 ## V11 Directional derivatives If you are on the side of a hill, the gradient depends on the direction you look. So the directional derivative is the gradient in a particular direction.

Learn how to find the directional derivative of a function of two variables f(x,y) or three variables g(x,y,z) at a point and in a given direction. This is very useful in engineering and especially computer graphics.

You will recall that the derivative of a function gives us the gradient or rate of change of the function. The rate of change of a function such as $$z=f\left(x,y\right)$$ can be found by $$\text{partial}$$ differentiation; $$\frac{\partial f}{\partial x}$$ gives the rate of change of the function $$f$$, with respect to $$x$$, that is, the gradient of the graph as we move in the $$x$$ direction. Keep in mind that the graph of $$z=f\left(x,y\right)$$ is a surface in three dimensional space and $$\frac{\partial f}{\partial y}$$ gives the rate of change of $$f$$ with respect to $$y$$, that is, the gradient of the graph as we move in the $$y$$ direction.

### Directional Derivatives

We can find the rate of change of a function in any direction (not just in the direction of the $$x-$$ axis or $$y-$$ axis) by finding the directional derivative.

The directional derivative of a function $$f$$ in the direction of a vector $$\vec{u}$$ is denoted by $$D_{u}$$ and is given by:1 Note that the function $$f$$ may be of $$2$$ or more variables. \begin{align*} D_{u} & =\nabla f\cdot\hat{u} \end{align*} where $$\hat{u}$$ is a unit vector in the direction of the vector $$\vec{u}.$$2 A unit vector is a vector of magnitude $$1.$$ For a vector $$\vec{u}$$ a unit vector in the direction of $$\vec{u}$$ is given by \begin{align*} \hat{u} & =\frac{1}{\left|\vec{u}\right|}\vec{u} \end{align*} where $$\left|\vec{u}\right|$$ is the magnitude of the vector $$\vec{u}$$. The following figure illustrates the meaning of the directional derivative: #### Example 1

Suppose that we wish to find the directional derivative of $$f\left(x,y\right)=x^{2}+y^{2}$$ at the point $$\left(2,3\right)$$ in the direction of the vector $$\vec{u}=4\vec{i}-3\vec{j}$$.

Solution:

The first step is to find the vector “grad f”, symbolised thus $$\nabla f$$.

\begin{align*} \nabla f=\frac{\partial f}{\partial x}\vec{i} & +\frac{\partial f}{\partial y}\vec{j}. \end{align*}

In this particular case \begin{align*} \nabla f & =2x\vec{i}+2y\vec{j}\\ & =4\vec{i}+6\vec{j} \end{align*} at the point where $$x=2$$ and $$y=3$$ .

The next step is to find the dot product of this vector $$\nabla f$$, and $$\hat{u}$$, the unit vector in the direction of $$\vec{u}$$.

In this case $$\vec{u}=4i-3j$$, therefore $$\hat{u}=\frac{1}{5}\left(4i-3j\right)$$

Hence the directional derivative of $$f$$ in the direction of $$u$$ is \begin{align*} D_{u} & =\left(4i+6j\right)\cdot\left(\frac{4}{5}i-\frac{3}{5}j\right)\\ & =\frac{16}{5}-\frac{18}{5}\\ & =-\frac{2}{5}. \end{align*} To generalise the above, the directional derivative of a function, $$f$$, in the direction of $$u$$ is $D_{u}=\nabla f\cdot\hat{u}$ where $\nabla f=\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j+\frac{\partial f}{\partial z}k.$ Since the directional derivative relies on a dot product, (remember $$\vec{a}.\vec{b}=\left|\vec{a}\right|\left|\vec{b}\right|\cos\theta$$), it will be a maximum when $$\cos\theta$$ is maximum (that is, $$\cos\theta=1$$), so the directional derivative will be maximum when $$\theta=0$$.

In other words, we can say that

• $$D_{u}$$ will be maximum when $$\vec{u}$$ and $$\nabla f$$ are in the same direction.
• $$D_{u}$$ will have its greatest negative value when $$u$$ and $$\nabla f$$ are in opposite directions (when $$\theta=\pi$$).
• $$D_{u}$$ will be zero when $$u$$ and $$\nabla f$$ are at right angles.

#### Example 2

If $$f\left(x,y,z\right)=x^{2}+y^{2}+xyz$$, find a unit vector $$\hat{u}$$ such that the rate of change of $$f$$ at $$\left(2,3,-1\right)$$ in the direction of $$u$$ is maximum.

Solution: \begin{align*} \nabla f= & \frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}\\ = & \left(2x+yz\right)\hat{i}+\left(2y+xz\right)\hat{j}+\left(xy\right)\hat{k}\\ = & \hat{i}+4\hat{j}+6\hat{k} \end{align*}

For the directional derivative to be maximum, $$u=\nabla f$$

Therefore $$u=\hat{i}+4\hat{j}+6\hat{k}$$ and $$\hat{u}=\frac{1}{\sqrt{53}}\left(\hat{i}+4\hat{j}+6\hat{k}\right)$$.

### Exercise

1. Find the directional derivative of the given function at the given point in the direction of the indicated vector:
$$\text{a) }$$ $$f\left(x,y\right)=xy^{2},\,\left(3,2\right),\,4\hat{i}+3\hat{j}$$
$$\text{b) }$$ $$f\left(x,y\right)=e^{xy},\,\left(0,2\right),\,\hat{i}$$
$$\text{c) }$$ $$f\left(x,y,z\right)=x^{2}y^{3}z,\,\left(2,-1,3\right),\,\hat{i}-2\hat{j}-2\hat{k}$$

$$\text{a) }$$ $$\frac{52}{5}\quad\text{b) 2 \quad\text{c) -\frac{76}{3} } }.$$
1. Find the unit vector in the direction in which $$f$$ increases most rapidly at $$P\left(1,\pi/2\right)$$ for $$f\left(x,y\right)=x^{2}+\cos xy$$.
$$\hat{u}=0.394\hat{i}-0.919\hat{j}.$$