#### xaktly | Vector calculus

Directional derivatives

#### Slope of a surface in an arbitrary direction

We've seen how a function of two variables, like $f(x, y)$ represents a surface in three dimensions — with two independent variables. We can write the $z$ coordinate as the dependent variable, $z = f(x, y)$.

We've studied partial derivatives:

• $\frac{\partial f}{\partial x}(a, b) = f_x(a, b)$ is the rate of change of $f(x,y)$ in the x direction with $y$ held constant $(y = b)$.

• $\frac{\partial f}{\partial y}(a, b) = f_g(a, b)$ is the rate of change of $f(x,y)$ in the $y$ direction with $x$ held constant $(x = a)$.

Now we'll ask whether we can find the slope of such a surface in any arbitrary direction. Say we have a point $P$ on $f(x, y)$. Now we can recall that the gradient vector is

$$\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$$

\begin{align} \nabla f \cdot \hat i &= f_x \\[5pt] \nabla f \cdot \hat j &= f_y \end{align}

are gradients (slopes) in the $x$- and $y$-directions, aligned with our coordinate system. What if we could replace those directions with some arbitrary direction vector, $\vec u$.

Here's a picture that illustrates the idea. The direction vector $\vec u$ is not along the $x$ or $y$ axes.

#### The directional derivative

The directional derivative of the function $f$ in the direction of $\vec u$, which we write as $D_{\vec u} f$, is

$$D_{\vec u} f = \frac{\nabla f \cdot \vec u}{|\vec u |}$$

Recall that $\frac{\vec u}{|\vec u|}$ is just the normalized version of $\vec u$, which is $\hat u$.

### Justification

Let's see if we can outline where that definition comes from. Take a function $z = f(x,y)$ that has a graph like the one shown below. Now let's take a direction vector $\vec u$ (red) that isn't parallel to the $x$- or $y$-axes. Let $P$ be a point on $\vec u$. Notice that both $\vec u$ and $P$ are in the $x-y$ plane, the plane of the independent variables. The red curve is meant to be the trace of the curve that would be the slice of our graph cut by the $u-z$ plane. Our aim will be to find the derivative of $f(x,y)$ in the direction of $\vec u$ at point $P$. We should also note that we'll be using the unit vector form of $\vec u$, which is $\hat u$ — because we only care about the direction of this vector.

First, let's call the components of $\hat u$ be $\vec u = (u_1, u_2)$. And let's parameterize our trip along $f(x,y)$ in the direction of $\hat u$ with a parameter $s$, which could loosely be speed or time — it doesn't really matter. That makes $f(x,y)$ into the compound function $f(x(s), y(s))$, which will beg the chain rule in a minute.

Then we can write

\begin{align} x(s) &= x_o + u_1 s \tag{1} \\[5pt] y(s) &= y_o + u_2 s \end{align}

Notice that $x'(s) = u_1$ and $y'(s) = u_2$. Then the derivative of our function in the direction of $\hat u$ is

$$D_{\hat u} f(x_o,y_o) = \lim_{s \to 0} \frac{f(x_o+s u_1, y_o +s u_2) - f(x_o, y_o)}{s}$$

This is the usual definition of the derivative, the slope (rise over run) of a secant line on our curve, taken as the distance between the two points that define the secant line goes to zero — that is, as $s \rightarrow 0$. But how do we actually compute it? First, we use the chain rule:

\begin{align} \frac{d}{ds} \left[ f(x(s), y(s)) \right] &= \left( \frac{\partial f}{\partial x}_{(x_o,y_o)} \frac{dx}{ds} \right) \\[5pt] &+ \left( \frac{\partial f}{\partial y}_{(x_o,y_o)} \frac{dy}{ds} \right) \end{align}

Recall from (1) above that $\frac{dx}{ds} = u_1$ and $\frac{dy}{ds} = u_2$, so we can write

$$\frac{df}{ds} = \left( \frac{\partial f}{\partial x}_{(x_o,y_o)} u_1 \right) + \left( \frac{\partial f}{\partial y}_{(x_o,y_o)} u_2 \right)$$

Well, there is a gradient and a dot product hidden in there:

$$\frac{df}{ds} = \nabla f |_{(x_o, y_o)} \cdot (u_1, u_2)$$

Let's streamline that a little more and write

$$\frac{df}{ds} = \nabla f |_{(x_o, y_o)} \cdot \hat u$$

Now notice that the directional derivative is just the dot product of the standard gradient vector of the function with the unit direction vector, $\hat u$ — pretty simple. And the calculation is equally simple. Here are a couple of examples.

### Example 1

Take the function in the graph above, and let $a = 9$, giving $f(x,y) = 9 - x^2 - y^2$. Find the directional derivative in the direction (1, 2) over the point (1, 1) in the $x-y$ plane.

Solution: First find $\nabla f$:

\begin{align} \nabla f &= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \\[5pt] &= (-2x, -2y) \\[5pt] \nabla f(1,1) &= (-2, -2) \end{align}

Let $\vec u = (1, 2) = \hat i + 2 \hat j$. Then the length of $\vec u$ is

$$| \vec u | = \sqrt{1^2 + 2^2} = \sqrt{5}$$

Now we can compute our directional derivative:

\begin{align} D_{\vec u} f &= \frac{\nabla f \cdot \vec u}{\vec u} \\[5pt] &= \frac{(-2,-2) \cdot (1, 2)}{\sqrt{5}} \\[5pt] &= \frac{-2(1) - 2(2)}{\sqrt{5}} \\[5pt] &= -\frac{6}{\sqrt{5}} = -\frac{6 \sqrt 5}{5} \end{align}

### Example 2

Find the directional derivative of $f(x,y) = (1+xy)^{3/2}$ at point $P = (3, 1)$ and in the direction $\vec u = (1, 1)$.

Solution: First find $\nabla f$:

\begin{align} \nabla f &= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \\[5pt] &= \left( \frac{3}{2} y \sqrt{1 + xy}, \frac{3}{2} x \sqrt{1 + xy} \right) \\[5pt] \nabla f(3,1) &= (3, 9) \end{align}

Let $\vec u = (1, 1) = \hat i + \hat j$. Then the length of $\vec u$ is

$$| \vec u | = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Now we can compute our directional derivative:

\begin{align} D_{\vec u} f &= \frac{\nabla f \cdot \vec u}{\vec u} \\[5pt] &= \frac{(3,9) \cdot (1, 1)}{\sqrt{2}} \\[5pt] &= \frac{3(1) + 9(1)}{\sqrt{2}} \\[5pt] &= \frac{12}{\sqrt{2}} = 6 \sqrt{2} \end{align}

### Example 3

Let's do an example with three independent variables — it's the same idea. Find the derivative of $f(x,y,z) = x^2+2y^2+3z^2$ at point $P_o = (3,2,1)$ along the direction $\vec u = (2, 1, 1)$.

Solution: First calculate $\hat u$:

$$\hat u = \frac{(2,1,1)}{\sqrt{2^2+1^2+1^2}} = \frac{1}{\sqrt{6}}(2,1,1)$$

The gradient of $f(x,y,z)$ is

$$\nabla f(x,y,z) = (2x,4y,6z)$$

Now the directional derivative is

\begin{align} \nabla f(x,y,z) \cdot \hat u &= (2x,4y,6z) \cdot (2,1,1) \frac{1}{\sqrt{6}} \\[5pt] &= \frac{1}{\sqrt{6}} (4x+4y+6z) \end{align}

Then we plug in our point, $P_o$:

\begin{align} D_{\hat u}(3,2,1) &= \frac{1}{\sqrt{6}} (4x+4y+6z) \\[5pt] &= \frac{1}{\sqrt{6}} [4(3)+4(2)+6(1)] \\[5pt] &= \frac{12}{\sqrt{6}} + \frac{8}{\sqrt{6}} + \frac{6}{\sqrt{6}} \\[5pt] &= \frac{26}{\sqrt{6}} \end{align}

### Practice problems

Find an equation of the plane tangent to the surface at the given point.

1.   For $f(x,y) = 4 - x^2 - y^2$, calculate the derivative of $f$ above point $(1, 1)$ in the $x-y$ plane and along the vector $\vec u = (1, 3)$. What does it mean that $\nabla f(0, 0) = \vec 0$ for this function?

Solution

First calculate the unit vector form of $\vec u$:

\begin{align} |\vec u| &= \sqrt{1^2 + 3^2} = \sqrt{10} \\[5pt] \hat u &= \frac{1}{\sqrt{10}} (1, 3) \end{align}

Now $\nabla f = (-2x, -2y)$, so

\begin{align} \nabla f \cdot \hat u &= (-2x, -2y) \cdot (1, 3) \frac{1}{\sqrt{10}} \\[5pt] &= \frac{-2x}{\sqrt{10}} - \frac{6y}{\sqrt{10}} \end{align}

Finally,

$$\nabla f \cdot \hat u(1, 1) = \frac{-2}{\sqrt{10}} - \frac{6}{\sqrt{10}} = \frac{-8}{\sqrt{10}}$$

This makes sense because the function is a downward-opening paraboloid, which we'd expect to have a negative slope in this direction. The gradient of the function, $\nabla f(0,0) = (0, 0)$, the zero vector. This is because $(0,0)$ is the global maximum value of the function.

2.   Find the directional derivative of $f(x,y) = sin(xy^2)$ above point $P = (\pi/4, 2)$ in the direction of $\vec u = (1, 1)$

Solution

The unit direction vector is

$$\hat u = \frac{1}{\sqrt{2}}(1, 1)$$

$$\nabla f = (y^2cos(xy^2), 2xy \, cos(xy^2))$$

Then we have

\begin{align} \nabla f &= (y^2cos(xy^2), 2xy \, cos(xy^2)) \cdot (1,1) \frac{1}{\sqrt{2}} \\[5pt] &= \frac{1}{\sqrt{2}} \left[ y^2 cos(xy^2) + 2xy \, cos(xy^2) \right] \end{align}

Finally,

\require{cancel} \begin{align} \nabla f &\left( \frac{\pi}{4}, 2 \right) \cdot \hat u \\[5pt] &= \frac{1}{\sqrt{2}} \left[ 4 cos(\pi) + \cancel{2} \left(\frac{\pi}{\cancel{4}} \right)\cancel{2} cos(\pi) \right] \\[5pt] &= \frac{1}{\sqrt{2}} \left[ (4 + \pi) cos(\pi) \right] \\[5pt] &= -\frac{1}{\sqrt{2}} (4 + \pi) \\[5pt] &= \frac{-(4+\pi)}{\sqrt{2}} \end{align}

3.   Find the derivative of $f(x,y) = x^2 + xy + y^2$ above point $P = (0, 1)$ in the direction of $-3 \hat i + 4 \hat j$.

Solution

The unit vector version of $\vec u$ is

\begin{align} |\vec u| &= \sqrt{(-3)^2+4^2} = \sqrt{25} = 5 \\[5pt] \rightarrow \; \hat u &= \frac{1}{5}(-3, 4) \end{align}

$$\nabla f(x,y) = (2x + y, 2y + x)$$

\begin{align} \nabla f \cdot \hat u &= -\frac{3}{5}(2x + y) + \frac{4}{5} (2y + x) \\[5pt] &= -\frac{6}{5} x - \frac{3}{5} y + \frac{8}{5} y + \frac{4}{5} x \\[5pt] &= -\frac{1}{5} x + y \end{align}

Then

$$\nabla f(0, 1) \cdot \hat u = -\frac{1}{5} (0) + 1 = 1$$

4.   Find the derivative of $f(x,y,z) = (x-y)(y-z)(z-x)$ above point $P = (1,1,1)$ in any direction.

Solution

It will be easier to find the gradient (take the derivatives) if we expand these binomials:

\begin{align} f(x,y,z) &= (x-y)(y-z)(z-x) \\[5pt] &= (xy - xz - y^2 + yz)(z-x) \\[5pt] &= \cancel{xyz} - x^2 y - xz^2 + x^2z \\[5pt] &\phantom{000} - y^2z + xy^2 + yz^2 - \cancel{xyz} \\[5pt] &= -x^2 y - xz^2 + x^2z - y^2z + xy^2 + yz^2 \end{align}

\begin{align} \nabla f &= (-2xy - z^2 + 2xz + y^2, \\[5pt] &\phantom{000} -x^2 - 2yz + x + 2yz, \\[5pt] &\phantom{000} -2xz + x^2 - y^2 + 2yz) \end{align}

and

\begin{align} \nabla f(1,1,1) &= (-2-1+2+1,\\[5pt] &-1-2+1+2, \\[5pt] &-2+1-1+2) \\[5pt] &= (0,0,0) \end{align}

Now the directional derivative in an unspecified direction, $\hat u = (u_1, u_2, u_3)$ is

\begin{align} \nabla f(1,1,1) &\cdot (u_1, u_2, u_3) \\[5pt] &= (0,0,0) \cdot (u_1, u_2, u_3) = 0 \end{align}

So the directional derivative at this point in any direction is zero.

Notice that in this problem you have to be meticulous about your algebra — there are a lot of steps and many opportunities to make a sign error.

5.   Suppose that $z = e^{xy+x-y}$.

1. How fast is $z$ changing when we move away from the origin toward $(2, 1)$ ?
2. In what direction should we move away from the origin for $z$ to change most rapidly? What would the maximum rate be?
Solution

Moving away from the origin toward $(2, 1)$ means moving in the direction of $\vec u = (2, 1)$. The unit vector is

$$\hat u = \frac{1}{\sqrt{5}}(2, 1)$$

The gradient of $f(x,y)$ is

\begin{align} \nabla f &= \left( e^{xy+x-y}(y+1), e^{xy+x-y}(x-1) \right) \\[5pt] \nabla f(2,1) &= \left( e^{2+2-1}(1+1), e^{2+2-1}(1) \right) \\[5pt] &= (2e^3, e^3) \\[5pt] &= e^3(2, 1) \end{align}

Now

\begin{align} \nabla f(2, 1) \cdot \hat u &= e^3(2, 1)\cdot (2, 1) \frac{1}{\sqrt{5}} \\[5pt] &= e^3[2(2)+1(1)] \frac{1}{\sqrt{5}} \\[5pt] &= e^3 \cdot 5 \sqrt{5} \end{align}

The direction of maximum slope is the gradient. The gradient we found above, evaluated at the origin is

\begin{align} \nabla f &= \left( e^{xy+x-y}(y+1), e^{xy+x-y}(x-1) \right) \\[5pt] \nabla f(0,0) &= \left( e^0(0+1), e^0(0-1) \right) \\[5pt] &= (1, -1) \end{align}

That is the direction of steepest ascent. The magnitude of the slope is the length of this vector:

$$|(1, -1)| = \sqrt{2}$$

6.   Consider the function $f(x,y) = \frac{x^2-y^2}{x^2+y^2}$. In what direction is the directional derivative of $f$ at $(1, 1)$ equal to zero?

Solution

We'll just call our unknown direction unit vector

$$\hat u = (u_1, u_2)$$

And we'll seek to find the $u_1$ and $u_2$ that yield

$$\nabla f \cdot \hat u = 0$$

Let's do the gradient in pieces:

\begin{align} \frac{\partial f}{\partial x} &= \frac{(x^2+y^2)(2x) - (x^2-y^2)(2x)}{(x^2+y^2)^2} \\[5pt] &= \frac{\cancel{2x^3}+2xy^2 - \cancel{2x^3} + 2xy^2}{(x^2+y^2)^2} \\[5pt] &= \frac{4xy^2}{x^2+y^2)^2} \\[5pt] \frac{\partial f}{\partial y} &= \frac{(x^2+y^2)(-2y) - (x^2-y^2)(2y)}{(x^2+y^2)^2} \\[5pt] &= \frac{-2x^2y - \cancel{-2y^3} - 2x^2y + \cancel{2y^3}}{(x^2+y^2)^2} \\[5pt] &= \frac{-4xy^2}{(x^2+y^2)^2} \\[5pt] \rightarrow \nabla f &= \left( \frac{4xy^2}{x^2+y^2)^2}, \frac{-4xy^2}{(x^2+y^2)^2} \right) \end{align}

Evaluated at $(1, 1)$, the gradient is

$$\nabla f(1, 1) = \left( \frac{4}{2^2}, \frac{-4}{2^2} \right) = (1, -1)$$

Now in order to have a directional derivative of zero, we require that

\begin{align} \nabla f(1, 1) \cdot (u_1, u_2) &= 0 \\[5pt] (1, -1) \cdot (u_1, u_2) &= 0 \\[5pt] u_1 - u^2 = 0 \end{align}

So the directional derivative is zero when $u_1 = u_2$, that is, directions specified by $(1, 1)$ and $(-1,-1)$.

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