Download Introduction to Calculus: Unit 1 - Algebra & Lines and more Lecture notes Calculus for Engineers in PDF only on Docsity!
Unit 1
Preliminaries
This is a brief overview of some topics from high-school algebra and geometry that are a
prerequisite for studying calculus. It is by no means a comprehensive summary of all that is needed; it focuses on the basic requirements for calculus.
Objectives
Upon completion of this unit you should be able to:
- perform algebraic manipulations, including factorization and rationalization;
- solve algebraic equations;
- solve algebraic inequalities;
- sketch and analyze the equations of lines; and
- understand the definitions of trigonometric functions.
Learning activities
- Read this unit. Pay attention to the boxed statements and definitions.
- If you need some additional explanations consult any high-school textbook.
- Browse through the list of typical errors and try to avoid making them.
- Do the diagnostic test and compare your answers with those provided.
Examples and comments
Module 0. Elementary algebra
Example 1. Write ax + 3 − x − 3 a as a product (factor as much as possible).
Solution:
The distributive law (of multiplication with respect to addition) tells us that a ( b + c )= ab + ac.
Going from to causes no trouble, but the other direction is not so obvious. Writing
as is sometimes called factoring. In the solution below we do it twice: in the
second and the fourth step.
a ( b + c )
a ( b + c )
ab + ac
ab + ac
a x
xa a xa a
ax x a ax x a
Example 2. Simplify (cancel as much as possible):
a) 3 2
2
x
x x ,
b) 3
2
x
x .
Solution:
In order to cancel a part c appearing both in the numerator and the denominator of a quotient, it should multiply all of the numerator and all of the denominator. Hence, to identify cancellable
Introduction to Calculus MATH 1500 Unit 1 1
expressions the first step should be to factor conveniently the numerator and the denominator of the
quotient.
a) x x
x x
x
x x =− −
2
There are some useful formulas (identities) that should be memorized. Here are three of them:
3 3 2 2
3 3 2 2
2 2
x y x y x xy y
x y x y x xy y
x y x y x y
b) Since , the first of the formulas in the above list is applicable, and we conclude
that
2 2 2 x − 9 = x − 3
2
x x
x x
x
x
Example 3. Rationalize the denominator (eliminate the roots in the denominator)
a) 3 5
,
b) 2 3
,
c) 5 1
3 −
.
Solution:
a) Recall that a ⋅ a = a = a 3 3 2 3 3 , for every number. So, in order to eliminate the root in the
denominator, we should multiply it by
a 3 2 5 , at the same time being careful not to change the
original number. So, we multiply both the numerator and the denominator by 3 2
3 2
3 3
3 2
3 2
3 2
3 3
b) Here we use the identity ( )( ). Applied to 2 2 x − y = x − y x + y x = 2 and y = 3 it yields
2 − 3 =( 2 − 3 )( 2 + 3 ). With this as a hint we proceed to rationalize the original expression.
Solution:
We use the symbol ⇔ again, meaning that the inequalities on both sides of it have the same solutions.
a) 2 x + 3 > 5 ⇔ 2 x > 5 − 3 ⇔ 2 x > 2 ⇔ x > 1
So, the solution is the set of all numbers x such that x > 1. ◊
Compare this solution with the one supplied for example 1(a). You notice that, except for = being
changed to <, there are no other changes. In the last step above we have divided both sides of the inequality by 2 to get the final answer. Now look at the solution for (b).
b) − 2 x + 3 > 5 ⇔ − 2 x > 5 − 3 ⇔ − 2 x > 2 ⇔ x <− 1.
Thus the solution is all numbers x such that x <− 1. ◊
Did you notice that the inequality has changed in the last step (after division by -2)?
If both sides of an inequality are multiplied (or divided) by a negative number then the inequality sign changes the orientation:
< changes to >, > changes to <,
≤ changes to ≥ and ≥ changes to ≤.
c) First of all, we notice that 2 3 5 2 x + x > ⇔. 2 3 5 0. Now we use the following fact. 2 x + x − >
If the solutions of the equality 0 are and , then 2 ax + bx + c = x 1 x 2
2 ax + bx + c = ax − x x − x.
We have established in example 1(b) that the solutions of 2 3 5 0 are and 2 x + x − = x 1 = 1
x 2 (^) =−. Hence, by the boxed formula, we have ) 2
2 3 x − 5 = 2 ( x − 1 )( x 2 x + + , so that
2 3 5 0 is the same as 2 x + x − > ) 0 2
2 ( x − 1 )( x + >.
Here we argue as follows (another procedure is given in the next subsection (module 2)).
The left hand side of the last inequality is > 0 only if
(i) both ( x − 1 )and ) 2
( x + are > 0, or if
(ii) both ( x − 1 )and ) 2
( x + are < 0.
Now case (i) yields x > 1 and 0 2
x + >. These two mean x > 1 , and, at the same time 2
x > −.
Hence, it has to be that x > 1. Case (ii), by the same argument leads us to the following: x < 1 ,
and, at the same time 2
x < −. We conclude that both inequalities happen only if 2
x < −. So we
summarize: the solution of the original inequality is the set of all numbers x , such that x > 1 or
x < −. ◊
Example 3. Solve
a) x − 1 = 1 ,
b) x − 1 < 1.
Solution:
First, recall that a = a , when a ≥ 0 and a = − a , when a < 0.
a) Assume first that x − 1 > 0 , that is, assume that x > 1. Then x − 1 = x − 1 so that x − 1 = 1
is equivalent to x − 1 = 1 , which in turn yields x = 2. Since 2 is indeed > 1, is a solution to the original equation.
x = 2
Now suppose that x − 1 < 0 , that is, suppose that x < 1. Then x − 1 =− ( x − 1 ), so that
x − 1 = 1 is equivalent to − ( x − 1 )= 1. The solution to the last equality is x = 0 , and since x < 1
in the case when x = 0 , we have that x = 0 is a solution to the original equation too.
So, the solutions are x = 2 and x = 0. ◊
b) Recall now that y < c means − c < y < c
Hence, x − 1 < 1 means − 1 < x − 1 < 1
1 1 1
, so that we need to solve these two inequalities
simultaneously. We have − < x − < ⇔ 0 < x < 2 , and thereby the solution is the set of all
numbers x in the interval ( 0 , 2 ). ◊
Recall that the open interval ( a , b ) is the set of all numbers x , such that ; the semiopen
interval is the set of all numbers
a < x < b
[ a , b ) x , such that a ≤ x < b ; are defined
similarly.
( a , b ] and [ a , b ]
Module 2. More on solving equations and inequalities
Example 1. Solve
a) ( x − 1 )( x + 1 )= 0 ,
b) ( 1 )( 1 ) 0 , 2 x − x + =
c) ( 1 )( 1 ) 1. 2 x − x + =−
Solution:
The following principle is called the Zero Product Principle (ZPP):
f ( x )⋅ g ( x )= 0 means that at least one of f ( x )and g ( x )is equal to 0.
a) According to ZPP, ( x − 1 )( x + 1 )= 0 means (is equivalent to) x − 1 = 0 or. The
former yields , while the latter yields
x + 1 = 0
x =− 1 x = 1. Hence, the solutions are x =− 1 and x = 1.
Introduction to Calculus MATH 1500 Unit 1 5
b) 2 x − x = 1
Solution:
a) First we note that x ≠ 0 or else x
has no clear meaning. Under that assumption, we have:
2 x +
x
(we have multiplied both sides of the first equation by 3 χ ). The last
equation is in turn equivalent to
2 x + 9 x − 9 = 0. Apply the quadratic formula to get
x =
or
x
b) There is a square root in the equation. Since a square root x^ is a real number only if^ ,
in this problem we deal only with numbers
x ≥ 0
x such that^ x ≥^0.
Notice that 2 x^^ −^ x =^1 ⇔^2 x −^1 = x. Now, the last equation implies (after squaring both sides
of it) that , which in turn yields. Apply the quadratic formula to get
the solution to the last equation:
4 x − 4 x + 1 = x
2 4 5 1 0
2 x − x + =
x = and x = 1. These are solutions of , but,
although they are both ≥0, they may not be solutions of
4 x − 4 x + 1 = x
2
2 x − x = 1.
The point to take a note is that although 2 x − x = 1 implies , it may be the case
that the latter does not imply the former. Indeed, we check both solutions in
4 x − 5 x + 1 = x
2
2 x − x = 1 , and
easily see that only satisfies the equation. Hence the only solution of the original equation is
x = 1
x = 1
Example 4. Solve the following inequalities.
a) ) 0 2
2 ( x − 1 )( x + > ,
b) ( 1 )( 2 ) 0 , 2 x + x − x <
c) x x 3 x 3 x. 4 3 2 − − >−
We outline a new procedure: suppose we need to solve , where
is a product of polynomials. We first find the solutions
of (listed in increasing order). Note that, by ZPP, the last amounts to finding
the solutions of all
f ( x )> 0
f ( x )= p 1 ( x )⋅ p 2 ( x )⋅...⋅ pk ( x )
x 1 (^) , x 2 ,..., x n f ( x )= 0
p 1 ( x )= 0 , p 2 ( )= 0 ,..., ( )= 0
x pk x
1 , ),( ,^ )
. Then we choose one number in each of
the intervals −∞ x x x xn − xn xn ∞
f ( x )
. Finally we test all , ,..., , at
each of these numbers, to get the signs of. The inequality is true on those intervals
for which the chosen number yields a positive value for. Let us now do this on the above
examples.
p 1 ( x )
x ) > 0
p 2 (^) ( x ) pk ( x )
f (
f ( x )
Introduction to Calculus MATH 1500 Unit 1 7
Solution:
a) This was solved in Example 2(c) in Module 1. Here we shall provide an alternative solution.
Denote ) 2
f ( x )= 2 ( x − 1 )( x +. The solution of 2 ( x − 1 )= 0 is 1 , while the solution of
( x + = is 2
− , so that , 1 2
x 1 (^) = − x 2 = is the list of the solutions of (as
established in 2(c). This splits the set of all real number into three intervals:
f ( x )= 0
( −∞, − − and ∞. We choose one number in each of these interval and then test
f on it. Let us pick the number− 3 from ) 2
( −∞ , −. We find that, for x =− 3 , we have that
2 ( x − 1 ) < 0 , and ( x + < so that, since in this case is a product of two negative
numbers, we have that is >0 on that interval. Now we choose
f ( x )
f ( x ) x^ =^0 from , 1 ) 2
(− ; here we
have, ( ) ⎟> 0
2 x − 1 < 0 , and ⎜ so, since “- times + is - ”, is < 0 on
x + f ( x ) , 1 ) 2
(−. Finally,
we choose x = 2 from (1, ∞) ; ) > 0 2
2 ( x − 1 ) > 0 , and ( x + , so that f ( x )is > 0 on (1, ∞).
Hence, the solution to (a) is the set of all numbers belonging to one of the intervals ) 2
( −∞ , − and
We may summarize all this in a table of signs:
Interval, number (^) 2 ( x − 1 ) ) 2
( x +
f ( x )
x =− 3 −^ − +
x = 0 − +^ −
(1, ∞ ), x = 2 +^ +^ +
f ( x )is > 0 for (^) x in ) 2
( −∞, − or in (1, ∞). ◊
b) Obviously, the solutions of ( 1 )( 2 ) 0 are 2 x + x − x = x = 0 and x = 2 ( is never equal to
0). This splits the real line into the intervals
2 x +
(−∞, 0 ), ( 0 , 2 )and( 2 ,∞). Note that is
always greater then 0. This implies that the sign of depends solely on
2 x +
x + 1 ) ( x − 2 ) x 2 g ( x )=(
slope of the corresponding line is defined to be the number b
a −. If the slope b
a − is 0, (that is, if
a = 0 ) then the equation of the line is b
c y = − and its graph is a horizontal line. On the other hand, if
b = 0 , then (*) becomes ax = c , i.e., a
c x = ; this determines a vertical line. Vertical lines have no
slope (though you may think of them as being lines with “infinite” slope).
Example 1. Find the slopes and graph the lines defined by the equations:
a) x + y = 1 ,
3 y
2 x
x +
b) − 3 = 0 ,
c) = 4.
Solution:
a) y = 1 ⇔ y =− x + 1 and we see that the coefficient in front of x in the last equation is -
1, so that the slope of the line is -1. To sketch the graph of the equation it suffices to plot two distinct points and then join them with a straight line. To find a point on the graph, choose any (^) x and compute (^) y from the given equation. For example, if x = 0 , then we compute y = 1 his
yields the point (^) (0. Similarly,
: t
,1) x^ =^1 implies^ y^ =^0 , hence the point^ (1,0 ).
x
y
(0,1)
(1,0)
b). This is a horizontal line; the last equation tells us that
irrespective of what the variable
3 y − 3 = 0 ⇔ y = 1 y = 1
x is. For example, (− 5 , 1 ),( 0 , 1 ),( 101. 5 , 1 )are all points on that line.
y
x
y=
c). This defines a vertical line (in a plane with standard coordinates). Vertical
lines have no slopes; informally, they have an infinite slope).
2 x = 4 ⇔ x = 2
x
y x=
(^12)
Let the line l1 be defined by the equation y = m 1 x + s 1 ( are constants), and let l2 be defined
by ( are constants).
m 1 , s 1
y = m 2 x + s 2 m 2 , s 2
Two non-vertical lines l1 and l2 are parallel if and only if they have the same slope (i.e., if and only if
m 1 (^) = m 2 ).
The lines l1 and l2 are perpendicular if and only if
2
1
m
m = −.
Introduction to Calculus MATH 1500 Unit 1 11
Example 3. Find the equation of the line l passing through the point (1,2) and perpendicular to the
line y = − x + 3.
Solution:
First we find the slope m of l. It is obvious that the slope of the line y = − x + 3 is -1. Since l is
perpendicular to that line, we have 1 1
m = −. Hence, according to (1), the equation of l is
y − 2 = x − 1. ◊
Module 4. Basic trigonometry
Degree is a measure of the magnitude of an angle. For example, the right angle has a measure of
, while the angle corresponding to a full circle has. An alternative measure is radian (rad):
1 radian is the measure of the central angle corresponding to a circular segment such that the length of the arc of that segment is equal to the radius of the circle.
o 360
o
The relation between degrees and radians is the following:
π rad = 180
o
Example 1
a) Change from radians into degrees:
b) Change from degrees into radians: 1.
o , 90
o
Introduction to Calculus MATH 1500 Unit 1 13
Solution:
The clue is given in the above boxed statement.
a)
rad =
o
o
rad =
o
o
o = 360
o
b) 1
o
rad
o = 90 ⋅
rad =
rad. ◊
If an explicit indication of the measure is missing, then you should assume that it is the radian measure that has been used.
The function sin x is defined as follows. Place a circle of radius 1 in a coordinate plane, so that the
center of the circle is at the origin. Let α be any number. Measure the angle of α radians such that
the positive part of the x^ -axis is its first leg: anticlockwise if α is positive, clockwise if α is negative.
Find the intersection of the second leg of the angle and the circle. Then sin α is defined to be
second coordinate of the intersection point. This is illustrated in the pictures below: for the indicated
angle, sin a is simply the second coordinate of the intersection point B.
The function cos x is defined similarly. The only difference is in the last step: cos x is the first
coordinate of the intersection point B (see the picture above).
The other significant trigonometric functions are defined in terms of the above two:
tan x =
sin x
cos x
, cot x =
cos x
sin x
, csc x =
sin x
and sec x =
cos x
.
Example 3
Find the values of all six trigonometric functions at
.
Solution:
, so this angle is made of full counter clockwise circle followed by three quarters of a
circle (see the picture below).
B
The intersection point B has coordinates (0, −1) , so that sin
= − 1 , cos
= 0. It follows that
tan
is not well defined (for tan x =
sin x
cos x
and the denominator is 0 when x =
). The same
argument holds for sec
. On the other hand cot
cos
sin
= 0 , while
csc
7 π
2
sin
7 π
2
A Gallery of common errors Comments
1.
x + 2
x + 3
ac
ab
c
b
: in order to cancel a number (or an
expression) a it has to multiply all of denominator and all of the numerator.
2. x^ + y = x + y 3. x^ ⋅^ y = x ⋅ y
This is subtler than the error 2. It is true that
x ⋅ y = x ⋅ y but ONLY if both x and y are ≥
4. x^ = x
2 x = x 2
. Thus, for example, (^3 )^3
2 − =. But
note though that x^ is equal to x
2 ( )
5. x^ y =^ xy Here is a typical argument with the same error (pay attention to the second step).
(^1) x
x
x
x
x x
x
6. − 2 x > 3, so x > −
− 2 x > 3, so x < −
since we have divided by a
negative number (-2).
7. ( x
2 − 1)( x + 1) = − 1 so
x
2 − 1 = − 1 or x + 1 = − 1
The ZPP principle is not extendible to expressions of type f ( x )⋅ g ( x ) = s , where s is not 0.
8. sin x
2 = sin
2 x sin
2 x is short for (sin x ) , while sin
2 x
2 stands for
sin( x.
2 )
Gems of humor
“What’s one and one and one and one and one and one and one and one and one and one and one and one?”
“I don’t know,” said Alice. “I lost count.”
“She can’t do addition”, said the Red Queen.
Lewis Carrol, Through the Looking Glass
Introduction to Calculus MATH 1500 Unit 1 17
Module 3.
- Sketch the graph of the line
a) 2 x + y = 3
b) 2 x + 1 = 3
- Find an equation of the line passing through the points (−1,2) and (2,. Is this line parallel or
perpendicular to the line
y − 1 = −
x?
Module 4.
- Change 30 into radians. Compute.
o sin
o
- Suppose sin x =
. Compute cos x , tan x and cot x.
[Hint: start with cos x .]
Answers
Module 0.
- a)
y 2 a
a + x
2
b)
x − 2
x
- a) 23
b) 4
3
(^233)
x
x x
Module 1.
- a) x = 1
b) x = 1, x = −
- a) x > − 1
b) −
< x <
Introduction to Calculus MATH 1500 Unit 1 19
Module 2.
- a) x =^1 ,^ x =^2 , x =−^2
b) x = 0
- a) x = 4, x =
b) The intervals ( − ∞,−2] and [0 ,2] ; that is, all x such that x ≤ − 2 or such that 0 ≤ x ≤ 2.
Module 3.
- a)
x
y
y=-2x+
b)
x
y
x=2X=
- y − 2 =
( x + 1). The two lines are perpendicular.
