Headline

Let's begin with our derivatives in the $x$- and $y$-directions — the normal partial derivatives:

$$ \begin{align} D_{x} f(x_o, \; y_o) = \lim_{h \to 0} \frac{f(x_o+h, \; y_o) - f(x_o, \; y_o)}{h} \\[5pt] D_{y} f(x_o, \; y_o) = \lim_{h \to 0} \frac{f(x_o, \; y_o+h) - f(x_o, \; y_o)}{h} \\[5pt] \end{align}$$

These are the rates of change of our function in the $x$- and $y$ directions, respectively. Now we'd like to find the rate of change of $z$ at point $P = (x_o, \; y_o)$ in the direction of some other unit vector $\hat u = \langle a, \; b \rangle$.

Let's let $S$ be a surface with $z = f(x, \; y)$ and $z = f(x_o, \; y_o)$. Then point $P = (x_o, \; y_o, \; z_o)$ lies on surface $S$. The vertical plane passing through $P$ in the direction of $\hat u$ intersects $S$ in a curve, $C$. Here's the basic geometry:

Here is a blowup of the geometry shown in yellow dashed lines in the previous figure. The unit direction vector $\hat u = \langle a, \; b \rangle$. Any vector of lenth $h$ (a scalar) can be written as $h\hat u$ or $h \langle a, \; b \rangle$.

Example 1

Calculate the derivative of the function $z = x^2+y^2$ at point $P = (0, 2)$ in the direction of $\vec u = \langle 1, \; 4 \rangle$.

First we'll need the length of $\vec u$. It's $|\vec u| = \sqrt{2}$. Then our unit direction vector is

$$\hat u = \bigg< \frac{1}{\sqrt{17}}, \; \frac{4}{\sqrt{17}} \bigg>$$

The derivative is

$$ \require{cancel} \begin{align} D_{\hat u}f(x,y) &= f_x(x,y) a + f_y(x,y) b \\[5pt] &= 2x \left( \frac{1}{\sqrt{2}} \right) + 2y \left( \frac{1}{\sqrt{2}} \right) \\[5pt] &= \cancel{2(0) \left( \frac{1}{\sqrt{17}} \right)} + 2(2) \left( \frac{1}{\sqrt{17}} \right) \\[5pt] &= \frac{4}{\sqrt{17}} \approx 3.88 \end{align}$$

The gradient vector

Let's re-express the directional derivative in the direction of $\hat u = \langle a, \; b \rangle $ slightly:

$$ \begin{align} D_{\hat u} f(x,y) &= f_x(x,y)a + f_y(x,y)b \\[5pt] &= \langle f_x(x,y), \; f_y(x,y) \rangle \cdot \langle a, \; b \rangle \\[5pt] &= \langle f_x(x,y), \; f_y(x,y) \rangle \cdot \hat u \end{align}$$

We'll call the vector $\langle f_x(x,y), \; f_y(x,y) \rangle$ the gradient vector. We'll further define an "operator" called "del", with symbol $\nabla$ (called "nabla") as a pseudo-vector:

$$\nabla = \bigg< \frac{\partial}{\partial x}, \; \frac{\partial}{\partial y}\bigg>$$

The three-dimensional version is

$$\nabla = \bigg< \frac{\partial}{\partial x}, \; \frac{\partial}{\partial y}, \; \frac{\partial}{\partial z} \bigg>$$

We can use $\nabla$ in two ways. One is as an operator, which is just a set of instructions for what to do with another vector of the same dimensions. That is,

Gradient: Vector of steepest ascent

The gradient vector always points in the direction of steepest ascent.

Theorem

If $f$ is a differentiable function of two or three variables (represented by $\vec x$), the maximum value of the directional derivative in the direction of $\hat u$ is $|\nabla \cdot f(\vec x)|$. It occurs when $\hat u$ has the same direction as $\nabla f(\vec x)$.

Example 2

Consider the function $f(x,y) = 9-x^2-y^2$. Find the directional derivative in the direction of $\vec u = \langle 1, \; 2 \rangle$ over the point $P = (1, 1)$ in the $xy$ plane.

Solution: First find $\nabla f$:

$$ \begin{align} \nabla f &= \bigg< \frac{\partial f}{\partial x}, \; \frac{\partial f}{\partial y} \bigg> \\[5pt] &= \langle -2x, \; -2y\rangle \\[5pt] \nabla f(1,1) &= \langle -2, \; -2\rangle \end{align}$$

The length of $\vec u$ is

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

Example 3

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

Solution: First find $\nabla f$:

$$ \begin{align} \nabla f &= \bigg< \frac{\partial f}{\partial x}, \; \frac{\partial f}{\partial y} \bigg> \\[5pt] &= \bigg< \frac{3}{2} y \sqrt{1 + xy}, \; \frac{3}{2} x \sqrt{1 + xy} \bigg> \\[5pt] \nabla f(3,1) &= \langle 3, \; 9 \rangle \end{align}$$

Let $\vec u = \langle 1, \; 1\rangle = \hat i + \hat j$. Then the length of $\vec u$ is

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

Example 4

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 = \langle 2, \; 1, \; 1\rangle $.

Solution: First calculate $\hat u$:

$$\hat u = \frac{\langle 2,1,1 \rangle}{\sqrt{2^2+1^2+1^2}} = \frac{1}{\sqrt{6}}\langle 2,1,1\rangle $$

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

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

Now the directional derivative is

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

Example 5

Suppose that the temperature, $T$ at point $(x,y,z)$ of a block of metal is given by

$$T = \frac{30}{1+3x^2+2y^2+z^2}$$

with $T$ in $^{\circ}C$, and $x, \; y, \; z$ in meters. In which direction does $T$ increase fastest at point $(1, \; 1, \; 2)$ ?

Solution: The gradient of the temperature function is

$$ \begin{align} \nabla T &= \bigg< \frac{\partial T}{\partial x}, \; \frac{\partial T}{\partial y}, \; \frac{\partial T}{\partial z} \bigg> \\[5pt] &= \bigg< \frac{-180x}{(1+3x^2+2y^2+z^2)^2}, \; \frac{-120y}{(1+3x^2+2y^2+z^2)^2}, \; \frac{-60x}{(1+3x^2+2y^2+z^2)^2} \bigg> \\[5pt] \nabla T_{(1, 1, -2)} &= \bigg< \frac{-180}{7}, \; \frac{-120}{7}, \; \frac{120}{25} \bigg> \\[5pt] &\approx \langle -25.7, -17.1, \; 4.8 \rangle \end{align}$$

The length of the gradient vector is the value of the slope, which is $31.24^{\circ}C/m$.

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 = \langle 1, 3 \rangle $. 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 = \langle 1, 1\rangle $

   Solution

   
   

   The unit direction vector is

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

   The gradient vector is

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

   Then we have

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

   
   

   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

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

   The gradient is

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

   
   

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

   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:

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

   Now the gradient is

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

   
   

   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 $\vec u = \langle 2, 1\rangle$. 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 $(-1,-1)$.