Directional Derivatives Examples 4

# Directional Derivatives Examples 4

Recall from the Directional Derivatives page that for a two variable real-valued function $z = f(x, y)$, the directional derivative of $f$ at a point $(x, y) \in D(f)$ in the direction of the unit vector $\vec{u} = (a, b)$ is given by the formula:

(1)
\begin{align} \quad D_{\vec{u}} \: f(x, y) = \frac{\partial z}{\partial x} a + \frac{\partial z}{\partial y} b \end{align}

For a three variable real-valued function $w = f(x, y, z)$, the directional derivative of $f$ at a point $(x, y, z) \in D(f)$ in the direction of the unit vector $\vec{u} = (a, b, c)$ is given by the formula:

(2)
\begin{align} \quad D_{\vec{u}} \: f(x, y, z) = \frac{\partial w}{\partial x} a + \frac{\partial w}{\partial y} b + \frac{\partial w}{\partial z} c \end{align}

## Example 1

Find the directional derivative of $f(x, y, z) = 2xe^y\cos z$ in the direction of the vector $(1, 2, 3)$.

We must first find a vector that points in the same direction as $(1, 2, 3)$ and has norm $1$. Let $\vec{u}$ be this unit vector. Then:

(3)
\begin{align} \quad \vec{u} = \frac{(1, 2, 3)}{\| (1, 2, 3) \|} = \left ( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right ) \end{align}

Now applying the formula above and we have that:

(4)
\begin{align} \quad D_{\vec{u}} \: f(x, y, z) = \left ( \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right ) \cdot \left ( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right ) \\ \quad D_{\vec{u}} \: f(x, y, z) = \left ( 2e^y \cos z, 2xe^y \cos z, -2xe^y \sin z \right ) \cdot \left ( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right ) \\ \quad D_{\vec{u}} \: f(x, y, z) = \frac{2e^y \cos z + 4xe^y \cos z - 6xe^y \sin z}{\sqrt{14}} \end{align}

## Example 2

Find the directional derivative of $f(x, y, z, w) = xy^2z^3w^4$ in the direction of the vector $(-1, -2, -3, -4)$.

We first need a unit vector that points in the direction of $(-1, -2, -3, -4)$. Let $\vec{u}$ be this vector. Then:

(5)
\begin{align} \quad \vec{u} = \frac{(-1, -2, -3, -4)}{\| (-1, -2, -3, -4) \|} = \frac{1}{\sqrt{30}} \left (-1, -2, -3, -4 \right ) \end{align}

Therefore we have that:

(6)
\begin{align} \quad D_{\vec{u}} \: f(x, y, z, w) = \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}, \frac{\partial f}{\partial w} \right ) \cdot \frac{1}{\sqrt{30}} (-1, -2, -3, -4) \\ \quad D_{\vec{u}} \: f(x, y, z, w) = \left ( y^2 z^2 w^4, 2xyz^3w^4, 3xy^2z^2w^4, 4xy^2z^3w^3 \right ) \cdot \frac{1}{\sqrt{30}} (-1, -2, -3, -4) \\ \quad D_{\vec{u}} \: f(x, y, z, w) = \frac{-y^2 z^2 w^4 -4xyz^3w^4 -9xy^2z^2w^4 -16xy^2z^3w^3}{\sqrt{30}} \end{align}