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How to find grad u of a function of three variables at a point

The rule: grad u = (u_x, u_y, u_z)

The gradient of a function $u (x, y, z)$ is the vector built from its three partial derivatives:

grad u = (\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z})

For a point you do it in two stages: first differentiate in general form, then substitute the coordinates of the point into each component. It points in the direction of the fastest growth of $u$ , and its length is the rate of that growth.

Problem

A function $u = \frac{z ^{2}}{x y ^{2}}$ is given. Find $grad u$ at the point $M (\frac{1}{3}, 2, \frac{2}{3})$ .

Answer

grad u_{M} = (- \frac{3}{2}, - \frac{1}{2}, \frac{6}{2}) = (- 1.5, - 0.5, 1.2247 \dots)

Checked by hand: all three components were recomputed from scratch.

Step-by-step solution

1. Differentiate in general form

Write the function as $u = z^{2} \cdot x^{- 1} \cdot y^{- 2}$ - in this shape every derivative is a power rule. With respect to $x$ (here $z^{2} y^{- 2}$ is a constant):

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

With respect to $y$ :

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

With respect to $z$ :

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

2. Substitute the point

At $M$ we have $x = \frac{1}{3}$ , $y = 2$ , $z = \frac{2}{3}$ , so $z^{2} = \frac{2}{3}$ - the root disappears in the first two components:

\frac{\partial u}{\partial x}_{M} = - \frac{\frac{2}{3}}{\frac{1}{9} \cdot 4} = - \frac{\frac{2}{3}}{\frac{4}{9}} = - \frac{2}{3} \cdot \frac{9}{4} = - \frac{3}{2}

\frac{\partial u}{\partial y}_{M} = - \frac{2 \cdot \frac{2}{3}}{\frac{1}{3} \cdot 8} = - \frac{\frac{4}{3}}{\frac{8}{3}} = - \frac{1}{2}

In the third component the root stays. Using $\frac{2}{3} = \frac{6}{3}$ :

\frac{\partial u}{\partial z}_{M} = \frac{2 \frac{2}{3}}{\frac{1}{3} \cdot 4} = \frac{3}{2} \frac{2}{3} = \frac{3}{2} \cdot \frac{6}{3} = \frac{6}{2}

3. Assemble the vector

grad u_{M} = (- \frac{3}{2}, - \frac{1}{2}, \frac{6}{2})

Its length, if the problem asks for the rate of the fastest growth, is

∣ grad u ∣ = \frac{9}{4} + \frac{1}{4} + \frac{6}{4} = 4 = 2

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