Three-Dimensional Gradients and Directional Derivatives

Learning Objectives

Calculate directional derivatives and gradients in three dimensions.

The definition of a gradient can be extended to functions of more than two variables.

Definition

Let $w=f(x, y ,z)$ be a function of three variables such that $f_x$ , $f_y$ , and $f_z$ exist. The vector $\nabla{f}(x,y,z)$ is called the gradient of $f$ and is defined as

$\nabla{f}(x,y,z)=f_x(x,y,z){\bf{i}}+f_y(x,y,z){\bf{j}}+f_z(x,y,z){\bf{k}}.$

$\nabla{f}(x,y,z)$ can also be written as $\text{grad }f(x,y,z)$ .

Calculating the gradient of a function in three variables is very similar to calculating the gradient of a function in two variables. First, we calculate the partial derivatives $f_x$ , $f_y$ , and $f_z$ , and then we use the Equation for $\nabla{f}(x,y,z)$ .

Example: finding gradients in three dimensions

Find the gradient $\nabla{f}(x,y,z)$ of each of the following functions:

a. $f(x,y,z)=5x^2-2xy+y^2-4yz+z^2+3xz$

b. $f(x,y,z)=e^{-2z}\sin 2x\cos 2y$

Show Solution

For both parts a. and b., we first calculate the partial derivatives $f_x$ , $f_y$ , and $f_z$ , then use the Equation for $\nabla{f}(x,y,z)$ .

$\hspace{1cm}\begin{align} f_x(x,y,z)&=10x-2y+3z,f_y(x,y,z)=-2x+2y-4z\text{ and }f_z(x,y,z)=3x-4y+2z\text{ so,} \\ \nabla{f}(x,y,z)&=f_x(x,y,z){\bf{i}}+f_y(x,y,z){\bf{j}}+f_z(x,y,z){\bf{k}} \\ &=(10x-2y+3z){\bf{i}}+(-2x+2y-4z){\bf{j}}+(-4x+3y+2z){\bf{k}} \end{align}$

$\hspace{1cm}\begin{align} f_x(x,y,z)&=-2e^{-2z}\cos2x\cos 2y,f_y(x,y,z)=-2e^{-2z}\sin 2x\sin2y\text{ and }f_z(x,y,z)=-2e^{-2z}\sin 2x\cos 2y\text{ so,} \\ \nabla{f}(x,y,z)&=f_x(x,y,z){\bf{i}}+f_y(x,y,z){\bf{j}}+f_z(x,y,z){\bf{k}} \\ &=(2e^{-2z}\cos2x\cos 2y){\bf{i}}+(-2e^{-2z}){\bf{j}}+(-2e^{-2z}){\bf{k}} \\ &=2e^{-2z}(\cos2x\cos 2y{\bf{i}}-\sin 2x\sin2y{\bf{j}}-\sin 2x\cos 2y{\bf{k}}). \end{align}$

Try it

Find the gradient $\nabla{f}(x,y,z)$ of $f(x,y,z)=\frac{x^2-3y^2+z^2}{2x+y-4z}$ .

Show Solution

$\nabla{f}(x,y,z)=\frac{2x^2+2xy+6y^2-8xz-2z^2}{(2x+y-4z)^2}{\bf{i}}-\frac{x^2+12xy+3y^2-24yz+z^2}{(2x+y-4z)^2}{\bf{j}}+\frac{4x^2-12y^2-4z^2+4xz+2yz}{(2x+y-4z)^2}{\bf{k}}$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 4.32” here (opens in new window).

>The directional derivative can also be generalized to functions of three variables. To determine a direction in three dimensions, a vector with three components is needed. This vector is a unit vector, and the components of the unit vector are called . Given a three-dimensional unit vector $\bf{u}$ in standard form (i.e., the initial point is at the origin), this vector forms three different angles with the positive $x-,y-$ and $z-$ axes. Let’s call these angles $\alpha,\beta$ and $\gamma$ . Then the directional cosines are given by $\cos\alpha,\cos\beta$ and $\cos\gamma$ . These are the components of the unit vector $\bf{u}$ ; since $\bf{u}$ is a unit vector, it is true that $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ .

Definition

Suppose $w=f(x,y,z)$ is a function of three variables with a domain of $D$ , Let $(x_0,y_0,z_0)\in{D}$ and let ${\bf{u}}=cos\alpha{\bf{i}}+\cos\beta{\bf{j}}+\cos\gamma{\bf{k}}$ be a unit vector. Then, the directional derivative of $f$ in the direction of $u$ is given by

$D_{\bf{u}}f(x_0,y_0,z_0)=\displaystyle{\lim_{t\to0}}\frac{f(x_0+t\cos\alpha,y_0+t\cos\beta,z_0+t\cos\gamma)-f(x_0,y_0,z_0)}t,$

providing the limit exists.

We can calculate the directional derivative of a function of three variables by using the gradient, leading to a formula that is analogous to the dot product definition of the Directional Derivative of a Function of Two Variables.

Theorem: Directional derivative of a function of three variables

Let $f(x,y,z)$ be a differentiable function of three variables and let ${\bf{u}}=\cos\alpha{\bf{i}}+\cos\beta{\bf{j}}+\cos\gamma{\bf{k}}$ be a unit vector. Then, the directional derivative of $f$ in the direction of ${\bf{u}}$ is given by

[latex]\hspace{6cm}\begin{align}

D_{\bf{u}}f(x,y,z)&=\nabla{f}(x,y,z)\cdot{\bf{u}} \\ &=f_x(x,y,z)\cos\alpha+f_y(x,y,z)\cos\beta+f_z(x,y,z)\cos\gamma. \end{align}[/latex]

The three angles $\alpha,\beta$ , and $\gamma$ determine the unit vector $\bf{u}$ . In practice, we can use an arbitrary (nonunit) vector, then divide by its magnitude to obtain a unit vector in the desired direction.

Example: finding a directional derivative in three dimensions

Calculate $D_{\bf{u}}f(1,-2,3)$ in the direction of $v=-{\bf{i}}+2{\bf{j}}+2{\bf{k}}$ for the function

$f(x,y,z)=5x^2-2xy+y^2-4yz+z^2+3xz$ .

Show Solution

First, we find the magnitude of ${\bf{v}}$ :

$\|{\bf{v}}\|=\sqrt{(-1)^2+(2)^2+(2)^2}=3$

Therefore $\frac{{\bf{v}}}{\|{\bf{v}}\|}=\frac{-{\bf{i}}+2{\bf{j}}+2{\bf{k}}}3=-\frac13{\bf{i}}+\frac23{\bf{j}}+\frac23{\bf{k}}$ is a unit vector in the direction of ${\bf{v}}$ , so $\cos\alpha=-\frac13$ , $\cos\beta=\frac23$ , and $\cos\gamma=\frac23$ . Next, we calculate the partial derivatives of $f$ :

$\hspace{7cm}\begin{align} f_x(x,y,z)&=10x-2y+3z \\ f_y(x,y,z)&=-2x+2y-4z \\ f_z(x,y,z)&=-4y+2z+3x, \end{align}$

Then substitute them into the Directional Derivative of a Function of Three Variables:

$\hspace{5cm}\begin{align} D_{\bf{u}}f(x,y,z)&=10x-2y+3z \\ f_y(x,y,z)&=f_x(x,y,z)\cos\alpha+f_y(x,y,z)\cos\beta+f_z(x,y,z)\cos\gamma \\ &=(10x-2y+3z)(-\frac13)+(-2x+2y-4z)(\frac23)+(-4y+2z+3x)(\frac23) \\ &=-\frac{10x}3+\frac{2y}3-\frac{3z}3-\frac{4x}3+\frac{4y}3-\frac{8z}3-\frac{8y}3+\frac{4z}3+\frac{6x}3 \\ &=-\frac{8x}3-\frac{2y}3-\frac{7z}3 \end{align}$

Last, to find $D_{\bf{u}}f(1,-2,3)$ , we substitute $x=1$ , $y=-2$ , and $z=3$ :

$\hspace{5cm}\begin{align} D_{\bf{u}}f(1,-2,3)&=-\frac{8(1)}3-\frac{2(-2)}3-\frac{7(3)}3 \\ &=-\frac83+\frac43-\frac{21}3 \\ &=-\frac{25}3. \end{align}$

Try it

Calculate $D_{\bf{u}}f(x,y,z)$ and $D_{\bf{u}}f(0,-2,5)$ in the direction of ${\bf{v}}=-3{\bf{i}}+12{\bf{j}}-4{\bf{k}}$ for the function $f(x,y,z)=3x^2+xy-2y^2+4yz-z^2+2xz$ .

Show Solution

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 4.33” here (opens in new window).

Module 4: Differentiation of Functions of Several Variables