Given \(u=x y+y z+z \sin x\), find (a) \(\mathbf{V}_{\mathrm{H}}\) at \((0,1,2) ;\) (b) the directional derivative of \(u\) at \((0,1,2)\) in the direction of \(2 \mathbf{i}+2 \mathrm{j}-\mathrm{k} ;\) (c) the equations of the tangent plane and of the normal line to the level surface \(u=2\) at \((0,1,2)\) (d) a unit vector in the direction of most rapid increase of \(u\) at \((0,1,2)\).

In vector calculus, the concept of partial derivatives helps us understand how a function changes as we vary one of its variables while keeping the others constant. For our function, \(u = xy + yz + z \sin(x)\), we compute the partial derivatives with respect to each variable (x, y, z). This gives us an array of derivatives: \(u_x\), \(u_y\), and \(u_z\). These derivatives measure the rate of change of the function as we change x, y, or z individually.
In this exercise, we found:

• \(u_x = y + z \cos(x)\)
• \(u_y = x + z\)
• \(u_z = y + \sin(x)\)

Evaluating these partial derivatives at the point (0,1,2), we get:

• \(u_x(0,1,2) = 3\)
• \(u_y(0,1,2) = 2\)
• \(u_z(0,1,2) = 1\)

The gradient vector, represented as \(abla u\), is an essential concept in vector calculus. It consists of the partial derivatives of the function and points in the direction of the steepest ascent. For our function, the gradient vector at any point (x, y, z) is:
\(abla u = \langle u_x, u_y, u_z \rangle\)
Thus, after evaluating the partial derivatives at (0,1,2), we get:
\(abla u(0,1,2) = \langle 3, 2, 1 \rangle\)
This vector tells us the direction and rate of the fastest increase of the function at the specified point.

The directional derivative measures how a function changes as we move in a specific direction. To find the directional derivative, we first need a unit vector in the desired direction. For the direction given by \(2 \mathbf{i} + 2 \mathbf{j} - \mathbf{k}\), we normalize it:
\(\mathbf{v} = \frac{1}{\sqrt{2^2 + 2^2 + (-1)^2}} \langle 2, 2, -1 \rangle = \frac{1}{3} \langle 2, 2, -1 \rangle\)
The directional derivative is then computed by the dot product of the gradient vector and the unit direction vector:
\(D_{\mathbf{v}}u = abla u \cdot \mathbf{v} = \langle 3, 2, 1 \rangle \cdot \frac{1}{3} \langle 2, 2, -1 \rangle = 3\)
This tells us how much the function changes in that specific direction.

The tangent plane to a surface at a given point provides an approximation of the surface near that point. For the surface defined by \(u = 2\) at the point (0,1,2), we use the gradient vector to find the equation of the tangent plane. The formula is:
\(abla u \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0\)
Plugging in our gradient \(\langle 3, 2, 1 \rangle\) and point (0,1,2), we get:
\(3(x - 0) + 2(y - 1) + 1(z - 2) = 0\)
Simplifying gives us:
\(3x + 2y + z - 4 = 0\)
This is the equation of the tangent plane at the given point.

The normal line is perpendicular to the tangent plane at a specific point on the surface. Its direction is along the gradient vector. For our function, the normal line at (0,1,2) with direction \(abla u = \langle 3, 2, 1 \rangle\) can be expressed as:
\(\frac{x - 0}{3} = \frac{y - 1}{2} = \frac{z - 2}{1}\)
This set of parametric equations describes the normal line, revealing the trajectory moving away from the point along the gradient vector.