Math 16b: Calculus and Analytic Geometry

http://www.math.berkeley.edu/~scanlon/m16bs04/index.html

Chapter 7: Functions of Several Variables

Section 7.1: Examples of Functions of Several Variables

• $f(x,y) = x + y^2$
• $g(\theta,\varphi) = \sin(4 \theta) \cos(\varphi^2)$
• $h(s,t,u) = \frac{e^{s t} + u}{t^2 + us}$

Graphs

A function need not be expressed in terms of a formula.

• The population $P$ of the state $s$ at the beginning of the year $y$ is a function of the variables $s$ and $y$.

Traditionally, the variables of a function of several variables are written as $x, y, z$ or if there are more than three variables, using subscripts $x_1, x_2, \ldots, x_n$.

The functions themselves are usually written using symbols such a $f, g, h, F, G, H$ and if it is important to list the variables as, eg $f(x,y)$, $g(x,y,z)$, or $H(x_1, x_2, x_3, x_4, x_5)$.

Graphing

For function of two variables, one may graph this function by plotting the solutions to $z = f(x,y)$ in three space.

Examples:

• $z = x + y^2$
• $z = \frac{1}{x + y} + y^2$
• $z = \sin(x y^2)$

Graphs

Level curves

One may produce a two-dimensional graph of a function of two variables by plotting the solutions to $f(x,y) = c$ for various constants $c$.

A contour map is precisely such a graph. Here the variables are the lattitude and longitude of a point on the Earth and the function gives the altitude.

(More on contour plots)

Functions of several variables in applied problems

Suppose that one wishes to produce a structure in the shape of a rectangular box. The material for the floor costs $\$ 7$ per square foot, the material for the walls costs $\$ 5$ per square foot, and the material for the roof costs $\$ 2$ per square foot. Write the total cost as a function of $w$, the width, $\ell$, the length, and $h$, the height, of the structure.

Solution

$$= w \ell$$ Area of the floor

$$= (2 w h) + (2 \ell h)$$ Area of walls

$$= w \ell$$ Area of roof

Solution (continued)

$$= + +$$ Total cost

$$= 7 + 5 + 3$$

$$= 7 w \ell + 5 ( (2w h) + 2(\ell h)) + 2 w \ell$$

$$= 9 w \ell + 10 w h + 10 \ell h$$

Section 7.2: Partial Derivatives

If $F(x,y,z)$ is a function of several variables, then for any fixed value of $x$ and $y$, say, $x=a$ and $y=b$, the function $f(z) := F(a,b,z)$ is a function of the single variable $z$.

As such, it makes sense to compute the derivative of $f$, or what is the same thing, the derivative of $f$ with respect to $z$:

$$\frac{\partial F}{\partial z} = f'(z)$$

Derivatives as limits

For a function of one variable, $f(x)$, the derivative of $f$ at $a$ is defined as a limit:

$$f'(a) = \frac{df}{dx} \vert_{x=a} := \lim_{\epsilon \to 0} \frac{f(a+\epsilon) - f(a)}{\epsilon}$$

Partial derivatives as limits

For a function of several variables, partial derivatives are defined by the same kind of limit.

$$\frac{\partial F}{\partial x}(x,y,z) := \lim_{\epsilon \to 0} \frac{F(x+\epsilon,y,z) - F(x,y,z)}{\epsilon}$$

Computing partial derivatives

In general, to compute the partial derivative of a function with respect to some variable, treat the function as a function of that single variable with all the other named variables regarded as constants.

Computing partial derivatives: Example 1

If $C$ is a constant and $n$ a natural number, then the formula

$$\frac{d}{dx}(C x^n) = C n x^{n-1}$$

is familiar to you.

Instead, we could consider this monomial as a function of three variables $f(x,y,z) = y x^z$ (at least for $z \geq 0$) and the above formula expresses

$$\frac{\partial f}{\partial x} = y z x^{z - 1}$$

Computing partial derivatives: Example 2

Let $F(x,y,z) = x \sin(y) + z^2$. Compute $\frac{\partial F}{\partial x}$, $\frac{\partial F}{\partial y}$, and $\frac{\partial F}{\partial z}$.

Solution

$$\frac{\partial F}{\partial x} = \sin(y)$$

$$\frac{\partial F}{\partial y} = x \cos(y)$$

$$\frac{\partial F}{\partial z} = 2 z$$

Computing partial derivatives: Example 3

Let $g(x,y) = x e^{x y^2}$. Compute $\frac{\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$.

Solution

$$\frac{\partial g}{\partial x} = \frac{d}{dx} (x) \cdot e^{x y^2} + x \frac{d}{dx} (e^{x y^2})$$

$$= e^{x y^2} + x (y^2 e^{x y^2})$$

$$= (1 + x y^2) e^{x y^2}$$

Solution continued

$$\frac{\partial g}{\partial y} = \frac{d}{dy} (x e^{x y^2})$$

$$= x \frac{d}{dy} (e^{x y^2})$$

$$= x \frac{d}{dt}(e^t)\vert_{t=x y^2} \frac{d}{dy}(x y^2)$$

$$= x e^{x y^2} (2xy)$$

$$= 2 x^2 y e^{x y^2}$$

Geometric interpretation

Unlike a curve, a surface has many tangent lines at each point. The partial derivatives give the slopes of the tangent lines at a point in a specific direction.

More precisely, the partial derivative at a point $P$ of a function $F$ with respect to $x$ is the slope of the tangent line to the graph of $F$ at $(P,f(P))$ along the direction where all coordinates save $x$ are held fixed.

Partial derivatives as rates of change

As with derivatives of a function of a single variable, partial derivatives may be interpreted as rates of change. In this case, $\frac{\partial f}{\partial x}$ is the rate at which $f$ changes relative to changes in the $x$-variable with all other variables held fixed.

Partial derivatives as rates of change: an example

Let $f(x,y) = \frac{x}{y}$. Compute and interpret $\frac{\partial f}{\partial x} \vert_{(1,2)}$ and $\frac{\partial f}{\partial y} \vert_{(1,2)}$.

Solution

$\frac{\partial f}{\partial x} \vert_{(1,2)} = \frac{1}{y} \vert_{(1,2)} = \frac{1}{2}$. So the slope of the tangent line in the $x$ direction at $(1,2,\frac{1}{2})$ is $\frac{1}{2}$.

$\frac{\partial f}{\partial y} \vert_{(1,2)} = \frac{-x}{y^2} \vert_{(1,2)} = \frac{-1}{4}$. That is, the slope of the tangent line in the $y$-direction is $\frac{-1}{4}$.

Notice that $f$ is increasing in the $x$-direction while it is decreasing in the $y$-direction.

Graph

Higher order derivatives

A partial derivative of a function is itself a function and may be differentiated again.

It is a non-trivial, though true, theorem that for a sufficiently smooth function the order of differentiation is immaterial. That is, $\frac{\partial}{\partial x} (\frac{\partial F}{\partial y}) = \frac{\partial}... ...}(\frac{\partial F}{\partial x}) =: \frac{\partial^2 F}{\partial x \partial y}$

Second partial derivatives: an example

Let $F(x,y) = x^2 y + y^3$. Compute $\frac{\partial^2 F}{\partial x^2}$, $\frac{\partial^2 F}{\partial x \partial y}$ and $\frac{\partial^2 F}{\partial y^2}$.

Solution

$\frac{\partial F}{\partial x} = 2 x y$ and $\frac{\partial F}{\partial y} = x^2 + 3 y^2$.

So, $\frac{\partial^2 F}{\partial x^2} = 2 y$ and $\frac{\partial^2 F}{\partial y^2} = 6 y$, while $\frac{\partial^2 F}{\partial x \partial y} = \frac{\partial}{\partial x}(x^2 + 3 y^2) = 2x$ (or we may compute $\frac{\partial^2 F}{\partial x \partial y} = \frac{\partial}{\partial y}(2 x y) = 2x$).