Contents / Calculus / Partial Derivatives

Chapter 11

Partial Derivatives

Functions of several variables, partial derivatives, the chain rule, gradients, and optimization with Lagrange multipliers.

Introduction

Single-variable calculus studies functions $f:\mathbb{R}\to\mathbb{R}$ : one input, one output, and a graph you can draw on a page. Almost nothing in the world is that simple. The temperature in a room depends on where you stand, the pressure of a gas depends on its volume and its temperature, the cost of a product depends on the price of every ingredient. These are functions of several variables, and this chapter builds their differential calculus.

The plan is the same as in one variable — limits, then derivatives, then tangent lines, then optimisation — but every idea acquires more room to move. A function of two variables can be approached from infinitely many directions, so a limit is a stronger claim than before. Its graph is a surface, so a tangent line becomes a tangent plane. Its rate of change depends on which way you walk, and one vector, the gradient, packages every such rate at once.

The chapter ends with the two tools the rest of applied mathematics leans on constantly: the second derivatives test for locating maxima, minima and saddle points, and the method of Lagrange multipliers for optimising under a constraint. Multiple integrals, vector fields and the big integral theorems live in the chapters that follow; everything here is differentiation.

11.1Functions of several variables

Definition 11.1 (Function of two variables). A function $f$ of two variables is a rule that assigns to each ordered pair $(x,y)$ in a set $D\subseteq\mathbb{R}^2$ a unique real number $f(x,y)$ . The set $D$ is the domain of $f$ and the set of values $\{f(x,y) : (x,y)\in D\}$ is its range. The graph of $f$ is the set of points $(x,y,z)$ in $\mathbb{R}^3$ with $z = f(x,y)$ and $(x,y)\in D$ .

We write $z = f(x,y)$ and call $x$ and $y$ the independent variables, $z$ the dependent variable. When a formula is given without a domain, the domain is understood to be every $(x,y)$ for which the formula makes sense — no square roots of negatives, no logarithms of non-positive numbers, no division by zero. Finding it is a two-dimensional exercise: each restriction cuts out a region of the plane, and the domain is what survives all of them.

The graph of $f$ is a surface sitting over the domain. A linear function $f(x,y) = ax + by + c$ has a plane for its graph; $f(x,y) = \sqrt{9 - x^{2} - y^{2}}$ has the upper hemisphere of radius $3$ ; $f(x,y) = x^{2} + y^{2}$ has a bowl called a paraboloid, and $f(x,y) = y^{2} - x^{2}$ has a saddle, curving up in one direction and down in the other. Surfaces are hard to draw, so the working tool is a flat picture instead.

Definition 11.2 (Level curves and contour maps). The level curves of $f$ are the curves in the plane with equations $f(x,y) = k$ , one for each constant $k$ in the range of $f$ . A collection of level curves for equally spaced values of $k$ is a contour map of $f$ .

A level curve is the set of inputs on which $f$ takes the value $k$ ; it is the shadow in the $xy$ -plane of the horizontal slice of the graph at height $z = k$ . Where the level curves crowd together the surface is steep, because a small step in the plane crosses many heights; where they are far apart the surface is nearly flat. Every idea in this chapter — partial derivatives, the gradient, constrained optimisation — has a picture on the contour map, and it is worth learning to read one.

Intuition. A walker's map of a mountain draws a line through every point at $100$ m, another through every point at $200$ m, and so on. No third dimension is printed, yet you can read the mountain off the map: contours packed tight mean a cliff, contours far apart mean a meadow, a closed ring means a summit or a hollow. A contour map of $f$ is exactly that map, with $f$ playing the role of altitude.

Everything extends to three or more inputs. A function $f(x,y,z)$ assigns a number to each point of a region in space — the temperature at each point of a room, say. Its graph would live in four dimensions, so we never draw it; instead we draw its level surfaces $f(x,y,z) = k$ , which are surfaces in space. For $f(x,y,z) = x^{2} + y^{2} + z^{2}$ the level surfaces are concentric spheres, and knowing them is knowing the function.

Example 11.3 (Finding a domain). Find the domain of $f(x,y) = \dfrac{\sqrt{x + y + 1}}{x - 1}$ and of $g(x,y) = \ln(9 - x^{2} - 9y^{2})$ .

Solution. For $f$ , the square root requires $x + y + 1 \ge 0$ and the denominator requires $x \neq 1$ . The domain is the closed half-plane on and above the line $y = -x - 1$ , with the vertical line $x = 1$ removed.

For $g$ , the logarithm requires $9 - x^{2} - 9y^{2} > 0$ , that is,

\frac{x^2}{9} + y^2 < 1.

The domain is the interior of the ellipse with semi-axes $3$ and $1$ , boundary excluded. As a check, $(0,0)$ satisfies the inequality and $g(0,0) = \ln 9$ is a perfectly good number, while $(3,0)$ lies on the ellipse and gives $\ln 0$ , which is undefined.□

Example 11.4 (Level curves of a paraboloid and of a saddle). Sketch the level curves of $f(x,y) = x^{2} + y^{2}$ and of $g(x,y) = y^{2} - x^{2}$ .

Solution. The level curves of $f$ are $x^{2} + y^{2} = k$ : for $k > 0$ a circle of radius $\sqrt{k}$ centred at the origin, for $k = 0$ the single point $(0,0)$ , and for $k < 0$ nothing at all. Since the radii $\sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots$ grow ever more slowly, the circles for equally spaced $k$ bunch together as they move outward — the bowl steepens away from its base.

The level curves of $g$ are $y^{2} - x^{2} = k$ . For $k = 0$ this is the pair of lines $y = \pm x$ . For $k > 0$ it is a hyperbola opening up and down, for $k < 0$ a hyperbola opening left and right. The two families are separated by the crossed lines, which is the fingerprint of a saddle: walk along the $y$ -axis and $g$ rises, walk along the $x$ -axis and it falls.□

Example 11.5 (Reading a contour map). The level curves of $h(x,y) = 4x^{2} + y^{2} + 1$ for $k = 2, 5, 10$ are drawn. Describe them, and say in which direction from the origin $h$ increases fastest.

Solution. Setting $4x^{2} + y^{2} + 1 = k$ gives $4x^{2} + y^{2} = k - 1$ , an ellipse for $k > 1$ :

\frac{x^2}{(k-1)/4} + \frac{y^2}{k-1} = 1.

For $k = 2$ the semi-axes are $\tfrac12$ (along $x$ ) and $1$ (along $y$ ); for $k = 5$ they are $1$ and $2$ ; for $k = 10$ they are $\tfrac32$ and $3$ . Every ellipse is twice as tall as it is wide.

Contours are closer together along the $x$ -axis: moving from $k = 2$ to $k = 10$ takes only $1$ unit in the $x$ -direction but $2$ units in the $y$ -direction. So $h$ climbs fastest along the $x$ -axis, which agrees with the formula — the coefficient $4$ on $x^{2}$ makes $x$ count double.□

Example 11.6 (Level surfaces of a function of three variables). Describe the level surfaces of $F(x,y,z) = x^{2} + y^{2} - z$ and find the domain of $G(x,y,z) = \ln(z - x^{2} - y^{2})$ .

Solution. The level surface $F = k$ is $z = x^{2} + y^{2} - k$ , a paraboloid with vertex at $(0,0,-k)$ . The level surfaces are all translates of a single bowl, sliding up as $k$ decreases.

For $G$ we need $z - x^{2} - y^{2} > 0$ , so the domain is the set of points strictly above the paraboloid $z = x^{2} + y^{2}$ — the inside of the bowl, its surface excluded. The point $(0,0,1)$ is in the domain and $G(0,0,1) = 0$ ; the point $(1,0,1)$ sits on the paraboloid and is not.□

Exercises for §11.1 →

11.2Limits and continuity

Definition 11.7 (Limit of a function of two variables). Let $f$ be defined on a set $D$ containing points arbitrarily close to $(a,b)$ . We write

\lim_{(x,y)\to(a,b)} f(x,y) = L

if for every $\varepsilon > 0$ there is a $\delta > 0$ such that $|f(x,y) - L| < \varepsilon$ whenever $(x,y)\in D$ and $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$ .

Compare this with the one-variable definition from Limits & Continuity. The expression $\sqrt{(x-a)^2 + (y-b)^2}$ is the distance from $(x,y)$ to $(a,b)$ , so the condition says: whenever the input is within $\delta$ of $(a,b)$ but not equal to it, the output is within $\varepsilon$ of $L$ . Nothing is said about how the input approaches; the disc of radius $\delta$ contains every direction, every spiral, every curve into $(a,b)$ , and $f$ must be near $L$ on all of them at once.

That is a far stronger demand than in one variable, where a point can only be approached from the left or from the right. On the line, the limit exists when two one-sided limits agree. In the plane there are infinitely many "sides", and a single disagreement kills the limit.

Theorem 11.8 (Two-path test). If $f(x,y)\to L_1$ as $(x,y)\to(a,b)$ along a path $C_1$ and $f(x,y)\to L_2$ along a path $C_2$ , with $L_1 \neq L_2$ , then $\lim_{(x,y)\to(a,b)} f(x,y)$ does not exist.

Proof. Suppose the limit existed and equalled $L$ . Points on $C_1$ close to $(a,b)$ are inside the $\delta$ -disc, so $f$ is within $\varepsilon$ of $L$ on them, forcing $L_1 = L$ . The same argument on $C_2$ forces $L_2 = L$ . Then $L_1 = L_2$ , contrary to assumption.∎

The test is the standard way to show a limit fails. Try the two axes first, then the line $y = x$ , then the family of lines $y = mx$ ; if the answer depends on $m$ , you are done. If every line gives the same value, try a parabola — the third example shows why.

Showing that a limit exists is harder, because you must control every path at once. Two tools do it. The first is a squeeze: bound $|f(x,y) - L|$ by something that visibly goes to zero. The second, which usually amounts to the same thing but organises it, is polar coordinates centred at $(a,b)$ : put $x = a + r\cos\theta$ , $y = b + r\sin\theta$ . Then $(x,y)\to(a,b)$ means exactly $r\to 0^+$ , with $\theta$ free to do anything, and if the resulting expression tends to $L$ uniformly in $\theta$ — bounded by a function of $r$ alone that goes to $0$ — the limit is $L$ .

Intuition. Picture a summit and hikers arriving from every direction — up the north ridge, the south face, a spiral trail. The altitude at the summit is well defined only if every hiker reads the same number on the final step. Two hikers reporting different altitudes at the same point is the two-path test. A single altimeter reading that is within $\varepsilon$ of $L$ for every hiker within $\delta$ of the summit is the definition of the limit.

Example 11.9 (Different limits along the axes). Show that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ does not exist.

Solution. Along the $x$ -axis, where $y = 0$ and $x \neq 0$ ,

f(x,0) = \frac{x^2}{x^2} = 1,

so $f\to 1$ as $x\to 0$ along this path. Along the $y$ -axis, where $x = 0$ ,

f(0,y) = \frac{-y^2}{y^2} = -1,

so $f \to -1$ . Two paths, two different limits: by the two-path test the limit does not exist. As a sanity check, along $y = x$ the function is identically $0$ , a third value again.□

Example 11.10 (Agreeing on the axes, disagreeing on a diagonal). Show that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{xy}{x^{2} + y^{2}}$ does not exist.

Solution. On the $x$ -axis $f(x,0) = 0$ and on the $y$ -axis $f(0,y) = 0$ , so both axes give the limit $0$ . That does not settle anything. Along the line $y = x$ ,

f(x,x) = \frac{x\cdot x}{x^2 + x^2} = \frac{1}{2},

so the limit along this path is $\tfrac12 \neq 0$ , and the limit does not exist. Along a general line $y = mx$ the value is $m/(1 + m^{2})$ , a different constant for every slope: the function takes every value between $-\tfrac12$ and $\tfrac12$ in every neighbourhood of the origin.□

Example 11.11 (Every line agrees, a parabola does not). Show that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{xy^{2}}{x^{2} + y^{4}}$ does not exist.

Solution. Along any line $y = mx$ through the origin,

f(x,mx) = \frac{x\cdot m^2 x^2}{x^2 + m^4 x^4} = \frac{m^2 x}{1 + m^4 x^2} \to 0 \quad\text{as } x \to 0,

and along the $y$ -axis $f(0,y) = 0$ as well. Every straight line gives $0$ . But along the parabola $x = y^{2}$ ,

f(y^2, y) = \frac{y^2\cdot y^2}{y^4 + y^4} = \frac{1}{2},

so the limit along this curve is $\tfrac12$ . Two paths disagree, and the limit does not exist. Lines are not enough; the definition demands every path.□

Example 11.12 (Proving a limit exists with polar coordinates). Show that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{3x^{2} y}{x^{2} + y^{2}} = 0$ .

Solution. Substituting $x = r\cos\theta$ , $y = r\sin\theta$ turns the denominator into $r^{2}$ :

\frac{3x^2 y}{x^2 + y^2} = \frac{3r^3\cos^2\theta\sin\theta}{r^2} = 3r\cos^2\theta\sin\theta.

Since $|\cos^{2}\theta\sin\theta| \le 1$ for every $\theta$ ,

\left|\frac{3x^2 y}{x^2 + y^2} - 0\right| \le 3r,

and $3r \to 0$ as $r \to 0^+$ regardless of $\theta$ . The limit is $0$ . In $\varepsilon$ – $\delta$ terms, $\delta = \varepsilon/3$ works. The same estimate can be reached without polar coordinates: $x^{2}/(x^{2} + y^{2}) \le 1$ gives $|f(x,y)| \le 3|y|$ , and $|y| \to 0$ .□

Example 11.13 (A logarithm tamed by polar coordinates). Evaluate $\displaystyle\lim_{(x,y)\to(0,0)} (x^{2} + y^{2})\ln(x^{2} + y^{2})$ .

Solution. In polar coordinates $x^{2} + y^{2} = r^{2}$ , so the expression is $r^{2}\ln(r^{2}) = 2r^{2} \ln r$ , which depends on $r$ alone. As $r \to 0^+$ this is a $0\cdot(-\infty)$ form; rewrite it as

\lim_{r\to 0^+} \frac{2\ln r}{1/r^2} = \lim_{r\to 0^+}\frac{2/r}{-2/r^3} = \lim_{r\to 0^+}(-r^2) = 0

by L'Hôpital's Rule. There is no $\theta$ to worry about, so the two-variable limit is $0$ .□

Pitfall. Substituting polar coordinates and letting $r\to 0$ is only valid when the result is controlled for every $\theta$ at once. If the expression after substitution still depends on $\theta$ in a way that does not vanish — for $xy/(x^{2} + y^{2})$ it becomes $\cos\theta\sin\theta$ , which does not go to $0$ at all — then the limit does not exist, and the polar form has just handed you the paths that prove it.

Definition 11.14 (Continuity). A function $f$ of two variables is continuous at $(a,b)$ if

\lim_{(x,y)\to(a,b)} f(x,y) = f(a,b).

It is continuous on $D$ if it is continuous at every point of $D$ .

Continuity means what it meant before: the limit exists, the value exists, and they agree. The practical content is a list of functions you never have to check. Polynomials in $x$ and $y$ are continuous everywhere, because limits of sums and products are sums and products of limits, and $\lim x = a$ , $\lim y = b$ come straight from the definition. Rational functions are continuous wherever the denominator is nonzero. And if $f$ is continuous at $(a,b)$ and $g$ is a continuous function of one variable at $f(a,b)$ , the composition $g\circ f$ is continuous at $(a,b)$ : near $(a,b)$ the value $f(x,y)$ is near $f(a,b)$ , and then $g(f(x,y))$ is near $g(f(a,b))$ . So $e^{x^{2} y}$ , $\sin(xy)$ , $\sqrt{x^{2} + y^{2}}$ and $\ln(1 + x^{2} + y^{2})$ are continuous on all of $\mathbb{R}^2$ without any further work.

Example 11.15 (Where is a function continuous?). Determine where $g(x,y) = \arctan(y/x)$ is continuous, and whether

f(x,y) = \begin{cases} \dfrac{3x^2 y}{x^2 + y^2}, & (x,y)\neq(0,0) \\[6pt] 0, & (x,y) = (0,0) \end{cases}

is continuous at the origin.

Solution. The quotient $y/x$ is a rational function, continuous wherever $x \neq 0$ , and $\arctan$ is continuous everywhere, so the composition $g$ is continuous on the set $\{(x,y) : x \neq 0\}$ . On the line $x = 0$ it is undefined, hence not continuous there. It cannot even be repaired: approaching $(0, 1)$ from $x > 0$ gives $\arctan(+\infty) = \pi/2$ , from $x < 0$ gives $-\pi/2$ .

For $f$ : away from the origin it is a rational function with nonzero denominator, so continuous. At the origin the limit was shown to be $0$ in the polar-coordinates example, and $f(0,0) = 0$ by definition, so $f$ is continuous at the origin too. The function is continuous on all of $\mathbb{R}^2$ .□

Exercises for §11.2 →

11.3Partial derivatives

Definition 11.16 (Partial derivatives). The partial derivatives of $f$ with respect to $x$ and to $y$ at $(a,b)$ are

f_x(a,b) = \lim_{h\to 0}\frac{f(a+h,\,b) - f(a,b)}{h}, \qquad f_y(a,b) = \lim_{h\to 0}\frac{f(a,\,b+h) - f(a,b)}{h},

whenever these limits exist.

In $f_x(a,b)$ the second input is frozen at $b$ and only the first moves; the limit is the ordinary derivative of the one-variable function $g(x) = f(x,b)$ at $x = a$ . That observation is also the rule for computing: to find $f_x$ , regard $y$ as a constant and differentiate with respect to $x$ by every rule you already know; to find $f_y$ , regard $x$ as a constant and differentiate with respect to $y$ .

Notation. For $z = f(x,y)$ the following all mean the same thing:

f_x = \frac{\partial f}{\partial x} = \frac{\partial z}{\partial x} = D_x f = f_1,

and likewise for $y$ . The curly $\partial$ is read "partial" and is never written $d$ : $\partial f/\partial x$ is not a quotient of two quantities $\partial f$ and $\partial x$ .

Geometrically, freezing $y = b$ slices the surface $z = f(x,y)$ with a vertical plane; the slice is a curve $C_1$ , and $f_x(a,b)$ is the slope of its tangent line at the point $(a,b,f(a,b))$ . Freezing $x = a$ gives a second curve $C_2$ whose tangent slope is $f_y(a,b)$ . Two slices, two slopes — and, as the next section shows, those two tangent lines span the tangent plane.

As rates of change, $f_x(a,b)$ is how fast $f$ changes per unit step in the $x$ -direction from $(a,b)$ , and $f_y(a,b)$ how fast it changes per unit step in the $y$ -direction. If $T(x,y)$ is the temperature of a plate, $T_x$ is the temperature gradient felt walking east and $T_y$ the one felt walking north.

Intuition. Stand on a hillside at a point where the ground is a plane tilted so that walking east climbs $3$ m per $100$ m and walking north climbs $-1$ m per $100$ m. Those two numbers, $0.03$ and $-0.01$ , are the partial derivatives of the altitude function at your feet. Neither is "the slope of the hill" — the hill has a different slope in every direction — but the section on the gradient shows that these two determine all the others.

Example 11.17 (Partial derivatives at a point). If $f(x,y) = x^{3} + x^{2} y^{3} - 2y^{2}$ , find $f_x(2,1)$ and $f_y(2,1)$ .

Solution. Holding $y$ constant and differentiating with respect to $x$ ,

f_x(x,y) = 3x^2 + 2xy^3, \qquad\text{so}\qquad f_x(2,1) = 3\cdot 4 + 2\cdot 2\cdot 1 = 16.

Holding $x$ constant and differentiating with respect to $y$ (the term $x^{3}$ is a constant and disappears),

f_y(x,y) = 3x^2 y^2 - 4y, \qquad\text{so}\qquad f_y(2,1) = 3\cdot 4\cdot 1 - 4 = 8.

Check $f_x(2,1)$ directly: with $y = 1$ , $g(x) = x^{3} + x^{2} - 2$ , and $g'(2) = 12 + 4 = 16$ .□

Example 11.18 (Partial derivatives with the chain rule inside). Find $\partial f/\partial x$ and $\partial f/\partial y$ for $f(x,y) = \sin\!\left(\dfrac{x}{1 + y}\right)$ .

Solution. With $y$ fixed, $\dfrac{x}{1+y}$ is a constant multiple of $x$ , so

\frac{\partial f}{\partial x} = \cos\!\left(\frac{x}{1+y}\right)\cdot\frac{1}{1+y}.

With $x$ fixed, $\dfrac{x}{1+y} = x(1+y)^{-1}$ has derivative $-x(1+y)^{-2}$ , so

\frac{\partial f}{\partial y} = -\cos\!\left(\frac{x}{1+y}\right)\cdot\frac{x}{(1+y)^2}.

□

Example 11.19 (Implicit partial derivatives). If $z$ is defined implicitly as a function of $x$ and $y$ by $yz - \ln z = x + y$ , find $\partial z/\partial x$ and $\partial z/\partial y$ .

Solution. Differentiate both sides with respect to $x$ , treating $y$ as a constant and $z$ as a function of $x$ :

y\frac{\partial z}{\partial x} - \frac{1}{z}\frac{\partial z}{\partial x} = 1 \quad\Longrightarrow\quad \frac{\partial z}{\partial x} = \frac{1}{y - 1/z} = \frac{z}{yz - 1}.

Differentiate with respect to $y$ , now treating $x$ as a constant; the product $yz$ needs the product rule:

z + y\frac{\partial z}{\partial y} - \frac{1}{z}\frac{\partial z}{\partial y} = 1 \quad\Longrightarrow\quad \frac{\partial z}{\partial y} = \frac{1 - z}{y - 1/z} = \frac{z(1 - z)}{yz - 1}.

The chain rule section turns this procedure into a formula.□

Functions of three or more variables are handled the same way: $f_z$ freezes $x$ and $y$ and differentiates in $z$ . For $f(x,y,z) = e^{xy}\ln z$ ,

f_x = y e^{xy}\ln z, \qquad f_y = x e^{xy}\ln z, \qquad f_z = \frac{e^{xy}}{z}.

Since a partial derivative is again a function of $x$ and $y$ , it can be differentiated again. Four second-order partial derivatives arise:

f_{xx} = \frac{\partial}{\partial x}\!\left(\frac{\partial f}{\partial x}\right) = \frac{\partial^2 f}{\partial x^2}, \qquad f_{xy} = \frac{\partial}{\partial y}\!\left(\frac{\partial f}{\partial x}\right) = \frac{\partial^2 f}{\partial y\,\partial x},

and similarly $f_{yx}$ and $f_{yy}$ . Read the subscripts left to right: $f_{xy}$ means differentiate in $x$ first, then in $y$ . In the $\partial$ notation the order is reversed, $\partial^{2} f/\partial y\,\partial x$ , because the operator nearest $f$ acts first. The two mixed partials $f_{xy}$ and $f_{yx}$ look as if they could differ, and the next example shows what happens in practice.

Example 11.20 (All four second partials). Find the second partial derivatives of $f(x,y) = x^{3} + x^{2} y^{3} - 2y^{2}$ .

Solution. From the first example, $f_x = 3x^{2} + 2xy^{3}$ and $f_y = 3x^{2} y^{2} - 4y$ . Differentiating each in both variables:

f_{xx} = 6x + 2y^3, \qquad f_{xy} = 6xy^2, \qquad f_{yx} = 6xy^2, \qquad f_{yy} = 6x^2 y - 4.

The mixed partials agree. That is not a coincidence.□

Theorem 11.21 (Clairaut's Theorem). Suppose $f$ is defined on a disc $D$ containing the point $(a,b)$ . If the functions $f_{xy}$ and $f_{yx}$ are both continuous on $D$ , then

f_{xy}(a,b) = f_{yx}(a,b).

The proof applies the Mean Value Theorem twice to the double difference $f(a+h,b+h) - f(a+h,b) - f(a,b+h) + f(a,b)$ , once in each order, and belongs to real analysis; the idea is that both mixed partials are limits of that same symmetric quantity divided by $h^{2}$ . What matters in practice is the hypothesis: continuity of the mixed partials. For every function built from polynomials, exponentials, logarithms and trigonometric functions on its natural domain it holds automatically, and you may compute whichever mixed partial is easier.

Remark. The continuity hypothesis is not decoration. The function

f(x,y) = \begin{cases}\dfrac{xy(x^2 - y^2)}{x^2 + y^2}, & (x,y)\neq (0,0) \\[6pt] 0, & (x,y) = (0,0)\end{cases}

has $f_x(0,y) = -y$ and $f_y(x,0) = x$ , so $f_{xy}(0,0) = -1$ while $f_{yx}(0,0) = 1$ . Its mixed partials exist everywhere but are discontinuous at the origin, and there they disagree.

Partial derivatives of order three and higher are defined the same way, and by Clairaut's Theorem applied repeatedly, $f_{xyy} = f_{yxy} = f_{yyx}$ whenever these are continuous. Only the number of differentiations in each variable matters.

Partial derivatives appear in the laws of physics chiefly through partial differential equations, equations relating an unknown function to its partial derivatives. Three of them describe most of classical physics. Laplace's equation

\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0

governs steady-state temperature, electrostatic potential and the flow of ideal fluids; its solutions are called harmonic functions. The wave equation

\frac{\partial^2 u}{\partial t^2} = a^2\frac{\partial^2 u}{\partial x^2}

describes a vibrating string of displacement $u(x,t)$ , with $a$ the wave speed. The heat equation

\frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2}

describes how temperature $u(x,t)$ along a rod evens out over time. Solving them is a later course; verifying that a given function solves one is a partial-derivative computation.

Example 11.22 (Verifying a harmonic function). Show that $u(x,y) = e^x\sin y$ satisfies Laplace's equation.

Solution. Differentiate twice in each variable:

u_x = e^x\sin y, \quad u_{xx} = e^x\sin y, \qquad u_y = e^x\cos y, \quad u_{yy} = -e^x\sin y.

Then $u_{xx} + u_{yy} = e^x\sin y - e^x\sin y = 0$ , so $u$ is harmonic. The same computation shows $e^x\cos y$ is harmonic too.□

Example 11.23 (Verifying a solution of the wave equation). Show that $u(x,t) = \sin(x - at)$ satisfies the wave equation $u_{tt} = a^{2} u_{xx}$ .

Solution. Treating $t$ as constant, $u_x = \cos(x - at)$ and $u_{xx} = -\sin(x - at)$ . Treating $x$ as constant, the chain rule contributes a factor $-a$ each time:

u_t = -a\cos(x - at), \qquad u_{tt} = -a^2\sin(x - at).

So $u_{tt} = a^{2}\cdot(-\sin(x - at)) = a^{2} u_{xx}$ . Physically, $\sin(x - at)$ is a sine wave travelling to the right at speed $a$ : at time $t$ the shape is the graph of $\sin x$ shifted $at$ units right.□

Example 11.24 (Verifying a solution of the heat equation). Show that $u(x,t) = e^{-kt}\sin x$ satisfies the heat equation $u_t = k\,u_{xx}$ .

Solution. $u_t = -k e^{-kt}\sin x$ , while $u_x = e^{-kt}\cos x$ and $u_{xx} = -e^{-kt}\sin x$ . Hence

k\,u_{xx} = -k e^{-kt}\sin x = u_t.

The picture: the initial temperature profile $\sin x$ keeps its shape and decays exponentially in amplitude, faster for larger $k$ — heat spreads from the warm crests into the cool troughs.□

Pitfall. The existence of both partial derivatives at a point does not make $f$ continuous there, let alone nice. The function $f(x,y) = xy/(x^{2} + y^{2})$ with $f(0,0) = 0$ has $f_x(0,0) = 0$ and $f_y(0,0) = 0$ , because it vanishes identically on both axes, yet it has no limit at the origin. Partial derivatives only look along two lines; the next section introduces the right notion of "differentiable", which looks in every direction.

Exercises for §11.3 →

11.4Tangent planes and linear approximations

In one variable, the tangent line at $a$ is the line through $(a, f(a))$ with slope $f'(a)$ , and zooming in on a differentiable graph makes it indistinguishable from its tangent line. The same picture holds for a surface: zoom in on a smooth surface and it flattens into a plane, the tangent plane.

Let $S$ be the surface $z = f(x,y)$ and $P = (x_0, y_0, z_0)$ a point on it, with $z_0 = f(x_0,y_0)$ . The tangent plane at $P$ should contain the tangent lines to the two slice curves $C_1$ (in the plane $y = y_0$ ) and $C_2$ (in the plane $x = x_0$ ). Any non-vertical plane through $P$ has an equation

z - z_0 = A(x - x_0) + B(y - y_0)

for some constants $A$ and $B$ . Intersect it with the plane $y = y_0$ : the second term vanishes and we get $z - z_0 = A(x - x_0)$ , a line of slope $A$ . For this to be the tangent line to $C_1$ we need $A = f_x(x_0, y_0)$ . Intersecting with $x = x_0$ in the same way forces $B = f_y(x_0,y_0)$ . Two conditions, two unknowns, and the plane is determined.

Theorem 11.25 (Equation of the tangent plane). If $f$ has continuous partial derivatives, the tangent plane to the surface $z = f(x,y)$ at the point $(x_0, y_0, z_0)$ is

z - z_0 = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0).

Example 11.26 (A tangent plane to a paraboloid). Find the tangent plane to the elliptic paraboloid $z = 2x^{2} + y^{2}$ at the point $(1,1,3)$ .

Solution. With $f(x,y) = 2x^{2} + y^{2}$ , $f_x = 4x$ and $f_y = 2y$ , so $f_x(1,1) = 4$ and $f_y(1,1) = 2$ . The tangent plane is

z - 3 = 4(x - 1) + 2(y - 1), \qquad\text{that is,}\qquad z = 4x + 2y - 3.

Check: the plane passes through $(1,1,3)$ , since $4 + 2 - 3 = 3$ . Near that point the surface and the plane are close: at $(1.1, 1.05)$ the surface gives $2(1.21) + 1.1025 = 3.5225$ and the plane gives $4.4 + 2.1 - 3 = 3.5$ .□

The last computation is the whole point. The right-hand side of the tangent plane equation is a linear function of $x$ and $y$ , easy to evaluate, and it approximates $f$ near $(x_0,y_0)$ .

Definition 11.27 (Linearization and linear approximation). The linearization of $f$ at $(a,b)$ is the linear function

L(x,y) = f(a,b) + f_x(a,b)(x - a) + f_y(a,b)(y - b),

and the approximation $f(x,y) \approx L(x,y)$ for $(x,y)$ near $(a,b)$ is the linear (or tangent plane) approximation of $f$ at $(a,b)$ .

Example 11.28 (Using a linear approximation). Find the linearization of $f(x,y) = xe^{xy}$ at $(1,0)$ and use it to approximate $f(1.1, -0.1)$ .

Solution. $f(1,0) = 1\cdot e^{0} = 1$ . By the product rule $f_x = e^{xy} + xy e^{xy}$ , so $f_x(1,0) = 1$ ; and $f_y = x^{2} e^{xy}$ , so $f_y(1,0) = 1$ . The linearization is

L(x,y) = 1 + 1\cdot(x - 1) + 1\cdot(y - 0) = x + y.

Hence $f(1.1, -0.1) \approx 1.1 - 0.1 = 1$ . The true value is $1.1\,e^{-0.11} \approx 0.985$ , so the approximation is off by about $1.5\%$ — reasonable for a step of size $0.14$ from the base point.□

There is a subtlety hiding here. The tangent plane formula uses only $f_x$ and $f_y$ , and the pitfall at the end of the previous section produced a function with $f_x(0,0) = f_y(0,0) = 0$ that is not even continuous at the origin. Its "tangent plane" $z = 0$ is a terrible approximation: the function takes the value $\tfrac12$ arbitrarily close to $(0,0)$ . So the mere existence of the partial derivatives does not guarantee that the tangent plane approximates the surface. We need a definition of differentiability that asks for exactly that guarantee.

Write $\Delta z = f(a + \Delta x,\, b + \Delta y) - f(a,b)$ for the change in $f$ . In one variable, differentiability at $a$ is equivalent to $\Delta y = f'(a)\Delta x + \varepsilon\Delta x$ with $\varepsilon \to 0$ as $\Delta x \to 0$ : the change is the tangent-line prediction plus an error that is small compared with $\Delta x$ . The two-variable version reads the same way.

Definition 11.29 (Differentiability). The function $z = f(x,y)$ is differentiable at $(a,b)$ if $\Delta z$ can be written as

\Delta z = f_x(a,b)\,\Delta x + f_y(a,b)\,\Delta y + \varepsilon_1\,\Delta x + \varepsilon_2\,\Delta y,

where $\varepsilon_1, \varepsilon_2 \to 0$ as $(\Delta x, \Delta y) \to (0,0)$ .

In words: $f$ is differentiable when the tangent plane approximation is good, with an error that is small compared with the distance moved. The definition is awkward to verify directly, but there is a clean sufficient condition that covers every function met in practice.

Theorem 11.30 (Continuous partials imply differentiability). If the partial derivatives $f_x$ and $f_y$ exist near $(a,b)$ and are continuous at $(a,b)$ , then $f$ is differentiable at $(a,b)$ .

Proof. Split the change into two one-variable steps, first in $y$ then in $x$ :

\Delta z = \big[f(a + \Delta x,\, b + \Delta y) - f(a,\, b + \Delta y)\big] + \big[f(a,\, b + \Delta y) - f(a,b)\big].

In the first bracket only $x$ varies, so the Mean Value Theorem gives a $u$ between $a$ and $a + \Delta x$ with the bracket equal to $f_x(u,\, b + \Delta y)\,\Delta x$ . In the second bracket only $y$ varies, and there is a $v$ between $b$ and $b + \Delta y$ with the bracket equal to $f_y(a, v)\,\Delta y$ . Now set

\varepsilon_1 = f_x(u,\, b + \Delta y) - f_x(a,b), \qquad \varepsilon_2 = f_y(a,v) - f_y(a,b).

Then $\Delta z = f_x(a,b)\Delta x + f_y(a,b)\Delta y + \varepsilon_1\Delta x + \varepsilon_2\Delta y$ exactly, and as $(\Delta x,\Delta y)\to(0,0)$ the points $(u, b + \Delta y)$ and $(a,v)$ tend to $(a,b)$ , so continuity of $f_x$ and $f_y$ at $(a,b)$ makes $\varepsilon_1,\varepsilon_2 \to 0$ .∎

Corollary 11.31 (Differentiability implies continuity). If $f$ is differentiable at $(a,b)$ , then $f$ is continuous at $(a,b)$ .

Proof. Every term on the right of the differentiability equation contains a factor $\Delta x$ or $\Delta y$ , so $\Delta z \to 0$ as $(\Delta x,\Delta y)\to(0,0)$ ; that is, $f(x,y)\to f(a,b)$ .∎

Polynomials, rational functions on their domains, and any combination of them with exponentials, logarithms and trigonometric functions have continuous partials, hence are differentiable wherever they are defined. The example $xy/(x^{2} + y^{2})$ has partials at the origin, but they are not continuous there, and the theorem does not apply — correctly, since the function is not differentiable there.

Intuition. Differentiability is the promise that the surface is locally flat. Rest a rigid sheet of card on a smooth hill at one point: it touches along the tangent plane and, in a small enough patch, the hill barely leaves the card. Now try the same on the surface $z = xy/(x^{2} + y^{2})$ near the origin, where it looks like a twisted ribbon rising to height $\tfrac12$ in some directions and dropping to $-\tfrac12$ in others: no card fits. Two slopes exist, but no plane.

Once $f$ is differentiable, the approximation $\Delta z \approx f_x\Delta x + f_y\Delta y$ is worth its own notation. For a differentiable function $z = f(x,y)$ , the differentials $dx$ and $dy$ are independent variables, and the total differential is

dz = f_x(x,y)\,dx + f_y(x,y)\,dy = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy.

Taking $dx = \Delta x$ and $dy = \Delta y$ , the differential $dz$ is the change in height of the tangent plane while $\Delta z$ is the change in height of the surface, and the definition of differentiability says exactly that $\Delta z \approx dz$ .

Example 11.32 (Comparing $\Delta z$ and $dz$ ). For $z = f(x,y) = x^{2} + 3xy - y^{2}$ , compare $dz$ and $\Delta z$ when $x$ changes from $2$ to $2.05$ and $y$ changes from $3$ to $2.96$ .

Solution. The differential is $dz = (2x + 3y)\,dx + (3x - 2y)\,dy$ . At $(2,3)$ with $dx = 0.05$ and $dy = -0.04$ ,

dz = (4 + 9)(0.05) + (6 - 6)(-0.04) = 0.65.

The actual change is

\Delta z = f(2.05, 2.96) - f(2,3) = (4.2025 + 18.204 - 8.7616) - (4 + 18 - 9) = 13.6449 - 13 = 0.6449.

The differential is easier to compute and is off by only $0.005$ .□

The typical use is error estimation: measured quantities carry errors, a computed quantity inherits an error from them, and $dz$ says how large.

Example 11.33 (Estimating the error in a computed volume). The base radius and height of a right circular cone are measured as $10$ cm and $25$ cm, each with a possible error of $0.1$ cm. Estimate the maximum error in the calculated volume, and the maximum relative error.

Solution. The volume is $V = \tfrac13\pi r^{2} h$ , so

dV = \frac{\partial V}{\partial r}dr + \frac{\partial V}{\partial h}dh = \frac{2\pi r h}{3}\,dr + \frac{\pi r^2}{3}\,dh.

The errors satisfy $|dr| \le 0.1$ and $|dh| \le 0.1$ ; the worst case is both at $0.1$ with the same sign:

|dV| \le \frac{2\pi(10)(25)}{3}(0.1) + \frac{\pi(100)}{3}(0.1) = \frac{50\pi}{3} + \frac{10\pi}{3} = 20\pi \approx 63\ \text{cm}^3.

That looks large, but $V = \tfrac13\pi(100)(25) = 2500\pi/3 \approx 2618\ \text{cm}^3$ , so the relative error is

\frac{dV}{V} = \frac{2\,dr}{r} + \frac{dh}{h} = 2(0.01) + 0.004 = 0.024,

about $2.4\%$ . Note how the exponent $2$ on $r$ doubles the relative contribution of the radius error.□

For three variables everything reads the same: $w = f(x,y,z)$ is linearised by $L = f(a,b,c) + f_x\,(x - a) + f_y\,(y - b) + f_z\,(z - c)$ , and $dw = f_x\,dx + f_y\,dy + f_z\,dz$ .

Example 11.34 (Approximating a value with three variables). Use differentials to approximate $\sqrt{(2.02)^2 + (1.97)^2 + (0.98)^2}$ without a calculator.

Solution. Take $w = f(x,y,z) = \sqrt{x^{2} + y^{2} + z^{2}}$ at the base point $(2,2,1)$ , where $w = \sqrt{9} = 3$ . Each partial has the form $f_x = x/w$ , so at the base point

f_x = \frac{2}{3}, \qquad f_y = \frac{2}{3}, \qquad f_z = \frac{1}{3}.

With $dx = 0.02$ , $dy = -0.03$ , $dz = -0.02$ ,

dw = \frac{2}{3}(0.02) + \frac{2}{3}(-0.03) + \frac{1}{3}(-0.02) = \frac{0.04 - 0.06 - 0.02}{3} = -\frac{0.04}{3} \approx -0.0133,

so the value is approximately $3 - 0.0133 = 2.9867$ . Squaring the true numbers gives $4.0804 + 3.8809 + 0.9604 = 8.9217$ , whose square root is $2.98692$ ; the estimate is correct to three decimals.□

Exercises for §11.4 →

11.5The chain rule

In one variable the chain rule says that if $y = f(x)$ and $x = g(t)$ then $dy/dt = (dy/dx)(dx/dt)$ : rates multiply along a chain. With several variables a quantity can depend on $t$ through more than one intermediate variable, and the rule acquires one term per route.

Theorem 11.35 (Chain Rule, case 1). Suppose $z = f(x,y)$ is a differentiable function of $x$ and $y$ , where $x = g(t)$ and $y = h(t)$ are differentiable functions of $t$ . Then $z$ is a differentiable function of $t$ and

\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}.

Proof. A change $\Delta t$ produces changes $\Delta x = g(t + \Delta t) - g(t)$ and $\Delta y = h(t + \Delta t) - h(t)$ , which produce a change $\Delta z$ in $f$ . Since $f$ is differentiable,

\Delta z = \frac{\partial f}{\partial x}\Delta x + \frac{\partial f}{\partial y}\Delta y + \varepsilon_1\Delta x + \varepsilon_2\Delta y,

with $\varepsilon_1,\varepsilon_2\to 0$ as $(\Delta x,\Delta y)\to(0,0)$ . Divide by $\Delta t$ and let $\Delta t \to 0$ . Then $\Delta x/\Delta t \to dx/dt$ and $\Delta y/\Delta t \to dy/dt$ by definition of the derivative. Because $g$ and $h$ are differentiable they are continuous, so $\Delta x,\Delta y\to 0$ and hence $\varepsilon_1,\varepsilon_2\to 0$ ; the last two terms therefore contribute $0\cdot dx/dt + 0\cdot dy/dt = 0$ , leaving the formula.∎

The formula is the one-variable chain rule applied once per route from $t$ to $z$ and summed. Think of a tree: $z$ at the top with branches down to $x$ and $y$ , each of which has a branch down to $t$ . Every root-to-leaf path is a product of derivatives along its edges, and $dz/dt$ is the sum over all paths. That description works for every case of the chain rule and is worth drawing whenever the bookkeeping gets heavy.

Example 11.36 (A rate along a curve). If $z = x^{2} y + 3xy^{4}$ , where $x = \sin 2t$ and $y = \cos t$ , find $dz/dt$ when $t = 0$ .

Solution. By case 1,

\frac{dz}{dt} = (2xy + 3y^4)(2\cos 2t) + (x^2 + 12xy^3)(-\sin t).

At $t = 0$ we have $x = \sin 0 = 0$ and $y = \cos 0 = 1$ , so

\frac{dz}{dt}\bigg|_{t=0} = (0 + 3)(2) + (0 + 0)(0) = 6.

Check by substituting first: $z(t) = \sin^{2} 2t\cos t + 3\sin 2t\cos^{4} t$ , and near $t = 0$ the dominant term is $3\sin 2t \approx 6t$ , whose derivative is $6$ .□

Example 11.37 (Related rates for a gas). The pressure $P$ (kPa), volume $V$ (L) and temperature $T$ (K) of a mole of ideal gas satisfy $PV = 8.31\,T$ . Find the rate at which the pressure is changing when $T = 300$ K and increasing at $0.1$ K/s while $V = 100$ L and increasing at $0.2$ L/s.

Solution. Write $P = 8.31\,T/V$ as a function of the two variables $T$ and $V$ , each a function of $t$ :

\frac{dP}{dt} = \frac{\partial P}{\partial T}\frac{dT}{dt} + \frac{\partial P}{\partial V}\frac{dV}{dt} = \frac{8.31}{V}\frac{dT}{dt} - \frac{8.31\,T}{V^2}\frac{dV}{dt}.

Substituting the given values,

\frac{dP}{dt} = \frac{8.31}{100}(0.1) - \frac{8.31(300)}{100^2}(0.2) = 0.00831 - 0.04986 \approx -0.042\ \text{kPa/s}.

The pressure is falling: the expansion of the volume outweighs the warming.□

When the intermediate variables themselves depend on two variables, the same tree has two leaves at the bottom, and we get a partial derivative for each.

Theorem 11.38 (Chain Rule, case 2). Suppose $z = f(x,y)$ is differentiable, where $x = g(s,t)$ and $y = h(s,t)$ have first partial derivatives. Then

\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}, \qquad \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}.

Proof. To compute $\partial z/\partial s$ we hold $t$ fixed, and then $x$ and $y$ are functions of the single variable $s$ . Case 1 applies verbatim, with the ordinary derivatives $dx/ds$ , $dy/ds$ replaced by the partials $\partial x/\partial s$ , $\partial y/\partial s$ because $t$ is being held constant. The formula for $\partial z/\partial t$ is the same argument with the roles of $s$ and $t$ exchanged.∎

Example 11.39 (Chain rule with two independent variables). If $z = e^x\sin y$ , where $x = st^{2}$ and $y = s^{2} t$ , find $\partial z/\partial s$ and $\partial z/\partial t$ .

Solution. The four partial derivatives of the intermediate variables are

\frac{\partial x}{\partial s} = t^2, \quad \frac{\partial x}{\partial t} = 2st, \quad \frac{\partial y}{\partial s} = 2st, \quad \frac{\partial y}{\partial t} = s^2.

Case 2 gives

\frac{\partial z}{\partial s} = e^x\sin y\cdot t^2 + e^x\cos y\cdot 2st = t^2 e^{st^2}\sin(s^2 t) + 2st\,e^{st^2}\cos(s^2 t),

\frac{\partial z}{\partial t} = e^x\sin y\cdot 2st + e^x\cos y\cdot s^2 = 2st\,e^{st^2}\sin(s^2 t) + s^2 e^{st^2}\cos(s^2 t).

□

Theorem 11.40 (Chain Rule, general version). Suppose $u$ is a differentiable function of the $n$ variables $x_1, \ldots, x_n$ , and each $x_j$ is a differentiable function of the $m$ variables $t_1,\ldots,t_m$ . Then $u$ is a function of $t_1,\ldots,t_m$ and, for each $i = 1,\ldots,m$ ,

\frac{\partial u}{\partial t_i} = \frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial t_i} + \frac{\partial u}{\partial x_2}\frac{\partial x_2}{\partial t_i} + \cdots + \frac{\partial u}{\partial x_n}\frac{\partial x_n}{\partial t_i}.

The tree diagram now has $u$ at the top, the $n$ intermediate variables in a row beneath it, and the $m$ independent variables in a row beneath each of those. To find $\partial u/\partial t_i$ , follow every path from $u$ down to a leaf labelled $t_i$ , multiply the derivatives written on the edges of that path, and add up over paths. There are $n$ such paths, one through each intermediate variable, which is why the formula has $n$ terms.

Example 11.41 (Three intermediate and three independent variables). If $u = x^{4} y + y^{2} z^{3}$ , where $x = rse^t$ , $y = rs^{2} e^{-t}$ and $z = r^{2} s\sin t$ , find $\partial u/\partial s$ when $r = 2$ , $s = 1$ , $t = 0$ .

Solution. At the given point $x = 2$ , $y = 2$ , $z = 0$ . The three paths from $u$ to $s$ give

\frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s} + \frac{\partial u}{\partial z}\frac{\partial z}{\partial s} = (4x^3 y)(re^t) + (x^4 + 2yz^3)(2rse^{-t}) + (3y^2 z^2)(r^2\sin t).

Substituting $x = 2$ , $y = 2$ , $z = 0$ , $r = 2$ , $s = 1$ , $t = 0$ :

\frac{\partial u}{\partial s} = (4\cdot 8\cdot 2)(2) + (16 + 0)(4) + (0)(0) = 128 + 64 = 192.

□

Example 11.42 (Rewriting an expression in polar coordinates). If $z = f(x,y)$ has continuous second partials and $x = r\cos\theta$ , $y = r\sin\theta$ , show that

\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 = \left(\frac{\partial z}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial z}{\partial\theta}\right)^2.

Solution. By case 2, with $x_r = \cos\theta$ , $y_r = \sin\theta$ , $x_\theta = -r\sin\theta$ , $y_\theta = r\cos\theta$ ,

z_r = z_x\cos\theta + z_y\sin\theta, \qquad z_\theta = -z_x\,r\sin\theta + z_y\,r\cos\theta.

Square and combine:

z_r^2 + \frac{1}{r^2}z_\theta^2 = z_x^2(\cos^2\theta + \sin^2\theta) + z_y^2(\sin^2\theta + \cos^2\theta) + 2z_x z_y(\cos\theta\sin\theta - \sin\theta\cos\theta) = z_x^2 + z_y^2.

The cross terms cancel and the Pythagorean identity does the rest. This identity is what lets physicists write the gradient's magnitude in polar coordinates.□

The chain rule also settles implicit differentiation, replacing the case-by-case procedure of the earlier chapters by a formula. Suppose an equation $F(x,y) = 0$ defines $y$ implicitly as a differentiable function of $x$ . Then $F(x, y(x)) = 0$ for all $x$ in the domain, and differentiating both sides with respect to $x$ by case 1 (the two intermediate variables being $x$ itself and $y$ ),

\frac{\partial F}{\partial x}\frac{dx}{dx} + \frac{\partial F}{\partial y}\frac{dy}{dx} = 0.

Since $dx/dx = 1$ , solving gives the first formula below; the same argument with $F(x,y,z) = 0$ and $z = z(x,y)$ , differentiating in $x$ with $y$ held fixed so that $\partial y/\partial x = 0$ , gives the second.

Theorem 11.43 (Implicit differentiation formulas). If $F(x,y) = 0$ defines $y$ implicitly as a differentiable function of $x$ and $F_y \neq 0$ , then

\frac{dy}{dx} = -\frac{F_x}{F_y}.

If $F(x,y,z) = 0$ defines $z$ implicitly as a differentiable function of $x$ and $y$ and $F_z \neq 0$ , then

\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}.

When does an equation actually define such a function? The Implicit Function Theorem, proved in real analysis, gives the answer: if $F$ has continuous partials on a disc containing $(a,b)$ , with $F(a,b) = 0$ and $F_y(a,b) \neq 0$ , then near $(a,b)$ the equation $F(x,y) = 0$ defines $y$ as a differentiable function of $x$ , and its derivative is given by the formula. The three-variable version is identical with $F_z(a,b,c)\neq 0$ . The condition $F_y \neq 0$ is exactly the condition that the formula makes sense — and geometrically, that the level curve is not vertical at the point.

Example 11.44 (Implicit differentiation, two variables). Find $y'$ if $x^{3} + y^{3} = 6xy$ .

Solution. Let $F(x,y) = x^{3} + y^{3} - 6xy$ . Then $F_x = 3x^{2} - 6y$ and $F_y = 3y^{2} - 6x$ , so

\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{3x^2 - 6y}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}.

This is the same answer the direct method from the derivatives chapter gives, with the algebra done once and for all. The formula fails where $y^{2} = 2x$ , which is where the curve has a vertical tangent.□

Example 11.45 (Implicit differentiation, three variables). Find $\partial z/\partial x$ and $\partial z/\partial y$ if $x^{3} + y^{3} + z^{3} + 6xyz = 1$ .

Solution. Let $F(x,y,z) = x^{3} + y^{3} + z^{3} + 6xyz - 1$ . Then

\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{3x^2 + 6yz}{3z^2 + 6xy} = -\frac{x^2 + 2yz}{z^2 + 2xy}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{y^2 + 2xz}{z^2 + 2xy}.

The symmetry of the original equation in $x$ and $y$ is reflected in the answers, which is a quick sanity check.□

Intuition. A bug crawls across a heated plate whose temperature is $T(x,y)$ . The bug's position at time $t$ is $(x(t),y(t))$ . How fast does the temperature it feels change? Two effects add up: it is moving east at speed $dx/dt$ through a temperature gradient $T_x$ , contributing $T_x\,dx/dt$ , and it is moving north at speed $dy/dt$ through a gradient $T_y$ , contributing $T_y\,dy/dt$ . The chain rule is the statement that these two effects simply add. If the plate's temperature also drifted with time, a third term $\partial T/\partial t$ would join them — one term per way the input can change.

Pitfall. In case 2, $\partial z/\partial s$ and $\partial z/\partial x$ are not interchangeable symbols. Both are partial derivatives of " $z$ ", but with respect to different sets of variables: $\partial z/\partial x$ holds $y$ fixed, while $\partial z/\partial s$ holds $t$ fixed, and $y$ generally moves when $s$ does. Keep the tree in front of you and never cancel a $\partial x$ against a $\partial x$ as if they were fractions.

Exercises for §11.5 →

11.6Directional derivatives and the gradient vector

The partial derivatives $f_x$ and $f_y$ give the rate of change of $f$ in the directions of the two coordinate axes. There is nothing special about those directions; a hiker can walk north-east as easily as north. The directional derivative measures the rate of change in an arbitrary direction, given by a unit vector $\mathbf{u} = \langle a, b\rangle$ .

Definition 11.46 (Directional derivative). The directional derivative of $f$ at $(x_0,y_0)$ in the direction of the unit vector $\mathbf{u} = \langle a,b\rangle$ is

D_{\mathbf{u}}f(x_0,y_0) = \lim_{h\to 0}\frac{f(x_0 + ha,\; y_0 + hb) - f(x_0,y_0)}{h},

if this limit exists.

The point $(x_0 + ha, y_0 + hb)$ is a step of length $h$ from $(x_0,y_0)$ in the direction $\mathbf{u}$ , so the quotient is the change in $f$ per unit distance travelled in that direction. Taking $\mathbf{u} = \mathbf{i} = \langle 1,0\rangle$ recovers $f_x$ , and $\mathbf{u} = \mathbf{j}$ recovers $f_y$ : the partial derivatives are the two special cases. The definition uses a unit vector so that $h$ really measures distance; with a longer vector the same limit would be scaled by its length.

Computing this limit from scratch every time would be painful. The chain rule turns it into a dot product.

Theorem 11.47 (Computing directional derivatives). If $f$ is differentiable at $(x_0,y_0)$ , then $f$ has a directional derivative there in every direction, and for a unit vector $\mathbf{u} = \langle a,b\rangle$ ,

D_{\mathbf{u}}f(x_0,y_0) = f_x(x_0,y_0)\,a + f_y(x_0,y_0)\,b.

Proof. Define the one-variable function $g(h) = f(x_0 + ha,\, y_0 + hb)$ . By the definition of the derivative,

g'(0) = \lim_{h\to 0}\frac{g(h) - g(0)}{h} = \lim_{h\to 0}\frac{f(x_0 + ha,\, y_0 + hb) - f(x_0,y_0)}{h} = D_{\mathbf{u}}f(x_0,y_0).

On the other hand $g(h) = f(x,y)$ with $x = x_0 + ha$ and $y = y_0 + hb$ , so by the chain rule (case 1),

g'(h) = \frac{\partial f}{\partial x}\frac{dx}{dh} + \frac{\partial f}{\partial y}\frac{dy}{dh} = f_x(x,y)\,a + f_y(x,y)\,b.

Setting $h = 0$ gives $x = x_0$ , $y = y_0$ , and the two expressions for $g'(0)$ agree.∎

The right-hand side is a dot product of $\mathbf{u}$ with the vector of partial derivatives, which deserves a name.

Definition 11.48 (Gradient). The gradient of $f$ is the vector function

\nabla f(x,y) = \langle f_x(x,y),\, f_y(x,y)\rangle = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j}.

With it, the directional derivative reads

D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u}.

Example 11.49 (Directional derivative in a given direction). Find the directional derivative of $f(x,y) = x^{3} - 3xy + 4y^{2}$ at $(1,2)$ in the direction making angle $\pi/6$ with the positive $x$ -axis.

Solution. The unit vector is $\mathbf{u} = \langle\cos\tfrac{\pi}{6}, \sin\tfrac{\pi}{6}\rangle = \langle\tfrac{\sqrt3}{2}, \tfrac12\rangle$ . The partials are $f_x = 3x^{2} - 3y$ and $f_y = -3x + 8y$ , so at $(1,2)$ , $\nabla f = \langle 3 - 6,\; -3 + 16\rangle = \langle -3, 13\rangle$ . Then

D_{\mathbf{u}}f(1,2) = \langle -3, 13\rangle\cdot\left\langle\frac{\sqrt3}{2}, \frac12\right\rangle = \frac{-3\sqrt3 + 13}{2} \approx 3.90.

The value is positive: heading at $30^\circ$ from (1,2), $f$ increases. It should, since the direction has a large component along $\mathbf{j}$ and $f_y = 13$ dominates.□

Example 11.50 (Direction given by a non-unit vector). Find the directional derivative of $f(x,y) = x^{2} y^{3} - 4y$ at $(2,-1)$ in the direction of $\mathbf{v} = 2\mathbf{i} + 5\mathbf{j}$ .

Solution. First the gradient: $\nabla f = \langle 2xy^{3},\; 3x^{2} y^{2} - 4\rangle$ , so $\nabla f(2,-1) = \langle -4, 8\rangle$ . The vector $\mathbf{v}$ is not a unit vector; $|\mathbf{v}| = \sqrt{29}$ , so the direction is $\mathbf{u} = \tfrac{1}{\sqrt{29}}\langle 2, 5\rangle$ . Then

D_{\mathbf{u}}f(2,-1) = \langle -4, 8\rangle\cdot\frac{\langle 2,5\rangle}{\sqrt{29}} = \frac{-8 + 40}{\sqrt{29}} = \frac{32}{\sqrt{29}}.

Forgetting to normalise would give $32$ , wrong by a factor of $\sqrt{29}$ .□

For a function of three variables the definition and theorem read the same with a third component: $\nabla f = \langle f_x, f_y, f_z\rangle$ and $D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u}$ for unit $\mathbf{u}$ in $\mathbb{R}^3$ .

Example 11.51 (A directional derivative in space). If $f(x,y,z) = x\sin(yz)$ , find the directional derivative of $f$ at $(1,3,0)$ in the direction of $\mathbf{v} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ .

Solution. $\nabla f = \langle\sin(yz),\; xz\cos(yz),\; xy\cos(yz)\rangle$ , and at $(1,3,0)$ this is $\langle 0, 0, 3\rangle$ . The unit vector is $\mathbf{u} = \tfrac{1}{\sqrt6}\langle 1, 2, -1\rangle$ , so

D_{\mathbf{u}}f(1,3,0) = \langle 0,0,3\rangle\cdot\frac{\langle 1,2,-1\rangle}{\sqrt6} = -\frac{3}{\sqrt6} = -\frac{\sqrt6}{2}.

□

The formula $D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u}$ answers a natural question: among all directions from a point, which one makes $f$ increase fastest, and how fast?

Theorem 11.52 (Maximising the directional derivative). Suppose $f$ is differentiable and $\nabla f(\mathbf{x}) \neq \mathbf{0}$ . The maximum value of $D_{\mathbf{u}}f(\mathbf{x})$ over all unit vectors $\mathbf{u}$ is $|\nabla f(\mathbf{x})|$ , attained when $\mathbf{u}$ points in the direction of $\nabla f(\mathbf{x})$ . The minimum value is $-|\nabla f(\mathbf{x})|$ , attained in the direction of $-\nabla f(\mathbf{x})$ .

Proof. By the geometric form of the dot product, $D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u} = |\nabla f|\,|\mathbf{u}|\cos\theta = |\nabla f|\cos\theta$ , where $\theta$ is the angle between $\nabla f$ and $\mathbf{u}$ . The factor $\cos\theta$ ranges over $[-1,1]$ , is $1$ exactly when $\theta = 0$ , i.e. $\mathbf{u}$ is parallel to $\nabla f$ , and is $-1$ exactly when $\theta = \pi$ .∎

So the gradient vector points uphill, and its length is the steepness of the steepest ascent. Perpendicular to it, $\cos\theta = 0$ and $f$ does not change at all — that is the direction along the level curve, a fact made precise below.

Intuition. On the hillside from the partial-derivatives section, with eastward slope $0.03$ and northward slope $-0.01$ , the gradient is $\langle 0.03, -0.01\rangle$ . It points somewhat south of east; that is the direction a ball would not roll — it is straight uphill — and the ball rolls the opposite way. The steepest slope is $|\nabla f| = \sqrt{0.0009 + 0.0001} \approx 0.032$ , a bit more than the eastward slope alone, because a little southward tilt adds to it. Walk perpendicular to the gradient and you stay on your contour line, neither climbing nor descending.

Example 11.53 (Steepest ascent). For $f(x,y) = xe^y$ , find the rate of change of $f$ at $P(2,0)$ in the direction from $P$ to $Q(\tfrac12, 2)$ . In what direction does $f$ increase fastest at $P$ , and what is that maximum rate?

Solution. $\nabla f = \langle e^y, xe^y\rangle$ , so $\nabla f(2,0) = \langle 1, 2\rangle$ . The vector from $P$ to $Q$ is $\langle -\tfrac32, 2\rangle$ , of length $\tfrac52$ , so the unit direction is $\mathbf{u} = \langle -\tfrac35, \tfrac45\rangle$ and

D_{\mathbf{u}}f(2,0) = \langle 1,2\rangle\cdot\left\langle -\frac35, \frac45\right\rangle = -\frac35 + \frac85 = 1.

The function increases fastest in the direction of $\nabla f(2,0) = \langle 1,2\rangle$ , and the maximum rate is $|\langle 1,2\rangle| = \sqrt5 \approx 2.24$ . The rate toward $Q$ , being $1$ , is indeed less than $\sqrt5$ .□

Example 11.54 (Temperature in space). The temperature at $(x,y,z)$ is $T = \dfrac{80}{1 + x^{2} + 2y^{2} + 3z^{2}}$ . In which direction does the temperature increase fastest at $(1,1,-2)$ , and what is the maximum rate of increase?

Solution. Write $T = 80\,w^{-1}$ with $w = 1 + x^{2} + 2y^{2} + 3z^{2}$ . By the chain rule,

\nabla T = -\frac{80}{w^2}\nabla w = -\frac{80}{w^2}\langle 2x, 4y, 6z\rangle = -\frac{160}{w^2}\langle x, 2y, 3z\rangle.

At $(1,1,-2)$ , $w = 1 + 1 + 2 + 12 = 16$ , so $-160/w^{2} = -160/256 = -\tfrac58$ and

\nabla T(1,1,-2) = -\frac58\langle 1, 2, -6\rangle = \frac58\langle -1, -2, 6\rangle.

The temperature increases fastest in the direction of $\langle -1,-2,6\rangle$ — toward the origin, where the denominator is smallest, as it should — and the maximum rate is $\tfrac58|\langle -1,-2,6\rangle| = \tfrac58\sqrt{41} \approx 4$ degrees per unit distance.□

The claim that the gradient is perpendicular to the level curve is now a two-line consequence of the chain rule, and it extends to level surfaces in space.

Theorem 11.55 (The gradient is normal to level sets). Let $F$ be differentiable and let $S$ be the level surface $F(x,y,z) = k$ . If $P = (x_0,y_0,z_0)$ lies on $S$ and $\nabla F(x_0,y_0,z_0)\neq\mathbf{0}$ , then $\nabla F(x_0,y_0,z_0)$ is perpendicular to the tangent vector of every smooth curve on $S$ through $P$ . The same holds in the plane: $\nabla f(x_0,y_0)$ is perpendicular to the level curve $f(x,y) = k$ through $(x_0,y_0)$ .

Proof. Let $\mathbf{r}(t) = \langle x(t), y(t), z(t)\rangle$ be any smooth curve lying on $S$ , with $\mathbf{r}(t_0) = P$ . Because the curve lies on $S$ , $F(x(t), y(t), z(t)) = k$ for every $t$ . Differentiate this identity with respect to $t$ by the chain rule:

F_x\frac{dx}{dt} + F_y\frac{dy}{dt} + F_z\frac{dz}{dt} = 0, \qquad\text{that is,}\qquad \nabla F\cdot\mathbf{r}'(t) = 0.

At $t = t_0$ this says $\nabla F(P)\cdot\mathbf{r}'(t_0) = 0$ : the gradient is perpendicular to the tangent vector at $P$ . The planar statement is the same argument with one fewer coordinate.∎

Since every tangent vector at $P$ is perpendicular to $\nabla F(P)$ , they all lie in one plane, the tangent plane to $S$ at $P$ , whose normal vector is $\nabla F(P)$ . Writing the plane through $P$ with that normal gives

F_x(x_0,y_0,z_0)(x - x_0) + F_y(x_0,y_0,z_0)(y - y_0) + F_z(x_0,y_0,z_0)(z - z_0) = 0,

and the normal line to $S$ at $P$ — the line through $P$ in the direction of $\nabla F(P)$ — has symmetric equations

\frac{x - x_0}{F_x(x_0,y_0,z_0)} = \frac{y - y_0}{F_y(x_0,y_0,z_0)} = \frac{z - z_0}{F_z(x_0,y_0,z_0)}.

If the surface is given as a graph $z = f(x,y)$ , take $F(x,y,z) = f(x,y) - z$ ; then $F_x = f_x$ , $F_y = f_y$ , $F_z = -1$ , and the tangent plane equation becomes $f_x(x - x_0) + f_y(y - y_0) - (z - z_0) = 0$ , exactly the formula derived from slices in the previous section. The two approaches agree, and the level-surface one is more general because it handles surfaces that are not graphs.

Example 11.56 (Tangent plane and normal line to an ellipsoid). Find the tangent plane and the normal line to the ellipsoid $\dfrac{x^{2}}{4} + y^{2} + \dfrac{z^{2}}{9} = 3$ at the point $(-2, 1, -3)$ .

Solution. The ellipsoid is the level surface $F = 3$ of $F(x,y,z) = \tfrac{x^{2}}{4} + y^{2} + \tfrac{z^{2}}{9}$ . Its gradient is

\nabla F = \left\langle\frac{x}{2},\; 2y,\; \frac{2z}{9}\right\rangle, \qquad \nabla F(-2,1,-3) = \left\langle -1,\; 2,\; -\frac23\right\rangle.

Tangent plane:

-(x + 2) + 2(y - 1) - \frac23(z + 3) = 0, \qquad\text{or, multiplying by } {-3},\qquad 3x - 6y + 2z + 18 = 0.

Normal line:

\frac{x + 2}{-1} = \frac{y - 1}{2} = \frac{z + 3}{-2/3}.

Check that the point lies on the plane: $3(-2) - 6(1) + 2(-3) + 18 = -6 - 6 - 6 + 18 = 0$ .□

Example 11.57 (Gradient perpendicular to a level curve). Verify directly that $\nabla f$ is perpendicular to the level curve of $f(x,y) = x^{2} + y^{2}$ through $(3,4)$ .

Solution. The level curve through $(3,4)$ is $x^{2} + y^{2} = 25$ , a circle of radius $5$ . The gradient is $\nabla f = \langle 2x, 2y\rangle$ , so $\nabla f(3,4) = \langle 6, 8\rangle$ , which points radially outward from the origin — visibly perpendicular to the circle. For a check by computation, the circle can be parametrised as $\mathbf{r}(t) = \langle 5\cos t, 5\sin t\rangle$ ; at the point $(3,4)$ the tangent vector is $\mathbf{r}'(t) = \langle -5\sin t, 5\cos t\rangle = \langle -4, 3\rangle$ , and

\langle 6, 8\rangle\cdot\langle -4, 3\rangle = -24 + 24 = 0.

□

Pitfall. The gradient of $f(x,y)$ is a vector in the plane, not on the surface $z = f(x,y)$ ; it lives in the domain and points along the direction of steepest ascent as read off the contour map. When you want a normal vector to the surface itself, use the three-dimensional gradient $\nabla F = \langle f_x, f_y, -1\rangle$ of $F = f(x,y) - z$ . Confusing the two produces a "normal" that is missing its $z$ -component.

Exercises for §11.6 →

11.7Maximum and minimum values

Definition 11.58 (Local and absolute extrema). A function of two variables has a local maximum at $(a,b)$ if $f(x,y)\le f(a,b)$ for all $(x,y)$ in some disc centred at $(a,b)$ ; the number $f(a,b)$ is a local maximum value. If the inequality holds for all $(x,y)$ in the domain of $f$ , then $f$ has an absolute maximum at $(a,b)$ . Local and absolute minima are defined with $\ge$ .

Theorem 11.59 (Fermat's Theorem for two variables). If $f$ has a local maximum or minimum at $(a,b)$ and the first-order partial derivatives of $f$ exist there, then

f_x(a,b) = 0 \qquad\text{and}\qquad f_y(a,b) = 0.

Proof. Let $g(x) = f(x,b)$ . If $f$ has a local maximum at $(a,b)$ then $g$ has a local maximum at $a$ , so by the one-variable Fermat's Theorem $g'(a) = 0$ ; but $g'(a) = f_x(a,b)$ . Applying the same argument to $h(y) = f(a,y)$ gives $f_y(a,b) = 0$ .∎

Geometrically, at a local extremum the tangent plane $z - z_0 = f_x(x - a) + f_y(y - b)$ is horizontal. As in one variable the theorem runs in one direction only: the partials vanishing is necessary for an extremum, not sufficient.

Definition 11.60 (Critical point). A point $(a,b)$ is a critical point of $f$ if $f_x(a,b) = 0$ and $f_y(a,b) = 0$ , or if one of these partial derivatives does not exist.

Example 11.61 (A critical point that is a minimum). Find the extreme values of $f(x,y) = x^{2} + y^{2} - 2x - 6y + 14$ .

Solution. $f_x = 2x - 2$ and $f_y = 2y - 6$ vanish only at $(1,3)$ : the single critical point. Completing the squares,

f(x,y) = (x - 1)^2 + (y - 3)^2 + 4,

and since squares are non-negative, $f(x,y)\ge 4 = f(1,3)$ for every $(x,y)$ . So $f(1,3) = 4$ is a local minimum and in fact the absolute minimum; the graph is a paraboloid with vertex $(1,3,4)$ . There is no maximum, since $f\to\infty$ far from the origin.□

Example 11.62 (A critical point that is neither). Find the extreme values of $f(x,y) = y^{2} - x^{2}$ .

Solution. $f_x = -2x$ and $f_y = 2y$ , so the only critical point is $(0,0)$ , where $f = 0$ . Along the $x$ -axis $f(x,0) = -x^{2} < 0$ for $x \neq 0$ , so $(0,0)$ is not a minimum; along the $y$ -axis $f(0,y) = y^{2} > 0$ , so it is not a maximum. The origin is a saddle point: a minimum along one line and a maximum along another. The graph is the hyperbolic paraboloid, shaped like a horse's saddle, and $f$ has no extreme values.□

Deciding which of these three things happens at a critical point is the job of the second derivatives test. It is the two-variable analogue of the second derivative test, and the quantity that decides is not $f_{xx}$ alone but a combination of all three second partials.

Theorem 11.63 (Second Derivatives Test). Suppose the second partial derivatives of $f$ are continuous on a disc centred at $(a,b)$ , and suppose $f_x(a,b) = 0$ and $f_y(a,b) = 0$ . Let

D = D(a,b) = f_{xx}(a,b)\,f_{yy}(a,b) - \big[f_{xy}(a,b)\big]^2.

If $D > 0$ and $f_{xx}(a,b) > 0$ , then $f(a,b)$ is a local minimum.
If $D > 0$ and $f_{xx}(a,b) < 0$ , then $f(a,b)$ is a local maximum.
If $D < 0$ , then $(a,b)$ is a saddle point of $f$ .
If $D = 0$ , the test gives no information.

Proof. Here is the idea, with the details of the error term left to real analysis. Write $A = f_{xx}(a,b)$ , $B = f_{xy}(a,b)$ , $C = f_{yy}(a,b)$ . Since the first partials vanish, the change in $f$ for a small step $(h,k)$ is governed by the second-order Taylor term, a quadratic form:

f(a + h,\, b + k) - f(a,b) \approx \frac12\left(Ah^2 + 2Bhk + Ck^2\right) = \frac12 Q(h,k),

with an error small compared with $h^{2} + k^{2}$ because the second partials are continuous. So the sign of $f(a+h,b+k) - f(a,b)$ for small steps is the sign of $Q$ , and the question is whether $Q$ is always positive, always negative, or takes both signs. Suppose $A\neq 0$ and complete the square in $h$ :

Q(h,k) = A\left(h + \frac{B}{A}k\right)^2 + \frac{AC - B^2}{A}k^2 = A\left(h + \frac{B}{A}k\right)^2 + \frac{D}{A}k^2.

If $D > 0$ , both coefficients $A$ and $D/A$ have the sign of $A$ , so $Q$ has the sign of $A$ for every $(h,k)\neq(0,0)$ : a minimum when $A > 0$ , a maximum when $A < 0$ . (When $D > 0$ we automatically have $A \neq 0$ , since $AC > B^{2} \ge 0$ .) If $D < 0$ , choose $k = 0$ : then $Q = Ah^{2}$ has the sign of $A$ ; now choose $h = -\tfrac{B}{A}k$ : then $Q = \tfrac{D}{A}k^{2}$ has the opposite sign. Two directions with opposite signs of $Q$ is a saddle. If $A = 0$ and $D < 0$ then $B\neq 0$ and $Q = k(2Bh + Ck)$ , which also changes sign. When $D = 0$ the quadratic form vanishes along some direction and the ignored higher-order terms decide, which is why the test is silent.∎

For a unit vector $\langle h,k\rangle$ the quantity $Q(h,k)$ is the second directional derivative of $f$ in that direction, so the argument says: at a minimum every second directional derivative is positive, at a maximum every one is negative, and at a saddle some are positive and some negative. The determinant $D$ is what makes the sign of $Q$ decidable from three numbers instead of infinitely many directions. Note the symmetry: when $D > 0$ , $f_{xx}$ and $f_{yy}$ have the same sign, so either may be used in the test.

Example 11.64 (Classifying several critical points). Find and classify the critical points of $f(x,y) = x^{4} + y^{4} - 4xy + 1$ .

Solution. Setting the partials to zero,

f_x = 4x^3 - 4y = 0, \qquad f_y = 4y^3 - 4x = 0,

so $y = x^{3}$ and $x = y^{3} = x^{9}$ . Then $x^{9} - x = x(x^{8} - 1) = 0$ gives $x = 0$ or $x = \pm 1$ (the other roots of $x^{8} = 1$ are not real). The critical points are $(0,0)$ , $(1,1)$ and $(-1,-1)$ .

The second partials are $f_{xx} = 12x^{2}$ , $f_{yy} = 12y^{2}$ , $f_{xy} = -4$ , so

D(x,y) = 144x^2 y^2 - 16.

At $(0,0)$ : $D = -16 < 0$ , a saddle point. At $(1,1)$ : $D = 128 > 0$ and $f_{xx} = 12 > 0$ , a local minimum with value $f(1,1) = 1 + 1 - 4 + 1 = -1$ . At $(-1,-1)$ : $D = 128 > 0$ and $f_{xx} = 12 > 0$ , another local minimum, also with value $-1$ . Since $f\to\infty$ as $|(x,y)|\to\infty$ , these two local minima are the absolute minimum.□

Example 11.65 (When $D = 0$ ). Show that $f(x,y) = x^{4} + y^{4}$ , $g(x,y) = -(x^{4} + y^{4})$ and $h(x,y) = x^{4} - y^{4}$ all have $D = 0$ at the origin, yet behave differently there.

Solution. For each, the first partials $\pm 4x^{3}, \pm 4y^{3}$ vanish at the origin, so it is a critical point. The second partials $\pm 12x^{2}$ , $\pm 12y^{2}$ and $0$ all vanish at $(0,0)$ , so $D = 0\cdot 0 - 0 = 0$ in each case, and the test says nothing.

Direct inspection settles it: $f(x,y) = x^{4} + y^{4} \ge 0 = f(0,0)$ , so the origin is a minimum of $f$ ; likewise it is a maximum of $g$ ; and $h(x,0) = x^{4} > 0$ while $h(0,y) = -y^{4} < 0$ , so it is a saddle of $h$ . Three functions with identical second-order data and three different answers — the $D = 0$ case genuinely cannot be decided by second derivatives.□

Example 11.66 (Shortest distance to a plane). Find the shortest distance from the point $(1,0,-2)$ to the plane $x + 2y + z = 4$ .

Solution. A point on the plane has $z = 4 - x - 2y$ , so its squared distance to $(1,0,-2)$ is

f(x,y) = (x - 1)^2 + y^2 + (6 - x - 2y)^2.

Minimising the square avoids a square root and gives the same minimiser. The partials are

f_x = 2(x - 1) - 2(6 - x - 2y) = 4x + 4y - 14, \qquad f_y = 2y - 4(6 - x - 2y) = 4x + 10y - 24.

Solving $4x + 4y = 14$ and $4x + 10y = 24$ : subtracting, $6y = 10$ , so $y = \tfrac53$ and $x = \tfrac{14 - 20/3}{4} = \tfrac{11}{6}$ . At this point $f_{xx} = 4$ , $f_{yy} = 10$ , $f_{xy} = 4$ , so $D = 40 - 16 = 24 > 0$ with $f_{xx} > 0$ : a local minimum, and by the geometry the absolute one. The squared distance is

f\!\left(\tfrac{11}{6}, \tfrac53\right) = \left(\tfrac56\right)^2 + \left(\tfrac53\right)^2 + \left(\tfrac56\right)^2 = \frac{25}{36} + \frac{100}{36} + \frac{25}{36} = \frac{150}{36} = \frac{25}{6},

so the distance is $\sqrt{25/6} = \dfrac{5\sqrt6}{6}$ . The point-to-plane formula $|1 + 0 - 2 - 4|/\sqrt{1 + 4 + 1} = 5/\sqrt6$ agrees.□

Example 11.67 (Largest box with a given surface area). A rectangular box without a lid is to be made from $12$ m $^2$ of cardboard. Find the maximum volume of such a box.

Solution. Let the box have base $x$ by $y$ and height $z$ . Then $V = xyz$ and the area of the base plus four sides is $xy + 2xz + 2yz = 12$ . Solving the constraint for $z = \dfrac{12 - xy}{2(x + y)}$ ,

V(x,y) = xy\cdot\frac{12 - xy}{2(x + y)} = \frac{12xy - x^2 y^2}{2(x + y)}, \qquad x > 0,\ y > 0,\ xy < 12.

By the quotient rule, after simplification,

V_x = \frac{y^2(12 - 2xy - x^2)}{2(x + y)^2}, \qquad V_y = \frac{x^2(12 - 2xy - y^2)}{2(x + y)^2}.

For $x, y > 0$ both vanish only when $12 - 2xy - x^{2} = 0$ and $12 - 2xy - y^{2} = 0$ . Subtracting, $x^{2} = y^{2}$ , so $x = y$ , and then $12 - 3x^{2} = 0$ gives $x = y = 2$ and $z = \dfrac{12 - 4}{8} = 1$ . The single critical point is $(2,2)$ with $V = 4$ m $^3$ . That it is a maximum is clear from the physical situation — $V\to 0$ as the box becomes very flat or very thin — and the second derivatives test would confirm it. The box is $2 \times 2 \times 1$ .□

Intuition. Water settles at the bottom of a bowl. Which way is "down" at a critical point? Nowhere — that is what the vanishing partials say. Whether you are at the bottom of a bowl, the top of a dome or the middle of a mountain pass is a question about curvature, and $D$ answers it. A pass has the fingerprint $D < 0$ : the ground curves up along the ridge and down along the road. A bowl and a dome both have $D > 0$ , curving the same way in every direction, and $f_{xx}$ says which.

Local extrema are found at critical points; absolute extrema need one more ingredient. In one variable a continuous function on a closed interval $[a,b]$ attains its extremes either at a critical point or at an endpoint. The two-variable analogue of a closed interval is a closed, bounded set: closed meaning it contains all its boundary points, bounded meaning it fits inside some disc. Discs, rectangles and triangles with their edges included are closed and bounded; the open disc $x^{2} + y^{2} < 1$ is not closed, and the half-plane $y\ge 0$ is not bounded.

Theorem 11.68 (Extreme Value Theorem for two variables). If $f$ is continuous on a closed, bounded set $D$ in $\mathbb{R}^2$ , then $f$ attains an absolute maximum value $f(x_1,y_1)$ and an absolute minimum value $f(x_2,y_2)$ at some points $(x_1,y_1)$ and $(x_2,y_2)$ in $D$ .

The proof is real analysis (compactness), but the consequence is a procedure. An absolute extremum occurring in the interior of $D$ is also a local extremum, hence at a critical point by Fermat's Theorem; otherwise it occurs on the boundary. So:

Method 11.69 (Absolute extrema on a closed bounded set).

Find the values of $f$ at the critical points of $f$ in the interior of $D$ .
Find the extreme values of $f$ on the boundary of $D$ . Parametrise each boundary piece as a curve, so that $f$ restricted to it becomes a function of one variable on a closed interval, and apply the one-variable closed-interval method.
The largest of the values from steps 1 and 2 is the absolute maximum; the smallest is the absolute minimum.

Example 11.70 (Absolute extrema on a rectangle). Find the absolute maximum and minimum values of $f(x,y) = x^{2} - 2xy + 2y$ on the rectangle $D = \{(x,y) : 0\le x\le 3,\ 0\le y\le 2\}$ .

Solution. Step 1. $f_x = 2x - 2y$ and $f_y = -2x + 2$ vanish together only at $(1,1)$ , which is inside $D$ , with $f(1,1) = 1$ .

Step 2. The boundary has four edges. On the bottom edge $y = 0$ , $0\le x\le 3$ : $f = x^{2}$ , increasing, with minimum $0$ at $(0,0)$ and maximum $9$ at $(3,0)$ . On the right edge $x = 3$ , $0\le y\le 2$ : $f = 9 - 6y + 2y = 9 - 4y$ , decreasing from $9$ at $(3,0)$ to $1$ at $(3,2)$ . On the top edge $y = 2$ , $0\le x\le 3$ : $f = x^{2} - 4x + 4 = (x - 2)^2$ , with minimum $0$ at $(2,2)$ and maximum $4$ at $(0,2)$ . On the left edge $x = 0$ , $0\le y\le 2$ : $f = 2y$ , from $0$ at $(0,0)$ to $4$ at $(0,2)$ .

Step 3. Collecting the candidates $1, 0, 9, 1, 0, 4, 0, 4$ : the absolute maximum is $9$ at $(3,0)$ and the absolute minimum is $0$ , attained at both $(0,0)$ and $(2,2)$ .□

Example 11.71 (Absolute extrema on a triangle). Find the absolute extrema of $f(x,y) = 4x^{2} + y^{2} - 4x - 2y$ on the closed triangular region $T$ with vertices $(0,0)$ , $(2,0)$ and $(0,2)$ .

Solution. Step 1. $f_x = 8x - 4 = 0$ and $f_y = 2y - 2 = 0$ give the single critical point $(\tfrac12, 1)$ . It lies inside $T$ , since $\tfrac12 + 1 < 2$ , and $f(\tfrac12, 1) = 1 + 1 - 2 - 2 = -2$ .

Step 2. The boundary has three edges. On the edge $y = 0$ , $0\le x\le 2$ : $g(x) = 4x^{2} - 4x$ , with $g'(x) = 8x - 4 = 0$ at $x = \tfrac12$ , so the candidates are $g(0) = 0$ , $g(\tfrac12) = -1$ , $g(2) = 8$ . On the edge $x = 0$ , $0\le y\le 2$ : $h(y) = y^{2} - 2y$ , with $h'(y) = 0$ at $y = 1$ , giving $h(0) = 0$ , $h(1) = -1$ , $h(2) = 0$ . On the hypotenuse $x + y = 2$ , parametrise by $y = 2 - x$ with $0\le x\le 2$ :

k(x) = 4x^2 + (2 - x)^2 - 4x - 2(2 - x) = 5x^2 - 6x,

with $k'(x) = 10x - 6 = 0$ at $x = \tfrac35$ , so the candidates are $k(0) = 0$ , $k(\tfrac35) = \tfrac{9}{5} - \tfrac{18}{5} = -\tfrac95$ , $k(2) = 8$ .

Step 3. The candidate values are $-2, 0, -1, 8, -1, 0, -\tfrac95$ . The absolute maximum is $8$ at the vertex $(2,0)$ ; the absolute minimum is $-2$ at the interior point $(\tfrac12, 1)$ . Sanity check: $f_{xx} = 8$ , $f_{yy} = 2$ , $f_{xy} = 0$ , so $D = 16 > 0$ and the interior critical point is indeed a local minimum.□

Example 11.72 (Absolute extrema on a disc). Find the absolute extrema of $f(x,y) = 2x^{2} + y^{2} - 2x$ on the closed disc $x^{2} + y^{2}\le 1$ .

Solution. Step 1. $f_x = 4x - 2 = 0$ and $f_y = 2y = 0$ give the critical point $(\tfrac12, 0)$ , inside the disc, with $f(\tfrac12,0) = \tfrac12 - 1 = -\tfrac12$ .

Step 2. Parametrise the boundary circle by $x = \cos t$ , $y = \sin t$ , $0\le t\le 2\pi$ :

g(t) = 2\cos^2 t + \sin^2 t - 2\cos t = \cos^2 t + 1 - 2\cos t = (1 - \cos t)^2,

using $\sin^{2} t = 1 - \cos^{2} t$ . Since $\cos t$ ranges over $[-1,1]$ , $(1 - \cos t)^2$ ranges over $[0, 4]$ : minimum $0$ at $t = 0$ , the point $(1,0)$ , and maximum $4$ at $t = \pi$ , the point $(-1,0)$ .

Step 3. Comparing $-\tfrac12, 0, 4$ : the absolute maximum is $4$ at $(-1,0)$ and the absolute minimum is $-\tfrac12$ at $(\tfrac12, 0)$ . The next section redoes the boundary step with Lagrange multipliers, which avoids the parametrisation.□

Pitfall. The second derivatives test classifies interior critical points only. A point on the boundary of a closed region can be an absolute extremum with no partial derivative vanishing at all — in the rectangle example the maximum $9$ sits at a corner where $f_x = 6$ and $f_y = -4$ . Never apply the test to boundary candidates, and never skip the boundary because the interior critical point "looks like" the answer.

Exercises for §11.7 →

11.8Lagrange multipliers

The box example of the previous section was a constrained problem: maximise $V = xyz$ subject to $xy + 2xz + 2yz = 12$ . We solved it by eliminating $z$ , which was possible only because the constraint could be solved for one variable and led to messy quotients even so. Lagrange's method handles constraints without solving them.

Consider maximising $f(x,y)$ subject to $g(x,y) = k$ . Draw the constraint curve $g = k$ and, on the same axes, the level curves $f = c$ for various $c$ . Maximising $f$ on the constraint means finding the largest $c$ for which the level curve $f = c$ still meets the curve $g = k$ . Picture $c$ increasing: the level curves sweep across the constraint curve, and at the last moment of contact the two curves touch rather than cross — they have a common tangent line. Their normal vectors are therefore parallel, and the normal vectors are the gradients: $\nabla f = \lambda\nabla g$ for some scalar $\lambda$ .

The same argument works in space, and there it can be made precise with the tools of the gradient section. Suppose $f(x,y,z)$ has an extreme value at $P = (x_0,y_0,z_0)$ on the level surface $S$ given by $g = k$ . Let $\mathbf{r}(t)$ be any smooth curve on $S$ passing through $P$ at $t = t_0$ , and let $h(t) = f(\mathbf{r}(t))$ . Since $f$ is extreme at $P$ among points of $S$ , the one-variable function $h$ has an extremum at $t_0$ , so $h'(t_0) = 0$ ; by the chain rule,

0 = h'(t_0) = \nabla f(P)\cdot\mathbf{r}'(t_0).

So $\nabla f(P)$ is perpendicular to the tangent vector of every curve on $S$ through $P$ — it is normal to $S$ at $P$ . But we know that $\nabla g(P)$ is normal to $S$ at $P$ too. Two vectors normal to the same surface at the same point are parallel, so if $\nabla g(P)\neq\mathbf{0}$ there is a number $\lambda$ with $\nabla f(P) = \lambda\nabla g(P)$ .

Method 11.73 (Method of Lagrange multipliers). To find the maximum and minimum values of $f(x,y,z)$ subject to the constraint $g(x,y,z) = k$ , assuming these exist and $\nabla g\neq\mathbf{0}$ on the constraint surface:

Find all values of $x, y, z$ and $\lambda$ such that

\nabla f(x,y,z) = \lambda\nabla g(x,y,z) \qquad\text{and}\qquad g(x,y,z) = k.

Evaluate $f$ at all the points found in step 1. The largest value is the maximum of $f$ on the constraint; the smallest is the minimum.

The number $\lambda$ is called a Lagrange multiplier. For two variables drop $z$ throughout.

Step 1 is a system of four equations, the three components of the vector equation and the constraint, in the four unknowns $x, y, z, \lambda$ . It is rarely linear; the usual technique is to eliminate $\lambda$ by dividing or cross-multiplying equations, watching for the cases where a divisor might be zero. The multiplier itself is often not needed, although it has a meaning: it is the rate at which the extreme value changes as the constraint level $k$ is relaxed.

Intuition. A fence runs across a hillside and you may walk only along the fence. Where is the highest point you can reach? Not necessarily where the fence crosses the summit — it may not — but at the spot where the fence runs along a contour line. Anywhere the fence cuts across contours you could gain height by walking on; only where fence and contour are tangent is there nowhere higher to go. Tangent curves share a normal direction, and the normals are $\nabla g$ (to the fence) and $\nabla f$ (to the contour). That is the equation $\nabla f = \lambda\nabla g$ .

Example 11.74 (A first Lagrange problem). Find the minimum value of $f(x,y) = x^{2} + y^{2}$ subject to $x + y = 1$ .

Solution. Take $g(x,y) = x + y$ . Then $\nabla f = \langle 2x, 2y\rangle$ and $\nabla g = \langle 1, 1\rangle$ , so the equations are

2x = \lambda, \qquad 2y = \lambda, \qquad x + y = 1.

The first two give $x = y$ , and the constraint then gives $x = y = \tfrac12$ . The only candidate is $(\tfrac12, \tfrac12)$ , with $f = \tfrac12$ . Since $f$ is the squared distance from the origin and grows without bound along the line, this candidate must be the minimum; there is no maximum. Geometrically, $\tfrac{1}{2}$ is the square of the distance $1/\sqrt2$ from the origin to the line, as it should be.□

Example 11.75 (The box again, by Lagrange multipliers). A rectangular box without a lid is to be made from $12$ m $^2$ of cardboard. Find the maximum volume by the method of Lagrange multipliers.

Solution. Maximise $V = xyz$ subject to $g(x,y,z) = xy + 2xz + 2yz = 12$ . The equations $\nabla V = \lambda\nabla g$ and $g = 12$ are

yz = \lambda(y + 2z), \qquad xz = \lambda(x + 2z), \qquad xy = \lambda(2x + 2y), \qquad xy + 2xz + 2yz = 12.

Multiply the first equation by $x$ , the second by $y$ and the third by $z$ ; each left side becomes $xyz$ :

xyz = \lambda(xy + 2xz) = \lambda(xy + 2yz) = \lambda(2xz + 2yz).

Now $\lambda\neq 0$ (otherwise $yz = xz = xy = 0$ , forcing two of the dimensions to be $0$ , which contradicts the constraint). So $xy + 2xz = xy + 2yz$ , giving $x = y$ since $z\neq 0$ , and $xy + 2xz = 2xz + 2yz$ , giving $xy = 2yz$ , so $x = 2z$ . Then $x = y = 2z$ , and the constraint reads $4z^{2} + 4z^{2} + 4z^{2} = 12$ , so $z = 1$ , $x = y = 2$ . The maximum volume is $V = 2\cdot 2\cdot 1 = 4$ m $^3$ , as before, with no quotient rule in sight.□

Example 11.76 (Extreme values on a circle). Find the extreme values of $f(x,y) = x^{2} + 2y^{2}$ on the circle $x^{2} + y^{2} = 1$ .

Solution. With $g(x,y) = x^{2} + y^{2}$ , the equations are

2x = 2\lambda x, \qquad 4y = 2\lambda y, \qquad x^2 + y^2 = 1.

The first says $x(1 - \lambda) = 0$ , so $x = 0$ or $\lambda = 1$ . If $x = 0$ , the constraint gives $y = \pm 1$ : the points $(0, \pm 1)$ . If $\lambda = 1$ , the second equation gives $4y = 2y$ , so $y = 0$ , and the constraint gives $x = \pm 1$ : the points $(\pm 1, 0)$ . Evaluating,

f(0,\pm 1) = 2, \qquad f(\pm 1, 0) = 1.

The maximum on the circle is $2$ at $(0,\pm 1)$ and the minimum is $1$ at $(\pm 1, 0)$ . On the contour map, the level curves of $f$ are ellipses that are wider than tall; the ellipse $f = 1$ touches the unit circle at $(\pm 1, 0)$ from inside and the ellipse $f = 2$ touches it at $(0,\pm 1)$ from outside.□

Example 11.77 (Combining an interior search with Lagrange multipliers). Find the extreme values of $f(x,y) = x^{2} + 2y^{2}$ on the closed disc $x^{2} + y^{2}\le 1$ .

Solution. Interior: $f_x = 2x$ and $f_y = 4y$ vanish only at $(0,0)$ , inside the disc, with $f(0,0) = 0$ .

Boundary: by the previous example, the extreme values of $f$ on the circle $x^{2} + y^{2} = 1$ are $2$ and $1$ .

Comparing $0, 1, 2$ : the absolute maximum on the disc is $2$ at $(0,\pm1)$ and the absolute minimum is $0$ at $(0,0)$ . The value $1$ is only the minimum on the circle — a candidate that lost. This is the pattern for any closed bounded region with a smooth boundary: critical points inside, Lagrange on the edge, compare.□

Example 11.78 (Closest and farthest points on a sphere). Find the points on the sphere $x^{2} + y^{2} + z^{2} = 4$ that are closest to and farthest from the point $(3,1,-1)$ .

Solution. The distance squared from $(x,y,z)$ to $(3,1,-1)$ is $f = (x - 3)^2 + (y - 1)^2 + (z + 1)^2$ , to be optimised subject to $g = x^{2} + y^{2} + z^{2} = 4$ . The equations are

2(x - 3) = 2\lambda x, \qquad 2(y - 1) = 2\lambda y, \qquad 2(z + 1) = 2\lambda z, \qquad x^2 + y^2 + z^2 = 4.

Solving each of the first three for the coordinate ( $\lambda\neq 1$ , since $\lambda = 1$ would force $-3 = 0$ ):

x = \frac{3}{1 - \lambda}, \qquad y = \frac{1}{1 - \lambda}, \qquad z = \frac{-1}{1 - \lambda}.

Substituting in the constraint, $\dfrac{9 + 1 + 1}{(1 - \lambda)^2} = 4$ , so $(1 - \lambda)^2 = \tfrac{11}{4}$ and $1 - \lambda = \pm\tfrac{\sqrt{11}}{2}$ . The two candidate points are

\left(\frac{6}{\sqrt{11}}, \frac{2}{\sqrt{11}}, -\frac{2}{\sqrt{11}}\right) \qquad\text{and}\qquad \left(-\frac{6}{\sqrt{11}}, -\frac{2}{\sqrt{11}}, \frac{2}{\sqrt{11}}\right).

The first is a positive multiple of $(3,1,-1)$ and so lies on the same side of the origin: it is the closest point, at distance $\sqrt{11} - 2$ . The second is the farthest, at distance $\sqrt{11} + 2$ . Both are on the line through the origin and $(3,1,-1)$ , which is what geometry predicts.□

A constraint curve in space is usually given as the intersection of two surfaces, $g = k$ and $h = c$ . At an extremum $P$ of $f$ on the curve, the same chain-rule argument shows $\nabla f(P)$ is perpendicular to the tangent vector of the curve. But $\nabla g(P)$ and $\nabla h(P)$ are both perpendicular to that tangent vector too, since the curve lies on both surfaces, so all three gradients lie in the plane normal to the curve at $P$ . If $\nabla g$ and $\nabla h$ are not parallel, they span that plane, and $\nabla f(P)$ is a combination of them:

\nabla f = \lambda\nabla g + \mu\nabla h.

Together with $g = k$ and $h = c$ this is five equations in the five unknowns $x, y, z, \lambda, \mu$ .

Example 11.79 (Two constraints). Find the maximum value of $f(x,y,z) = x + 2y + 3z$ on the curve of intersection of the plane $x - y + z = 1$ and the cylinder $x^{2} + y^{2} = 1$ .

Solution. Take $g = x - y + z$ and $h = x^{2} + y^{2}$ . Since $\nabla f = \langle 1, 2, 3\rangle$ , $\nabla g = \langle 1, -1, 1\rangle$ and $\nabla h = \langle 2x, 2y, 0\rangle$ , the vector equation $\nabla f = \lambda\nabla g + \mu\nabla h$ reads

1 = \lambda + 2\mu x, \qquad 2 = -\lambda + 2\mu y, \qquad 3 = \lambda.

The third gives $\lambda = 3$ at once; then $2\mu x = -2$ and $2\mu y = 5$ , so $x = -1/\mu$ and $y = 5/(2\mu)$ (note $\mu\neq 0$ ). The cylinder constraint gives

\frac{1}{\mu^2} + \frac{25}{4\mu^2} = 1 \quad\Longrightarrow\quad \mu^2 = \frac{29}{4}, \quad \mu = \pm\frac{\sqrt{29}}{2}.

So $x = \mp\dfrac{2}{\sqrt{29}}$ , $y = \pm\dfrac{5}{\sqrt{29}}$ , and from the plane $z = 1 - x + y = 1 \pm\dfrac{7}{\sqrt{29}}$ . The corresponding values are

f = x + 2y + 3z = \mp\frac{2}{\sqrt{29}} \pm\frac{10}{\sqrt{29}} + 3 \pm\frac{21}{\sqrt{29}} = 3 \pm\frac{29}{\sqrt{29}} = 3\pm\sqrt{29}.

The intersection curve is an ellipse, closed and bounded, so the Extreme Value Theorem guarantees both extremes exist: the maximum is $3 + \sqrt{29}$ and the minimum is $3 - \sqrt{29}$ .□

Pitfall. Lagrange's equations produce candidates, not classifications, and the recipe only works when an extreme value is known to exist — on a closed bounded constraint set by the Extreme Value Theorem, or by a geometric argument otherwise. On an unbounded constraint such as a line or a hyperbola, a lone candidate may be a minimum, a maximum, or neither, and the values must be compared with the behaviour of $f$ far out. Also check any point of the constraint where $\nabla g = \mathbf{0}$ separately, since the method's derivation assumed $\nabla g\neq\mathbf{0}$ there.

Summary. A limit $\lim_{(x,y)\to(a,b)} f(x,y) = L$ demands the same value along every approach, so two paths with different limits prove it does not exist. If $f_{xy}$ and $f_{yx}$ are continuous on a disc containing $(a,b)$ , then $f_{xy}(a,b) = f_{yx}(a,b)$ (Clairaut). If $f_x$ and $f_y$ exist near $(a,b)$ and are continuous there, $f$ is differentiable at $(a,b)$ , hence continuous, and

L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)

is both the tangent plane and the linear approximation. The chain rule chains one factor per intermediate variable, $\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$ , and implicitly $\frac{dy}{dx} = -F_x/F_y$ and $\frac{\partial z}{\partial x} = -F_x/F_z$ where the denominators are nonzero. For differentiable $f$ and a unit vector $\mathbf{u}$ , $D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u}$ ; this is largest, equal to $|\nabla f|$ , in the direction of $\nabla f$ , and $\nabla F$ is normal to the level set $F = k$ . Extrema with existing partials occur only at critical points, and where the second partials are continuous on a disc about a critical point, $D = f_{xx}f_{yy} - f_{xy}^2$ decides: $D>0$ with $f_{xx}>0$ a local minimum, $D>0$ with $f_{xx}<0$ a local maximum, $D<0$ a saddle, $D=0$ no information. A continuous $f$ on a closed bounded set attains an absolute maximum and minimum, found by comparing interior critical values with the boundary; under the constraint $g = k$ with $\nabla g \neq \mathbf{0}$ , the candidates solve $\nabla f = \lambda\nabla g$ together with $g = k$ .

Exercises for §11.8 →

Common Mistakes

Checking a limit along the two axes, finding they agree, and concluding the limit exists. Two paths agreeing proves nothing; two paths disagreeing proves non-existence. To prove existence, bound $|f - L|$ by something in $r = \sqrt{(x-a)^2 + (y-b)^2}$ alone.
Trusting straight lines in the two-path test. The function $xy^{2}/(x^{2} + y^{4})$ tends to $0$ along every line through the origin and to $\tfrac12$ along the parabola $x = y^{2}$ .
Forgetting to hold the other variable constant, or forgetting the chain rule inside a partial derivative: $\partial/\partial y$ of $\sin(xy)$ is $x\cos(xy)$ , not $\cos(xy)$ .
Reading $f_{xy}$ in the wrong order. Subscripts read left to right ( $x$ first), the operator form $\partial^{2} f/\partial y\,\partial x$ reads right to left; both mean the same thing. By Clairaut's Theorem it rarely matters, but it does at points where the mixed partials are discontinuous.
Believing that existence of $f_x$ and $f_y$ makes $f$ differentiable, or even continuous. The right sufficient condition is *continuity* of the partials, and $xy/(x^{2} + y^{2})$ is the standard counterexample.
Using a non-unit vector in $D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u}$ . Normalise first, or the answer is scaled by the vector's length.
Treating $\nabla f(x,y)$ as a vector in three-dimensional space lying on the surface. It is a vector in the domain plane. The normal vector to the surface $z = f(x,y)$ is $\langle f_x, f_y, -1\rangle$ .
In the chain rule, dropping a term because a variable "does not appear": if $z = f(x,y)$ with $x = g(s,t)$ and $y = h(s,t)$ , then $\partial z/\partial s$ has two terms even when only one of them ends up nonzero.
Applying the second derivatives test with $D = 0$ and reporting an answer. The test is silent; inspect the function directly, as with $x^{4} + y^{4}$ versus $x^{4} - y^{4}$ .
Concluding "local minimum" from $f_{xx} > 0$ alone. The sign of $f_{xx}$ is only decisive once $D > 0$ ; with $D < 0$ the point is a saddle regardless.
Finding absolute extrema on a closed region from interior critical points only. The boundary must be parametrised and searched separately, endpoints of each edge included, and boundary candidates are never fed into the second derivatives test.
Dividing a Lagrange equation by $x$ (or by $\lambda$ ) without treating the case $x = 0$ (or $\lambda = 0$ ) separately. Solutions lost this way are often exactly the extremes.
Reporting the points found by Lagrange multipliers as "the maximum and the minimum" without evaluating $f$ at each and comparing, and without confirming an extreme value exists at all on an unbounded constraint.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

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