Activity 7.5.1. Verifying 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 7.17, where \(\mathbf u\) can be dragged around the circle of directions.
(a)
Beginning with \(D_{\mathbf u} f = \nabla f \cdot \mathbf u\text{,}\) explain why the dot product can be rewritten as \(|\nabla f|\cos\theta\text{.}\) Where is the assumption \(|\mathbf u| = 1\) used?
Answer.
For any two vectors, \(\mathbf a \cdot \mathbf b = |\mathbf a||\mathbf b|\cos\theta\) with \(\theta\) the angle between them. Applying this to \(\nabla f\) and \(\mathbf u\text{,}\)
\begin{align*}
D_{\mathbf u} f = \nabla f \cdot \mathbf u\\
\amp= |\nabla f|\,|\mathbf u|\cos\theta\\
\amp= |\nabla f|\cos\theta ,
\end{align*}
where the last step uses \(|\mathbf u| = 1\text{.}\) Without that assumption the factor \(|\mathbf u|\) would survive, and rescaling \(\mathbf u\) would change \(D_{\mathbf u} f\) without changing the direction being measured.
(b)
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 \(D_{\mathbf u} f = 0\text{?}\)
Answer.
Since \(-1 \le \cos\theta \le 1\text{,}\) the quantity \(|\nabla f|\cos\theta\) ranges between \(-|\nabla f|\) and \(|\nabla f|\text{.}\) The largest value \(|\nabla f|\) occurs at \(\theta = 0\text{,}\) with \(\mathbf u\) pointing along \(\nabla f\text{;}\) the smallest, \(-|\nabla f|\text{,}\) occurs at \(\theta = \pi\text{,}\) with \(\mathbf u\) pointing along \(-\nabla f\text{.}\) The directional derivative vanishes when \(\cos\theta = 0\text{,}\) that is at \(\theta = \pi/2\) and \(\theta = 3\pi/2\text{,}\) the two directions orthogonal to \(\nabla f\text{.}\)
(c)
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.
Answer.
Here \(\nabla f = 2x\,\mathbf i + 2y\,\mathbf j\text{,}\) so \((\nabla f)_{(1,1)} = 2\,\mathbf i + 2\,\mathbf j\) and \(|\nabla f| = 2\sqrt 2\text{.}\) Therefore
\begin{gather*}
D_{\mathbf u} f = 2\sqrt 2 \,\cos\theta .
\end{gather*}
The fastest increase happens along \(\mathbf u = \frac{1}{\sqrt 2}\mathbf i + \frac{1}{\sqrt 2}\mathbf j\text{,}\) giving \(D_{\mathbf u} f = 2\sqrt 2 \approx 2.83\text{,}\) which is the value computed in Example 7.6. The fastest decrease happens along \(-\frac{1}{\sqrt 2}\mathbf i - \frac{1}{\sqrt 2}\mathbf j\text{,}\) giving \(-2\sqrt 2\text{.}\) There is no change along \(\pm\left(\frac{1}{\sqrt 2}\mathbf i - \frac{1}{\sqrt 2}\mathbf j\right)\text{,}\) the two directions orthogonal to \(\nabla f\text{.}\)
(d)
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?
Instructions.
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).
Answer.
The three buttons should give \(2\sqrt 2 \approx 2.83\text{,}\) \(-2\sqrt 2 \approx -2.83\text{,}\) and \(0\text{.}\) The reading changes fastest where \(\left|\frac{d}{d\theta}\cos\theta\right| = |\sin\theta|\) is largest, namely at \(\theta = \pi/2\) and \(3\pi/2\text{,}\) the very directions where \(D_{\mathbf u} f\) itself is zero. It is momentarily stationary at \(\theta = 0\) and \(\theta = \pi\text{,}\) where \(D_{\mathbf u} f\) is largest and smallest. So a small change of heading matters least when you are already pointing straight uphill, and most when you are moving along a level direction.
