The surface is plotted in Figure 12.6.4, where the point \(P=(1,2)\) is indicated in the \(x,y\)-plane as well as the point \((1,2,9)\) which lies on the surface of \(f\text{.}\) We find that \(f_x(x,y) = -2x\) and \(f_x(1,2) = -2\text{;}\) \(f_y(x,y) = -2y\) and \(f_y(1,2) = -4\text{.}\)

1. Let \(\vec u_1\) be the unit vector that points from the point \((1,2)\) to the point \(Q=(3,4)\text{,}\) as shown in the figure. The vector \(\vv{PQ} = \la 2,2\ra\text{;}\) the unit vector in this direction is \(\vec u_1=\la 1/\sqrt{2}, 1/\sqrt{2}\ra\text{.}\) Thus the directional derivative of \(f\) at \((1,2)\) in the direction of \(\vec u_1\) is \begin{equation*} D_{\vec u_1}f(1,2) = -2(1/\sqrt{2}) +(-4)(1/\sqrt{2}) = -6/\sqrt{2}\approx -4.24. \end{equation*} Thus the instantaneous rate of change in moving from the point \((1,2,9)\) on the surface in the direction of \(\vec u_1\) (which points toward the point \(Q\)) is about \(-4.24\text{.}\) Moving in this direction moves one steeply downward.

2. We seek the directional derivative in the direction of \(\la 2,-1\ra\text{.}\) The unit vector in this direction is \(\vec u_2 = \la 2/\sqrt{5},-1/\sqrt{5}\ra\text{.}\) Thus the directional derivative of \(f\) at \((1,2)\) in the direction of \(\vec u_2\) is \begin{equation*} D_{\vec u_2}f(1,2) = -2(2/\sqrt{5})+(-4)(-1/\sqrt{5}) = 0. \end{equation*} Starting on the surface of \(f\) at \((1,2)\) and moving in the direction of \(\la 2,-1\ra\) (or \(\vec u_2\)) results in no instantaneous change in \(z\)-value. This is analogous to standing on the side of a hill and choosing a direction to walk that does not change the elevation. One neither walks up nor down, rather just “along the side” of the hill. Finding these directions of “no elevation change” is important.

3. At \(P=(1,2)\text{,}\) the direction towards the origin is given by the vector \(\la -1,-2\ra\text{;}\) the unit vector in this direction is \(\vec u_3=\la -1/\sqrt{5},-2/\sqrt{5}\ra\text{.}\) The directional derivative of \(f\) at \(P\) in the direction of the origin is \begin{equation*} D_{\vec u_3}f(1,2) = -2(-1/\sqrt{5}) + (-4)(-2/\sqrt{5}) = 10/\sqrt{5} \approx 4.47. \end{equation*} Moving towards the origin means “walking uphill” quite steeply, with an initial slope of about \(4.47\text{.}\)

We begin by finding the gradient. \(f_x = \cos(x) \cos(y)\) and \(f_y = -\sin(x) \sin(y)\text{,}\) thus \begin{equation*} \nabla f = \la \cos(x) \cos(y) ,-\sin(x) \sin(y) \ra \text{ and, at \(P\), } \nabla f\left(\frac{\pi}3,\frac{\pi}3\right) = \la\frac14,-\frac34\ra. \end{equation*}

Thus the direction of maximal increase is \(\la 1/4, -3/4\ra\text{.}\) In this direction, the instantaneous rate of \(z\) change is \(\norm{\la 1/4,-3/4\ra} = \sqrt{10}/4 \approx 0.79\text{.}\)

Figure 12.6.9 shows the surface plotted from two different perspectives. In each, the gradient is drawn at \(P\) with a dashed line (because of the nature of this surface, the gradient points “into” the surface). Let \(\vec u = \la u_1, u_2\ra\) be the unit vector in the direction of \(\nabla f\) at \(P\text{.}\) Each graph of the figure also contains the vector \(\la u_1, u_2, \norm{\nabla f\,}\ra\text{.}\) This vector has a “run” of 1 (because in the \(x\)-\(y\) plane it moves 1 unit) and a “rise” of \(\norm{\nabla f\,}\text{,}\) hence we can think of it as a vector with slope of \(\norm{\nabla f\,}\) in the direction of \(\nabla f\text{,}\) helping us visualize how “steep” the surface is in its steepest direction.

The direction of minimal increase is \(\la -1/4,3/4\ra\text{;}\) in this direction the instantaneous rate of \(z\) change is \(-\sqrt{10}/4 \approx -0.79\text{.}\)

Any direction orthogonal to \(\nabla f\) is a direction of no \(z\) change. We have two choices: the direction of \(\la 3,1\ra\) and the direction of \(\la -3,-1\ra\text{.}\) The unit vector in the direction of \(\la 3,1\ra\) is shown in each graph of the figure as well. The level curve at \(z=\sqrt{3}/4\) is drawn: recall that along this curve the \(z\)-values do not change. Since \(\la 3,1\ra\) is a direction of no \(z\)-change, this vector is tangent to the level curve at \(P\text{.}\)

We find \(\nabla f = \la -2x+2, -2y+2\ra\text{.}\) At \(P\text{,}\) we have \(\nabla f(1,1) = \la 0,0\ra\text{.}\) According to Theorem 12.6.7, this is the direction of maximal increase. However, \(\la 0,0\ra\) is directionless; it has no displacement. And regardless of the unit vector \(\vec u\) chosen, \(D_{\vec u\,}f = 0\text{.}\)

Figure 12.6.13 helps us understand what this means. We can see that \(P\) lies at the top of a paraboloid. In all directions, the instantaneous rate of change is 0.

So what is the direction of maximal increase? It is fine to give an answer of \(\vec 0 = \la 0,0\ra\text{,}\) as this indicates that all directional derivatives are 0.

Let \(\vrt = \la x(t), y(t)\ra\) be the vector–valued function describing the path of the water in the \(x\)-\(y\) plane; we seek \(x(t)\) and \(y(t)\text{.}\) We know that water will always flow downhill in the steepest direction; therefore, at any point on its path, it will be moving in the direction of \(-\nabla f\text{.}\) (We ignore the physical effects of momentum on the water.) Thus \(\vrp(t)\) will be parallel to \(\nabla f\text{,}\) and there is some constant \(c\) such that \(c\nabla f = \vrp(t) = \la x'(t), y'(t)\ra\text{.}\)

We find \(\nabla f = \la -2x, -4y\ra\) and write \(x'(t)\) as \(\frac{dx}{dt}\) and \(y'(t)\) as \(\frac{dy}{dt}\text{.}\) Then \begin{align*} c\nabla f \amp = \la x'(t), y'(t)\ra\\ \la -2cx, -4cy \ra \amp = \la \frac{dx}{dt}, \frac{dy}{dt}\ra. \end{align*}

This implies \begin{equation*} -2cx = \frac{dx}{dt} \text{ and } -4cy =\frac{dy}{dt}, \text{ i.e., } \end{equation*} \begin{equation*} c = -\frac{1}{2x}\frac{dx}{dt} \text{ and } c =-\frac{1}{4y}\frac{dy}{dt}. \end{equation*}

As \(c\) equals both expressions, we have \begin{equation*} \frac{1}{2x}\frac{dx}{dt} =\frac{1}{4y}\frac{dy}{dt}. \end{equation*}

To find an explicit relationship between \(x\) and \(y\text{,}\) we can integrate both sides with respect to \(t\text{.}\) Recall from our study of differentials that \(\ds \frac{dx}{dt}dt = dx\text{.}\) Thus: \begin{align*} \int \frac{1}{2x}\frac{dx}{dt}\ dt \amp =\int \frac{1}{4y}\frac{dy}{dt}\ dt\\ \int \frac{1}{2x}\ dx \amp =\int\frac{1}{4y}\ dy\\ \frac 12\ln\abs{x} \amp = \frac14\ln\abs{y} +C_1\\ 2\ln\abs{x} \amp = \ln\abs{y} +C_1\\ \ln\abs{x^2} \amp = \ln\abs{y}+C_1\\ \end{align*} Now raise both sides as a power of \(e\text{:}\) \begin{align*} x^2 \amp = e^{\ln\abs{y}+C_1}\\ x^2 \amp = e^{\ln\abs{y}}e^{C_1}\qquad \text{(Note that \(e^{C_1}\) is just a constant.)}\\ x^2 \amp = yC_2\\ \frac1{C_2}x^2 \amp =y \qquad \text{ (Note that \(1/C_2\) is just a constant.) }\\ Cx^2 \amp = y. \end{align*}

As the water started at the point \((1,1/4)\text{,}\) we can solve for \(C\text{:}\) \begin{equation*} C(1)^2 = \frac14 \Rightarrow C = \frac14. \end{equation*}

Thus the water follows the curve \(y=x^2/4\) in the \(x\)-\(y\) plane. The surface and the path of the water is graphed in Figure 12.6.15(a). In part (b) of the figure, the level curves of the surface are plotted in the \(x\)-\(y\) plane, along with the curve \(y=x^2/4\text{.}\) Notice how the path intersects the level curves at right angles. As the path follows the gradient downhill, this reinforces the fact that the gradient is orthogonal to level curves.