$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

## Directional Derivative

In the previous section, we found that if a surface is described by $$z=f(x,y)$$, the partial derivatives $$\partial z/\partial x$$ and $$\partial z/\partial y$$ describe the rate of change of z when we travel parallel to the x axis and parallel to the y axis, respectively. The question now is, what if we travel in a different direction, say along an arbitrary vector $$\vec{v}$$? The answer is that we need to use the directional derivative. This is nothing more than a weighted average of the partial derivatives, where the weights are the components of the direction vector (for which we'll need to use a unit vector; this is important).

#### Definition

If $$f(x,y)$$ is a differentiable function, then the directional derivative of $$f$$ in the direction of a unit vector $$\hat{v} = \langle v_1,v_2 \rangle$$ is given by $\ans{D_v\ f(x,y) = f_x(x,y) v_1 + f_y (x,y) v_2.}$ At a point $$(a,b,f(a,b))$$: $D_v\ f(a,b) = f_x(a,b) v_1 + f_y (a,b) v_2$

Find the directional derivative of $$f(x,y)=\dfrac{x^2}{4}+\dfrac{y^2}{9}$$ at $$(2,3,2)$$ in the direction of $$\vec{v}=\langle 3,4 \rangle$$.

#### Solution

First, find the unit vector in the direction of $$\vec{v}$$: $\hat{v} = \dfrac{\langle 3,4 \rangle}{5} = \left\langle \dfrac{3}{5},\dfrac{4}{5} \right\rangle$

Next, find the partial derivatives:

\begin{align} f_x &= \dfrac{x}{2} \longrightarrow f_x (2,3) = 1\\ f_y &= \dfrac{2y}{9} \longrightarrow f_y (2,3) = \dfrac{2}{3} \end{align}

Finally, plug these into the directional derivative formula:

$D_v = 1\left(\dfrac{3}{5}\right) + \dfrac{2}{3}\left(\dfrac{4}{5}\right) = \ans{\dfrac{17}{15}}$

The interpretation of this is that for every unit you travel in the direction of this vector, z increases by 17/15.

#### Try it yourself:

(click on a problem to show/hide its answer)

1. Find the directional derivative of $$f(x,y) = \dfrac{1}{4}(x^2+2y^2)+2$$ at $$(3,2)$$ in the direction of $$\vec{u}=\langle 1,1 \rangle$$ and $$\vec{v}=\langle 1,-\sqrt{3} \rangle$$.
2. \begin{align} D_u &= \dfrac{7}{2\sqrt{2}} \approx 2.47\\ D_v &= \dfrac{3}{4}-\sqrt{3} \approx -0.98 \end{align}

3. Find the directional derivative of $$f(x,y) = 3x^2+y^3$$ at $$(3,2)$$ in the direction of $$\vec{u}=\langle 5,12 \rangle$$.
4. $$D_u = 18$$

The gradient of $$f(x,y)$$, denoted by the "del" operator $$\nabla$$, is a vector that points in the direction of steepest ascent on the surface $$z=f(x,y)$$. The magnitude of this vector is the slope in that direction (i.e. the largest slope at that point).

#### Definition

The gradient of $$f(x,y)$$ is \begin{align} \nabla f(x,y) &= \langle f_x (x,y) , f_y (x,y) \rangle\\ &= f_x (x,y) \hat{\imath} + f_y (x,y) \hat{\jmath} \end{align} At a point $$(a,b,f(a,b))$$: $\nabla f(a,b) = f_x (a,b) \hat{\imath} + f_y (a,b) \hat{\jmath}$

#### Note

$D_v\ f(x,y) = \nabla f(x,y) \cdot \hat{v}$

Find the gradient of $$f(x,y)=\dfrac{x^2}{4}+\dfrac{y^2}{9}$$ at $$(2,3,2)$$.

#### Solution

\ans{\begin{align} \nabla f(x,y) &= \dfrac{x}{2} \hat{\imath} + \dfrac{2y}{9} \hat{\jmath}\\ \nabla f(a,b) &= \hat{\imath} + \dfrac{2}{3} \hat{\jmath} \end{align}}

## Examples

A hill has a shape given by $$f(x,y)=1000-0.01x^2-0.02y^2$$.

1. In which direction should you travel from $$(60,100,764)$$ to reach the top the fastest?
2. What is the slope in that direction?

#### Solution

1. Find the gradient: \begin{align} \nabla f(x,y) &= -0.02x \hat{\imath} -0.04y \hat{\jmath}\\ \nabla f(60,100) &= -1.2 \hat{\imath} - 4 \hat{\jmath} \end{align} Thus, to reach the top the fastest, you should travel in the direction of $$\ans{\langle -1.2,-4 \rangle}$$.
2. The slope in that direction is the magnitude of the gradient: \begin{align} |\nabla f(60,100)| = \sqrt{1.2^2+4^2} = \ans{4.18} \end{align}

Note that if you traveled in the opposite direction ($$\langle 1.2,4 \rangle$$), the slope would be $$-4.18$$.

#### Note:

$$\nabla f(x,y)$$ points to the steepest ascent, $$-\nabla f(x,y)$$ points to the steepest descent, and the directional derivative is zero in any direction orthogonal to $$\nabla f(x,y)$$.

For instance, using the hill from the previous example, if you travel in the direction of $$\langle -4,-1.2 \rangle$$ from $$(60,100,764)$$, you'll keep the same elevation.

Consider the function $$f(x,y)=3x^2-2y^2$$.

1. Compute $$\nabla f(x,y)$$ and $$\nabla f(2,3)$$.
2. Consider the unit vector $$\hat{u} = \langle \cos \theta, \sin \theta \rangle$$. At $$(2,3)$$, for what values of $$\theta$$ between $$0$$ and $$2\pi$$ does the directional derivative have its maximum and minimum values?

#### Solution

1. Find the gradient: \ans{\begin{align} \nabla f(x,y) &= 6x \hat{\imath} - 4y \hat{\jmath}\\ \nabla f(2,3) &= 12 \hat{\imath} - 12 \hat{\jmath} \end{align}}
2. Find the directional derivative in the direction of $$\hat{u}$$: $D_u\ f(x,y) = \nabla f(x,y) \cdot \hat{u} \longrightarrow D_u\ f(2,3) = 12\cos \theta - 12\sin \theta$ To find the maximum and minimum values for this, take its derivative with respect to $$\theta$$ and set the derivative to 0: $D'_u\ f(2,3) = -12\sin\theta - 12\cos\theta = 0 \longrightarrow \sin\theta = -\cos\theta \longrightarrow \ans{\theta = \dfrac{3\pi}{4}, \dfrac{7\pi}{4}}$ Note that the slope is zero if $$\theta = \dfrac{\pi}{4}, \dfrac{5\pi}{4}$$.

The gradient of a function of three variables is similar, with a third component: $$f_z (x,y,z) \hat{k}$$.

Consider the function $$f(x,y,z)=x^2+2y^2+4z^2-1$$.

1. Find the gradient of $$f$$ at $$(2,0,0)$$.
2. Find the directional derivative of $$f$$ at $$(2,0,0)$$ in the direction of $$\hat{u} = \left\langle \dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}} \right\rangle$$.

#### Solution

1. Find the gradient: \begin{align} \nabla f(x,y,z) &= f_x (x,y,z) \hat{\imath} + f_y (x,y,z) \hat{\jmath} + f_z (x,y,z) \hat{k}\\ &= 2x \hat{\imath} + 4y \hat{\jmath} + 8z \hat{k}\\ \nabla f(2,0,0) &= 4 \hat{\imath} = \ans{\langle 4,0,0 \rangle} \end{align}
2. Find the directional derivative in the direction of $$\hat{u}$$: $D_u\ f(2,0,0) = \nabla f(2,0,0) \cdot \left\langle \dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}} \right\rangle = \ans{\dfrac{4\sqrt{3}}{3}}$

#### Try it yourself:

(click on a problem to show/hide its answer)

1. Find the gradient of $$f(x,y)=4x^2-2xy+y^2$$.
2. $$\nabla f(x,y) = \langle 8x-2y , -2x+2y \rangle$$

3. Find the gradient of $$f(x,y)=\sqrt{12-4x^2-y^2}$$.
4. $$\nabla f(x,y) = \dfrac{1}{2\sqrt{12-4x^2-y^2}} \left\langle -8x , -2y \right\rangle$$

5. Find the gradient of $$f(x,y)=xe^{2xy}$$.
6. $$\nabla f(x,y) = \left\langle e^{2xy}+2xye^{2xy} , 2x^2e^{2xy} \right\rangle$$

7. Find the unit vector that gives the direction of steepest ascent for the surface $$z=x^2+4xy-y^2$$ at $$(2,1)$$.
8. $$\vec{u} = \nabla f(2,1) = \langle 8,6 \rangle \longrightarrow \hat{u} = \left\langle \dfrac{4}{5},\dfrac{3}{5} \right\rangle$$