calculus ii fall 2014 - university of colorado...

Calculus II – Fall 2014

Eitan Angel

University of Colorado

Tuesday, November 18, 2014

E. Angel (CU) Calculus II 18 Nov 1 / 30

Introduction

We will gain intuition into functions of two variables graphically,numerically, and analytically.

A two variable function z = f(x, y) has two independent inputs, xand y, which may be varied separately.

We will discuss three dimensional coordinate systems in order tounderstand two variable functions.

Functions of two variables and their domains. Cross sections and levelcurves can be used to graph two variable functions.

Introduction

Graphical Example: A Weather Map

The weather map below displays temperature, T . What are theindependent variables? We may consider T (x, y).

Graphical Example: A Weather Map

The weather map below displays temperature, T . What are theindependent variables? We may consider T (x, y).

Numerical Example: Veggie Burger Consumption

Suppose you distribute veggie burger patties and you want to understandyour market. The quantity of burger patties bought by a household in amonth, C, depends on the price of patties, p, as well as the householdincome, I. So some values of C(I, p) (in lbs.) are

Household income I (in $1000)

Price of patties, p ($ / lb.)

3.00 3.50 4.00 4.50

20 2.65 2.59 2.51 2.43

40 4.14 4.05 3.94 3.88

60 5.11 5.00 4.97 4.84

80 5.35 5.29 5.19 5.07

100 5.79 5.77 5.60 5.53

We can visualize this data with a bar graph

Transparency Master for bar graph of Table 12.1 in the text

income

pounds

A three-dimensional bar graphE. Angel (CU) Calculus II 18 Nov 5 / 30

Given more refined data:

income

pounds

More and more refined beef dataE. Angel (CU) Calculus II 18 Nov 6 / 30

Analytic Examples

If M is the amount of money in a bank account t years after aninvestment of P dollars, and if interest is accrued at a rate of 5% peryear compounded annually, then

M = f(P, t) = P (1.05)t.

If P is the population of rabbits whose growth rate is a continuous3% per year, and P0 is the initial population of rabbits, then

P = f(P0, t) = P0e0.03t

A cone has radius r and height h. If its volume is V we can give aformula for the volume by

V (r, h) =1

3πr2h

Analytic Examples

M = f(P, t) = P (1.05)t.

P = f(P0, t) = P0e0.03t

V (r, h) =1

3πr2h

Analytic Examples

M = f(P, t) = P (1.05)t.

P = f(P0, t) = P0e0.03t

V (r, h) =1

3πr2h

Coordinate Systems

Given a fixed point O, a pair of realnumbers describes a point on a plane.The x-axis and y-axis are called thecoordinate axes.

A triple of real numbers describes apoint in space. The x-axis, y-axis, andz-axis are called the coordinate axes.

Coordinate Systems

Given a fixed point O, a pair of realnumbers describes a point on a plane.The x-axis and y-axis are called thecoordinate axes.

A triple of real numbers describes apoint in space. The x-axis, y-axis, andz-axis are called the coordinate axes.

Handedness

Left-handed system Right-handed system

We will only use right-handed coordinate systems in this course.

Coordinate Planes

The three illustratedcoordinate planes dividespace into eight parts, calledoctants. The first octant, inthe foreground, is determinedby the positive axes.

As mentioned before, a point P isspecified by a triple of real numbers,(a, b, c). To locate P ,

First travel a distance a along thex-axis.

Next travel a distance b parallel to they-axis.

Finally travel a distance c parallel tothe z-axis.

We call a the x-coordinate of P , b the y-coordinate of P , and c thez-coordinate of P .

Points and Space

The Cartesian product

R× R× R = {(x, y, z) : x, y, z ∈ R}

is the set of all ordered triples of real numbers and is denoted by R3.

We have described a one-to-one correspondence between points P inspace and ordered triples (a, b, c) in R3.

This is called a three dimensional rectangular coordinate system.

Notice that the first octant can be described as points with allpositive coordinates.

Points and Space

R× R× R = {(x, y, z) : x, y, z ∈ R}

Points and Space

R× R× R = {(x, y, z) : x, y, z ∈ R}

Points and Space

R× R× R = {(x, y, z) : x, y, z ∈ R}

Distance

Theorem (Distance Formula)

The distance between a point P1(x1, y1, z1) and P2(x2, y2, z2) is given by

d =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

This formula can be proven by repeated application of the Pythagoreantheorem.

Distance

Draw a box where P1 and P2

form the main diagonal.

Label A(x2, y1, z1),B(x2, y2, z1), a = |x2 − x1|,b = |y2 − y1|, c = |z2 − z1|.|P1B|2 = a2 + b2 (Pythagoras)

Using Pythagoras again:

d2 = |P1P2|2

= |P1B|2 + c2

= a2 + b2 + c2

= (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

Distance

d2 = |P1P2|2

= |P1B|2 + c2

= a2 + b2 + c2

= (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

Distance

d2 = |P1P2|2

= |P1B|2 + c2

= a2 + b2 + c2

= (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

Distance

d2 = |P1P2|2

= |P1B|2 + c2

= a2 + b2 + c2

= (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

Distance

What is the distance between (2, 7, 3) and (1, 9, 5)?

d =√(1− 2)2 + (9− 7)2 + (5− 3)2 = 3

Sphere

What is the equation of a sphere with radius r and center C(x0, y0, z0)?

A sphere is the set of all pointsP (x, y, z) whose distance from the Cis r, i.e., |PC| = r. Upon squaringboth sides,

r2 = |PC|2

= (x− x0)2 + (y − y0)2 + (z − z0)2

Distance

d =√(1− 2)2 + (9− 7)2 + (5− 3)2 = 3

Sphere

r2 = |PC|2

= (x− x0)2 + (y − y0)2 + (z − z0)2

Distance

d =√(1− 2)2 + (9− 7)2 + (5− 3)2 = 3

Sphere

r2 = |PC|2

= (x− x0)2 + (y − y0)2 + (z − z0)2E. Angel (CU) Calculus II 18 Nov 16 / 30

Distance

d =√(1− 2)2 + (9− 7)2 + (5− 3)2 = 3

Sphere

r2 = |PC|2

= (x− x0)2 + (y − y0)2 + (z − z0)2E. Angel (CU) Calculus II 18 Nov 16 / 30

Spheres

The previous example implies a standard way to describe a sphere.

The standard equation of a sphere with radius r and centerC(x0, y0, z0) is

r2 = (x− x0)2 + (y − y0)2 + (z − z0)2.

If the terms the standard equation of a sphere are multiplied out and liketerms are collected, we arrive at the general equation of the sphere

x2 + y2 + z2 +Ax+By + Cz = D.

Spheres

Example

Show thatx2 + y2 + z2 + 4x− 2y + 6z − 2 = 0

is the equation of a sphere, and find the center and radius.

Complete the square (thrice!)

(x2 + 4x+ 4) + (y2 − 2y + 1) + (z2 + 6z + 9)− 2 = 4 + 1 + 9

(x+ 2)2 + (y − 1)2 + (z + 3)2 = 16

which implies the radius is 4 and the center is (−2, 1,−3).

Q: Could the equation fail to describe a sphere?A: Sure. What if the right hand side is not positive?

Spheres

Example

Show thatx2 + y2 + z2 + 4x− 2y + 6z − 2 = 0

(x2 + 4x+ 4) + (y2 − 2y + 1) + (z2 + 6z + 9)− 2 = 4 + 1 + 9

(x+ 2)2 + (y − 1)2 + (z + 3)2 = 16

Spheres

Example

Show thatx2 + y2 + z2 + 4x− 2y + 6z − 2 = 0

(x2 + 4x+ 4) + (y2 − 2y + 1) + (z2 + 6z + 9)− 2 = 4 + 1 + 9

(x+ 2)2 + (y − 1)2 + (z + 3)2 = 16

Spheres

Example

Show thatx2 + y2 + z2 + 4x− 2y + 6z − 2 = 0

(x2 + 4x+ 4) + (y2 − 2y + 1) + (z2 + 6z + 9)− 2 = 4 + 1 + 9

(x+ 2)2 + (y − 1)2 + (z + 3)2 = 16

Spheres

Thus the relevant theorem:

Theorem

An equation of the form

x2 + y2 + z2 +Ax+By + Cz = D

represents a sphere, a point, or has no graph.

So just because an equation has the general form of a sphere, it may bedegenerate or have no solution.

Functions of Two Variables

The temperature T at a point on the Earth’s surface depends on thelatitude x and the longitude y of that point. So we can think oftemperature as a function of two variables, T (x, y).

The volume Vcyl of a cylinder depends on its radius r and its heighth. We say Vcyl is a function of r and h, and is given by the equationVcyl(r, h) = πr2h.

Definition

Let D ⊂ R2. A function f of two variables is a rule f : D → R thatassigns to each ordered pair (x, y) in D a unique real number f(x, y)denoted (x, y) 7→ f(x, y). The set D is called the domain of f and itsrange is the set of values that f takes on, i.e., {f(x, y)|(x, y) ∈ D}.

We often write z = f(x, y) to make explicit the value taken on by f at thepoint (x, y). The variables x and y are independent variables and z isthe dependent variable.

Definition

Domain of Functions of Two Variables

Example

Find the domain of the following and evaluate f(3, 2).

f(x, y) =

√x+ y + 1

x− 1

Example

Find the domain of the following and evaluate f(3, 2).

f(x, y) =

√x+ y + 1

x− 1

f(3, 2) =

√3 + 2 + 1

3− 1=

The expression for f makes sense ifthe denominator is not 0 and thequantity under the square root sign isnonnegative. So the domain of f is

D = {(x, y) |x+ y + 1 ≥ 0, x 6= 1}

Example

Find the domain of the following.

f(x, y) = x ln(y2 − x)

Example

Find the domain of the following.

f(x, y) = x ln(y2 − x)

Since ln(y2 − x) is defined only wheny2 − x > 0, that is, x < y2, thedomain of f is

D = {(x, y) |x < y2}

In other words, the set of points tothe left of the parabola x = y2.

Graphs of Functions of Two Variables

Definition

If f is a function of two variables with domain D, the graph of f is the set

S = {(x, y, z) ∈ R3 | z = f(x, y), (x, y) ∈ D}

The graph of a function f of onevariable is a curve C with equationy = f(x). The graph of a function fof two variables is a surface S withequation z = f(x, y). We canvisualize the graph S of f as lyingdirectly above or below its domain Din the xy-plane.

Definition

If f is a function of two variables with domain D, the graph of f is the set

S = {(x, y, z) ∈ R3 | z = f(x, y), (x, y) ∈ D}

The graph of a function f of onevariable is a curve C with equationy = f(x). The graph of a function fof two variables is a surface S withequation z = f(x, y). We canvisualize the graph S of f as lyingdirectly above or below its domain Din the xy-plane.

There are a couple of basic techniques to graph a two variable functionf(x, y).

We can numerically plot points. This consists of choosing a list ofvalues for x and y, computing f(x, y) for these chosen values, thenplotting in 3-space.

We can take cross-sections. This consists of holding one dependentvariable constant while allowing the other to vary.

We will give an example of both techniques.

There are a couple of basic techniques to graph a two variable functionf(x, y).

We can numerically plot points. This consists of choosing a list ofvalues for x and y, computing f(x, y) for these chosen values, thenplotting in 3-space.

We can take cross-sections. This consists of holding one dependentvariable constant while allowing the other to vary.

We will give an example of both techniques.

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.Let’s compute f(x, y) for some points.

-3 -2 -1 0 1 2 3

-3 45 40 37 36 37 40 45

-2 25 20 17 16 17 20 25

-1 13 8 5 4 5 8 13

0 9 4 1 0 1 4 9

1 13 8 5 4 5 8 13

2 25 20 17 16 17 20 25

3 45 40 37 36 37 40 45

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.Now let’s plot these points:

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.We can roughly guess how they connect with a wireframe:

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.We can roughly guess how they connect with a wireframe:

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.The other strategy is to take cross sections, that is fix one variable whileletting the other vary.

x = −2 h(−2, y) = 16 + y2

x = −1 h(−1, y) = 4 + y2

x = 0 h(0, y) = y2

x = 1 h(1, y) = 4 + y2

x = 2 h(2, y) = 16 + y2

y = −2 h(x,−2) = 4x2 + 4

y = −1 h(x,−1) = 4x2 + 1

y = 0 h(x, 0) = 4x2

y = 1 h(x, 1) = 4x2 + 1

y = 2 h(x, 2) = 4x2 + 4

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.We obtain families of parabolas which we can graph

Example

Sketch the graph of the function h(x, y) = 4x2 + y2.Filling in we obtain the graph of an elliptic paraboloid:

Example

Find the domain and range and sketch the graph of

g(x, y) =√9− x2 − y2

Example

Find the domain and range and sketch the graph of

g(x, y) =√9− x2 − y2

The domain of g is

D = {(x, y) | 9− x2 − y2 ≥ 0} = {(x, y) |x2 + y2 ≤ 9}

which is the disk with radius 3 and center (0, 0). The range of g is

{z | z =√

9− x2 − y2, (x, y) ∈ D}

Since z is a positive square root, z ≥ 0. Also,√

9− x2 − y2 ≤ 3. So therange is

{z | 0 ≤ z ≤ 3} = [0, 3]

Cross-sections off(x, y) = sinx+ sin y

f(x, y) = sinx+ sin y

f(x, y) = (x2 + 3y2)e−x2−y2

calculus ii fall 2014 - university of colorado...

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