holding the barometer steady means circling the storm at a fixed radius of \(\sqrt{500} \approx 22.4\) miles.
Use the interactive figure to explore the problem further geometrically, switching between the two headings and watching how the barometer responds as the ship moves.
Figure 7.28. The pressure field \(P(x,y) = 988 + 0.004(x^2+y^2)\) around a low. One heading circles the eye at constant \(990\) mb; the other runs outward along \(\nabla P\text{.}\) The distance readout shows why \(|\nabla P| = 0.008r\) grows with radius.
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