Gradient of a scalar field at M(1;1;1)

This problem asks to find the gradient of a scalar field at a given point.

Problem

Find the gradient of the scalar field:

u = x^{2} y - 2 y^{3} z + 3 z^{2} x, M (1; 1; 1)

Solution

Step 1

Write the formula for the gradient of the scalar field u(x, y, z):

grad u = \frac{\partial u}{\partial x} \vec{i} + \frac{\partial u}{\partial y} \vec{j} + \frac{\partial u}{\partial z} \vec{k}

Step 2

Compute the partial derivatives of the scalar field with respect to each variable:

How does the gradient differ from divergence and curl?

The gradient applies to a scalar field and yields a vector field. Divergence applies to a vector field and yields a scalar. Curl applies to a vector field and yields a vector. These are three different first-order differential operators.

\frac{\partial u}{\partial x} = 2 x y + 3 z^{2}

\frac{\partial u}{\partial y} = x^{2} - 6 y^{2} z

\frac{\partial u}{\partial z} = - 2 y^{3} + 6 z x

Step 3

Assemble the gradient as a vector from the found partial derivatives:

grad u = (2 x y + 3 z^{2}) \vec{i} + (x^{2} - 6 y^{2} z) \vec{j} + (- 2 y^{3} + 6 z x) \vec{k}

Step 4

Substitute the coordinates of the point M(1; 1; 1) into each of the partial derivatives:

{(\frac{\partial u}{\partial x})}_{M} = 2 + 3 = 5

{(\frac{\partial u}{\partial y})}_{M} = 1 - 6 = - 5

{(\frac{\partial u}{\partial z})}_{M} = - 2 + 6 = 4

Step 5

Write the final value of the gradient at M:

grad u (M) = 5 \vec{i} - 5 \vec{j} + 4 \vec{k} = (5; - 5; 4)

Answer

grad u(M) = (5; −5; 4).

Geometric meaning

The gradient points in the direction of the fastest increase of the scalar field at a given point, and its magnitude equals the maximum rate of that increase. At M(1;1;1) the gradient is directed along (5; −5; 4), with magnitude √(25 + 25 + 16) = √66.

Frequently asked questions

How do I find the gradient of a scalar field?

The gradient of a scalar field u(x, y, z) is computed by: grad u = (∂u/∂x) ī + (∂u/∂y) j̄ + (∂u/∂z) k̄. Find three partial derivatives and assemble them into a vector. The result is a vector field.

What is the geometric meaning of the gradient?

The gradient points in the direction of the fastest growth of the scalar field at a given point. It is perpendicular to the level line (or level surface) of the field passing through this point. Its magnitude equals the maximum rate of growth of the field.

How does the gradient differ from the directional derivative?

The directional derivative is the rate of change of the field in a given direction — a scalar. The gradient is a vector pointing in the direction of maximum growth. The directional derivative equals the scalar product of the gradient and the unit direction vector.

How do I compute the gradient at a specific point?

First find the general expression for the gradient as a vector function of coordinates (x, y, z). Then substitute the coordinates of the point M into this expression. In our case grad u(M) = (5; −5; 4).