Lecture Notes
for
Transition to Calculus
Department of Mathematics
University of North Dakota
Grand Forks, ND 58202
Fall 2003
Preface
This collection of lecture notes is designed for use in Arts and Sciences 250 - Transition to
Calculus, at the University of North Dakota. This course is not a comprehensive preparation
for calculus. It is designed for either the student who needs a few gaps filled in order to be
ready for calculus, or for the student who is in calculus, but needs to refresh their precalculus
skills.
c
°2003
University of North Dakota Mathematics Department
Permission is granted to copy, distribute and/or modify this document under the terms of
the GNU Free Documentation License, Version 1.2 or any later version published by the Free
Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts. A copy of the license is included at http://www.gnu.org/copyleft in the section
entitled ”GNU Free Documentation License.”
i
TABLE OF CONTENTS
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1. Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2. Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3. Rational and Algebraic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4. Partial Fraction Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
5. Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
6. Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
7. Conic Sections. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23
8. Basic Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
9. Special Values of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
10. General Sinusoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
11. Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
12. Applications of Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
13. Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
14. Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Appendices
A. The Real Number Line and the Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
B. Lines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .66
C. Factoring Polynomials and Complex Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
D. Solving Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . on line
E. Rotation of Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
ii
Introduction
We denote the set of real numbers by IR, and the Cartesian Plane by IR2 = IR × IR. For
more information on these sets see appendix A. Information on lines is contained in appendix
B. These appendices should be read carefully before proceeding.
A relation of real numbers is a subset of IR × IR. Des Cartes’ invention allows us to
represent any relation of real numbers graphically. That is, the graph of a relation, R, consists
of all ordered pairs (a, b) ∈ IR2 , where (a, b) ∈ R. Given a relation, R, its inverse relation,
denoted by R−1 is the relation R−1 = {(b, a)|(a, b) ∈ R}. The graph of R−1 will therefore be
the graph of R reflected across the line y = x.
Commonly a relation will be describable by a system of equations and inequalities. Thus
we may think of a relation as the set of pairs (a, b) for which the equations and inequalities
E1 (a, b), E2 (a, b), . . . , En (a, b) are all true. So for example E1 (x, y) may be x2 = y 2 , and
E2 (x, y) might be |x| ≤ 1. The relation defined by these is S = {(a, b)|a = ±b and − 1 ≤ a ≤
1}. If we put R1 = {(a, a)|a ∈ IR} and R2 = {(a, −a)|a ∈ IR}, then R = R1 ∪ R2 is the relation
defined by E1 . Moreover if we set T = {(x, y)| − 1 ≤ x ≤ 1}, then T is the relation defined by
E2 , and S = R ∩ T . Notice that S = S −1 .
In an attempt to deal with the general case we see that we must at least be able to
determine relations described by a single equation, rule, or formula. Moreover it will be
advantageous if one variable in the equation is completely determined by the other, since in
this case every single input will yield exactly one ordered pair as output. This explains why
we will spend most of our time considering functions.
Section 1: Functions
Given two sets of real numbers A and B, a function, f , is a relation of real numbers with
the property that for every a ∈ A, there is exactly one b ∈ B, so that (a, b) is in the relation.
Notationally we write f : A −→ B. Also when (a, b) is in the relation we write b = f (a). As
alluded to above, we will usually be able to describe f by an equation in x and y. But since
y is determined, given x, we can rewrite the equation to be y − f (x) = 0, where f (x) is some
formula, or rule dependent on x. Equivalently we write y = f (x).
Given a function y = f (x), where f : A −→ B, we call A the domain of f and B the
codomain. We write A = dom f . Graphically a ∈ dom f if and only if the vertical line x = a
1
Section 1: Functions
intersects the graph of f . In fact this vertical line test allows us to separate those relations
which are functions, from those which are not. A relation f is a function only if every vertical
line x = r intersects the graph of f at most once. The domain of a function can be equated
with those values of r for which the intersection is non-empty.
We will suppose that B = IR. However we are interested in the subset C of B consisting
of all b ∈ B so that b = f (a) for at least one a ∈ A. This subset is the range of f . We write
C = ran f . A number s ∈ IR is in the range of f only if the horizontal line y = s intersects
the graph of f at least once.
The end behavior of the functions describes how the output values tend as the inputs
become large in absolute value. We write x −→ ∞ to mean x grows without bound positively,
and x −→ −∞ to mean that x grows without bound negatively. In general the notation
E −→ F means that the expression E is becoming arbitrarily close to the expression F . A
line ax + by + c = 0 (not both a = 0 and b = 0) is an asymptote for a function if the graph of
the function becomes arbitrarily close to the graph of the line.
If a ∈ IR and f (a) ≤ f (b) for all b ∈ dom f , (a, f (a)) is a global minimum. If a ∈ IR
and f (a) ≤ f (b) for all b in an interval I containing a, then (a, f (a)) is a relative minimum.
Global maximum and relative maximum are defined similarly.
A function for which f (−x) = f (x) will have a graph which is symmetric about the y-axis.
Such a function is called an even function. A function for which f (−x) = −f (x) will have a
graph which is symmetric about the origin. Such a function is called odd.
An intersection of the graph of a function with a coordinate axis is called an intercept. A
function may have many x-intercepts, but has at most one y-intercept.
Given a function as a rule or formula we can use dom f , ran f , its end behavior, any
asymptotes it has, any symmetry it has, and its x-intercepts and y-intercept to help sketch
the graph of f . Conversely, given the graph of f we can describe each of the characteristics
above.
As the example in the introduction points out, we also need to be able to determine
how functions interact. To begin with, given real functions f : A −→ IR and g : C −→ IR
their sum is f + g : A ∩ C −→ IR defined by (f + g)(x) = f (x) + g(x). The difference, and
g
g
product are defined similarly. The ratio of g by f is denoted by . The domain of consists
f
f
2
Section 1: Functions
1
.
f
Finally, the composition of g by f is denoted by (f ◦ g)(x) = f (g(x)). The domain of f ◦ g is
of all x ∈ A ∩ C so that f (x) 6= 0. A special type of ratio is the reciprocal, denoted
{x ∈ dom g|g(x) ∈ dom f }.
An important case of the above operations occurs when y = g(x) = c, and c is a positive
real number. In this case dom g = IR and if f : A −→ IR, then A∩ dom g = A = dom f . Now
the function f1 (x) = (f +g)(x) has the same domain as f and its graph is {(x, f (x)+c)|x ∈ A}.
Which is to say that we may graph f1 by translating every point of the graph of f up c units
vertically. Similarly the graph of f2 = f − g can be made by translating every point of the
graph of f down c units. The graph of f3 = (f g) can be formed by scaling every second
coordinate of a point on the graph of f by a factor of c. Similarly the ratio of f by g can
be graphed by scaling all second coordinates of points on the graph of f by a factor of 1/c.
When g(x) = −1, the product f g may be graphed as the reflection of the graph of f across the
x-axis. More generally if g(x) = d 6= 0 we can graph the function f g by reflecting the graph
of f across the x-axis if necessary, and then scaling second coordinates by a factor of |d|.
Another special case occurs when g(x) = dx, d 6= 0, or g(x) = x − c, where c > 0. In the
second case when we compose g by f we have
dom (f ◦ g) = {x ∈ dom g|g(x) ∈ dom f } = {x|x − c ∈ dom f } = {x + c|x ∈ dom f }.
So the graph of this composition consists of {(x + c, f (x))|x ∈ dom f }. Which means that
to graph the composition, we simply translate the graph of f to the right c units. Similarly if
g(x) = x + c, where c > 0, then the graph of f ◦ g would be the graph of f shifted left c units.
In the first case when d > 0 the graph of f ◦ g is {( xd , f (x))|x ∈ A}. So this composition can
be graphed by compressing the scale of the x-axis by a factor of d. Notice that if 0 < d < 1,
this compression turns out to be expansion by a factor of 1/d. When g(x) = −1, the graph of
f ◦ g is the reflection of the graph of f across the y-axis. In general we reflect across the y-axis
if d < 0 and compress the scale of the x-axis by a factor of |d|.
Of very great interest is the possibility that g = f −1 . To begin with we must have
that f −1 is a function, and not simply a relation. This means that the graph of f −1 must
pass the vertical line test. Since the graph of f −1 is the graph of f reflected across the
line x = y we have that the graph of f must satisfy the horizontal line test. That is, every
horizontal line y = b intersects the graph of f at most once. In this case we say that f is a
3
Section 1: Functions
one-to-one function. Now for every point (a, b) on the graph of f , so that b = f (a), we have
a = f −1 (b). Consequently (f −1 ◦ f )(a) = f −1 (f (a)) = f −1 (b) = a. Also for b ∈ dom f −1 ,
(f ◦ f −1 )(b) = f (f −1 (b)) = f (a) = b.
One last characteristic of a function we will investigate is its difference quotient, which
f (x + h) − f (x)
is found by forming the expression
, where h is an infinitessimally small, but
h
nonzero, real number. The denominator is actually (x + h) − x, so the difference quotient is
the slope of the line through the points (x, f (x)), and (x + h, f (x + h)). This slope describes
how y = f (x) changes when x changes by h units.
To illustrate the ideas we have developed let us consider a function of the form y = f (x) =
xn , where n is a non-negative whole number. We call such a function a natural power function.
Henceforth we will dispense with the case n = 0, since we get f (x) = 1. So in the sequel unless
otherwise specified, natural power function will denote a power function f (x) = xn with n > 0.
Notice that xn = x · x · . . . · x, with n factors makes sense for any x ∈ IR. We can therefore
take dom f to be any subset of IR. Most often we will take dom f = IR.
Next, if n = 2m, where m is a positive whole number, then f (x) = xn = x2m = (x2 )m ≥ 0,
since x2 ≥ 0. Thus a function of this form has ran f ⊂ [0, ∞). Indeed graphing a few of these
leads us to conclude that ran f = [0, ∞). On the other hand when n = 2k + 1, where k is
a non-negative whole number, this is no longer true. In fact as x −→ ∞, xn −→ ∞ whether
n is even or odd. But as x −→ −∞, x2m −→ ∞ while x2k+1 −→ −∞. So when n is odd,
ran f = IR. Moreover we see that the end behavior of a natural power function is determined
by whether n is odd or even. You might also have guessed that f (x) = x2m is even while
f (x) = x2k+1 is odd.
Every natural power function passes through (0, 0). In fact if xn = 0 we deduce that
x = 0. So each natural power function f (x) = xn has exactly one x-intercept which coincides
with its y-intercept.
By definition, the reciprocal, g(x), of a natural power function f (x) = xn has domain
1
1
= n , which we also write as x−n . When n is even
IR − {0}. A formula for g is g(x) =
f (x)
x
ran g = (0, ∞). When n is odd, ran g = IR − {0}. As x −→ ∞, xn −→ ∞, when n > 0. The
reciprocals of arbitrarily large positive numbers are infinitessimally small positive numbers.
So as x −→ ∞, g(x) −→ 0. Likewise, the reciprocals of arbitrarily large negative numbers
are infinitessimally small negative numbers. Thus as x −→ −∞, g(x) −→ 0. So g(x) has the
4
Section 1: Functions
x-axis y = 0 as a horizontal asymptote. When n is even, g(x) gets arbitrarily close to 0 but
is positive so we write g(x) −→ 0+ . When n is odd, the values of g(x) are negative when x is
so as x −→ −∞, g(x) −→ 0− . By reversing our reasoning we realize that x = 0 is a vertical
asymptote of g(x). We now write g(x) −→ ∞, as x −→ 0+ . When n is even we get g(x) −→ ∞
as x −→ 0− . When n is odd we have g(x) −→ −∞ as x −→ 0− .
The functions of the form f (x) = xl where l is a whole number are the integral power
functions. From the preceding paragraphs we see that the integral power functions consist of
the natural power functions and their reciprocals.
Next, when n is odd, any integral power function passes both the vertical and horizontal
√
line tests. Therefore f −1 is a function. We use the formula f −1 (x) = n x. Alternatively we
1
have f −1 (x) = x n , from the multiplicative law of exponents. However when n 6= 0 is even,
f (x) fails the horizontal line test. In this case f −1 is a relation but not a function. In order
to partially correct this problem we restrict the domain of f to a set A for which the graph of
f will pass the horizontal line test. By convention for n even we take A = [0, ∞), for n > 0
and A = (0, ∞), for n < 0. For this restricted function, f −1 is now a function and we use the
same notation as the odd case.
When we repeatedly combine natural power functions (including f (x) = 1) using sums,
multiples, and differences we get the class of polynomial functions. When we compose integral
power functions and inverses of integral power functions we get the rational power functions.
√
For example if we compose f (x) = x3 by g(x) = x1/2 we get h(x) = (x3 )1/2 = x3/2 = ( x)3 .
Lastly, when f is a natural power function the difference quotient for f takes the form
(x + h)n − xn
xn + nxn−1 h + h2 k(x, h) − xn
f (x + h) − f (x)
=
=
= nxn−1 + h(k(x, h))
h
h
h
as long as h 6= 0, where h2 k(x, h) is the rest of the expansion of (x + h)n in terms of x and
h. So as h −→ 0, the difference quotient of f approaches nxn−1 . We can also algebraically
simplify difference quotient formulas for more general power functions.
Exercises:
1. Find a simplified formula for the difference quotient of
1
a) f (x) =
x
√
b) f (x) = x
c) f (x) = x2 − 4x + 6
5
Section 1: Functions
2. Read Section 2 and appendix C.
3. Let f (x) = ax2 + bx + c, with a 6= 0 and a, b, c ∈ IR.
a) Complete the square on x.
b) Use part a) to show that f is a shift and/or scale of y = g(x) = x2 .
c) Use part a) to show that f has either one double zero, two real zeroes, or no real zeroes.
6
Section 2: Polynomial Functions
To help you grasp the importance of this little section we encourage you to borrow a
non-reform calculus text from another student, or the local library. There should be a chapter
titled “Sequences and Series”. In this chapter will be a section titled “Taylor Polynomials”.
This section is not normally covered in calculus until the second semester, so you’ll have to
wait quite a while to appreciate the contents. The gist of the section is that any function can
be approximated relatively well using polynomials. This is enormously important, since polynomials require no division, and completely accurate division is not easy to do on a computer.
Also since polynomials are nothing more than linear combinations of natural power functions
they are quite simple to utilize.
Starting with a polynomial p(x) = an xn + an−1 xn−1 + . . . + a1 x + a0 , where an 6= 0 and
a0 , a1 , . . . , an ∈ IR are constants, it is fairly clear that dom p is any subset of IR. We usually
choose dom p = IR.
Computing the range of an arbitrary polynomial can be more of a problem. Sometimes
we can do it by inspection. The general computation often requires calculus.
Since 1 ≤ x implies 0 < x we can successively multiply both sides of 1 ≤ x by x without
changing the direction of the inequality. We arrive at the string of inequalities 1 ≤ x ≤ x2 ≤
x3 ≤ . . . ≤ xn . The point is that as x gets large, the leading term, an xn of a polynomial will
dominate all other terms. So the end behavior of a polynomial is the same as the end behavior
of its leading term. Thus if n is odd, the range of a polynomial will be IR. If n is even we may
need calculus to specify the range, but at least we know a polynomial’s end behavior. Since
n is an important parameter of a polynomial we give it a name, to wit the degree. Strictly
speaking the degree of the polynomial p(x) = 0 is not defined. Some texts will set the degree
of the zero to be −1, others will define its degree to be negative infinity.
The graph of a polynomial of degree n will “waggle” no more than n − 1 times. Peaks of
waggles are relative (sometimes global) maximums. Valleys of waggles are relative (sometimes
global) minimums. This behavior is another feature of polynomials describable after just a
modicum of calculus. So is the fact that polynomials have no asymptotes.
A polynomial is even if and only if its only nonzero coefficients have even subscripts. A
polynomial is odd if and only if its only nonzero coefficients have odd subscripts.
In between the ends, the behavior of a polynomial is mostly dictated by the fact that its
7
Section 2: Polynomial Functions
graph has no breaks, and that it can be factored into the form (see appendix C)
p(x) = an (x−r1 )e1 (x−r2 )e2 (. . .)(x−rk )ek (x2 +b1 x+c1 )f1 (x2 +b2 x+c2 )f2 . . . (x2 +bj x+cj )fj ,
where n = e1 + e2 + . . . + ek + 2(f1 + f2 + . . . + fj ), none of the quadratic factors crosses
the x-axis (so b2i − 4ci < 0), and the values r1 < r2 < . . . < rk indicate the x-coordinates of
p(x)’s x-intercepts. The values r1 , r2 , . . . , rk are the roots of the polynomial. The y-intercept
is (0, p(0)) = (0, a0 ).
When we let qi (x) = (x − ri )ei , for i = 1, 2, . . . , k, and pi (x) = p(x)/qi (x), we see that
“near x = ri ” p(x) will behave like a shift ri units right (left if ri is negative) of the function
axei , where a = pi (ri ). This phenomenon is called local behavior.
A fairly decent sketch of the graph of a polynomial can be constructed by piecing together
the local behavior (including the end-behavior) so that the end result is a smooth curve with
no breaks (this is using what is known as the Intermediate Value Theorem from calculus).
The reciprocal of a polynomial is a special kind of rational function. Rational functions
are the topic of the next section.
A polynomial will be invertible as a function only if n is odd and the graph has no
waggles. So a polynomial often has only a local inverse. In fact it can be somewhat difficult to
algebraically find a formula for the local inverse. Try for example q(x) = x4 − 4x2 + 2. Some
√
examples are easier, such as p(x) = x5 − 2. Here if y = x5 − 2, y + 2 = x5 . So x = 5 y + 2.
√
Therefore p−1 (x) = 5 x + 2.
The difference quotient of a polynomial is not too hard to compute given that the difference
quotient of a sum is the sum of the difference quotients, and that the difference quotient of
a constant times f is the constant times the difference quotient of f . So for example if
p(x) = 3x2 − 4x + 5 its difference quotient is
[3(x + h)2 − 4(x + h) + 5] − [3x2 − 4x + 5]
p(x + h) − p(x)
=
h
h
2
2
3(x + 2xh + h ) − 4x − 4h + 5 − 3x2 + 4x − 5
=
h
3x2 + 6xh + 3h2 − 4x − 4h + 5 − 3x2 + 4x − 5
=
h
2
6xh + 3h − 4h
= 6x + 3h − 4, as long as h 6= 0
=
h
8
Section 2: Polynomial Functions
Exercises:
1. For each polynomial function sketch a graph of the function by transforming the graph of
an integral power function.
a) f (x) = x3 − 2
b) f (x) = −3x4 + 6
c) f (x) = 2(x − 3)5 − 1
d) f (x) = −(2x − 4)2
2. Use the zeroes and end behavior of each polynomial function to sketch an approximation
of its graph.
a) f (x) = (x + 1)2 (x − 2)2
b) f (x) = x(x + 2)2
c) f (x) = −x(x − 1)(x + 2)
d) f (x) = (x + 1)(x − 2)2
3. A square box without a lid is to be constructed by cutting out four squares of side x from
the corners of a 30in. by 30 in. piece of cardboard. Determine the surface area of the box as
a function S of the variable x. Compute the surface area when x = 4. What is dom S?
9
Section 3: Rational and Algebraic Functions
A rational function is one which is defined as the ratio of polynomials. So if r(x) is a
p(x)
, where p(x) = pn xn + pn−1 xn−1 + . . . + p1 x + p0 and q(x) =
rational function, r(x) =
q(x)
qm xm + qm−1 xm−1 + . . . + q1 x + q0 6= 0 are polynomials of degree n and m respectively.
Write p(x) = pn (x − r1 )e1 (x − r2 )e2 . . . (x − rk )ek (x2 + a1 x + b1 )f1 . . . (x2 + aj x + bj )fj and
q(x) = qm (x − s1 )g1 (x − s2 )g2 . . . (x − sl )gl (x2 + c1 x + d1 )h1 . . . (x2 + ci x + di )hi as per the real
factors theorem. If p and q have a common irreducible quadratic factor, we may cancel it since
it is never zero for real inputs. If p and q have a common zero, a, then r(a) is not defined and
otherwise for x 6= a, r(x) is the same as the ratio of p(x)/(x − a) by q(x)/(x − a), continuing to
cancel linear factors of the form (x − a) until at most one of the numerator and denominator
has x − a as a factor. In the sequel we can therefore focus our attention on rational functions
where the numerator and denominator have no common factor (so r(x) is in lowest terms so
to speak).
If r(x) is a rational function in lowest terms, then r(x) is defined, unless q(x) = 0. So
the domain of r(x) is all real numbers which are not roots of q(x). The range of a rational
function cannot in general be computed without the aid of graphing or calculus.
Supposing that we’ve already canceled common factors and p(x) and q(x) are as in the
pn n−m
x
as |x| −→ ∞. So if n < m r(x) −→ 0 as
first paragraph, r(x) will behave like
qm
x −→ ±∞. Thus the graph has y = 0 as a horizontal asymptote in this case. If n = m
pn
pn
as |x| −→ ∞. So now the graph has the line y =
as horizontal asymptote.
r(x) −→
qm
qm
If n > m, by the division algorithm we can write p(x) = d(x)q(x) + s(x), where the degree
of s(x) is strictly less than the degree of q(x). (s(x) is not zero here since we have already
s(x)
s(x)
canceled common factors.) Now we write r(x) = d(x) +
. Since
−→ 0 as |x| −→ ∞,
q(x)
q(x)
r(x) has the same end behavior as d(x). The graphs of r(x) and d(x) will become arbitrarily
close as |x| −→ ∞. If n = m+1, then d(x) takes the form d1 x+d0 i.e. d(x) is a linear function.
The line y = d1 x + d0 will be an asymptote for the graph of r(x). We call such an asymptote
a slant asymptote to distinguish it from the vertical and horizontal asymptotes. In general a
rational function r will have a vertical asymptote whenever c is a root of its denominator. To
help draw the graph of r(x) we determine the behavior of r(x) as x −→ c+ and as x −→ c− .
Only in special cases can we determine maxima and minima for a rational function.
A rational function is even iff both its numerator and denominator are even, or both its
10
Section 3: Rational and Algebraic Functions
numerator and denominator are odd. A rational function is odd iff the numerator is odd and
the denominator is even or vice versa. A rational function in reduced form has x-intercepts
wherever its numerator is zero. The y-intercept, if it exists, is r(0). If we have compiled the
previous information for a rational function, we can usually present a good sketch of its graph.
x2 − 2x − 3
, we have p(x) = x2 − 2x − 3 = (x − 3)(x + 1) and q(x) =
x3 − 4x
x3 − 4x = x(x − 2)(x + 2). So dom r = (−∞, −2) ∪ (−2, 0) ∪ (0, 2) ∪ (2, ∞). Since the degree
Example 1 If r(x) =
of p is strictly less than the degree of q, r(x) has y = 0 as a horizontal asymptote. The lines
x = −2, x = 0 and x = 2 are vertical asymptotes for the graph of r(x).
As x −→ −2− , p(x) −→ 5 and q(x) −→ 0− so r(x) −→ −∞. As x −→ −2+ , p(x) −→ 5
and q(x) −→ 0+ so r(x) −→ ∞. As x −→ 0− , p(x) −→ −3 and q(x) −→ 0− so r(x) −→ −∞.
Similarly as x −→ 0+ , r(x) −→ ∞. Finally as x −→ 2− , r(x) −→ ∞ and as x −→ 2+ , r(x) −→
∞.
The function r(x) is neither even nor odd. It has x-intecepts when x = 3 and x = −1.
The graph does not intersect the y-axis since r(0) is not defined. To facilitate sketching a
graph of r(x) we also compute values for r(x) near the interesting values of x on the graph.
x
−3
r(x) −4/5
−3/2
6/7
−1/2
−14/15
1
4/3
5/2
−14/45
7/2
6/77
Not only do these function values help us decide how to scale our y-axis, they tell us that there
is basically no hope of showing these points exactly. So the scale on our y-axis needs to be
fine enough to approximate −14/45 as about −1/3, but no finer.
4
2
-4
-2
2
-2
-4
11
4
Section 3: Rational and Algebraic Functions
Rational functions are interesting in that the sum, difference, product, quotient and composition of rational functions are again rational functions. In fact rational functions are linear
combinations of horizontal shifts of integral power functions. The topic of the next section is
the decomposition of a given rational function into a sum of simpler rational functions.
In general the inverse (or partial inverse) of a rational function, is a rational function.
2x + 1
2y + 1
. Then to solve for f −1 algebraically we set x = f (y) =
.
x−1
y−1
This means x(y − 1) = 2y + 1. So xy − x = 2y + 1. If we collect all terms with y’s in them on
Example 2: Let f (x) =
the right hand side and move all other terms to the left hand side we arrive at xy − 2y = x + 1.
x+1
.
So y(x − 2) = x + 1. Therefore y = f −1 (x) =
x−2
Difference quotients of rational functions tend to get pretty messy, which motivates generating more powerful methods in calculus.
Linear combinations of rational power functions form the set of algebraic functions. After
some algebra an algebraic function can be place in the form f (x) = g(x)/h(x). As with rational
functions we will emphasize the analysis of algebraic functions where g and h have no common
factors. The domain of an algebraic function is then {x ∈ dom g ∩ dom h|h(x) 6= 0}. As
before we usually need more power to compute the range of a general algebraic function. End
behavior of an algebraic function is determined by the dominate terms from its numerator
and denominator. An algebraic function f intersects the x-axis only when g(x) is zero, and
x ∈ dom h. Also f will have a vertical asymptote when h(x) = 0, and x ∈ dom g.
√
(x − 1)
Example 3 Let f (x) = √
. So g(x) = x − 1 with dom g = IR, and h(x) = x2 − 4x,
x2 − 4x
√
with dom h = {x ∈ dom p(x)|p(x) ∈ dom s(x)} where p(x) = x2 − 4x and s(x) = x and
dom s = [0, ∞). So x ∈ dom h iff x2 − 4x ≥ 0. Which by means of a sign graph we find to
be equivalent to x ∈ (−∞, 0] ∪ [4, ∞). So dom g ∩ dom h = dom h, and
dom f = {x ∈ dom h|h(x) 6= 0} = (−∞, 0) ∪ (4, ∞).
x
x
As x −→ ±∞, f (x) behaves like √ =
. So f (x) −→ 1 as x −→ ∞, and f (x) −→ −1
2
|x|
x
as x −→ −∞. As x −→ 0− , g(x) −→ −1 and h(x) −→ 0+ so f (x) −→ −∞. As x −→
4+ , g(x) −→ 3 and h(x) −→ 0+ , so f (x) −→ ∞.
It is again true that we cannot determine extreme values without graphing/calculus.
Our function f is neither even, nor odd.
The graph of f neither intersects the x-axis, nor intersects the y-axis. After computing a
12
Section 3: Rational and Algebraic Functions
few function values we sketch the following graph.
6
4
2
-2
2
4
6
-2
-4
You might suppose that as the complexity of a function increases (polynomial to rational
function to algebraic function) so does the computation for finding f −1 algebraically, if it’s
even possible.
As with rational functions, algebraic functions have difference quotients that in general
are pretty nasty. Some are not too bad.
√
√
Example 4: Let f (x) = x + 3. Then f (x + h) = x + h + 3, and f (x + h) − f (x) =
√
√
√
x + h + 3 − x + 3. The trick to proceeding is to multiply by 1 in the form ( x + h + 3 +
√
√
√
x + 3)/( x + h + 3 + x + 3), since the resulting numerator is the difference of squares,
1
√
namely (x + h + 3) − (x + 3) = h. So the difference quotient reduces to √
.
x+h+3+ x+3
Exercises:
x2 − 1
. Find the domain of f . Then determine its end behavior,
x3 + 3x2 − 4x − 12
and all asymptotes to its graph. Next find f ’s intercepts. Use all of this information to sketch
1. Let f (x) =
and label a graph of f .
2. Determine a rational function which has vertical asymptotes at x = −1 and x = 2, a
horizontal asymptote at y = 2, and x-intercepts at 0 and 4.
x2 − 3x
3. Repeat exercise 1 for f (x) =
.
x+4
x−1
, or show it can’t be done.
4. Algebraically compute the inverse function for f (x) = √
x+1
1
.
5. Find a simplified form for the difference quotient of f (x) = 2
x −4
13
Section 3: Rational and Algebraic Functions
6. A rectangular region of pasture land is to be fenced using 2, 400 linear feet of fence. Say
that the length of the region is l and the width of the region is w. Express the relation between
l, w and 2, 400. Then express the area, A, of the region as
a) a function of l only.
b) a function of w only.
14
Section 4: Partial Fraction Expansion
As alluded to in the previous section, every rational function is writable as the sum of
p(x)
simpler rational functions. Given a rational function, r(x) =
, where p(x) and q(x) 6= 0
q(x)
are polynomials of degree n and m respectively, we first reduce to the case that the numerator
and denominator have no common factor (so r(x) is in lowest terms).
If n ≥ m, by the division algorithm we can write p(x) = d(x)q(x) + s(x), where the degree
of s(x) is strictly less than the degree of q(x). (s(x) is not zero here since we have already
s(x)
. So every rational function is
canceled common factors.) Now we write r(x) = d(x) +
q(x)
the sum of a polynomial and a proper rational function which is one where the degree of the
numerator is strictly less than the degree of the denominator.
The partial fraction expansion of a proper rational function with non-zero, non-constant
denominator q(x) = (x − s1 )g1 (x − s2 )g2 . . . (x − sl )gl (x2 + c1 x + d1 )h1 . . . (x2 + ci x + di )hi
factored as per the real factors theorem, is the expression of r as a sum of proper rational
functions whose denominators are powers of irreducible factors of q(x).
A theorem, which requires linear algebra to prove, is that there are k + 2l real constants
which make this expression unique, where k is the total number of real zeroes of q(x) and l is
the total number of irreducible quadratic factors of q(x), both numbers reflecting multiplicities.
If (x − s)e is one of the factors of q(x), then the partial fraction expansion of r will have e
Ai
, as i = 1, 2, . . . , e. For each factor (x2 + bx + c)f with b2 − 4c < 0
terms of the form
(x − s)i
Bj x + Cj
the partial fraction expansion of r will have f terms 2
as j = 1, 2, . . . , f .
(x + bx + c)j
x2 − 2x − 3
, we have p(x) = x2 − 2x − 3 and q(x) = x3 − 4x = x(x −
x3 − 4x
2)(x + 2). So r is a proper rational function whose denominator has three real zeroes all with
Example 1 If r(x) =
multiplicity one. So there are three real constants, call them A, B and C so that
r(x) =
A
B
C
+
+
x
x−2 x+2
To find these constants we clear denominators on both sides of the equation by multiplying
through by r(x)’s denominator q(x). So
hA
B
C i
x2 − 2x − 3 =
x(x − 2)(x + 2)
+
+
x
x−2 x+2
Ax(x − 2)(x + 2) Bx(x − 2)(x + 2) Cx(x − 2)(x + 2)
+
+
=
x
x−2
x+2
=A(x − 2)(x + 2) + Bx(x + 2) + Cx(x − 2)
15
Section 4: Partial Fraction Expansion
We then can equate coefficients to determine A, B and C, or we can substitute in three distinct
values for x to get a system of three linear equations in the three unknowns A, B and C. For
example if we set x = 0 in x2 − 2x − 3 = A(x − 2)(x + 2) + Bx(x + 2) + Cx(x − 2), then on
the left hand side we get −3 and on the right hand side we get A(−2)(2) = −4A. We deduce
that A = 3/4. If we next set x = 2 (is there a pattern here?) then on the left hand side we
get 4 − 4 − 3 = −3 and on the right hand side we get B · 2(4). So B = −3/8. Finally we set
x to be the root of x + 2, which is to say x = −2 and get 4 + 4 − 3 = 5 on the left hand side
and C(−2)(−4) = 8C on the right. So C = 5/8. We can check that when we start with
3/4
3/8
5/8
−
+
x
x−2 x+2
and put everything back over the common denominator x(x − 2)(x + 2), we get r(x).
If instead we equate coefficients we first expand the left hand side to
A(x2 − 4) + B(x2 + 2x) + C(x2 − 2x) = (A + B + C)x2 + (2B − 2C)x − 4A. So we arrive at
the three linear equations A + B + C = 1, 2B − 2C = −2, and −3 = −4A. Which is simply an
equivalent system of equations.
x2
, then there are four real numbers A, B, C and D so
(x − 2)2 (x2 + 3)
A
B
Cx + D
that r(x) =
+
+ 2
. Clearing denominators yields x2 = A(x − 2)(x2 +
2
x − 2 (x − 2)
x +3
3) + B(x2 + 3) + (Cx + D)(x − 2)2 Equating coefficients yields the system of equations
Example 2: Let r(x) =
0 =A + C
1 = − 2A + B − 4C + D
0 =3A + 4C − 4D
0 = − 6A + 3B + 4D
which can be solved by substitution, or elimination.
Equivalently we can set x = 2 to obtain the equation 4 = 7B. which means B = 4/7.
√
√
√
√
When x = 3i, we get (C 3i + D)( 3i − −2)2 = . . . = (D + 12C) + i 3(C − 4D) = −3.
So C − 4D = 0 and D + 12C = −3. Which by substitution gives 49D = −3. So D =
−3/49 and C = 4D = −12/49. Finally setting x = 1 gives 1 = −4A + 4B + (C + D).
Substituting B = 4/7, C = −12/49 and D = −3/49 gives 1 = −4A + 16/7 − 15/49. So
A = (1 − 16/7 + 15/49)/ − 4 = 12/49.
16
Section 4: Partial Fraction Expansion
Partial fraction expansions are useful as a tactical device in integrating rational functions
in the second semester of calculus, they are used in circuit analysis when network outputs are
expressed as products of transfer functions and inputs under the Laplace transform. They
are also used in combinatorics/discrete mathematics/computer science to solve recurrence
relations by the method of generating functions. (In fact the last application is simply the
discrete version of the Laplace transform procedure.)
Exercises: Find the partial fraction expansion of each rational function.
1. f (x) =
2. f (x) =
3. f (x) =
4. f (x) =
5. f (x) =
x2 − 1
x3 + 3x2 − 4x − 12
x+4
x2 − 3x
x−1
(x + 1)2
1
2
x −4
3x4 − 2x2 + 4
(x2 + 1)2 (x − 2)
17
Section 5: Exponential and Logarithmic Functions
If a is a positive real number we can use the laws of exponents, and the rules for rational
p
power functions to compute ap/q , where
∈ Q. It can be proven that for any irrational
q
number s there is a sequence r1 , r2 , . . . , rn , . . . of rational numbers with the property that as
m approaches ∞, rm approaches s. We then define the quantity as as the value which arm
approaches as m approaches ∞. The result is the base a exponential function y = f (x) = ax .
Using a sequence of rational values to compute the value of as , where s is irrational is
tantamount to plotting rational points on the graph and filling in the remainder of the graph
by connecting the dots with a smooth curve. This is actually how we plot these curves and
their shifts/scales.
For an exponential function the value a = 1 is uninteresting. Moreover the function
1
1
y = ax , where 0 < a < 1 takes the form y = ( )x = x , where b > 1 and a = 1/b. So any such
b
b
function can be analyzed as the reciprocal of an exponential function whose base is greater
than one.
Of all the exponential functions with base greater than one, the most special is the
natural exponential function y = ex , where e is an irrational number approximately equal
to 2.718281828. The actual value of e is the value (1 + h)1/h approaches as h approaches ∞.
This particular value is very special because the slope of a line tangent to the curve y = ex
at the point (r, er ) is simply the function value er . In fact this is the only base for which the
base a exponential function has this property.
Every exponential function y = ax , where a > 0, and a 6= 1 is one-to-one with domain
IR and range (0, ∞). If a > 1 as x −→ −∞, ax −→ 0+ , while as x −→ ∞, ax −→ ∞. So
every exponential function with base a > 1 has the y-axis as horizontal asymptote and no
vertical asymptote. The reciprocal of such an exponential function has bx −→ 0 as x −→ ∞,
and bx −→ ∞ as x −→ −∞. So it has opposite end behavior in one sense, and similar end
behavior in another.
Since the exponential functions with non-trivial base are all one-to-one, they have no
relative extrema. In fact their function values either always increase as x increases, or always
decrease as x increases. Therefore none of the basic exponential functions are even or odd.
The basic exponential functions are never zero, so the graphs never intersect the x-axis.
They all have y-intercept (0, 1). They are only distinguished from each other by their slopes.
18
Section 5: Exponential and Logarithmic Functions
The difference quotient of y = f (x) = ax reduces as follows
f (x + h) − f (x)
ax+h − ax
ax ah − ax
ah − 1
=
=
= ax
h
h
h
h
Set Ca to be the value so that (ah − 1)/h −→ Ca as h −→ 0. It is shown in calculus that
Ce = 1, moreover e is the only value for which this is true.
The inverse function of ax is the base a log function, denoted loga x, except when a = e
in which case we denote it ln x and call it the natural log function.
As inverses each pair of functions y = ax and y = loga x satisfy x = aloga x = loga ax .
Especially for each positive number a we have a = eln a . Thus by the multiplicative law of
exponents ax = ex ln a . So the base a exponential function is a horizontal scaling of ex by a
factor of 1/ ln a. So if we understand the natural exponential function, we should be able to
understand any other exponential function.
ln x
. So every logarithm function is a
ln a
vertical scaling of the natural log function. Thus it will suffice to understand the natural log
In a similar fashion we can show that loga x =
function. The natural log function has domain (0, ∞) and range IR. The end behavior and
asymptotes of ln x are found using the fact that its graph is the reflection of the graph of ex
across y = x. So as x −→ 0+ , ln x −→ −∞, and as x −→ ∞, ln x −→ ∞.
The natural log function has no relative extrema nor any symmetry for essentially the same
reason the function ex doesn’t. The natural log function has no y-intercept and x-intercept at
(1, 0). We have already discussed the inverse of the natural log. We choose not to discuss its
reciprocal or difference quotient.
However there is another reason the natural log function is so “natural.” For t > 1, ln t
can be defined as the area bounded by the curve y = 1/x, the x-axis, and the two vertical lines
x = 1 and x = t. This is sometimes shown in the second semester of calculus.
Finally we remark that the laws of exponents translate into the arithmetic properties of
the log functions. For example
Fact: For positive numbers a and b, ln(ab) = ln a + ln b.
Proof: We write ab = eln(ab) on the one hand. We write ab = eln a eln b = eln a+ln b by the
additive law of exponents on the other. Since ab = ab and ex is one-to-one, we are done.
The other arithmetic properties of ln x are proven similarly. They include ln ab = b ln a,
and ln ab = ln a − ln b.
19
Section 5: Exponential and Logarithmic Functions
Exercises:
1. Sketch and label a graph of each function
a) y = f (x) = 4x−2 + 1
1
b) y = f (x) = ( )x − 2
3
c) y = f (x) = 3e−2x
e) y = f (x) = ln(x + 1) − 2
d) y = f (x) = ex − e−x
2. Determine the value of a CD in the amount of $3000 that matures in ten years and pays
6% per year
a) twice a month, and b) once every 2 months.
3. Give the exact value of a) log27 3, and b) log3 27.
4. Use the properties of logarithms to write the expression ln
√
3x2
so that the result does
(x − 2)7
not contain logarithms of products, quotients, or powers.
1
5. Rewrite the expression ln(2x) − 2 ln(x + 1) + ln(x − 2) as a single logarithm.
3
6. Solve the equation ln(x − 2) + ln(2x − 3) = ln 3 for x.
20
Section 6: Exponential Growth and Decay
This section concerns the behavior of functions of the form y = ekx , where k ∈ IR − {0}.
If k < 0, then because ln x has range IR, we know that k = ln b, for some real number b.
From the graph of y = ln x we see that 0 < b < 1. So y = ekx = bx , where 0 < b < 1. Such
an exponential function is the reciprocal of an exponential function of the form y = ax , where
a = 1/b > 1. In a twist of fate if k > 0, then k = ln a for some a since ln x has range IR, and
from the graph we have a > 1.
This yields another explanation of why y = ex is the natural exponential function to work
with: The horizontal scale y = ekx of the natural exponential function y = ex decreases if
k < 0 and increases if k > 0. This feature allows us to easily model certain behaviors. The
two cases we model correspond to k > 0 which we call exponential growth, and k < 0 which
we call exponential decay.
For exponential growth we start with a quantity Q0 of some material (Q0 > 0). If the
amount of material grows exponentially as a function of time we get Q(t) = Q0 ekt , where the
value of k tells us how fast Q is growing. The function Q(t) is completely determined by Q0 and
k. In fact k can be determined by any two functional values Q(c) and Q(d), where c < d. Since
Q0 ekd
Q(d)
=
= ekd−kc . Applying ln to both
Q(c) = Q0 ekc and Q(d) = Q0 ekd we get that
Q(c)
Q0 ekc
Q(d)
ln Q(d) − ln Q(c)
sides gives ln
= ln Q(d) − ln Q(c) = kd − kc = k(d − c). So k =
. We
Q(c)
d−c
Q(c)
can also find Q0 since Q0 = kc . These simple functions can be used to model populations,
e
and principal in monetary accounts drawing interest compounded continuously.
Exponential decay works similarly except we model quantities which are decreasing. Common examples include modeling the quantity of radioactive substances, and Newton’s Law of
Cooling.
In the case of modeling the quantity of some radioactive substance, the constant k is
related to the half-life of the substance. The half-life, h, of a radioactive substance is the
1
amount of time it takes for a quantity measuring Q0 units to decay to Q0 units. This means
2
1
1
kh
kh
that Q0 = Q(h) = Q0 e . Which implies that = e . Applying ln to both sides gives
2
2
1
ln = ln 1 − ln 2 = − ln 2 = kh. So given one of h or k, we can compute the other.
2
In the case of exponential growth the corresponding idea to half-life, is called the doubling
time. In this case if d is the doubling time we get dk = ln 2.
21
Section 6: Exponential Growth and Decay
Exercises:
1. A biologist counts 6, 000 bacteria in a culture at 8:00 am. Eight hours later they count
18, 000 bacteria in the culture. Find an expression for the number of bacteria (in thousands)
as a function of time t (in hours), where t = 0 corresponds to 8:00 am. How long, in hours,
will it take for the population to reach 35, 000?
2. A 5kg mass of pretendium decays to 2.2 kg in 3 years. What is the half-life of pretendium
in years?
3. A bottle of water at 68◦ F is placed in a freezer whose temperature is 26◦ F. The temperature
of the water is measured as 54◦ F an hour later. From the time the bottle was initially placed
in the freezer, how long, in hours, will it take for the water to be frozen?
22
Section 7: Conic Sections
The general quadratic equation in x and y has the form Ax2 +Bxy+Cy 2 +Dx+Ey+F = 0,
where A, B, C, D, E and F are real numbers. One goal of this section is to determine what the
graph of such an equation looks like. We easily see that if all three of A, B and C are zero,
then we really have a linear equation as in appendix B. So we begin by supposing that not all
three of A, B and C are zero.
Next we realize that if there is no cross term (i.e. B=0), we can complete the squares on
x and y to simplify our task. If there is a cross term (B 6= 0), we can change our coordinate
system to one where there is no cross term. This involves rotating axes, a trick that the ancient
Greeks discovered. The details of this trick require some trigonometry. This is contained in
appendix E.
As soon as we know that it suffices to consider quadratic equations with no cross terms
we can begin looking at specific cases.
Subsection 1: Parabolas The first case is an equation of the form Ax2 +Cy 2 +Dx+Ey +F = 0,
where exactly one of A and C is zero.
Up to symmetry of argument we suppose that A 6= 0. Thus we have Ax2 +Dx+Ey+F = 0.
D
We next collect x-terms on one side of the equation to get Ey + F = −A(x2 + x). We
A
complete the square on x and simplify to
D
D2
D2
x+
−
)
A
4A2
4A2
D 2 D2
) +
= −A(x +
2A
4A
Ey + F = −A(x2 +
D2
D
and rewrite Ey + F −
= −A(x − h)2 .
2A
4A
If E = 0 we divide through by −A to get
We set h = −
D2
F
−
2
4A
A
2
D − 4AF
=
4A2
(x − h)2 =
√
D2 − 4AF
D2 − 4AF
Thus x − h = ±
=
±
. If D2 − 4AF = 0 we get the “double” line
4A2
2A
(x−h)2 = 0, which is x = h twice. If D2 −4AF < 0 we get no graph. Otherwise D2 −4AF > 0
r
and we have
√
√
D2 − 4AF
D2 − 4AF
and x = h −
x=h+
2A
2A
23
Section 7: Conic Sections
The graph of which is two distinct vertical lines. These three situations are what we call
degenerate cases.
In the nondegenerate case (E 6= 0) we move on by dividing through by E to write
y+
F
D2
−A
−
=
(x − h)2
E
4AE
E
−A
D2 − 4AF
and write y − k =
(x − h)2 . Finally we set c to be the value for
We set k =
4AE
E
1
1
−A
which
=
and get the standard equation of the parabola y − k = (x − h)2 . We can
4c
E
4c
graph this equation by shifting an appropriate scale of y = x2 . The point (h, k) is called the
vertex. The line x = h is called the axis, short for axis of symmetry. Due to an application in
optics the point (h, k + c) is called the focus. The line y = k − c is called the directrix.
A nondegenerate parabola, P may be defined as the set of points equidistant from the
focus and directrix. After translation we may suppose that the focus has coordinates (0, c)
and the directrix, l, has equation y = −c. So if P = {(x, y)|d((x, y), (0, c)) = d((x, y), l)}, we
p
p
have x2 + (y − c)2 = |y − (−c)| = (y + c)2 . So x2 + y 2 − 2cy + c2 = y 2 + 2cy + c2 . Which
becomes x2 = 4cy.
A parabola is also a conic section, the result of intersecting a plane with a double-napped
cone. Here we suppose that the plane intersects only one of the two nappes. Moreover the
plane is parallel to a generator of the cone.
The parabolas we derived above are functions. The graphs open up and down. If we had
worked on the case C 6= 0 instead we would have arrived eventually at the degenerate cases of
a double horizontal line, no graph, and two horizontal lines. We would have then arrived at the
1
the standard equation x − p = (y − q)2 . This is a relation whose graph is a nondegenerate
4c
parabola which opens left or right. Such a parabola has vertex at (p, q), focus at (p + c, q),
axis with equation y = q and x = p − c as directrix.
Subsection 2: Ellipses The second special case of Ax2 + Cy 2 + Dx + Ey + F = 0 to consider
is that neither of A and C is zero and that they are both positive or both negative. If both
are negative we have
−Ax2 − Cy 2 − Dx − Ey − F = 0,
with both −A and −C positive so let us suppose that A, C > 0.
24
Section 7: Conic Sections
We first complete the squares on x and y to write
A(x2 +
D
D2
E
E2
D2
E2
2
x+
)
+
C(y
+
y
+
)
+
(F
−
−
)=0
A
4A2
C
4C 2
4A 4C
D
E
D2
E2
,k=−
, and R = −F +
+
. Thus A(x − h)2 + C(y − k)2 = R.
2A
2C
4A 4C
Since A and C are strictly positive and the quantities (x−h)2 and (y −k)2 are nonnegative
We set h = −
the left hand side of the previous equation is nonnegative. So if R is negative we have a relation
with an empty graph. Also if R = 0 we must have that both (x − h)2 = 0 and (y − k)2 = 0.
This means that the graph of the relation consists of the single point (h, k). If R > 0 and
R
A = C then our equation becomes (x − h)2 + (y − k)2 = . This is the equation of a circle,
A
R
centered at (h, k) with radius squared . These are the degenerate cases.
A
In the nondegenerate case we have R > 0 and A 6= C. Up to symmetry of argument we
can suppose that A < C. Now we let a and b be the real positive numbers with
1
1
A
C
= , and 2 =
2
a
R
b
R
Our choice of A < C forces a > b, which is an important convention. Our equation now reads
(x − h)2
(y − k)2
+
=1
a2
b2
This is the standard form of the equation. Its graph is called an ellipse.
Similar to a parabola an ellipse can be defined as the set of points with sum of distances to
two foci (= focal points) a fixed constant. To verify this, again suppose that (h, k) = (0, 0) and
that the foci are at (±c, 0). Then if E = {(x, y) ∈ IR2 |d((x, y), (c, 0))+d((x, y), (−c, 0)) = k} we
p
p
have (x − c)2 + y 2 + (x + c)2 + y 2 = k. Notice that k > 2c which is the distance between
the foci. Also the points on the x-axis with coordinates (±a, 0) are on E iff a−c+a+c = 2a = k.
p
p
So now we can write (x − c)2 + y 2 = 2a − (x + c)2 + y 2 . Which by squaring both sides
p
gives (x − c)2 + y 2 = 4a2 − 4a (x + c)2 + y 2 + (x + c)2 + y 2 . We cancel the y 2 terms and
p
expand the terms (x ± c)2 to get x2 − 2xc + c2 = 4a2 − 4a (x + c)2 + y 2 + x2 + 2xc + c2 . Which
p
p
reduces to 0 = 4a2 − 4a (x + c)2 + y 2 + 4xc. We rewrite this as a (x + c)2 + y 2 = a2 + xc.
Squaring both sides gives a2 (x + c)2 + a2 y 2 = a4 + 2a2 xc + x2 c2 . Expanding the term (x + c)2
gives a2 x2 + 2a2 xc + a2 c2 + a2 y 2 = a4 + 2a2 xc + x2 c2 . Collecting x terms and cancelling the
2a2 xc terms leaves (a2 − c2 )x2 + a2 y 2 = a4 − a2 c2 = a2 (a2 − c2 ). Here we use the fact that
a > c > 0 and write b2 = a2 − c2 . Now b2 x2 + a2 y 2 = a2 b2 , which reduces to x2 /a2 + y 2 /b2 = 1.
25
Section 7: Conic Sections
In general the graph of an ellipse is an elongated circle. The graph has two axes of
symmetry = lines about which the graph is symmetric. These have equations x = h and y = k
respectively. The line y = k joins the two points (h − a, k) and (h + a, k) on the graph. These
are at distance 2a from each other. The other axis joins the two points (h, k − b) and (h, k + b)
which are at distance 2b from each other. So in the nondegenerate case we call y = k the
major axis, and x = h the minor axis.
If C < A our convention puts our standard equation in the form
(x − h)2
(y − k)2
+
=1
b2
a2
1
C
1
A
=
and 2 = . Now x = h is the major axis and y = k is the minor axis. The
2
a
R
b
R
foci are at (h, k ± c), where again a2 = b2 + c2 .
where
In either case the constant c is less than a. The ratio of c to a is a measure of how
noncircular the ellipse is. This measure is the eccentricity of the ellipse.
Finally we can realize an ellipse as a conic section where the plane intersects one nappe
and is not parallel to a generator of the cone.
Subsection 3: Hyperbolas The last case to consider is Ax2 + Cy 2 + Dx + Ey + F = 0 where one
of A and C is positive and the other is negative. We will start by supposing that we have an
equation of the form Ax2 −Cy 2 +Dx−Ey +F = 0, where A, C > 0. As in the previous section
we complete squares on x and y and simplify our equation by renaming certain quantities. We
arrive at A(x − h)2 − C(y − k)2 = T .
If T = 0 we get A(x − h)2 − C(y − k)2 = 0. We solve this for y in terms of x to get
C
C
y = k + (x − h) or y = k − (x − h). The graph is the union of the two intersecting lines.
A
A
This is the degenerate case.
1
A
1
C
If T > 0 we set a and b to be the positive constants so that 2 =
and 2 =
and get
a
T
b
T
the standard equation
(y − k)2
(x − h)2
−
=1
a2
b2
The line y = k serves as an axis of symmetry. The lines y = k ± a/b(x − h) serve as slant
asymptotes. The foci have coordinates (h ± c, k), where c2 = a2 + b2 .
1
C
1
A
If T < 0 we set a and b to be the positive constants so that 2 =
and 2 =
. The
a
−T
b
−T
standard equation now reads
(x − h)2
(y − k)2
−
=1
a2
b2
26
Section 7: Conic Sections
The graph of the equation is symmetric about x = h and asymptotically approaches the lines
y = k ± b/a(x − h). The foci have coordinates (h, k ± c), where c2 = a2 + b2 .
These are both examples of a hyperbola. A hyperbola is also defined as the set of points so
that the magnitude of the difference of distances to two foci is a fixed constant. For hyperbolas
the ratio of c to a is strictly larger than 1. This ratio is the eccentricity of the hyperbola. The
derivation of the equation of the hyperbola is left to the interested reader.
A hyperbola is also a conic section formed as the intersection of a double napped cone
with a plane which cuts both nappes.
Note on Conic Sections: The intent of this note is to describe the nomenclature used in the
section. To make a right circular cone we use two coplanar lines. We call one, a, the axis
and the other, l, the generator. We do not consider the case that a and l are perdendicular
nor the case that a = l. A right circular cone, C,is generated when l is revolved about a in
a circular fashion. If a and l are parallel, the result is degenerate. We call C a right circular
cylinder. Otherwise a and l intersect in a unique point f called the fulcrum. The result is a
double-napped cone. A conic section is the intersection of a plane P and a right circular cone
C.
If C is degenerate, one possibility is that the intersection of P and C is empty. (This is
either the empty parabola or the empty ellipse.) Another possibility is that P is tangent to
C, in which case the result is the “double” line parabola. When P is not tangent to C but
contains two generators for C we get the degenerate parabola consisting of two parallel lines.
If P is perdendicular to a the result is a circle. Otherwise P ∩ C is a nondegenerate ellipse.
If C is nondegenerate we also get a degenerate section if P contains the fulcrum f . If such
a plane misses both nappes we get a “point” ellipse. When P contains exactly one generator
we get the “double” line parabola. Finally, if P contains two generators (and a), we get the
degenerate hyperbola consisting of two intersecting lines.
If C is nondegenerate and P does not contain f , the only degenerate section occurs when
P is perpendicular to a. The result is a circle.
The last three cases occur when P is not perpendicular to a, does not contain f and C is
nondegenerate. These are the special conic sections which tend to have applications involving
their reflective properties.
We now see that every quadratic equation in x and y must represent a conic section and
27
Section 7: Conic Sections
vice-versa. Hence y = 1/x when written xy − 1 = 0 is recognized as a hyperbola with major
axis the line y = x. A rotation of axes by 45◦ will clear the cross term. So in retrospect we’ve
been discussing conic sections since day one.
Exercises:
1. Find the standard form of the equation for the parabola x2 + 4x − 4y + 8 = 0.
2. Sketch and label a graph of each parabola clearly indicating the vertex, focus, directrix and
axis.
1
(x + 1)2
2
1
b) y − 3 = (x + 1)2
4
1
c) y − 3 = (x + 1)2
8
3. Find the standard equation of a parabola with focus (−1, 2) and directrix x = 1.
a) y − 3 =
4. Sketch and label the graph of the parabola with equation 4x = (y − 1)2 clearly indicating
the vertex, focus, directrix and axis.
5. Sketch and label a graph of each conic section clearly indicating axes of symmetry, foci,
vertices and asymptotes if any.
a) 9x2 + 25y 2 + 36x − 50y − 164 = 0
b) 9x2 − 16y 2 + 36x + 32y − 124 = 0
c) −9x2 + 16y 2 − 36x − 32y − 164 = 0
6. Find the standard equation of an ellipse with foci at (−1, 14) and (−1, −10) which also has
one vertex at (−1, 15).
28
Section 8: Basic Trigonometric Functions
Our goal in this section is to define and develop the six basic trigonometric functions. We
start with two of the more important transcendental functions, namely the sine function and
the cosine function. These are universally denoted by y = sin x and y = cos x respectively.
In order to define sin x and cos x, we first need to agree on how we are going to measure
angles. An angle, θ, consists of two rays emanating from a common point called the vertex.
One ray is called the initial side, the other is the terminal side. Because we can choose where
to put the origin in the Cartesian plane and we can decide where the positive x-axis goes, we
will assume that the vertex of our angle is at the origin and that the initial side is along the
positive x-axis. This is called putting the angle in standard position. Finally, since only the
direction of the rays matters, we can chop off both rays at length one. Our angle theta is now
determined by the two points at the end of these line segments of length 1, both of which lie
on the unit circle whose equation is x2 + y 2 = 1. The point on the initial side is simply (1, 0).
Here we will denote the point on the terminal side by P (θ).
All we need to determine how to measure θ is to know how we generated the angle. That
is we need to know how we go from (1, 0) to P (θ). If we generate θ by moving from (1, 0)
counterclockwise (CCW), we say θ has positive measure. If we generate θ by moving from
(1, 0) clockwise (CW), we say that θ has negative measure. The radian measure of θ is the
length of the arc on the unit circle we travel from (1, 0) to P (θ) when generating θ. We use a
minus sign in front when θ is generated CW.
Since the unit circle has circumference 2π · 1 = 2π, once around the circle CCW is 2π
radians. Halfway around the circle CCW is therefore π radians. One fourth the way around
π
2π
= radians. So we see that P (θ) = P (ϕ) any time that θ and ϕ have
the circle CCW is
4
2
measures which differ by a multiple of 2π radians. So we write P (θ) = P (θ + 2kπ) for all
k ∈ ZZ. Two such angles are called coterminal.
We call the x-coordinate of P (θ) the cosine of θ and denote it by cos θ. The y-coordinate
of P (θ) is the sine of θ and is denoted by sin θ. It is fairly clear that these are functions of
θ where θ, as a length, can be any real number. Therefore both the sine function and cosine
function have domain IR, unless restricted for some reason.
Since all the points on the unit circle have coordinates bounded between 1 and −1, we
have that −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1. By the horizontal line test every one of these
29
Section 8: Basic Trigonometric Functions
values is taken on, so the sine function and the cosine function both have range [−1, 1]. These
serve as our first examples of functions with bounded range. Meaning there is a number M
with f (x) ≤ M , for all x ∈ dom f .
Since P (θ) = P (θ + 2kπ) for all k ∈ ZZ, we have sin(θ) = sin(θ + 2kπ), and cos(θ) =
cos(θ + 2kπ) for all k ∈ ZZ. A function which repeats its values regularly is called periodic.
The period of a periodic function is the length of the smallest continuous input interval between
repetitions. Here we see that sin x and cos x have period 2π. Thus we would not be able to
describe the behavior of sin x as x −→ ∞ except to say that the values oscillate.
Neither the sine function nor the cosine function have any asymptote.
The cosine has maximum value 1 and attains this value at θ = 0 + 2kπ, where k ∈ ZZ. The
cosine has minimum value −1 and attains this at θ = π + 2lπ = (2l + 1)π, where l ∈ ZZ. The
sine function attains maximum value of 1 when θ = π/2 + 2kπ = (4k + 1)π/2, where k ∈ ZZ.
The minimum value of the sine function is −1 and occurs when θ = 3π/2 + 2lπ = (4l + 3)π/2,
where l ∈ ZZ.
For a given point P (θ) on the unit circle there are several other points which are natural
to consider. The first is the antipodal point on the opposite side of the unit circle. This
point has coordinates (− cos θ, − sin θ). It is also P (θ + π). We thus derive the relationship
π
sin(θ + π) = − sin θ. Another point to consider is radians CW from P (θ). It has coordinates
2
(sin θ, − cos θ) and is P (θ − π/2). Thus cos(θ − π/2) = sin θ, which means that the graph of
the sine function is the graph of the cosine function shifted right π/2 units. Equivalently the
graph of the cosine function is the graph of the sine function shifted left π/2 units and we have
that cos θ = sin(θ + π/2).
In short to determine the graphs of these functions, it suffices to graph the sine function
for 0 ≤ θ ≤ π and apply all the relations we found above. This is made even easier when we
use the fact that the unit circle is symmetric about the y-axis, giving us sin θ = sin(π − θ). So
we only need to determine values for 0 ≤ θ ≤ π/2.
Before we do so we remark that the x-intercepts of the sine function have x-coordinates
nπ, where n ∈ ZZ. The x-coordinates of the intercepts of the cosine function are (2m + 1)π/2,
where m ∈ ZZ.
Now sin 0 = 0, and sin π/2 = 1. Also as θ increases from 0 to π/2, the value of the sine
√
function will increase. In the next section it is derived that sin π/6 = 1/2, sin π/4 = 2/2,
30
Section 8: Basic Trigonometric Functions
and sin π/3 =
√
3/2. With just these five values (which you should memorize), we can sketch
a remarkably good graph of z = f (θ) = sin θ. We can also therefore graph its shift left by π/2
units, z = cos θ.
The reciprocal of the sine function is the cosecant function, and is denoted csc x. It has a
vertical asymptote when sin x = 0. It has a relative minimum when sin x = 1, and a relative
maximum when sin x = −1. It is also periodic with period 2π.
The reciprocal of the cosine function is the secant function, and is denoted sec x. It has
period 2π, and is a shift left π/2 units of the cosecant function.
The ratio of the sine function to the cosine function is the tangent function, denoted
tan x = sin x/ cos x. This function is zero when sin x = 0, i.e. x = nπ, where n ∈ ZZ. It has a
vertical asymptote whenever cos x = 0, which is when x = (2m + 1)π/2, where m ∈ ZZ. Since
−/− is +, the tangent function has period π. As the ratio of an odd function and an even
function the tangent function is odd. Finally as x −→ (π/2)− , sin x −→ 1, and cos x −→ 0+ ,
so tan x −→ ∞. On the other hand as x −→ (π/2)+ , tan x −→ −∞. The reciprocal of the
tangent function is the cotangent function, denoted cot x. Collectively these make up the six
basic trigonometric functions. We normally concentrate on sin x, cos x, tan x, and sec x.
We will discuss inverse trigonometric functions, and the difference quotients of trigonometric functions in later sections.
Several last points need to be made here. First of all, for any θ ∈ IR, there is a value r ∈ IR,
called the reference angle, so that 0 ≤ r ≤ π/2, with | sin(θ)| = sin r and cos r = | cos(θ)|. For
r 6= 0, π/2, there is therefore a unique right triangle of hypotenuse 1 which has base cos r and
height sin r. This is the reference triangle for θ. We investigate this connection with right
triangles further in a later section. Second, since the point P (θ) lies on the unit circle we have
that (sin θ)2 + (cos θ)2 = 1, for all θ ∈ IR. Since this follows from the Pythagorean Theorem,
this is called the Pythagorean Identity. This is one of the more important of hundreds of
trigonometric identities.
Exercise: Read sections 9 and 10.
31
Section 9: Special Values of Trigonometric Functions
A crucial theorem in trigonometry is Euclid’s Theorem on similar triangles. The triangle
with vertices at A, B and C is denoted by 4ABC. Two triangles 4ABC and 4DEF are
similar provided that their interior angles have the same measure. So the angle with vertex
at A has the same measure as the angle with vertex at D, etc. We state the theorem without
proof.
Theorem: (Euclid) 4ABC with side lengths a = |AB|, b = |BC|, and c = |AC| is similar to
b
c
a
4DEF with side lengths d = |DE|, e = |EF | and f = |DF | if and only if = = .
d
e
f
This theorem allows us to define the trigonometric functions using P (θ). First we draw
a line which touches the unit circle only at P (θ). Such a line is said to be tangent to the
curve at the point P (θ). (You might be asking how this can be done.) Except when the line
is horizontal or vertical (i.e. θ is a multiple of π/2) it has both an x-axis intercept D, and a
y-intercept E. Call the origin O, and put B = (cos θ, 0). Then 4OP B is similar to 4ODP .
|OP |
|OD|
1
|OD|
=
. That is,
=
. So since sec θ and cos θ have the same sign
Therefore
|OP |
|OB|
1
| cos θ|
the coordinates of D are (sec θ, 0). Similarly the coordinates of E are (0, csc θ).
A vertical line is tangent to the circle only when θ = kπ. For these values of θ the sine
function is zero. Hence the cosecant is undefined. Notice though that the coordinates of D
are still (sec θ, 0). The anagolous result holds true for horizontal tangents to the unit circle.
|P D|
|P B|
|P D|
| sin θ|
To define the tangent we note that
=
. Which is to say
=
. So
|OP |
|OB|
1
| cos θ|
the length from P to D measures the absolute value of the tangent of θ. In fact we see that
the tangent line to the curve at P (θ) is perpendicular to the line, l, through O and P , so the
slope of the tangent line is the negative reciprocal of the slope of l. But the slope of l is simply
sin θ
(rise over run)
. So the value of tan θ tells us how to draw the tangent line to the unit
cos θ
circle at P (θ)! The point is that the definitions we made for trigonometric functions in section
8 did not fall out of a hat, they have geometric significance.
Another application of Euclid’s Theorem is that given any right triangle with an included
angle θ, where 0 < θ < π/2, we can define the six trigonometric functions as ratios of the
lengths of the sides of the triangle. To do this draw the given triangle with the right angle in
the lower right hand corner and θ in the lower left corner. Let the length of the base be a,
the height be b, and the length of the hypotenuse be c. The other triangle is drawn with θ in
32
Section 9: Special Values of Trigonometric Functions
standard position in IR2 with vertices (0, 0), (cos θ, 0), and (cos θ, sin θ). This triangle has base
of length cos θ, height sin θ, and hypotenuse of length 1. By Euclid’s Theorem the ratios of
corresponding side lengths are equal so we have
1
sin θ
cos θ
=
=
c
b
a
So given two sides, for example a and c, we have an equation relating the two sides which can
rewritten as an equation describing their ratio. In this case 1/c = cos θ/a implies a/c = cos θ.
Equivalently c/a = 1/ cos θ = sec θ. The other formulas are b/c = sin θ, which is equivalent to
c/b = csc θ, and b/a = tan θ, which is equivalent to a/b = cot θ.
The famous mnemonic here is “soh-cah-toa” which is an acronym for “sine is opposite
over hypotenuse–cosine is adjacent over hypotenuse– tangent is opposite over adjacent.”
The sum of the measures of the angles of a triangle is 180◦ , which is π radians. So for a
right triangle with included angle θ the other nonright angle ϕ must have measure (π/2) − θ
radians. We call the two angles θ and ϕ complementary. Notice that the value of f (ϕ) = cof (θ)
for any of the six trigonometric functions f , when defined as ratios of lengths of sides. Since
the complement of the complement of θ is simply θ we get the mnemonic device “co-co=no-co”.
The fact that a triangle has three sides explains why there are three pairs of complementary
trigonometric functions.
Finally, to demonstrate the utility of these concepts we derive some special values of the
basic trigonometric functions.
√
First take an isosceles right triangle with side s. The hypotenuse has length s2 + s2 =
√
√
2s2 = s 2. Also the included acute angles must be equal and have measure π/4. Therefore
√
√
√
√
cos π/4 = s/s 2 = 1/ 2 = 2/2 = sin π/4. Also tan π/4 = 1 = cot π/4, and sec π/4 = 2 =
csc π/4. Moreover any value of θ with reference angle π/4 has |f (θ)| = f (π/4), where f is any
one of the six trigonometric functions. So to determine the actual value of f (θ) we simply need
to know whether its positive or negative, which can be done since we know what quadrant the
point P (θ) lies in, so we know the sign of its coordinates sin θ and cos θ. This is sufficient since
all of the other basic trigonometric functions are defined in terms of sine and cosine.
Second take an equilateral triangle of side s, drawn with horizontal base. The interior
angles are all π/3. If we drop a perpendicular from the top vertex to the base, we split the
triangle into two right triangles. The one on the left has base s/2 and hypotenuse s. By the
33
Section 9: Special Values of Trigonometric Functions
√
Pythagorean theorem the height is s 3/2. The complementary angle has measure π/6, so we
√
have cos π/6 = sin π/3 = 1/2, sin π/6 = cos π/3 = 3/2. The values of the other four basic
trigonometric functions are determined as before.
Identical to the first case, we can now find exact values of trigonometric functions for any
angle which has reference angle π/6, or reference angle π/3.
Example 1: Let’s find the exact value of some trigonometric functions at 4π/3. The reference
angle is π/3 and P (4π/3) is in the third quadrant where both x and y coordinates are negative.
√
Therefore cos 4π/3 = − cos π/3 = −1/2, and sin 4π/3 = − sin π/3 = − 3/2. Then for example
√
tan 4π/3 = 3, since the ratio of two negative numbers is positive.
Example 2: If we have a right triangle with base of length 12 and height 5, the Pythagorean
theorem gives 13 as the length of the hypotenuse. In this case if θ denotes the angle adjacent to
the base and hypotenuse, sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, sec θ = 13/12
and csc θ = 13/5. Moreover we know how to find lots points on the unit circle which have a
similar reference triangle.
Example 3: If we have an isosceles right triangle of side 4, then if h denotes the length of the
√
√
hypotenuse we have 4/h = sin(π/4) = 2/2. Here we write 8 = h 2 by cross multiplying and
√
√
can solve for h = 8/ 2 = 4 2.
It seems reasonable that we should be able to perform exercise 3 for any angle, θ, where
0 < θ < π/2. And of course it is possible, if we have inverse trigonometric functions. As with
even power functions, due to their periodic nature trigonometric functions are not one-to-one.
Therefore if we wish to invert them we must restrict their domain to a subset where they are
one-to-one. In order to complete exercises like example 3 we want this restricted domain to
include (0, π/2) and to be of most value it would be as large as possible.
For y = sin x we use the convention that its domain is [−π/2, π/2] when discussing its
inverse function denoted arcsin x. Formally the function is defined by the equations y = arcsin x
iff x = sin y. Which means that we can draw a reference triangle with enclosed angle |y|,
√
opposite side of length |x| and hypotenuse of length 1. The remaining side has length 1 − x2
by the Pythagorean theorem. This construct allows us to compute tan arcsin x = tan y =
√
x/ 1 − x2 .
When we start instead with y = cos x we use the convention that its domain is [0, π] and
34
Section 9: Special Values of Trigonometric Functions
denote its inverse by arccos x.
The conventional domain for y = tan x as an invertible function is (−π/2, π/2). Now we
have y = arctan x iff x = tan y. So we can draw a reference triangle with enclosed angle |y|,
opposite side of length |x| and adjacent side of length 1. The hypotenuse therefore has length
√
√
x2 + 1, and we can write for example cos arctan x = cos y = 1/ x2 + 1. Notice that here the
cosine function always has positive value which agrees with our convention about the domain
of a composition of functions.
One last picture to consider is the one flowing from considering sec x as an invertible
function with inverse denoted arcsec x. Now we suppose that the domain of the secant function
is limited to [0, π/2) ∪ (π/2, π]. By definition y = arcsec x iff x = sec y. So we can draw a
right triangle with included angle |y|, hypotenuse of length x, adjacent side of length 1, and
√
√
opposite side of length x2 − 1. Now for example csc arcsec x = csc y = x/ x2 − 1.
The restricted domain for cot x as an invertible function is (0, π). For y = csc x to be
considered invertible our convention is to limit its domain to [−π/2, 0) ∪ (0, π/2].
Exercises:
1. Find the exact value of each of the following:
µ ¶
³π´
2π
a) sin
b) tan
3
3
µ
¶
µ ¶
−π
7π
d) csc
e) cos
6
6
µ ¶
3π
g) sin (π)
h) cot
4
j) cos (3π)
k) tan (π)
2. Find the exact value of each of the following:
a) tan arccos (−3/5)
b) sin arccot (1/2)
µ
¶
−5π
c) cos
4
µ
¶
−π
f) sin
3
µ ¶
3π
i) sec
2
³π ´
l) csc
6
c) sec arccsc (2/3)
3. Sketch the picture described in the second paragraph of this section. Label the points O,
P , B, D, and E.
4. Sketch and label a triangle in which:
a) x = sin y
b) x = tan y
35
Section 10: General Sinusoids
A general sinusoid is a function of the form y = k + A sin(B(x − c)) or y = k + A cos(B(x − c)),
where k, A, B and c are real numbers. Since the sine function and cosine function are horizontal
shifts of each other we will only consider the form y = k + A sin(B(x − c)). Such a function is a
shift right by c units (left if c < 0), of a compression horizontally by a factor of 1/B (expanded
if B < 1), of a vertical scale by A units, of a shift up by k units (down if k < 0) of the basic
function y = sin x. For such a function |A| is the amplitude. About the central line y = k
the sine wave will fluctuate up |A| units and down |A| units. All function values are bounded
below by k − |A| and above by k + |A|. The bounds are obtained when sin(B(x − c)) = ±1,
which means that B(x − c) = (2n + 1)π/2, where n is an integer. This is equivalent to
x = c + [(2n + 1)π/2B]. The general sinusoid is periodic with period 2π/B.
π
Example 1: Let’s sketch a graph of y = f (x) = 2 − 3 sin(3x − ). First we put the function
2
π
in standard form y = f (x) = 2 − 3 sin(3(x − )). Then it’s a matter of determining where
6
f takes on maximal and minimal values, since half-way between it will take value 2. The
extreme values happen when 3(x −
π
6)
= (2n + 1)π/2. So x −
π
6
= (2n + 1)π/6. Thus
6x − π = (2n + 1)π = 2nπ + π. So 6x = (2n + 2)π, or x = (n + 1)π/3. We see that f (π/3) = −1
and f (2π/3) = 5. So f (π/2) = 2, and the rest is determined by symmetry. Of course we need
to mark our x-axis off in units of π/6 to be able to clearly indicate where f takes on its extreme
values and its central values. The y-axis needs to be marked so that all values between −1
and 5 can be seen. Elongating the x-axis a little gives a sketch which is smoother – more like
the basic sine wave. We can also shorten the y-scale somewhat to acheive the same effect.
6
4
2
−π
− 2π
3
− π3
π
3
-2
36
2π
3
π
Section 10: General Sinusoids
It is also useful to be able to combine two (or more) general sinusoids arithmetically.
Given two sinusoids y = k1 + A1 sin(B1 (x − c1 )) and y = k2 + A2 sin(B2 (x − c2 )), their sum
is is shift up k1 + k2 units of y = A1 sin(B1 (x − c1 )) + A2 sin(B2 (x − c2 )). This function is
bounded since −(|A1 | + |A2 |) ≤ y ≤ |A1 | + |A2 |. The bounds may or may not be met. To
graph this function we find scales for the x-axis and y-axis that allow both simpler functions
to be graphed simultaneously. Then we add ordinates. Ordinate is a fancy name for the
y-coordinate.
Example 2: Sketch a graph of y = 3 sin x + sin(3x). First of all the outputs are bounded
between −4 and 4. Second the function y = 3 sin x has period 2π, and is simply a vertical
scale of the regular sine function. Third the function sin 3x has period 2π/3. So it is a
horizontal compression of the sine wave by a factor of 3. To show 2 periods of each component
function simultaneously we need an interval of x-values of length 4π. We might as well take
x ∈ [−2π, 2π] since the overall function is odd. We do need to mark off multiples of π/6 on the
x-axis in order to see all zeroes, and extreme values of sin 3x. We can also make two copies of
our axes, one for scratch work, the other for the final product (neatness counts).
4
2
π
−π
−2π
2π
-2
-4
Exercises:
1. For each of the following general sinusoidal functions identify the amplitude A, the vertical
shift k, the period p, and the phase shift c. Then sketch and label a graph of the function
on the given interval. In each instance choose scales for your axes which allow all important
details to be clearly shown. At least two periods of the function should be shown.
a) y = 2 − 3 cos(2πx +
π
)
2
37
Section 10: General Sinusoids
b) y = 1 + 2 sin(
x π
− )
2
2
c) y = −3 + sin(3x)
2. Sketch and label a graph of each of the functions below on the specified interval. In each
instance choose scales for your axes which allow all important details to be clearly shown. At
least two periods of the resulting function should be graphed.
a) y = 3 sin πx + 2 sin(2πx)
b) y = 2 cos(2x) + cos(4x)
38
Section 11: Trigonometric Identities
We’ve already seen many basic identities which trigonometric functions satisfy. For example if f is a basic trigonometric function, and g is its co-function, then f (x) = g(π/2−x), as long
as both sides of the equation make sense. These are called the co-function identities. Another
example are what are called the reciprocal identities, such as csc x = 1/ sin x. A third example is the quotient identity, tan x = sin x/ cos x. Finally, consider the Pythagorean identity
sin2 x + cos2 x = 1.
For each of these identities there is usually an equivalent way of stating the identity. For
example we may rewrite the Pythagorean identity as sin2 x = 1−cos2 x, or as cos2 x = 1−sin2 x.
We can even divide the equation through by cos2 x to obtain tan2 x+1 = sec2 x, after appealing
to algebra, the quotient identity, and a reciprocal identity. This latest version of the identity is
not true for all real x, but is true as long as cos2 x 6= 0, which means x 6= (2n+1)π/2. One game
we might play is to continue generating equivalent identities from old using other identities.
The result is a very impressive list of identities, too many for most people to memorize. So
instead of recommending that people memorize every form of an identity, we recommend that
they remember one form of the identity and develop the ability to re-derive the rest. This is
very much like the difference between giving someone a fish, and teaching them how to fish.
To build this skill a common practice is to use an exercise such as the following.
Example 1: Verify that 1 + cot2 x = csc2 x is an identity.
Solution :
cos2 x
sin2 x
sin2 x + cos2 x
=
sin2 x
1
=
sin2 x
= csc2 x
1 + cot2 x = 1 +
(quotient identity)
(algebra)
(Pythagorean identity)
(reciprocal identity)
Notice in our solution we did not start by affirming the conclusion. That is we did not
suppose that the identity was true. We simply started with the left hand side and continued
to apply identities until we arrived at the right hand side. Not only is this process logically
proper, it is clearly reversible. So when we verify an identity we will use this process. Next
39
Section 11: Trigonometric Identities
realize that every step was justified as either algebra or as the application of an identity. While
this is not mandatory it is suggested. Third, observe that we did not show every single step of
algebra. The real challenge is for each individual to learn to take as many steps as they need
to verify the identity. Some people are more adroit than others combining algebraic steps. So
two valid solutions might have a different number of delineated algebraic steps.
Now to demonstrate the power of this process let’s derive one identity and develop a
variety of other identities from it.
Example 2: cos(θ − φ) = cos θ cos φ + sin θ sin φ, for θ, φ ∈ IR.
Proof: Let θ, φ ∈ IR. On the unit circle the square of the distance from P (θ) to P (φ) is the
same as the square of the distance from P (θ − φ) to P (0). Therefore by the distance formula
we have
(cos θ − cos φ)2 + (sin θ − sin φ)2 = (cos(θ − φ) − 1)2 + (sin(θ − φ) − 0)2
After expansion the left hand side is cos2 θ − 2 cos θ cos φ + cos2 φ + sin2 θ − 2 sin θ sin φ + sin2 φ.
By applying the Pythagorean identity twice we get 2 − 2 cos θ cos φ − 2 sin θ sin φ. Meanwhile
the right hand side becomes cos2 (θ −φ)−2 cos(θ −φ)+1+sin2 (θ −φ). Which after applying the
Pythagorean identity once becomes 2 − 2 cos(θ − φ). Cancelling the 20 s and dividing through
by −2 yields the identity. This is called the difference identity for cosine.
From the difference identity for cosine we can derive the sum identity for cosine,
Theorem: cos(θ + φ) = cos θ cos φ − sin θ sin φ.
Proof: Here we simply write cos(θ + φ) = cos(θ − (−φ))
= cos θ cos(−φ) + sin θ sin(−φ)
= cos θ cos φ − sin θ sin φ
since cos(−φ) = cos φ and sin(−φ) = − sin φ.
Similarly we can prove the sum identity for sine.
Theorem: sin(θ + φ) = sin θ cos φ + sin φ cos θ
40
Section 11: Trigonometric Identities
Proof: We use sin((π/2) − x) = cos x, sin(−z) = − sin z and cos(y − (π/2)) = sin y.
π
sin(θ + φ) = cos((θ + φ) − )
2
π
= cos(θ + (φ − ))
2
π
π
= cos θ cos(φ − ) − sin θ sin(φ − )
2
2
π
= cos θ sin φ + sin θ sin( − φ)
2
= cos θ sin φ + sin θ cos φ
The rest is a matter of rearranging terms.
The difference identity for sine is sin(θ − φ) = sin θ cos φ − sin φ cos θ. We leave the verification of this identity as an exercise.
As interesting corollaries we get the double angle formulas which are
2 tan x
. For example
cos 2x = cos2 x − sin2 x, sin 2x = 2 sin x cos x, and tan 2x =
1 − tan2 x
Theorem: cos 2x = cos2 x − sin2 x
Proof: cos(2x) = cos(x + x) = cos x cos x − sin x sin x = cos2 x − sin2 x.
Continuing, we can produce cos 2x = cos2 x − sin2 x = cos2 x + sin2 x − sin2 x − sin2 x =
1 − 2 sin2 x. This is an equivalent version of the double angle formula for cosine. One can also
show that cos 2x = 2 cos2 x − 1.
Each of the last two versions of the double angle formula for cosine can be rewritten. For
1 − cos 2x
. Which means that the function sin2 x can be analyzed via the
example sin2 x =
2
cosine function using shifting and scaling. Or if we wish we can derive one of the half-angle
formulas.
r
1 − cos x
x
Theorem: sin = ±
2
2
Proof: Set y = x/2, so 2y = x. Then sin2 (x/2) = sin2 y = (1 − cos 2y)/2 = (1 − cos x)/2. The
result follows by taking square roots of both sides, realizing that the answer may be positive
or negative.
As a nice application of these formulas we have the following examples.
Example 3: Find the exact value of sin(5π/12).
41
Section 11: Trigonometric Identities
8π 3π
2π π
5π
= sin(
−
) = sin(
− )
12
12
12
3
4
2π
π
π
2π
= sin
cos − sin cos
3
4
4
3
√ √
√
3 2
21
=
−
2 2
2 2
√
√
6− 2
=
4
7π
Example 4: Find the exact value of cos
.
8
Solution: sin
Solution: Since 7π/8 is in the second
s quadrant
s
√
p
√
1 + cos 7π
1 + 22
2+ 2
7π
4
cos
=−
=−
=−
8
2
2
2
Example 5: If we know that cos θ = 4/5 and 3π/2 ≤ θ ≤ 2π then cos 2θ = 2 cos2 θ − 1 =
2(16/25) − 1 = 7/25. While because θ is in quadrant four the Pythagorean identity implies
sin θ = −3/5. So sin 2θ = 2 sin θ cos θ = −24/25. Therefore tan 2θ = −24/7, and we can also
compute exact values for csc 2θ and cot 2θ.
The previous example can be inverted by giving exact value of cos x, for example, and
quadrantal information about x, to enable using the half-angle formulas to compute exact
values of any of the trigonometric functions at
x
2.
Identities are also important since they may be used to simplify a problem. For example
a3
we put u = a tan θ. We find that a2 + u2 =
if we want to analyze the formula z = 2
(a + u2 )3/2
a2 + a2 tan2 θ = a2 (1 + tan2 θ) = a2 sec2 θ. So (a2 + u2 )3/2 = (a2 sec2 θ)3/2 = a3 sec3 θ,
and z = cos3 θ. That is, trigonometric identities and composition of functions may allow us
to change venue to a friendlier setting. To be sure no one has all trigonometric identities
memorized, and many of them are not initially obvious. However we can for instance go one
step further and write z = (cos 3θ + 3 cos θ)/4 which is a restatement of one of the homework
exercises. This version of z as a function of θ can be visualized using addition of ordinates.
Yet another application of trigonometric identities is in simplifying difference quotients of
sin h
−→ 1,
trigonometric functions. For example by graphing we can observe that as h −→ 0,
h
cos h − 1
while
−→ 0. We then have
h
sin(x + h) − sin x
sin x cos h + sin h cos x − sin x
=
h
h
³ cos h − 1 ´
³ sin h ´
= sin x
+ cos x
h
h
42
Section 11: Trigonometric Identities
So as h −→ 0, the difference quotient of the sine function approaches cos x.
As a final demonstration of the power of these identities you may now read appendix E
on rotation of axes to clear cross terms in general quadratic equations in two variables.
Exercises:
1. Verify the following trigonometric identities.
a) sin(θ − φ) = sin θ cos φ − sin φ cos θ
tan θ + tan φ
b) tan(θ + φ) =
1 − tan θ tan φ
c) sin 2x = 2 sin x cos x
2 tan x
d) tan 2x =
1 − tan2 x
1 + cos 2x
e) cos2 x =
2
3
f) cos 3x = 4 cos x − 3 cos x
r
x
1 + cos x
g) cos = ±
2
2
1 − cos x
x
h) tan =
2
sin x
1
i) sin α cos β = (sin(α − β) + sin(α + β))
2
1
j) cos α cos β = (cos(α − β) + cos(α + β))
2
1
k) sin α sin β = (cos(α − β) − cos(α + β))
2
2. Use trigonometric identities to find the exact value of each of the following.
π
12
7π
b) sec
12
5π
c) csc
8
a) cot
−4
and that θ is in quadrant III. Find the exact values of the
5
other five trigonometric functions at θ.
3
4. Given that θ is in quadrant I and that cos 2θ = − , find the exact values of the six
5
trigonometric functions at θ.
3. Using the fact that sin θ =
5. Read appendix E.
43
Section 12: Applications of Trigonometry
Whenever a section in a math book has applications in the title it means “word exercises.”
So now you know what you will mostly face in this section. The reason that many people have
difficulty with word problems is that their solutions tend to be more involved than the “plug
and chug” exercises otherwise found in text books. In fact the next to last part of a word
problem usually is completing a particular type of “plug and chug” exercise. The difficulty is
arriving at that stage. Here follows a short description of what is supposed to happen.
Step 1: Read the problem carefully, several times. Make sure you understand what is being
asked.
Step 2: Draw a picture or diagram.
Step 3: Label all known quantities including units, such as ft., lbs. etc. Usually word problems
do not contain superfluous information. Thus any information given is vital.
Step 4: Label all unknown quantities with distinct variables. In contrast to step 3, some
unknown quantities may be irrelevant. With practice should come some facility in deciding
which unknown quantities you can omit. Any unknown quantity which you are being asked
to find must be labeled.
Step 5: This is the hard part. Write down as many equations/relations/identities as you can
which relate the known and unknown quantities. Some of these will be irrelevant. As before,
with practice will come some ability to determine which equations are to be used. The hint is
that the equations usually have something to do with the title of the section.
Step 6: After eliminating as many variables as possible, hopefully you have reduced the word
problem to a “plug and chug” exercise.
Step 7: When you have solved the appropriate exercise, you should carefully and neatly
re-write your solution eliminating all irrelevant unknown quantities,labels,equations,relations
and identities. The finished product should look like the examples, possibly minus some of the
prose.
Before we get to the fun we need to develop a few more tools.
First consider a triangle drawn with a horizontal base and one vertex labelled B above it.
Label the left end point of the base A, the right end point C. Label the angle with vertex A,
α, the angle with vertex B, β, and the angle with vertex C, γ. Finally label the length of the
44
Section 12: Applications of Trigonometry
side opposite A, a, the length of the side opposite B, b, and the length of the third side c.
Drop a perpendicular from B to the line the base lies in. Call the height of the perpendicular h, and the intersection of the perpendicular with the base P . If the perpendicular coincides
with the side CB we have a right triangle and c2 = a2 + b2 . Otherwise the perpendicular splits
the triangle into two right triangles AP B and BP C.
The length from A to P is |c cos α|. Therefore the length from P to C is b − c cos α. Also
we have c2 = h2 + c2 cos2 α and a2 = h2 + (b − c cos α)2 . From the first equation we have
h2 = c2 − c2 cos2 α so by substitution
a2 = h2 + (b − c cos α)2
= c2 − c2 cos2 α + b2 − 2bc cos α + c2 cos2 α
= c2 + b2 − 2bc cos α
This is the law of cosines. Since the parts of the triangle can be rearranged we also have
b2 = a2 + c2 − 2ac cos β, and c2 = a2 + b2 − 2ab cos γ.
Another tool arises from this scenario since in the triangles above we have h = a sin γ =
sin γ
sin β
sin α
=
. In fact whatever this common ratio is it must equal
.
c sin α. Therefore
a
c
b
This is the law of sines.
Though this law is simpler, and therefore more easy to use, it is more dangerous to use
too. If we are only given a, c and α, then if a < c sin α, there is no triangle with these
properties. If a = c sin α , then we get a unique right triangle where γ is the right angle. When
a ≥ c(≥ c sin α) we again get only one triangle with these properties. For this triangle γ must
be acute. However when c sin α < a < c, there are two angles γ1 and γ2 , where γ1 + γ2 = π
and γ1 is obtuse while γ2 is acute. So when applying the law of sines we must be careful to
consider whether our answer makes sense physically.
Now for the fun. A fairly representative exercise whose solution uses trigonometry is the
following.
Example 1: An executive from a window washing company is putting in a bid for the weekly
washing of the windows of a skyscraper. To set his price he needs to know approximately how
tall the building is. His ex-wife works for the city planning office, so he doesn’t dare call them
to ask for the information. Instead he moves away from the building 101 feet and measures
the angle of elevation from that point to the roof of the building as 73◦ = 73π/180 radians.
45
Section 12: Applications of Trigonometry
Thus the height of the building h satisfies h/101 = tan(73◦ ). So h is about 101 tan(73◦ ) ≈ 330
feet.
Example 2: Fred and George are playing wallyball. George wants to pass the ball to Fred but
a very tall opposing player is in the way. Fred is standing 4 feet away from a wall which George
√
is standing 3 feet from. Fred and George are standing 145 feet apart and George decides
that he wants to make a chest-high carom-pass off the wall to Fred. How far along the wall
between them should George pass the ball.
Solution: Let the point where George is standing be labelled A, and the point where Fred is
standing B. Drop perpendiculars from A and B to the wall and label these points C and D
respectively. Let P be the point on the wall which George should hit. Denote the length from
P to C as x, so the length from P to D is 12 − x. Let θ be the angle with vertex P , initial side
along the wall, and terminal side P A. Basic physics tells us that θ should also be the angle with
x
12 − x
vertex P , one side along the wall, and the other side P B. Therefore = cot θ =
. Thus
3
4
1
4x = 3(12 − x) = 36 − 3x, which means x = 36/7 = 5 feet. By the way x/3 = 12/7 = cot θ,
7
so θ = arccot (12/7) ≈ 30.2◦ .
Example 3: A car leaves a little town on a road traveling due South at 75 mph. Twenty minutes
later a truck heads out of town traveling due East at 55 mph. How far apart in miles are the
vehicles forty-five minutes after the truck departed?
Solution: Let t = 0 correspond to when the truck departed and let us measure t in hours. Let
a(t) denote the distance in miles of the car to the town as a function of t, and b(t) denote the dis75
+ 75t = 25 + 75t
tance of the truck in miles to the town as a function of time. Then a(t) =
3
and b(t) = 55t. By the Pythagorean theorem the distance between them in miles as a function
of time t in hours is
d(t) =
So d(3/4) = 5
miles.
p
p
p
(25 + 75t)2 + (55t)2 = 625 + 3750t + 8650t2 = 5 25 + 150t + 346t2
p
√
√
25 + 150(3/4) + 346(9/16) = (5/4) 400 + 1800 + 3114 = (5/4) 5314 ≈ 91.1
Example 4: A passenger jet leaves an airport at 12:00 pm traveling due west at 430 mph. A
private jet leaves the same airport at 12:30 pm traveling 65◦ N of West at 552 mph. How far
apart in miles are the planes at 2:00 pm?
46
Section 12: Applications of Trigonometry
Solution: Let t denote time in hours and let t = 0 correspond to 12:30 pm. Let a(t) denote
the distance of the passenger plane to the airport in miles. Let b(t) denote the distance of
the private jet to the airport in miles. So similar to the previous example a(t) = 215 + 430t
and b(t) = 552t. By the law of cosines the distance squared between them at time t is
d2 (t) = (215 + 430t)2 + (552t)2 − 2(215 + 430t)(552t) cos(65◦ ). So when t = (3/2), the distance
squared is d2 (3/2) = (4 · 215)2 + (828)2 − 2(860)(828) cos(65◦ ). Therefore
p
d(3/2) = (4 · 215)2 + (828)2 − 2(860)(828) cos(65◦ ) ≈ 907 miles.
Example 5: A drug smuggler is traveling due North at 35 knots in a small boat. A coast guard
cutter is 125 nautical miles away on a heading 22◦ East of North from the smuggler. What
heading should the cutter take to intercept the smuggler, supposing that the cutter will run
at its top speed of 40 knots? How long will it take?
Solution: We could use the law of cosines here, but the law of sines is easier. Let A denote the
position of the smuggler, P denote the position of the cutter and I denote the point of intercept.
Let θ be the angle with vertex P and sides AP and IP . Let t denote time to intercept in hours.
sin θ
sin(22◦ )
=
So sin θ = (35/40) sin(22◦ ) = (7/8) sin(22◦ ). Which
Then by the law of sines
40t
35t
means that θ = arcsin((7/8) sin(22◦ )) ≈ 19◦ . So the cutter should head roughly 49◦ South of
sin(139◦ )
sin(22◦ )
West to intercept the smuggler. The time of intercept can be found using
=
125
35t
125 sin(22◦ )
≈ 2.04 hours.
So t =
35 sin(139◦ )
Example 6: A pilot is trying to get to an airport 362 miles away which lies on a major highway
that runs East and West. The control tower tells them to proceed on a heading 35◦ South
of West. The pilot knows the compass in the plane is not reliable so they make a guess at
the correct direction and proceed at 156 mph. Two hours later the pilot reaches the major
highway. How far from the airport is she?
Solution: Label the airport point A, the starting point of the plane P and the point where
the airplane reaches the highway C. Then the length of AP is 362 miles, and the length
of CP is 2 · 156 = 312 miles. The angle with vertex A and sides AP and AC has measure
35◦ . Since 362 sin(35◦ ) < 312 < 362 this is the ambiguous case of the law of sines. Still
if we let γ denote the angle with vertex C and sides CA and CP , the law of sines implies
sin γ = (362/312) sin(35◦ ) So γ = arcsin((181/156) sin(35◦ )) ≈ 41.7◦ or 138.3◦ . Unless the
pilot has a completely lousy sense of direction, γ is the larger angle. Therefore the distance
47
Section 12: Applications of Trigonometry
sin(6.7◦ )
sin(35◦ )
sin(6.7◦ )
=
. Which means d = 312
≈ 63.4 miles. If
d
312
sin(35◦ )
we choose the other value of γ we find d ≈ 529.4 miles!
d from A to P satisfies
As an example of a possible non-word exercise we have
Example 7: Show that for each A, B ∈ IR there is C, δ ∈ IR so that A sin x + B cos x =
C sin(x + δ).
Solution: Let C satisfy C 2 = A2 + B 2 , and δ = arctan(B/A). Draw a right reference triangle
with sides of length |A|, |B| and |C| and included angle |δ|. Then A = C cos δ and B = C sin δ
so that A sin x + B cos x = C cos δ sin x + C sin δ cos x = C sin(x + δ).
Exercises: Give exact answers if possible.
1. The Butler Office Building (BOB) is 847 feet high. An apartment building stands across
Main Avenue from BOB. Someone standing on the edge of the roof of the apartment building
closest to BOB measures the angle of elevation to the top of BOB as 45◦ . They measure the
angle of depression from their location to the foot of BOB as 60◦ . How far apart are the two
buildings, and how tall is the apartment building?
2. An Army Reserve Squad is dropped off at point B and ordered to march on a heading of
42◦ North of East through a dense forest to a bivouac site at point A, 23 miles away on a
North-South road. During the drop all their compasses are damaged. They take a guess at
their proper direction and march off. Marching at 6 miles per hour they arrive at the road in
1
4 hours. How far are they from the bivouac site? How far off was their estimated heading?
2
3. Two jets depart LAX simultaneously at midnight. The first flies on a heading 32◦ North of
East at 350 mph. The second flies on a heading 13◦ South of East at 467 mph. How far apart
are the planes at 2:20 am?
4. Show that for each A, B ∈ IR there is C, δ ∈ IR so that A sin x + B cos x = C cos(x − δ)
5. John has a tree in his back yard that he wants removed. The tree removal company charges
$25 a foot of height to remove it. John wants to know if he can afford it so he needs to know
the height of the tree. He stands 21 feet away from the tree on level ground and measures the
angle of elevation from himself to the top of the tree as 75◦ . About how much will it cost John
to have the tree removed?
48
Section 13: Polar Coordinates
A polar coordinate system is one in which there is a unique point in the plane called
the pole and a ray through the pole called the polar axis. A point, P , in the plane can be
coordinatized by its distance |r| to the pole and the measure of the angle θ from the polar axis
to the ray through the pole and P . Normally we take the pole to lie at (0, 0) in the Cartesian
coordinate system, and take the polar axis to be the positive x-axis. It is therefore common to
measure θ in radians. Of course if θ and φ are coterminal angles, then the polar coordinates
(r, θ) and (r, φ) describe the same point. So polar coordinates are not unique! In fact if we
take the convention that when r is negative we move in the antipodal direction of θ, then the
polar coordinates (r, θ) and (−r, θ ± π) represent the same point. To make our life easier we
normally restrict the values of θ to lie in the range (−π, π].
With these facts in mind we seek a way to convert polar coordinates and Cartesian (or rectangular) coordinates. The point P (θ) lies on the line through the origin and (r, θ). Therefore
by Euclid’s Theorem on similar triangles the Cartesian coordinates of the point are x = r cos θ,
and y = r sin θ. Very simple!
The conversion from rectangular coordinates to polar coordinates is not so smooth. When
x = 0, the polar coordinates of the point are (y, π/2). When y = 0, the polar coordinates
of the point are (x, 0). Otherwise x 6= 0 and the point with Cartesian coordinates (x, y) is a
vertex of the nondegenerate right triangle whose other two vertices are (0, 0) and (x, 0). By
p
the Pythagorean theorem the distance from (x, y) to the origin is r = x2 + y 2 . Let θ denote
the angle with vertex at the origin, initial side the positive x-axis and terminal side the line
segment between the origin and (x, y). If α is the reference angle for this value of θ, then
we know that tan α = |y|/|x| and in fact tan θ = y/x. So if −π/2 < θ < π/2, we can take
θ = arctan(y/x). In general as long as θ is coterminal with an angle in the domain of the
arctangent function, we have no problem. When θ is not coterminal with such and angle,
θ + (2n + 1)π is for some integer n. We can therefore describe the point with rectangular
coordinates (x, y), by the polar coordinates (−r, θ + (2n + 1)π).
A major application of polar coordinates is that the description of certain graphs in polar
coordinates lets us analyze them as functions, rather than relations. For example, x2 + y 2 = a2
has graph a circle with radius a centered at the origin. In polar coordinates x2 + y 2 = r2
(squaring the top equation on the right) so we can describe this circle as r = a, a very simple
49
Section 13: Polar Coordinates
formula in polar coordinates.
By convention we do try to write polar equations in the form r = f (θ). So r is the
dependent variable and θ is independent. In order to graph polar equations we use T-tables.
π π π π
The values we normally plug-in for θ are multiples of 0, , , , etc. That is values of θ
6 4 3 2
for which we can compute exact values for r. Standardly we keep the θ values in increasing
order and after connecting our dots we use arrows to display the behavior of the curve as θ
increases. This is called orienting the graph. This is a crucial part of the graph. If a > 0,
the polar equations r = a and r = −a both represent the circle centered at the origin with
radius a. The curve r = a is oriented positively, meaning that the motion along the curve as θ
increases is counter-clockwise. The curve r = −a is oriented negatively, since the motion along
the curve as θ increases is clockwise. These are therefore different curves in polar coordinates,
though they transform to the same curve in rectangular coordinates.
There is therefore some inherent danger when we attempt to convert equations in one
coordinate system to the other. When we convert from polar to rectangular we lose orientation.
When we convert from rectangular to polar we must often choose an orientation. The loose
convention we use is to opt for the positively oriented polar curve.
Example 1: Transform the rectangular equation y = 2x to a polar equation.
Solution: By substitution we have r sin θ = 2r cos θ. So if r 6= 0, sin θ = 2 cos θ. Equivalently
tan θ = 2, or θ = arctan 2. This works when r = 0 too, since we are at (0, θ) which is (0, 0)
in rectangular coordinates. This kind of polar curve is not orientable, since θ is fixed. This is
the polar equivalent of the vertical line in rectangular coordinates which has no slope.
Example 2: Transform the polar equation r = 2 sin θ to a rectangular equation.
Solution: Supposing that r 6= 0, we can multiply both sides by r to obtain r2 = 2r sin θ. By
substitution we have x2 + y 2 = 2y. By completing the square on y we get x2 + (y − 1)2 = 1.
Thus this polar equation describes a circle of radius 1 tangent to the pole, oriented positively.
Example 3: Transform the polar equation r = sin 2θ to a rectangular equation.
Solution: Again supposing that r 6= 0 we first write r3 = r2 sin 2θ = r2 (2 sin θ cos θ). Which
we rewrite as r3 = 2(r sin θ)(r cos θ). So by substitution (x2 + y 2 )3/2 = 2xy. Therefore
(x2 + y 2 )3 = 4x2 y 2 . But unless you know about the classification of general expressions of
total degree 6 in x and y, this equation will not help you sketch a graph of the curve. It does
50
Section 13: Polar Coordinates
enable us to realize that the un-oriented curve is symmetric with respect to the y-axis, the
x-axis, the line y = x, and with respect to the origin, since substituting in ±x for x and/or
±y for y leaves the equation unchanged. So if we figure out what the graph looks like in the
first quadrant, we can apply symmetry to determine the rest.
The values of θ we want to use to build a T-table are those for which we can compute
exact values of sin 2θ. So for example 2θ = π/6, 2θ = π/4,, 2θ = π/3, etc. So we build the
T-table
θ 0
r(θ) 0
π/12 √
π/8
1/2
2/2
π/6
√
3/2
π/4
1
Thus the graph of this polar equation is
θ=
π
3
θ=
π
6
1
We don’t always need to transform a polar equation to rectangular form to investigate
symmetry. If replacing θ with −θ leaves a polar equation unchanged, then its oriented graph
will have polar axis (x-axis) symmetry. If replacing θ by π − θ leaves a polar equation unchanged, then the oriented graph will have symmetry with respect to the line θ = π/2 (the
y-axis). Finally if replacing r by −r leaves a polar equation unchanged, then its oriented graph
will have polar (original) symmetry.
Example 3 indicates that these tests for polar symmetry are not as easy to implement
as tests for rectangular symmetry. So there can be value to transforming polar equations to
rectangular ones, even if the rectangular equations cannot be easily graphed.
But the real problem persists, what does the graph of a polar equation look like? Aside
from symmetry many polar equations have graphs which are periodic. This is because the
most common functions to plug θ into are trigonometric. In such cases we are interested to
51
Section 13: Polar Coordinates
know the range of θ values we must run through before the graph begins to repeat itself. In
the example above, whereas sin 2θ has period π viewing it as a rectangular equation relating
r and θ, the polar point (sin 2θ, θ) is antipodal to the polar point (sin(2(θ + π)), θ + π), since
sin 2(θ + π) = sin(2θ + 2π) = sin 2θ. So in this case the period of the polar curve turns out to
be 2π.
Example 4: Sketch a graph of the polar equation r = sin 3θ.
Solution: One can check that this graph will have y-axis symmetry. To determine further
symmetries we plot r versus θ rectangularly.
1
−π
− 2π
3
− π3
π
3
2π
3
π
-1
From this we observe that the polar graph will be symmetric about θ = π/6, about θ = π/2
and about θ = 5π/6. Also we note that the period of the curve will be π since sin(3(θ + π)) =
sin(3θ+3π) = sin 3θ cos 3π+sin 3π cos 3θ = − sin 3θ and therefore (sin 3θ, θ) and (− sin 3θ, θ+π)
are the same point. Moreover the part of the polar curve about π/6 in shape will be the same
as the part about π/2 with negative radii, and will be identical to that about 5π/6, back to
positive radii.
52
Section 13: Polar Coordinates
θ=
π
3
θ=
π
6
1
A general analysis leads one to recognize that a curve with polar equation r = a cos 2kθ,
where k is a whole number, will have a graph which is a rose with 4k petals of length a, period
2π, and x-axis symmetry. The graph of r = a sin 2kθ, is virtually identical except that it has
y-axis symmetry. In contrast the graph of the curve with polar equation r = a cos(2l + 1)θ,
where l is a whole number, will be a rose with 2l + 1 petals of length a, each traversed twice
as θ runs from 0 to 2π. This graph therefore has period π and also has x-axis symmetry. So
rose curves might be analogous to quadratic equations in two variables, in that their behavior
is completely describable by the form of the equation after it is put in standard form.
Another class of polar curves with this property are the limaçons. A limaçon is a polar
curve with equation of the form r = a + b cos θ, or r = a + b sin θ, where a and b are constants.
The second class behave like the first except that they have y-axis symmetry instead of x-axis
symmetry.
For the first class there are three distinct possibilities: 1) a < b, 2) a = b, and 3) a > b.
In each case the curve has period 2π. In the first case r = 0 twice in each period, since r = 0
implies that cos θ = −a/b < −1. Which means θ = arccos(−a/b), and 2π − arccos(−a/b). Also
the sign of r changes at θ and again at 2π − θ. Consequently the graph has an inner loop. We
distinguish this case by saying that the curve is a limaçon with an inner loop. In the second
case r = 0 only once per period, when θ is an odd multiple of π. The result is what we called a
cardioid. In the third case r is never zero and the graph looks rather like an egg, or a blunted
ellipse.
53
Section 13: Polar Coordinates
a<b
a=b
a>b
Another application of polar coordinates is that they allow us to easily define and describe
nondegenerate conic sections. This is a two-step process.
First we define a conic as the set of points in the plane so that the distance from any
point to a fixed point F , called the focus, is e times the distance from the point to a fixed line
l, called the directrix. When e = 1 we have by definition a parabola. It turns out that e is
exactly the eccentricity of the conic as discussed in section 7.
We can then derive that if F is the pole, any conic has equation of the form
r=
ed
,
1 + e cos θ
r=
ed
,
1 − e cos θ
r=
ed
,
1 + e sin θ
or r =
ed
1 − e sin θ
where d is the distance from F to l.
Exercises:
1. Plot the point with polar coordinates (5, 5π/6). b) Convert the polar coordinates to
rectangular coordinates. c) Give two other pairs of polar coordinates which represent the
same point, one pair with r > 0, the other with r < 0.
2. Convert the rectangular equation to a polar equation
a) y = x
b) x2 + y 2 = 4x
c) y = x2
3. Sketch and label a graph of each polar equation clearly indicating orientation.
a) r = 3 cos 3θ
b) r = −3 cos 3θ
c) r = 1 − 2 sin θ
d) r = −1 − 2 sin θ
e) r = 4 sin θ
4. Find a polar equation for the ellipse with focus at the pole, directrix with rectangular
1
equation x = 2 and eccentricity e = .
2
54
Section 14: Parametric Equations
The last topic for us to treat is parametric equations, so called because both rectangular
coordinates of a point on a curve are described by another quantity called the parameter. We
normally take our third variable to be t and think of t as denoting time. This is the situation
in many physical applications. A parametric curve is the set of all ordered pairs (x, y) where
x = g(t) and y = h(t), where g(t) and h(t) describe x and y respectively in terms of t.
An easy way to build a set of parametric equations is to start with y = f (x) and let
x = g(t) = t be the parameter so that y = f (t) = h(t). Any function can be parametrized in
this manner. There is a change in the graph of the curve in this case. As with polar curves,
every parametric curve is oriented. We place arrow marks on the curve to denote the motion
along the curve as the parameter t increases. Similarly any relation of the form x = k(y) can
be parametrized by y.
Every polar curve of the form r = f (θ) gives rise to a parametric curve parametrized
by θ. The parametric equations are x = r cos θ = f (θ) cos θ = g(θ), and y = r sin θ =
f (θ) sin θ = h(θ). This includes r = a, a circle with radius |a| centered at (0, 0). The standard
parametric equations for this curve are x = a cos θ and y = a sin θ. This works because
x2 + y 2 = a2 cos2 θ + a2 sin2 θ = a2 .
This construction generalizes to x = h + a cos t, y = k + b sin t which are parametric
equations for the ellipse (x − h)2 /a2 + (y − k)2 /b2 = 1. The sign of a and b don’t have to be
positive so there are really four possible sets of parametric equations of this form for this ellipse
x = h ± a cos t, and y = k ± b sin t. Since the crucial fact here is that sin2 t + cos2 t = 1, we can
also parametrize this ellipse via x = h ± a sin t, y = k ± b cos t. These eight sets of parametric
equations parametrize the same curve. They are distinct as is seen by considering which point
corresponds to (x(0), y(0)) and whether the curve is oriented positively or negatively.
If we tweak the Pythagorean identity to be in the form 1 = sec2 t − tan2 t, we can
parametrize the hyperbola (x − h)2 /a2 − (y − k)2 /b2 = 1 via x = h ± a sec t, and y =
k ± b tan t. In this case x = h ± a tan t and y = k ± b sec t parametrize the conjugate hy-
perbola (y − k)2 /b2 − (x − h)2 /a2 = 1.
et − e−t
et + e−t
e2t + 2 + e−2t
Define sinh t =
and cosh t =
, then cosh2 t =
, and
2
2
4
e2t − 2 + e−2t
sinh2 t =
. So cosh2 t − sinh2 t = 1. We can therefore also use these functions,
4
which are the hyperbolic sine and cosine functions, to parametrize a hyperbola.
55
Section 14: Parametric Equations
Since every parabola is either a function of the form y = k + (1/4c)(x − h)2 , or a relation
of the form x = h + (1/4c)(y − k)2 every parabola is parametrizable.
To summarize, every function is parametrizable, nearly every polar curve is parametrizable, and every conic section is parametrizable.
As we see with ellipses and hyperbolas, there are often many ways to parametrize a curve.
In fact if we have a set of parametric equations for a curve x = g(t) and y = h(t), then we can
reparametrize part or all of the curve by s where t = f (s), and f is a function of s. The new
parametric equations are x = g(t) = g(f (s)), and y = h(t) = h(f (s)).
Example 1: Let x = t and y = x2 = t2 be a set of parametric equations describing the parabola
y = x2 . If we set t = sin s, then x = t = sin s and y = t2 = sin2 s. Which means that the points
on the curve x = sin s, y = sin2 s are on the curve y = x2 . However not every point of y = x2
is on this parametric curve since −1 ≤ x = sin s ≤ 1, and 0 ≤ sin2 s = y ≤ 1. This example is
the first of a kind where we have the motion of the curve oscillating. If we want more or less
of the curve we can use x = c sin u, and y = c2 sin u. This parametric curve describes the path
of a particle that oscillates between (−c, c2 ) and (c, c2 ) along the path y = x2 .
12
10
8
6
4
2
-4
-2
2
4
x = 3 sin t, y = 9 sin2 t
Example 2: Let C denote the part of the ellipse x2 /4 + y 2 /9 = 1 in the first quadrant. We can
find sets of parametric equations for C in various ways.
First we can solve for y in terms of x, since this part of the ellipse satisfies the horizontal
line test. We get y 2 /9 = 1 − (x2 /4) = (4 − x2 )/4, so y 2 = (9/4)(4 − x2 ). Thus we can set x = t
√
and y = (3/2) 4 − t2 , where t ∈ [0, 2].
56
Section 14: Parametric Equations
We can reverse orientation by setting x = t = 2 − s. Now 4 − t2 = 4 − (2 − s)2 =
√
4 − (4 − 4s + s2 ) = 4s − s2 , so y = (3/2) 4s − s2 , and s ∈ [0, 2].
√
We can increase speed as it were by setting x = t = 2s, and y = 3 1 − s2 , where s ∈ [0, 1].
√
We can decrease speed by setting x = t = (1/2)s and y = (3/4) 16 − s2 , where s ∈ [0, 4].
p
√
We can call x = t = 2 sin θ, then y = (3/2) 4 − 4 sin2 θ = (3/2) 4 cos2 θ = 3 cos θ. Now
we take θ ∈ [0, π/2].
Example 3: Let C be the line segment between (1, −2) and (3, 6). The line through these points
has rectangular equation y = 4x − 6. We can therefore parametrize C by x = t, y = 4t − 6,
where t ∈ [1, 3].
If we set t = s + 1, then x = s + 1, and y = 4(s + 1) − 6 = 4s − 2, where s ∈ [0, 2].
If we then set s = 2u we get x = 2u + 1, and y = 8u − 2, where u ∈ [0, 1].
If we set u = 1 − v we reverse the orientation and have x = 3 − 2v, and y = 6 − 8v, where
v ∈ [0, 1].
If we have an object for which we know rectangular or polar equations, we can therefore
parametrize the curve fairly easily. One challenge may be to find the best parameterization
with respect to a given set of criteria. It might be instead that we want a parameterization
with a specific orientation, and with a specified domain for the parameter. The tricks above
work as long as we have at least one parameterization.
A problem that arises is that we won’t always be given the rectangular or polar equations
which a curve, or part of a curve satisfy. Instead we’ll begin with just a set of parametric
equations. In order to plot the graph of a set of parametric equations we can use a three
row/column table. The first row is labeled t and contains inputs. The second row consists of
the outputs x = g(t). The third column consists of the outputs y = h(t). The ordered pairs
(x, y) from the second and third columns belong on the graph. When we connect these dots
it is again important to orient the graph to indicate the behavior as t increases. The number
of values in the table varies from person to person. One should be careful to have a sufficient
number of values. Also the t-values should be carefully chosen so that, if possible, exact values
can be calculated for both x and y.
Even when one is careful mistakes can easily be made. Eliminating the parameter to find
an equation relating x and y is therefore a useful skill. We can then use what we know about
this relation to ensure we’ve got a proper graph.
57
Section 14: Parametric Equations
Example 4: Given the parametric equations x = tan θ + sec θ, and y = tan θ − sec θ, where
−π/2 < θ < π/2 let us sketch a graph of the curve. The first thing we might do is to compute
a three-row T-table where the input values of θ are ±π/6, ±π/4, ±π/3, and 0.
θ
−π/3
√
x(θ) 2 − √3
y(θ) −2 − 3
√−π/4
2−1
√
− 2−1
−π/6
√
3/3
√
− 3
0
1
−1
π/6
√
√3
− 3/3
√π/4
2+1
√
− 2+1
π/3√
2 + √3
−2 + 3
Plotting these points by hand and trying to connect the dots is dangerous. Much safer is to
realize that xy = tan2 θ − sec2 θ = −1 by a variation of the Pythagorean identity. So the points
of the parametric curve lie on the hyperbola xy = −1.
5
-5
5
-5
We may also need to remember that the parametric equations do not need to define the
whole rectangular curve.
√
Example 5: Given x = t and y = 1 − t, describe the curve parametrized.
Solution: To begin with the domain for the parameter is at most [0, ∞). Secondly we can solve
for t as a function of x to get t = x2 . So this set of parametric equations describes the right
half of the downward opening parabola y = 1 − x2 . The orientation has (0, 1) = (x(0), y(0))
and as t increases the point moves toward (∞, −∞).
Exercises:
1. Let x(t) = 1 + 4 sin t and y = −2 + 3 cos t. a) Find a rectangular equation for this relation.
b) Sketch and label a graph of the curve clearly indicating orientation.
2. Let x(t) = t + 2 and y(t) = 3t − 5. a) Find a rectangular equation for this relation. b)
Sketch and label a graph of the curve clearly indicating orientation.
58
Section 14: Parametric Equations
3. Find a set of parametric equations for the line through P = (1, −1) and Q = (−2, 4) so
that P = (x(0), y(0)) and Q = (x(1), y(1)).
4. Find a set of parametric equations for the line through P = (2, 1) and Q = (−2, 2) so that
Q = (x(1), y(1)) and P = (x(−1), y(−1)).
5. Find a set of parametric equations for the ellipse
(x(0), y(0)) and (0, −4) = (x(π/2), y(π/2)).
59
x2
(y + 1)2
+
= 1 so that (4, −1) =
16
9
Appendix A: The Real Line and the Cartesian Plane
The main purpose of this appendix is to present many properties of the real numbers you
hopefully already know. A second goal is to introduce you to specific notation. A third is to
give you a sense of just how highly developed the real number system is. Of course the same
is true of the Cartesian Plane.
Subsection 1: The Real Line
A natural number is a non-negative whole number. We use the set of natural numbers, IN,
for counting. It is useful to know that the sum or product of two natural numbers is again a
natural number. We say that IN is closed with respect to addition and multiplication. However
IN is not closed with respect to subtraction, since for example 2 − 4 = −2 and −2 is not a
non-negative whole number.
An integer is a whole number. The set of integers, ZZ, contains the set of natural numbers.
It is also closed with respect to addition and multiplication, and has the added advantage of
being closed with respect to subtraction. It is not closed with respect to division, since for
example 12 ÷ 5 = 2 52 is not a whole number.
The set of rational numbers, Q (q for quotient), consists of all ratios of integers with
non-zero denominators. Q is closed with respect to addition, multiplication, and subtraction.
Also as long as we don’t divide by zero, the ratio of two rational numbers is again a rational
number. In fact Q is the smallest set containing IN where we can perform all four basic binary
operations, barring only dividing by zero. End of story? NO.
The ancient Greeks knew that certain necessary numbers were not rational.
√
Fact: 2 is not rational.
√
√
p
Proof: Suppose 2 ∈ Q. Then 2 = , where p, q ∈ ZZ, and q 6= 0. We may also suppose
q
p
that p and q have no common prime divisor. We say that is in lowest terms. By squaring
q
p2
we obtain 2 = 2 . Clearing the denominator gives 2q 2 = p2 . From which we deduce that p2 is
q
even (it is divisible by 2). If p were odd, then p2 would also be odd. We therefore conclude that
p is even. Thus p = 2k for some whole number k. But this implies 2q 2 = p2 = (2k)2 = 4k 2 .
Which reduces to q 2 = 2k 2 . As above we deduce that q is even. But this is a contradiction
since now 2 is a common prime divisor of p and q. Since we arrived at a contradiction our
√
original hypothesis must be false. Therefore 2 6∈ Q.
60
Appendix A: The Real Line and the Cartesian Plane
To form the real numbers, IR, it is customary to consider a continuous horizontal line of
infinite length. One point on the line is designated as 0. Another point to its right is designated
as 1. The distance from 1 to 0 forms a unit by which all other distances are measured. For
any point x on the line exactly one of the following statements is true: 1) x is to the left of
0 (x is negative), 2) x is to the right of 0 (x is positive), or 3) x = 0. This property is called
the trichotomy law. We denote the distance from x to 0 by |x|. When x is to the left of
zero we have x = −|x|. Otherwise x = |x|. This basically defines the absolute value function
√
f (x) = |x| = x2 .
This model also gives rise to the notion of order on the line. We write a < b, for real
numbers a and b, when a − b = −|a − b|. Also a ≤ b is equivalent to b − a = |b − a|. The
trichotomy law generalizes in that if a and b are real then exactly one of a < b, b < a and a = b
is true. In fact one can prove that the order relations have the following properties.
Theorem: (Inequality Properties) For real numbers a, b and c
1) if a < b, then a + c < b + c
2) if a < b and 0 < c, then ac < bc
3) if a < b and c < 0, then bc < ac
To begin to understand the importance of this construction we first state and prove the
Pythagorean Theorem. The reader is strongly encouraged to draw the necessary figures to
illustrate the proof.
Theorem: For a right triangle with hypotenuse of length c and sides of lengths a and b, we
have a2 + b2 = c2 .
Proof: Let A = (a + b)2 be the area of a square of side a + b. Let B denote the area of a right
triangle with dimensions as per the theorem statement.
1) Draw a square S with side a + b and two sides horizontal. Let C1 denote the point on the
left hand side which is b units down from the top left corner. Let C2 be the point on the
bottom side which is b units to the right of the lower left corner. Let C3 be the point on the
right hand side which is b units up from the lower right hand corner. Let C4 denote the point
on the top side which is b units to the left of the upper right hand corner. Connect C1 and C2
by a line segment. Repeat by connecting C2 to C3 , C3 to C4 and C4 to C1 . S now contains
four non-overlapping right triangles whose sides measure a, b and c as above. The remaining
61
Appendix A: The Real Line and the Cartesian Plane
area inside S is the area of a square of side c. Therefore A = 4B + c2 .
2) Draw another square S 0 of side a + b and two sides horizontal. Let D1 = C1 and D2 = C2 .
Let D3 be the point on the right side of S 0 which lies b units down from the top right hand
corner. Let D4 be the point on the upper side of S 0 which lies b units to the right of the
upper left hand corner. With line segments connect D1 to D2 , D1 to D3 , D4 to D2 , and D4
to D3 . Now S 0 contains four non-overlapping right triangles with sides of lengths a, b and c.
The remaining area inside S 0 is the area of two squares, one of side a, and the other of side b.
Therefore A = 4B + a2 + b2 .
Since S and S 0 have the same area we deduce a2 + b2 = c2 .
Because
√
2 is the length of the hypotenuse of an isosceles right triangle with sides of unit
length, it is a real number. Real numbers which are not rational are called irrational. The set
√ √
of irrational numbers is not closed with respect to multiplication since for example 2 2 = 2.
Consequently we have no standard symbol for this set.
The absolute value function also plays a role in the definition of distance on the real
line. The distance from a to b, denoted d(a, b) equals |a − b|. Naturally d(a, b) = d(b, a) and
d(a, b) ≥ 0 with equality if and only if a = b. Seemingly innocuous is the statement that the
midpoint of the line segment joining a and b is (a + b)/2. This fact is used repeatedly in section
7.
Finally, the absolute value function satisfies the following properties.
Theorem: (Absolute Value Properties) For real numbers a, b and c
1) |ab| = |a||b|
¯ a ¯ |a|
¯ ¯
2) If b 6= 0, ¯ ¯ =
b
|b|
3) |c| < a if and only if −a < c < a
4) |c| ≤ a if and only if −a ≤ c ≤ a
3) |c| > a ≥ 0 if and only if c < −a or c > a
3) |c| ≥ a ≥ 0 if and only if c ≤ −a or c ≥ a
In turn the order relations are used to define intervals. For real numbers a and b, [a, b]
denotes all real numbers x with a ≤ x ≤ b. Similarly [a, b) = {x ∈ IR|a ≤ x < b}, (a, b] =
{x ∈ IR|a < x ≤ b}, and (a, b) = {x ∈ IR|a < x < b}. We use the infinity symbol, ∞ when
we have an interval of the real line which extends indefinitely in the positive direction. So
62
Appendix A: The Real Line and the Cartesian Plane
(a, ∞) = {x ∈ IR|a < x}. Also (−∞, b] = {x ∈ IR|x ≤ b}. It is to be understood that ∞ is a
symbol, not a real number.
Intervals are special subsets of the real line. Whenever we have sets A and B we denote
their union by A ∪ B. The union of two sets is the set whose elements are in at least one of
the two original sets. Also the intersection of two sets A ∩ B is the set whose elements are in
both A and in B. We can build more general subsets of IR using unions and intersections of
intervals.
The last computational tool we introduce in this subsection is the sign graph. The premise
behind this is contained in part 3 of the properties of inequalities. To paraphrase “minus times
minus is plus”, “minus times plus is minus”, and “plus times plus is plus”. So for example we
can determine all x ∈ IR so that x3 − 4x ≥ 0. To proceed we factor x3 − 4x = x(x2 − 4) =
x(x + 2)(x − 2). For each linear factor the trichotomy law allows us to determine intervals
where the factor is positive and negative. We combine these subsolutions according to the
rules above.
x
–
–
–
–
–
–
–
–
0
+ + + + + + + +
x−2
–
–
–
–
–
–
–
–
–
–
x+2
–
–
–
–
–
0
+
+
+ + + + + + + + +
x3 − 4x
–
|
–
–
|
–
0
|
+
+
|
+
0
|
-4
-3
-2
-1
–
0
–
–
|
1
–
–
0
0
|
2
+ + + +
+ + + +
|
|
3
4
So x3 − 4x ≥ 0 when x ∈ [−2, 0] ∪ [2, ∞).
Subsection 2: The Cartesian Plane
One of the greatest mathematical ideas of all time is due to René DesCartes. Given
an infinite flat plane choose a point O. Draw two perpendicular lines concurrent at O, one
horizontal and the other vertical. Call the horizontal line the x-axis and the vertical line the
y-axis. On the x-axis pick a point P to the right of O and let the distance from P to O
serve as 1 unit of length in the horizontal direction. Pick another point Q above O on the
y-axis and use the distance from O to Q as a unit of vertical distance. Any point in the
plane can then be assigned a unique ordered pair of Cartesian coordinates. To label a point
63
Appendix A: The Real Line and the Cartesian Plane
(a, b) means that it lies |a| units from O horizontally (to the left if a = −|a|, and to the
right if a = |a|.) and |b| units from O vertically (down if b = −|b|, and up otherwise). The
points with (a, b) = (|a|, |b|) lie in the first quadrant. Those with (a, b) = (−|a|, |b|) are in the
second quadrant. The third quadrant consists of those points with (a, b) = (−|a|, −|b|). The
points with (a, b) = (|a|, −|b|) make up the fourth quadrant. In essence each coordinate axis
is a copy of the real line. We denote the Cartesian plane by IR2 = IR × IR.
Starting with (x1 , y1 ) and (x2 , y2 ) in the Cartesian plane form a third point (x2 , y1 ). These
points form the vertices of a (possibly degenerate) right triangle. Measured against the x-axis
p
we see that the horizontal side of the triangle has length |x2 − x1 | = (x2 − x1 )2 . The vertical
p
side has length |y2 − y1 | = (y2 − y1 )2 . By the Pythagorean Theorem the distance between
p
the two points is d((x1 , y1 ), (x2 , y2 )) = (x2 − x1 )2 + (y2 − y1 )2 .
A circle is defined to be the set of points all of which have a prescribed distance from a
given point called the center. The equation for a circle follows from the distance formula. If
the center is the point (h, k) and the distance from any point on the circle to the center is |r|,
then the circle consists of all points (x, y) so that (x − h)2 + (y − k)2 = r2 .
In practice the equation of a circle rarely comes in this standard form. The standard
form can be recaptured by completing squares. The basic formula to remember is (z + a2 )2 =
z 2 + az + ( a2 )2 . Given an equation x2 + ax + y 2 + by = c, we strategically add zero two times.
The first time we add zero in the form ( a2 )2 − ( a2 )2 . The second time we add zero in the
form ( 2b )2 − ( 2b )2 . The result via substitution transforms the equation x2 + ax + y 2 + by = c
first to x2 + ax + ( a2 )2 − ( a2 )2 + y 2 + by = c. Which after adding ( a2 )2 to both sides becomes
(x+( a2 ))2 +y 2 +by = c+( a2 )2 . At this point we have completed the square in x. After completing
the square in y and simplifying we arrive at (x + ( a2 ))2 + (y + ( 2b ))2 = c + ( a2 )2 + ( 2b )2 .
We close this appendix with another seemingly innocent remark that the midpoint of the
line segment joining the points with coordinates (x1 , y1 ) and (x2 , y2 ) has coordinates
((x1 + x2 )/2, (y1 + y2 )/2).
Exercises:
1. Use interval notation to describe the values of x which satisfy x2 − 3x − 4 > 0.
2. Solve the inequality
1
2
≤
. Write your answer using interval notation.
x+2
x−3
3. a) Find the distance between the points (2, 1) and (−1, −3).
64
Appendix A: The Real Line and the Cartesian Plane
b) Find the midpoint of the line segment joining (2, 1) and (−1, −3).
4. a) Complete the squares to find the center and radius of the circle with equation x2 − 2x +
y 2 + 4y = 11.
b) Sketch a graph of the circle in part a). Choose equal scales for both axes.
c) Repeat part b) but choose the y-scale to be twice as long as the x-scale.
5. Find an equation of the circle with center (−3, 4) which is tangent to the y-axis.
65
Appendix B: Lines
A linear equation in x and y is of the form Ax + By = C, where we take A, B, C ∈ IR.
If A = B = 0, the equation is either inconsistent (like 2 = 0) and has empty graph, or is the
trivial equation 0 = 0 whose graph is all of IR2 . So let us suppose that at least one of A or B
is non-zero.
Thus if B = 0, we must have A 6= 0 and our equation is equivalent to x = C/A. The
graph of this contains all points (C/A, y) as y ∈ IR. So the graph is a vertical line.
Otherwise B 6= 0 and we have the equivalent equation y = (−A/B)x + (C/B). In which
case we set m = −A/B and b = (C/B). When x = 0 we see that y = b is the y-intercept
of the graph. Too if (x1 , y1 ) is on the graph we have y1 = mx1 + b. Therefore for w ∈ IR,
y1 + mw = mx1 + mw + b = m(x1 + w) + b which is to say that every point of the form
(x1 + w, y1 + mw) as w ∈ IR is on the graph. In fact if (x2 , y2 ) is a point on the graph other
than (x1 , y1 ), then simultaneously we have y1 = mx1 + b and y2 = mx2 + b. Subtracting the
first equation from the second gives y2 − y1 = mx2 − mx1 = m(x2 − x1 ). In this case when we
set w = x2 − x1 so that x2 = x1 + w, we have y2 − y1 = mw which simplifies to y2 = y1 + mw.
So the graph of the equation consists exactly of all points of the form (x1 + w, y1 + mw) as
w ∈ IR. We say that the equation is determined by one point and the value of m which we call
the slope. Moreover from y2 − y1 = m(x2 − x1 ) with (x1 , y1 ) 6= (x2 , y2 ) and B 6= 0 we have
that x2 6= x1 wherefore m = (y2 − y1 )/(x2 − x1 ). So the graph is also determined by any two
distinct points. Given two distinct points on the graph their midpoint is
(
x2 − x1
y 2 − y1
x2 − x1
x2 − x1
x1 + x2 y1 + y2
,
) = (x1 +
, y1 +
) = ((x1 +
, y1 + m
)
2
2
2
2
2
2
And since the shortest distance between points in IR2 is a straight line, the graph must contain
the line through the two points. Of course every point on the graph is also on the line as
argued above.
So every line in IR2 has an equation of the form Ax + By = C, where not both A and B
are zero. Every non-vertical line in IR2 also has an equation of the form y = mx + b. This is
the slope-intercept form. Equivalently we can describe the line as all points (x, y) satisfying
an equation of the form (y − y1 ) = m(x − x1 ). This is called the point-slope form.
We use the convention that a vertical line has infinite slope. A horizontal line has slope
0. This allows us to define parallel lines as those whose graphs have the same slope when
66
Appendix B: Lines
sketched on the same coordinate axes.
Two lines are perpendicular if they meet at right angles. Since the lines are not parallel,
they intersect at a point (x0 , y0 ). If one line is horizontal the other is vertical and vice versa.
When neither is horizontal, then both lines have non-zero slope. Suppose that the first line has
slope m and the second has slope n. For w = 1 we have that the first line contains the point
(x1 , y1 ) = (x0 + 1, y0 + m), while the second line contains the point (x2 , y2 ) = (x0 + 1, y0 + n).
The three points (x0 , y0 ), (x1 , y1 ), and (x2 , y2 ) form the vertices of a right triangle. Therefore
by the Pythagorean Theorem we have that
[(x1 − x0 )2 + (y1 − y0 )2 ] + [(x2 − x0 )2 + (y2 − y0 )2 ] = [(x2 − x1 )2 + (y2 − y1 )2 ]
But x1 − x0 = 1 = x2 − x0 , y1 − y0 = m, y2 − y0 = n, x2 − x1 = 0, and y2 − y1 = n − m. So
after substitution we have 12 + m2 + 12 + n2 = (n − m)2 = n2 − 2mn + m2 . Which implies
that mn = −1. So a line is perpendicular to a line with non-zero slope if and only if its slope
is the negative reciprocal of the first line’s slope.
Exercises:
1. Find the equation in slope-intercept form of the line
a) through (1, 1) and (2, 3)
b) through (0, −2) and (−3, −6)
c) through (1, 2) with slope 5
d) through (−1, 2) parallel to y = 3x − 4
e) through (0, 1) perpendicular to y = 3x + 4
2. Convert every equation from exercise 1 to point-slope form.
3. Graph the lines whose equations are given. Scales for your axes should be clearly labelled
and chosen so that all interesting features of the graph can be clearly seen.
a) 4x + 3y = 12
b) y = −2x + 4
c) x = 3
d) (y − 3) = 2(x − 1)
e) 4x − 3y = 12
67
Appendix C: Factoring Polynomials and Complex Arithmetic
This appendix deals with finding factors of polynomials. The first major theorem is the
division algorithm for polynomials which mirrors the division algorithm for integers.
Theorem: (The Division Algorithm) If a(x) and b(x) are polynomials with real coefficients,
then there exist unique polynomials q(x) and r(x) with real coefficients, so that
a(x) = b(x)q(x) + r(x) and either r = 0 or the degree of r is strictly less than the degree of b.
The polynomial q(x) is called the quotient and the polynomial r(x) is the remainder.
When b(x) is a non-zero polynomial and a(x) = b(x)q(x) (with remainder zero) we say that
b(x) divides a(x). We call b(x) a factor of a(x).
Theorem: (The Factor Theorem) A polynomial p(x) = an xn + an−1 xn−1 + . . . + a1 x + a0 ,
where a0 , a1 , . . . , an ∈ IR and an 6= 0 has p(c) = 0 if and only if x − c divides p(x).
Proof: By the division algorithm p(x) = (x − c)q(x) + r(x), where r(x) = 0 or its degree is
strictly less that the degree of (x − c), which is 1. So r(x) is zero, or a non-zero constant.
Now if we set x = c, we get p(c) = (c − c)q(c) + r(c) = r(c). So r is the constant p(c).
Thus r = 0 if and only if p(c) = 0 if and only if x − c divides p(x).
Of course we are interested in such linear factors of p(x), since c is a root of p(x). We
therefore would like to how many possible values of c there are. For this we need to know
Theorem: (The Fundamental Theorem of Algebra) A polynomial p(x) = an xn + an−1 xn−1 +
. . . + a1 x + a0 , where a0 , a1 , . . . , an ∈ IR and an 6= 0 can be written uniquely (up to reordering
terms) as p(x) = an (x − α1 )(x − α2 ) . . . (x − αn ), where the roots α1 , α2 , . . . , αn are complex
numbers.
So a polynomial of degree n has n complex roots counting multiplicity. To access the
theorem we therefore need some knowledge of the complex numbers. The complex numbers
are denoted by C and consist of all objects of the form z = a+bi, where i2 = −1. The real part
of z is a and we denote it by Re(z). The imaginary part of z is b and we denote it by Im(z).
So every complex number corresponds to an ordered pair (a, b). We can therefore think of
the complex numbers forming a plane where the x-axis is the real axis, and the y-axis is the
imaginary axis.
68
Appendix C: Factoring Polynomials and Complex Arithmetic
Complex numbers are operated on by addition, subtraction, or multiplication as if they
are linear polynomials. The reduction i2 = −1 is used, if necessary, after the operation. So
for example (2 + 3i) + (1 − 4i) = 3 − i, (−2 + i) − (3 − 2i) = 1 + 3i, and (2 + 3i)(−1 − i) =
−2 + (−2 + (−3))i − 3i2 = −2 − 5i − 3(−1) = 1 − 5i. Dividing a complex number by a nonzero
complex number can also be done. To do this we first introduce the conjugate of z = a + ib,
which is the complex number a − ib, and is denoted by z. Next we observe that zz reduces to
a2 + b2 , which we call the norm of z, and denote |z|. It is critical that |z| = 0 if and only if
z = 0. This follows since |z| is the square of the distance of the point z to the origin in the
complex plane. Also z = z if and only if z = Re(z). At any rate, if z 6= 0, we have 1/z = z/|z|.
So to divide by z, we multiply by its conjugate and divide by its norm (which is a real number).
Complex conjugation has several other important properties. If z = a + bi and w = c + di,
then z + w = (a + c) + (b + d)i = (a + c) − (b + d)i = (a − bi) + (c − di) = z + w. The conjugate
of a sum is the sum of the conjugates. Similarly one can show that the conjugate of a product
is the product of the conjugates. This applies to sums and products with more than two terms.
Especially we have that the conjugate of a power of z is the power of the conjugate, z n = (z)n .
Finally z + z = 2a ∈ IR for any z ∈ C.
Consequently we arrive at the following fact.
Theorem: If p(x) = an xn + an−1 xn−1 + . . . + a1 x + a0 is a degree n polynomial with
a0 , a1 , . . . , an ∈ IR and z ∈ C is a root of p(x), then so is z.
Proof: We have
0 = 0 = p(z) = an z n + an−1 z n−1 + . . . + a1 z + a0
= an z n + an−1 z n−1 + . . . + a1 z + a0
= an z n + an−1 z n−1 + . . . + a1 z 1 + a0
= an (z)n + an−1 (z)n−1 + . . . + a1 (z)1 + a0
= an (z)n + an−1 (z)n−1 + . . . + a1 (z)1 + a0 , since a0 , . . . , an ∈ IR
= p(z)
Thus if z ∈ C−IR is a root of a polynomial p(x) with real coefficients, then both (x−z) and
(x−z) are factors of p(x). Therefore p(x) has as factor the quadratic polynomial (x−z)(x−z) =
x2 − (z + z)x + zz = x2 − 2ax + |z|. This quadratic has real coefficients and no real roots. This
fact allows us to move from the Fundamental Theorem of Algebra to the theorem we need.
69
Appendix C: Factoring Polynomials and Complex Arithmetic
Theorem: (Real Factors Theorem) If p(x) = pn xn + pn−1 xn−1 + . . . + p1 x + p0 is a degree n polynomial with p0 , p1 , . . . , pn ∈ IR, then there exist k real numbers r1 , r2 , . . . , rk , not
necessarily distinct, and m quadratic polynomials with no real roots x2 + al x + bl , al , bl ∈
IR, a2l − 4bl < 0, l = 1, . . . , m, not necessarily distinct, so that n = k + 2m and p(x) =
pn (x − r1 )(x − r2 ) . . . (x − rk )(x2 + a1 x + b1 )(x2 + a2 x + b2 ) . . . (x2 + am x + bm ).
Proof: Since IR ⊂ C, by the Fundamental Theorem of Algebra p(x) has n complex roots if we
count multiplicities. For each non-real root z ∈ C − IR, its conjugate is also a root. Thus we
can combine these two linear factors to form a quadratic factor of the form described in the
theorem. After we have done this for all properly complex roots, the remaining roots must be
real. This forces n = k + 2m as in the theorem.
After collecting multiple occurences of linear and quadratic factors together we have
p(x) = pn (x−r1 )e1 (x−r2 )e2 (. . .)(x−rk )ek (x2 +a1 x+b1 )f1 (x2 +a2 x+b2 )f2 . . . (x2 +aj x+bj )fj ,
where n = e1 + e2 + . . . + ek + 2(f1 + f2 + . . . + fj ), and a2l − 4bl < 0, for l = 1, . . . .j. In this
form the linear and irreducible quadratic factors are distinct.
A proof of the fundamental theorem of algebra is beyond the scope of this course. In
fact most undergraduate math majors never see a proof. However it lies behind a common
application in calculus.
Fact: A polynomial p(x) with real coefficients and odd degree must have at least one real root.
Proof: If the degree of p(x) is n = 2r + 1 for some integer r, then 2r + 1 = k + 2m, where k is
the number of real roots p(x) has. Thus k = 2(r − m) + 1 is odd. In particular it cannot be
zero.
So the graph of any odd degree polynomial crosses the x-axis an odd number of times.
Also the graph of a polynomial with even degree intersects the x-axis an even number of times.
Unfortunately the theorem does not help us find these.
As an aid for picking out likely rational roots we have the rational root test.
Theorem: (Rational Root Test) If the rational number p/q in lowest terms is a root of the
polynomial p(x) = an xn + . . . + a1 x + a0 with real coefficients, and an 6= 0, then p divides a0
and q|an .
70
Appendix C: Factoring Polynomials and Complex Arithmetic
For fun, you might try to prove this theorem. This test alone does not necessarily solve
our problem. For example if p(x) = x2 − 30 the rational root test says that the possible
rational roots of p(x) are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30. However we can easily factor
p(x) and see that neither of its roots is rational. So testing p(r) for each of the above values of
r would prove fruitless, except to provide points on the graph of the function. Similarly if we
attempted to synthetically divide p(x) by x − r for the values of r we would be largely wasting
our time. So the rational root test should be employed very carefully, hopefully utilizing
additional theorems.
One additional theorem is DesCartes’ Rule of Signs. If we take the form of p(x) as given
by the real factors theorem and expand it then we see that the coefficients of p(x) change
sign for each time p has a positive real root, plus possibly two more times for each irreducible
quadratic factor. That is, the number of positive real roots of p either equals the number of
sign changes in coefficients of p or less this by an even number. Also the number of negative
real roots of p either equals the number of sign changes in p(−x), or is less than this number
by an even number. This rule can therefore help us be a bit more selective as to which and
how many values from the rational root test we check.
Exercises:
1. Find the quotient q(x) and the remainder r(x) when a(x) = 3x3 − x2 + 4x + 2 is divided by
b(x) = 2x2 + x − 1.
2. Given that p(x) = x6 − 8x4 + 6x3 + 7x2 − 6x has x − 2 and x − 1 as factors, factor p(x)
completely as per the real factor theorem.
3. List all possible rational roots of p(x) = 2x3 − 5x2 + 7x − 21.
4. Find all rational zeroes of the polynomial p(x) = x4 − 3x3 − 2x2 + 5x + 3. Then determine
any irrational zeroes and factor p(x) completely.
5. Find a third degree polynomial with zeroes at x = 2, x = 0 and x = −4 whose x2 term has
coefficient 7.
6. If z = 1 − 5i and w = −2 + 2i, write z 3 − 2zw + w2 in the form a + bi.
7. Given that x = 2 + 3i is a solution of x4 − 4x3 + 14x2 − 4x + 13 = 0, find all solutions.
8. Find the polynomial with real coefficients and leading coefficient 1 which has roots at −1, 2
and (1 + i) and degree 4.
71
Appendix E: Rotation of Axes
Given a non-degenerate quadratic of the form Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 we
can rotate axes to get a new coordinate system, where the corresponding quadratic equation
in these variables has no cross term.
Essentially we want to hold the origin fixed and tilt our heads counter-clockwise at
π
an angle θ where 0 < θ < . The Greeks determined that the appropriate angle to use is
2
A−C
A−C
1
, which means θ satisfies cot 2θ =
.
θ = arccot
2
B
B
To see that this works, with our heads tilted at such an angle we draw in new axes, call
them the z and w-axes. So θ is the angle between the positive x-axis and the positive z-axis and
also the angle bewteen the positive y-axis and the postive w-axis. Given a point P = (x, y) let
p
r = x2 + y 2 denote its distance to the origin. Let α denote the angle between the positive
z-axis and the line, l, joining the origin and P . Note that θ + α is the angle between the
positive x-axis and l. From right triangle trigonometry we know that x = r cos(θ + α) and
y = r sin(θ + α). With respect to our second set of axes P has coordinates z = r cos α and
w = r sin α.
w
y
....
... ...
.......
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...
...
.
...
...
...
...
...
...
...
.
.
...
...
...
.........
...
...
.........
...
...
...... .
.
.
...
.
......
.
.
.
.
.
.
.
.
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(x, y)
z
α
θ
x
We use the sum formulae for sine and cosine to write
x = r cos(θ + α) = r(cos θ cos α − sin θ sin α) = z cos θ − w sin θ
y = r sin(θ + α) = r(sin θ cos α + cos θ sin α) = w cos θ + z sin θ
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Appendix E: Rotation of Axes
We can temporarily let c = cos θ and s = sin θ, so x = zc − ws and y = wc + zs. Substitution
into our general equation yields
0 = Ax2 + Bxy + Cy 2 + Dx + Ey + F
= A(zc − ws)2 + B(zc − ws)(wc + zs) + C(wc + zs)2 + D(zc − ws) + E(wc + zs) + F
which is a general quadratic equation in z and w of the form Gz 2 + Hzw + Iw2 + Jz + Kw + F .
By expanding the expression and collecting coefficients we find that
G = Ac2 + Bsc + Cs2
I = As2 − Bsc + Cc2
J = Dc + Es
K = −Ds + Ec
H = −2(A − C)cs + B(c2 − s2 )
But since θ satisfies cot 2θ =
(A − C)
we have by the double angle identities
B
cot 2θ =
c2 − s2
(A − C)
cos 2θ
=
=
sin 2θ
2cs
B
Cross-multiplying gives us 2(A − C)cs = B(c2 − s2 ). Thus H = 0 as desired.
Exercises:
1. Find the angle θ by which rotation of axes will clear the cross term from the equation
x2 − 2xy + y 2 + 6 = 0.
2. Find the angle θ by which rotation of axes will clear the cross term from the equation
√
x2 + 3xy + 2y 2 + 14 = 0.
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