Question

Given \(u=x y+y z+z \sin x,\) find (a) \(\quad \nabla u\) at (0,1,2) (b) the directional derivative of \(u\) at (0,1,2) in the direction \(2 \mathbf{i}+2 \mathbf{j}-\mathbf{k}\) (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).

Short answer

(a) \(abla u = (2, 2, 1)\); (b) \(\frac{7}{3}\); (c) Tangent plane: \(2x + 2y + z - 4 = 0\); Normal line: \(x=2t, y=1+2t, z=2+t\); (d) \(\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\)

Step-by-step solution

Find the Gradient abla u

The gradient of a scalar function is the vector of its partial derivatives with respect to each variable. For \(u = xy + yz + z \sin x\), calculate each partial derivative:\[ \frac{\partial u}{\partial x} = y + z \cos x \]\[ \frac{\partial u}{\partial y} = x + z \]\[ \frac{\partial u}{\partial z} = y + \sin x \]Evaluate these partial derivatives at the point (0,1,2):\[ \frac{\partial u}{\partial x} \Bigg|_{(0,1,2)} = 2 \cos 0 = 2 \]\[ \frac{\partial u}{\partial y} \Bigg|_{(0,1,2)} = 0 + 2 = 2 \]\[ \frac{\partial u}{\partial z} \Bigg|_{(0,1,2)} = 1 + \sin 0 = 1 \]Thus, \(abla u = (2, 2, 1)\).

Find the Directional Derivative

To find the directional derivative, first normalize the direction vector (2, 2, -1) to get a unit vector. The magnitude of (2, 2, -1) is: \[ |(2, 2, -1)| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{9} = 3 \]So the unit vector is: \[ \mathbf{v} = \left( \frac{2}{3}, \frac{2}{3}, -\frac{1}{3} \right) \]The directional derivative is given by the dot product of \(abla u\) and the unit vector: \[ D_\mathbf{v} u = abla u \cdot \mathbf{v} = (2, 2, 1) \cdot \left( \frac{2}{3}, \frac{2}{3}, -\frac{1}{3} \right) = \frac{4}{3} + \frac{4}{3} - \frac{1}{3} = \frac{7}{3} \]

Find the Equations of the Tangent Plane and Normal Line

For the tangent plane at (0,1,2) on the level surface \(u=2\), use: \[ u_x(x-x_0) + u_y(y-y_0) + u_z(z-z_0) = 0 \]With \(abla u = (2, 2, 1)\) at (0,1,2), the equation is: \[ 2(x-0) + 2(y-1) + 1(z-2) = 0 \]Which simplifies to: \[ 2x + 2(y-1) + (z-2) = 0 \]\[ 2x + 2y - 2 + z - 2 = 0 \]\[ 2x + 2y + z - 4 = 0 \]For the normal line, parametrize the point (0,1,2) with the direction vector (2,2,1): \[ x = 0 + 2t \]\[ y = 1 + 2t \]\[ z = 2 + t \]

Find the Unit Vector in the Direction of Most Rapid Increase

The direction of most rapid increase of \(u\) is given by the gradient vector \(abla u\). Normalize the gradient (2,2,1) to find the unit vector: \[ |abla u| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3 \]So the unit vector is: \[ \mathbf{e} = \left( \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right) \]

Key concepts

Gradient

The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. For a function of three variables, like \( u = xy + yz + z \sin x \), the gradient is comprised of the partial derivatives with respect to each variable: \( \frac{\partial u}{\partial x} \), \( \frac{\partial u}{\partial y} \), and \( \frac{\partial u}{\partial z} \).

To find the gradient \(abla u\), we calculate each partial derivative. Substituting the given point \((0,1,2)\), we get:

• \( \frac{\partial u}{\partial x} = y + z \cos x \), evaluated as \( 1 + 2 \cdot \cos(0) = 2 \).
• \( \frac{\partial u}{\partial y} = x + z \), evaluated as \( 0 + 2 = 2 \).
• \( \frac{\partial u}{\partial z} = y + \sin x \), evaluated as \( 1 + \sin(0) = 1 \).

Therefore, the gradient at \((0, 1, 2)\) is \((2, 2, 1)\).

Partial Derivatives

Partial derivatives represent the rate of change of a function with respect to one variable while holding others constant. In our example \( u = xy + yz + z \sin x \), partial derivatives with respect to \( x \), \( y \), and \( z \) are calculated as follows:

• \( \frac{\partial u}{\partial x} = y + z \cos x \)
• \( \frac{\partial u}{\partial y} = x + z \)
• \( \frac{\partial u}{\partial z} = y + \sin x \)

These derivatives help form the gradient and are crucial for finding the directional derivative and equations of the tangent plane and normal line.

Level Surfaces

A level surface is a set of points where a function has a constant value. For our function \( u = xy + yz + z \sin x \), one such surface is \( u = 2 \). The points \((x, y, z)\) satisfying this equation form a surface in three-dimensional space.

The tangent plane to this level surface at a specific point can be found using the gradient vector. The equation of the tangent plane provides a linear approximation of the surface at that point.

Tangent Plane

The tangent plane to a level surface at a point provides a plane that is tangent to the surface at that point. The equation of the tangent plane at a point on the surface is given by:

\( abla u \cdot ((x, y, z) - (x_0, y_0, z_0)) = 0 \).
Substitute \( u (0, 1, 2)\) and the gradient \( abla u = (2, 2, 1) \) to get the plane:

\( 2(x - 0) + 2(y - 1) + (z - 2) = 0 \), which simplifies to \( 2x + 2y + z - 4 = 0 \).

Normal Line

The normal line to a surface at a point is a line that is perpendicular to the tangent plane at that point. It follows the direction of the gradient vector. For the level surface at the point \((0, 1, 2)\) with the gradient \( (2, 2, 1) \), the parametric equations of the normal line are:

• \( x = 0 + 2t \)
• \( y = 1 + 2t \)
• \( z = 2 + t \)

Here, \( t \) is a parameter. As \( t \) varies, the line traces out the path of the normal vector starting from the point \((0, 1, 2)\).