where \(\theta\) is the angle between \(\nabla f\) and the unit vector \(\mathbf u\text{.}\) The first line is (7.5), obtained from the chain rule in Section 7.2; the second is the geometric form of the dot product; and the third uses \(\|\mathbf u\| = 1\text{.}\)
Remark7.18Three Consequences of \(D_{\mathbf u} f = \|\nabla f\|\cos\theta\).
Only the factor \(\cos\theta\) depends on the direction, and it runs over \([-1,1]\) as \(\mathbf u\) turns through the circle of directions at \(P_0\text{.}\) Its three extreme values single out three directions, drawn in Figure 7.19:
When \(\nabla f\) and \(\mathbf u\) are in the same direction (\(\cos\theta = 1\)), \(f\) increases most rapidly and \(D_{\mathbf u} f = \|\nabla f\|\text{.}\)
When \(\nabla f\) and \(\mathbf u\) have opposite directions (\(\cos\theta = -1\)), \(f\) decreases most rapidly and \(D_{\mathbf u} f = -\|\nabla f\|\text{.}\)
Figure7.19.Directions around \(P_0\text{:}\)\(f\) increases fastest along \(\nabla f\text{,}\) decreases fastest along \(-\nabla f\text{,}\) and does not change in the two directions orthogonal to \(\nabla f\text{.}\)
Activity7.20Verifying the Three Special Directions.
The three cases listed above are all consequences of a single formula, \(D_{\mathbf u} f = \|\nabla f\|\cos\theta\text{.}\) In this activity you will derive them yourself and then check your answers against Figure 7.21, where \(\mathbf u\) can be dragged around the circle of directions.
Beginning with \(D_{\mathbf u} f = \nabla f \cdot \mathbf u\text{,}\) explain why the dot product can be rewritten as \(\|\nabla f\|\cos\theta\text{.}\)
Treat \(\|\nabla f\|\) as fixed and let \(\theta\) run over \([0, 2\pi)\text{.}\) What are the largest and smallest values of \(D_{\mathbf u} f\text{,}\) and at which angles do they occur? For which angles is \(D_{\mathbf u} f = 0\text{?}\)
Now take \(f(x,y) = 1 + x^2 + y^2\) at \(P_0(1,1)\text{,}\) the function of Example 7.6. Compute \(\nabla f\) and \(\|\nabla f\|\) at \(P_0\text{,}\) write \(D_{\mathbf u} f\) as a function of \(\theta\text{,}\) and give the three special unit vectors explicitly.
Check your work against the figure below. Drag \(\mathbf u\) around the circle and confirm that the meter agrees with \(2\sqrt 2\cos\theta\) at several angles. Then use the three buttons and compare the readings with your answers to the previous task. As you drag, at what angle does the reading change most quickly, and at what angle is it momentarily stationary?
Drag the blue unit vector \(\mathbf u\) around the circle of directions and watch the meter display \(D_{\mathbf u} f = \|\nabla f\|\cos\theta\text{.}\) Use the buttons to snap \(\mathbf u\) to the three special cases: the same direction as \(\nabla f\) (fastest increase), the opposite direction (fastest decrease), and a direction orthogonal to \(\nabla f\) (no change).
Note that we need to divide \(\left(\nabla f\right)_{(1,1)} = \mathbf i + \mathbf j\) by its length to find the unit vector corresponding to the direction of most rapid increase:
Note that \(\tilde{\mathbf u} = -\mathbf u\text{.}\) All four directions are shown in Figure 7.23: the directions of zero change are tangent to the level curve of \(f\) through \(P_0(1,1)\text{,}\) while the directions of most rapid increase and decrease are perpendicular to it.
Figure7.23.At \(P_0(1,1)\text{:}\) fastest increase along \(\nabla f/\|\nabla f\|\text{,}\) fastest decrease along \(-\nabla f/\|\nabla f\|\text{,}\) and zero change along \(\pm\mathbf u\text{,}\) tangent to the level curve \(f = 1\text{.}\)
Press “Play” to watch \(\mathbf u\) turn once around \(P_0(1,1)\text{.}\) The reading at the lower left follows \(D_{\mathbf u} f = \nabla f\cdot\mathbf u = \|\nabla f\|\cos\theta\) as it sweeps from \(\|\nabla f\|\) down to \(-\|\nabla f\|\) and back: it is largest along \(\nabla f\) (A), smallest along \(-\nabla f\) (B), and zero in the two directions tangent to the highlighted level curve \(f = 1\) (C), where the second vector \(\tilde{\mathbf u} = -\mathbf u\) appears. Stop the animation at any time and drag \(\mathbf u\) on the picture, or use the slider.
Figure7.24.The same three directions, animated: \(D_{\mathbf u} f\) sweeping between \(\|\nabla f\|\) and \(-\|\nabla f\|\) as \(\mathbf u\) turns around \(P_0(1,1)\text{.}\)
Press “Play” to watch the vertical plane through \(P_0(1,1)\) turn once around. The gold curve is where it cuts the surface \(z = \frac{x^2}{2} + \frac{y^2}{2}\text{,}\) and the straight line through \((1,1,1)\) is the tangent to that curve, whose slope is \(D_{\mathbf u} f\text{:}\) it is green and steepest upwards along \(\nabla f\) (A), red and steepest downwards along \(-\nabla f\) (B), and yellow and level in the two directions tangent to the pink level curve \(f = 1\) (C). Stop the animation at any time and drag the \(\varphi\) slider, drag the figure to change the viewpoint, or press “Spin”.
Figure7.25.The same 3D view, animated: the tangent to the sliced surface at \((1,1,1)\) tipping from steepest ascent through level to steepest descent as \(\mathbf u\) turns.
