Calculus I: Differential Calculus
A Companion Guide
Brenda Davison & Jamie Mulholland
Department of Mathematics
Simon Fraser University
c Draft date February 16, 2015
Contents
Contents
i
Preface
iii
Greek Alphabet
v
1 Preparation for Limits
1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2 Algebraic Simplification of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
2 Problem Solving and Rates of Change
13
2.1 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
2.2 Problem solving - skills . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
2.3 Average rate of change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
3 Preparation for Related Rates
21
3.1 Problem Solving - Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
3.2 Related Rates - Setting up the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
Bibliography
27
Index
28
i
ii
CONTENTS
Preface
This booklet consists of some review material that is used in Math150: Calculus I with Review.
The header of each page contains an image
which takes you right back to the table of contents.
These notes are meant to be a companion to Stewart’s Calculus [2]:
No project such as this can be free from errors and incompleteness. I will be grateful to everyone who points
out any typos, incorrect statements, or sends any other suggestion on how to improve this manuscript.
Brenda Davison
bdavison@sfu.ca
Jamie Mulholland
j mulholland@sfu.ca
Simon Fraser University
February 16, 2015
iii
iv
Greek Alphabet
lower capital
case
name
pronunciation
lower capital
case
name
pronunciation
α
β
γ
δ
ε
ζ
η
θ
ι
κ
λ
µ
alpha
beta
gamma
delta
epsilon
zeta
eta
theta
iota
kappa
lambda
mu
(al-fah)
(bay-tah)
(gam-ah)
(del-ta)
(ep-si-lon)
(zay-tah)
(ay-tah)
(thay-tah)
(eye-o-tah)
(cap-pah)
(lamb-dah)
(mew)
ν
ξ
o
π
ρ
σ
τ
υ
φ
χ
ψ
ω
nu
xi
omicron
pi
rho
sigma
tau
upsilon
phi
chi
psi
omega
(new)
(zie)
(om-e-cron)
(pie)
(roe)
(sig-mah)
(taw)
(up-si-lon)
(fie)
(kie)
(si)
(oh-may-gah)
A
B
Γ
∆
E
Z
H
Θ
I
K
Λ
M
v
N
Ξ
O
Π
P
Σ
T
Υ
Φ
X
Ψ
Ω
Chapter 1
Preparation for Limits
1.1
Introduction
The idea of a limit or of taking a limit is of fundamental importance in the study of calculus. In fact,
it is possible to think of calculus as the systematic study of limits. In order to understand the limiting
behaviour of any type of phenomena, the underlying physical reality is first represented by a function.
As such, it is important to have a solid understanding of functions and possess the skills to algebraically
simplify them.
1.2
Algebraic Simplification of Functions
There are many ways to simplify a function. The simplification methods below are key to being able to
quickly and accurately evaluate function limits.
1.2.1
Factoring Polynomials
We begin by highlighting some important things to remember when factoring polynomials.
Fact 1.1: ( Difference of Two Square) Factoring the difference of two squares:
x2 − a2 = (x − a)(x + a).
Example 1.1: x2 − 25 = (x − 5)(x + 5)
Example 1.2: x2 − 17 = (x −
√
17)(x +
√
17)
√ √
√
√
Example 1.3: x − 3 = ( x − 3)( x + 3)
The last example can sometimes be very useful and it sometimes may not immediately occur to you that
x − 3 can be factored.
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Fact 1.2: (Sum/Difference of Two Cubes) Factoring the difference of two cubes:
x3 − a3 = (x − a)(x2 + ax + a2 ).
Factoring the sum of two cubes:
x3 + a3 = (x + a)(x2 − ax + a2 ).
Example 1.4: x3 − 8 = (x − 2)(x2 + 2x + 4)
Example 1.5: x3 + 27 = (x + 3)(x2 − 3x + 9)
You can always check that you have factored correctly by multiplying the factored expression to ensure
that get the original expression back.
Fact 1.3: Factoring a polynomial of the form x2 + (a + b)x + ab:
x2 + (a + b)x + ab = (x + a)(x + b)
When you see a polynomial of the form x2 + (a + b)x + ab, check if the last term, ab, is the product of two
numbers such that the coefficient of the x term, (a + b), is the sum of the those same two numbers.
Example 1.6: x2 + 5x + 6 = (x + 2)(x + 3)
Example 1.7: x2 − 2x − 35 = (x − 7)(x + 5)
1.2.2
Note: 6 = 2 · 3 and 5 = 2 + 3
Note: 35 = −7 · 5 and −2 = −7 + 5
The Factor Theorem and Polynomial Division
Factoring polynomials can quickly become tricky, fortunately there is a connection between linear factors
of a polynomial and its roots. For example, since f (x) = x2 + 5x + 6 has x = −2 as a root (by which we mean
f (−2) = 0) then we know that f (x) has x + 2 as a factor. It remains to determine the other factor of f to
get a complete factorization: x2 + 5x + 6 = (x + 2)(x + 3). To find the other factor, which is x + 3, we could
look for another root of f , which would be x = −3, or we could divide x2 + 5x + 6 by x + 2 to get the cofactor
x + 3:
x+3
x+2
2
x + 5x + 6
− x2 − 2x
3x + 6
− 3x − 6
0
← multiple x, x + 2 and −1
← add previous two lines
← multiple 3, x + 2 and −1
← add previous two lines
Of course this example may be simple enough for you to see the factors without having to go to this much
trouble, however these are precisely the techniques that we will find useful for more complicated examples.
First we begin we a result that connects factors with roots:
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Theorem 1.1: Let P (x) be a polynomial and r a real number. If P (r) = 0, then x − r is a factor of
P (x), i.e. P (x) = (x − r)Q(x) for some polynomial Q(x). Also, if x − r is a factor of P (x), then P (r) = 0.
Example 1.8: Since f (x) = x3 − 2x2 + 4x + 7 has x = −1 as a root (because f (−1) = −1 − 2 − 4 + 7 = 0) then
x + 1 is a factor. So we know that f (x) = (x + 1)Q(x) for some polynomial Q(x), which would have degree 2.
Furthermore, since f (2) = 15 6= 0 then we know x − 2 is not a factor of x3 − 2x2 + 4x + 7.
For quadratic polynomials (that is, polynomials of degree 2) the Factor Theorem comes in really handy
since it tells us that we can look for the roots of the polynomial which can be found using the quadratic
formula (if necessary).
Theorem 1.2: (Quadratic Formula) The quadratic polynomial ax2 + bx + c has roots
√
−b ± b2 − 4ac
.
x=
2a
Note: If b2 − 4ac < 0 then the polynomial has no roots in the real numbers. The number b2 − 4ac is
called the discriminant of the polynomial.
2
Example 1.9: 2x − 3x + 1 has roots x =
(x − 1)(2x − 1).
3±
√
9−4(2)
4
= 1, 1/2 so it factors as 2x2 − 3x + 1 = 2(x − 1)(x − 1/2) =
Example 1.10: Since for 3x2 − x + 5 we have b2 − 4ac = 1 − 4(3)(5) < 0 then 3x2 − x + 5 has no roots, and
therefore doesn’t factor. We call 3x2 − x + 5 irreducible.
For some higher degree polynomials the Factor Theorem is great for allowing us to quickly spot factors,
however once we spot a factor we still need to determine the other cofactors to get a complete factorization.
In example 1.8 above we know that x3 − 2x2 + 4x + 7 = (x + 1)Q(x), but we still need to determine Q(x).
How do we do this?
This can be found by dividing x3 − 2x2 + 4x + 7 by x + 1, recall that we have an algorithm for doing this
(the Polynomial Long Division algorithm):
x2 − 3x + 7
x + 1)
x3 − 2x2 + 4x + 7
− x3 − x2
← multiple x2 , x + 1 and −1
− 3x2 + 4x + 7
2
← add previous two lines
← multiple −3x, x + 1 and −1
3x + 3x
7x + 7
← add previous two lines
7x + 7
← multiple 7, x + 1 and −1
0
← add previous two lines
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Therefore, x3 − 2x2 + 4x + 7 = (x + 1)(x2 − 3x + 7). Since the discriminant of x2 − 3x + 7 is 9 − 4(1)(7) < 0
then it doesn’t factor any further.
Polynomial Long Division Algorithm Uncovered:
if you’re at all curious as to why this algorithm actually works, it really isn’t all that deep. Let’s pretend
we don’t know anything about this algorithm, but we still want to divide x + 1 into x3 − 2x2 + 4x + 7, how
should we proceed?
Start by writing x3 − 2x2 + 4x + 7 = (x + 1)Q(x) and think to ourselves: What must the leading term
of Q(x) be? Well, it must multiply to x to give x3 , therefore it must be x2 . So we know Q(x) = x2 +
[smaller degree terms]. This means we can write:
x3 − 2x2 + 4x + 7 = (x + 1)x2 + [some polynomial of degree 2]
To figure out the remaining polynomial just add −(x + 1)x2 to both side of the equation to get x3 − 2x2 +
4x + 7 − (x + 1)x2 = −3x2 + 4x + 7.
x3 − 2x2 + 4x + 7 = (x + 1)x2 + (−3x2 + 4x + 7).
Now do the same this with this quadratic, write −3x2 + 4x + 7 = (x + 1)(−3x + [some constant]), so
x3 − 2x2 + 4x + 7 = (x + 1)x2 + (x + 1)(−3x) + [some polynomial of degree 1]
To figure out the remaining polynomial just add −(x + 1)(−3x) to −3x2 + 4x + 7 (which is what x3 − 2x2 +
4x + 7 − (x + 1)x2 is equal to) to get 7x + 7. This gives us
x3 − 2x2 + 4x + 7 = (x + 1)x2 + (x + 1)(−3x) + (7x + 7)
Now do the same for the last term in the sum:
x3 − 2x2 + 4x + 7 = (x + 1)x2 + (x + 1)(−3x) + (x + 1)7
Therefore, we can factor out the x + 1 from each term to get
x3 − 2x2 + 4x + 7 = (x + 1)(x2 − 3x + 7).
Can you see how the Polynomial Long Division algorithm is exactly these steps but put in a table-like
format?
1.2.3
Function Simplification by canceling a common factor from the numerator and denominator
Example 1.11: Consider
f (x) =
x2 − 4
.
x−2
Step 1: Factor both the numerator and denominator:
f (x) =
(x − 2)(x + 2)
.
x−2
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Step 2: Cancel any factors that are common to the numerator and denominator:
f (x) = x + 2
(for x 6= 2).
is a simplified version of the original function.
Note that just writing f (x) = x + 2 (that is, omitting the statement ”for x 6= 2”) would be incorrect since it
is not exactly the same as the original function. Using the convention that if the domain is not explicitly
stated then we assume it is all the values of x that make sense to plus into the formula then f (x) = x + 2
2
−4
(division
would imply x = 2 is in the domain of f . But x = 2 is not in the domain of f since f (x) = xx−2
by zero). However, for all numbers except for 2, the simplified function is identical to the original function,
which is why we have added the statement ”for x 6= 2”.
In other words, the function g(x) = x + 2 is the simplified form of f on the domain of f : f (x) = g(x) for
x 6= 2. Think of simplifying f as trying to come up with a function g which is equal to f on the domain of f ,
but which is defined at more values (in this case g is defined at x = 2). See Figure 1.1.
(a) f
(b) g
Figure 1.1: g is the simplified form of f . The only difference is the hole at x = 2.
A more complicated example:
Example 1.12: Simplify the function
f (x) =
2x3 − 54
.
2x3 − 18x
Step 1: Note the common factor of 2 in the numerator:
f (x) =
2(x3 − 27)
.
2x3 − 18x
Step 2: Note the common factor of 2x in the denominator:
f (x) =
2(x3 − 27)
.
2x(x2 − 9)
Step 3: Note and factor the difference of cubes in the numerator and the difference of squares in the
denominator:
f (x) =
2(x − 3)(x2 + 3x + 9)
2x(x − 3)(x + 3)
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Step 4: Cancel factors common to the numerator and denominator:
f (x) =
x2 + 3x + 9
,
x(x + 3)
for x 6= 3.
is a simplified version of the original function. See Figure 1.2.
(a) f
(b) g
Figure 1.2: g is the simplified form of f . The only difference is the hole at x = 3.
1.2.4
Function Simplification involving radicals
√
√
If a function contains a radical ( a ± b) in the numerator
or denominator,
the radical can be removed by
√
√
√
√
a− b
a+ b
√ or √
√ .
multiplying by 1 written in the particular form √
a− b
a+ b
√
9+h−3
.
h
Note: m(h) cannot be directly evaluated when h = 0. When we try to evaluate m at h = 0 both the
numerator and denominator are 0. It is clear why the denominator is 0, there is a factor of h there. We can
also view the numerator as having a hidden factor of h which is the reason it is 0. We now try to get that
hidden factor of h in the numerator to show itself.
√
9+h+3
Step 1: Multiply the function by 1, written as √
, and expand the numerator.
9+h+3
√
√
9+h−3
9+h+3
9+h−9
h
m(h) =
·√
= √
= √
.
h
9+h+3
h( 9 + h + 3)
h( 9 + h + 3)
Example 1.13: Simplify the function m(h) =
Note: We shouldn’t expand the denominator since our goal is to try to get a hidden factor of h in the
numerator to show itself, that way we can cancel it with the factor of h in the denominator. This means we
want to keep the denominator factored.
Step 2: Simplify algebraically by cancelling the common factor of h in the numerator and denominator.
m(h) = √
1
,
9+h+3
for h 6= 0.
Important Note: You can now evaluate this simplified function when h = 0. This allows us to determine
what m is doing near h = 0, even though m is not defined at h = 0. This will be a very useful way to
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calculate limits when we get to that concept.
√
9+h−3
was deficient in
Another way to think about this is as follows: the original function m(h) =
h
1
that sense that h = 0 is not in the domain. The function √
is better since it is equal to m on the
9+h+3
domain of m, but it has a larger domain. It seems to have filled in the holes in m. See Figure 1.3.
(a) m
(b) n
Figure 1.3: n is the simplified form of m. The only difference is the hole at x = 0.
The “simplified” function in this case will most often still contain a radical but it will be the conjugate of the
original radical. Simplified in this context may more accurately be described as “a different version” but
this different version will be one on which certain calculations will be easier to perform. In other words, it
is a ”different version” but one with a larger domain.
√
√
Fact 1.4: a − b is rationalized by multiplying by
√
√
√
√
√b .
a + b is rationalized by multiplying by √a−
a− b
Example 1.14: Rationalize √
√ √
a+ b
√ √ .
a+ b
1
√ .
4+ 3
√
√
4− 3
√ .
Step 1: Multiply the expression by 1, written as √
4− 3
√
√
√
√
1
1
4− 3
4− 3
√
√ =√
√ ·√
√ =
4−3
4+ 3
4+ 3
4− 3
Step 2: Arithmetic simplification in the denominator results in
√
1.2.5
√
√
1
√ = 4− 3
4+ 3
Function Simplification involving absolute value.
The absolute value function gives the the magnitude of a signed quantity. So, for example, velocity is
a signed quantity (you can be traveling in the positive or the negative direction) whereas speed is an
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unsigned (or scalar) quantity. Speed tells you how fast you are going without regard to which direction you
are traveling. The absolute value function is represented by two vertical bars.
Example 1.15: |10| = 10, | − 10| = 10
In both cases above the absolute value of the number is telling us how far the number is from the origin
without regard to which direction we would travel from the origin to get there. It is tempting to think of
the absolute value function as “removing the negative sign”. However, it is important not to do so. For
example, | − a| is not a (think about the case when a is a negative number.) The following definition should
be kept in mind every time you are simplifying a function with an absolute value function.
Definition 1.1: |x| = x when x ≥ 0 and |x| = −x when x < 0. This can also be written as
(
x
if x ≥ 0
|x| =
−x if x < 0
Consider the example of the absolute value of 10 and −10 again, in light of the above definition:
Example 1.16: What is |10|? The thought process is as follows: 10 is greater than or equal to zero so |10|
equals the number between the absolute value signs, 10.
What is | − 10|? The thought process is as follows: −10 is less than zero so | − 10| equals the negative of the
number between the absolute value signs, −(−10) = 10.
Typically, to simplify a function or solve an equation containing an absolute value, you will need to break
the function or equation into two cases and handle each case separately.
Example 1.17: Solve |2x − 3| = 7.
Step 1: Consider the values of x for which 2x − 3 ≥ 0 or 2x ≥ 3 or x ≥ 23 .
For the values of x ≥ 23 , 2x − 3 is greater than or equal to zero. Therefore |2x − 3| = 2x − 3.
And then, solve the equation in the normal matter: 2x − 3 = 7 or 2x = 10 or x = 5.
Check the solution but substituting into the original equation |2(5) − 3| = |10 − 3| = |7| = 7.
Step 2: Consider the values of x for which 2x − 3 < 0 or 2x < 3 or x < 23 .
For the values of x < 23 , 2x − 3 is less than zero. Therefore |2x − 3| = −(2x − 3).
And then, solve the equation in the normal matter: −(2x − 3) = 7 or −2x + 3 = 7 or −2x = 4 x = −2.
Check the solution but substituting into the original equation |2(−2) − 3| = | − 7| = −(−7) = 7
Therefore the two solutions to the original equation are x = 5 or x = −2.
Example 1.18: Simplify h(x) =
|3 − 6x|
.
2x − 1
Step 1: Consider case when 3 − 6x ≥ 0. This occurs when −6x ≥ −3 or x ≤ 12 .
Therefore |3 − 6x| = 3 − 6x for x ≤ 12 . Now simplify the function in the normal manner:
h(x) =
|3 − 6x|
3 − 6x
−3(2x − 1)
=
=
= −3.
2x − 1
2x − 1
2x − 1
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Step 2: Consider the case when 3 − 6x < 0. This occurs when −6x < −3 or x > 12 .
Therefore |3 − 6x| = −(3 − 6x) for x > 12 . Now simplify the function in the normal manner:
h(x) =
|3 − 6x|
−(3 − 6x)
3(2x − 1)
=
=
= 3.
2x − 1
2x − 1
2x − 1
Therefore the function is simplified as follows: h(x) =
when x >
1
2
|3 − 6x|
= −3 when x ≤
2x − 1
1
2
and h(x) =
|3 − 6x|
=3
2x − 1
Alternatively, you could simplify h, using piecewise function notation, as follows.

 3(2x − 1)
|3 − 6x|
3|2x − 1|  2x − 1
h(x) =
=
= −3(2x − 1)

2x − 1
2x − 1

2x − 1
if 2x − 1 > 0
(
=
if 2x − 1 < 0
−3
3
if x <
if x >
1
2
1
2
Now we have no problem defining a function g on all real numbers which is equal to h on the domain of h.
For example, take g(1/2) = −3.
(
−3
g(x) =
3
if x ≤
if x >
1
2
1
2
(a) h
(b) g
Figure 1.4: g is the simplified form of h. The only difference is that g is defined at x = 1/2.
1.2.6
Function Simplification involving fractions
If a function is the sum or difference of two fractions, it can be simplified by finding a common denominator
and performing the addition or subtraction.
Definition 1.2:
a
c
ad + bc
+ =
b
d
bd
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Example 1.19: Simplify f (x) =
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x
5
+
.
x2 + 1 x − 3
Step 1: Note that this is the addition of two fractions and find the common denominator.
The common denominator is (x2 + 1)(x − 3).
Step 2: Write both fractions using the common denominator.
f (x) =
(x2
5(x2 + 1)
x(x − 3)
+
.
+ 1)(x − 3) (x − 3)(x2 + 1)
Step 3: Perform the addition.
f (x) =
x(x − 3) + 5(x2 + 1)
.
(x2 + 1)(x − 3)
Step 4: Algebraically simplify the numerator.
f (x) =
6x2 − 3x + 5
.
(x2 + 1)(x − 3)
If the function is the quotient of two fractions, perform the division by multiplying the numerator by the
reciprocal of the denominator.
Definition 1.3:
( ab )
ad
=
( dc )
bc
2
Example 1.20: Simplify f (x) =
x
( 3−x
)
x
( x+1
)
.
Step 1: Note that this is compound fraction and perform the corresponding multiplication.
f (x) =
x+1
x2
·
3−x
x
Step 2: Algebraically simplify if possible.
f (x) =
x(x + 1)
.
3−x
(a) f
(b) g
Figure 1.5: g is the simplified form of f . The only difference is that g is defined at x = −1 and x = 0.
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Exercises
1. In this exercise you will verify the factorization of the sum/difference of cubes in Fact 1.2. This will
help you remember these factorization formulas.
(a) Use the Factor Theorem to show x − a is a factor of x3 − a3 .
(Hint: Show x = a is a root of the cubic polynomial.)
(b) Use long division to find the polynomial Q(x) such that x3 − a3 = (x − a)Q(x).
(c) Use the Factor Theorem to show x + a is a factor of x3 + a3 .
(d) Use long division to find the polynomial Q(x) such that x3 + a3 = (x + a)Q(x).
2. Simplify the following functions and give the domain of each function.
(a) f (x) =
(b) g(x) =
x2 − x − 6
x2 − 6x + 9
2
2
x+3 − 3
x
− 21
x−2
| − 3x + 2|
(d) f (x) =
3x − 2
(c) h(x) =
x
4
(e) g(x) =
x+3
x2 − 9
3. Rationalize the numerator in each of the following expressions and simplify:
√
√
√
√
x− 2
3−h− 3
(a)
(c)
x−2
h
√1 − 1
x
x
(b)
x−5
√
√
4. Sketch the graph of y = x. Sketch the line through the points (1, 1) and (x, x) for several different
choices of x. What can you say about the slope of these secant lines?
5. Rationalize the denominator in each of the following expressions:
(a)
x−9
√
3− 3
(b) √
5−x
√
5− x
(c)
x−4
√
−2 + x
6. Give the domain and a simplified expression for each of the following functions. Graph each function.
(a) f (x) =
x2 − 7x + 10
x−2
(b) g(x) =
|2 − x|
x−2
(c) h(x) =
x−2
x2 − 7x + 10
7. Let f (x) = x4 − 81. Factor f into a factor of the form x − a times a polynomial of degree 3.
8. Give the domain of each function and find a simpler expression for the function.
x4 − 1
(a) f (x) =
x−1
− x13
(b) g(x) =
x−2
1
8
√
3− x+4
(c) h(x) =
x−5
9. Give the domain and find a simpler expression for the function. Graph each function.
(a) f (x) =
x2 + 2x − 3
1−x
(b) g(x) =
|x − 2|
3x − 6
(c) h(x) =
3−
−x
√
x+9
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Chapter 2
Problem Solving and Rates of Change
2.1
Problem Solving
“If you can’t solve a problem, then there is an easier problem you can solve: find it.”
(George Polya , Hungarian Mathematician, 1887- 1985)
Sometimes the hardest part of solving a problem is getting started doing something. Often, once one thing
has been done, even if it is not obvious how it helps, it sheds light on something else that can be done, and
then a series of steps, not visible at the outset, can be taken which lead to the solution.
The quote above, by Geoge Polya, author of the highly recommended book “How to solve it”, indicates one
way. If the problem under consideration is too difficult, then removing some of the details or variables can
result in a problem that can be solved. With the solution to the easier problem in hand, the complicating
details can slowly be added back in and a solution to the original more difficult problem can be built.
Polya’s problem solving approach consists of four broad steps, each of which may be done differently depending on the type of problem. Those four steps are:
(a) Understand the stated problem.
(b) Make a plan for solving the problem.
(c) Carry out the plan.
(d) Test and evaluate the solution obtained.
For example: Here is a problem:
Example 2.1: A 10-ft wall stands 5 ft from a building. Suppose a ladder is placed on the ground so that it
leans against the building and touches the top of the 10-ft wall. How much does the length of the ladder
need to be adjusted by if the distance from the base of the ladder to the wall increases from 5 ft to 7 ft?
Step 1: Understand the stated problem:
One of the most useful things at this point is to draw a diagram. This will nearly always apply if the
problem, like this one, describes a physical situation that it simple to visualize. Take considerable care
and attention when drawing a diagram and, after it is drawn, check it carefully to see if it accurately
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reflects what the words in the problem describe. Often doing this step deliberately will make it obvious
what to do to solve the problem. Your diagram, to be useful, should be clear and neat and labelled with
all of the information that is given. Quantities that are unchanging (here the height of the wall and how
far the wall is from the building) should be labelled with numbers along with their units. Quantities that
are varying should be labelled with a variable name which you identify with a short statement. For this
problem, perhaps the following:
Let x be the distance the ladder is place from the wall (this will change from 5 ft to 7 ft).
Let L be the length of the ladder (this is not known and indeed the change in L is what we are trying to
find).
Step 2: Make a plan:
Consider the question being asked carefully. Here we want to know how much longer to make the ladder
so it will still reach over the wall and touch the building if we move it further from the wall. Look at the
diagram; and use it to devise a relationship between the things the are known and the things that are not.
In the above diagram, it is apparent that there are triangles which define the relationship between 5, 10, x,
L. Exploit those relationships and write them out algebraically. If at this point, you cannot remember how
to write out the relationships between the knowns and the unknowns because, for example in this case,
you have forgotten some of trigonometry previously learned, then NOW is the time to go and review that.
Each time a skill is mastered or remastered, you have one more tool that will make subsequent problems
easier to solve. The more tools you have, the more complicated solutions you can build. In the solving of
the most complicated problems, the solution will sometimes lie in devising new tools or in using existing
tools in novel ways.
In this case:
Consider the angle, call it θ, between x and L:
x+5
10
and tan θ =
.
L
x
If we know x, then we can find θ from the second formula and once we know θ we can find L from the first
formula. Since we know x is first 5 ft, we can find L, call it L1 when the ladder if 5 ft from the wall. Then
when the ladder moves to 7 ft from the wall, we can do the same calculation, finding L2 . The difference
between L2 and L1 is the amount the ladder will need to be adjusted.
Then cos θ =
Step 3: Carry out the plan:
1
10
= 2. Therefore cos θ = √ . This last step is a trigonometry skill, draw a right
5 √
5
triangle with sides 1, 2 and 5 and make sure you understand this step.
When x is 5 ft, tan θ =
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√
1
x+5
5+5
Then cos θ = √ =
=
and L1 = 10 5.
L
L
5
1
1
10
7
.
When x is 7 ft, tan θ =
. Therefore cos θ = √
7
149
√
7
x+5
7+5
12 149
Then cos θ = √
=
=
and L2 =
.
L2
L2
7
149
√
√
12 149
The change in the ladder length is L2 − L1 and this is
− 10 5 ≈ −1.435 ft
7
Step 4: Test and/or evaluate the solution:
In some cases it is possible to put the answer back into the problem and see if everything works out
according to the problem description. This is the best way to check an answer since you get independent
(from the way that the problem was solved) confirmation that the solution is correct. So, for example
had our problem been to find two numbers that when added total 12 and when multiplied gave 35 and
the solution was determined to 5 and 7, then we could check the solution. Does 5 + 7 = 12 and does
5 · 7 = 35? Since the answer to both questions is yes, we have tested the solution independently how those
two numbers were obtained.
In other cases, like this one, that may not be possible. Instead, the answer should be looked at for reasonableness. We have moved the ladder 2 ft from the wall and have determined that it needs to be approximately 1.435 shorter. The decrease in ladder length appears to be comparable to the amount of movement.
If the answer obtained was 143 ft shorter for example, this would not seem reasonable and then the solutions method and computations should be checked carefully, specifically with the idea of detecting an error
in mind.
2.2
Problem solving - skills
As you are solving problems, you want to make sure that your skills (tools) are available and ready for use.
As such, when you encounter a new technique, take the time to fully understand and master it.
You also want to be flexible. Sometimes, part way through solving a problem, a better (or maybe just
another) method will occur to you. It is not a waste of time to abandon previous work (particularly if you
seemed to be bogged down) and start again down a different path. The time spent on the wrong path gives
you experience and sometimes provides the inspiration for the better route.
Practice staying focussed on the problem and persisting for increasing lengths of time. This is hard to do
but both will improve with practice. A large part of the learning and improvement in your problem solving
skills come during this time that sometimes feels wasted or is frustrating.
The following quote from Dwight D. Eisenhower about warfare could equally well apply to problem solving:
“In preparing for battle, I have found that planning is essential, but plans are useless”
2.2.1
Some Strategies
Here is a summary of some strategies to try:
(a) Try a simpler problem of the same type.
(b) Solve a specific case of the general problem.
(c) Take a guess at the solution and see if you can check if it is correct. Or try working backwards.
(d) Draw a diagram or graph.
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(e) Add to your diagram to see if that helps relationships between different parts become more apparent.
(f) Name all of the variable quantities.
(g) Identify relationships between variable quantities and constant quantities.
(h) Consider if the formulas you have learned previously help you relate the quantities in the problem.
(i) Look for patterns.
(j) Ask if you have seen this type of problem before. If yes, review how that problem was solved.
2.2.2
Some 2-dimensional formulas you must know
Formulas for area A and perimeter P (circumference C).
Rectangle
A = bh
P = 2(b + h)
2.2.3
Triangle
A = 12 bh
P =a+b+c
Trapezoid
A = 12 (b + d)h
P =a+b+c+d
Circle
A = πr2
C = 2πr
Some 3-dimensional formulas you must know
Formulas for volume V and surface area A.
Rectangular prism
V = xyz
A = 2(xy + xz + yz)
2.2.4
Cylinder
V = πr2 h
A = 2πr2 + 2πrh
Other useful formulas worth knowing
(a) Pythagorean Theorem: a2 + b2 = c2
Right circular cone
V = 31 πr2 h
√
A = πr r2 + h2
Sphere
V = 43 πr3
A = 4πr2
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(c) Similar triangles:
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sin A
sin B
sin C
=
=
a
b
c
a
s
=
b
t
(d) Definitions of the trigonometric functions sine, cosine and tangent: referring to the diagram below,
the terminal line segment for the angle θ intersects the unit circle at the point (cos θ, sin θ), and the
slope of this line segment is tan θ.
(e) Displacement d (net distance)
d = st
where s is the average velocity (speed) and t is the time travelled.
(f) Production cost C
C = ax
where a is the cost per item and x is the number of items produced.
(g) Total cost T
T =C +F
where F is the fixed costs of production and C is the production cost.
(h) Revenue R
R = px
where p is the unit price and x is the number sold.
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(i) Profit P
P =R−T
where R is the revenue and T is the total cost.
2.3
Average rate of change
A rate of change is when one quantity changes with respect to another. For example, velocity is a rate of
change. It indicates how distance changes as time changes. Another example is acceleration. Acceleration
is how fast velocity changes as time changes. There are two types of rates of change - instantaneous and
average. Calculus will allow us to calculate instantaneous rates of change. For now, we will just focus on
the average rate of change. For example:
Example 2.2: It is 30km from downtown Vancouver to SFU. It takes me 45 min to drive. What is my
average velocity?
The average velocity is the distance travelled divided by the time taken.
45 min =
v=
45
= 0.75 hours.
60
30
= 40 km/hr (average velocity = the change in the distance/the time taken to travel that distance).
0.75
Definition 2.1: The average rate of change of a function f over the interval [x1 , x2 ] is defined as
f (x2 ) − f (x1 )
.
x2 − x1
Notice that this can be interpreted as the slope of the secant line through the points (x1 , f (x1 )) and
(x2 , f (x2 )).
In examples of business and economics the average rate of change in cost if production is increased by one
at a certain level of production is often called the marginal cost at that level. For example, see Exercise 4c.
2.4
Exercises
1. Jayden has a rectangular yard that is 30ft by 34ft. He would like to add a sidewalk along the outside
of two adjacent sides of the yard. The contractor quotes a price of $20 per square foot to put in the
sidewalk.
(a) Draw a picture of the situation that shows the yard and adjoining sidewalk.
(b) If the sidewalk is only 1 ft wide, how many square feet of sidewalk would be have?
(c) What is the cost of a 1 ft wide sidewalk?
(d) Write a function A(x) for the area (the number of square feet) of sidewalk installed if the sidewalk is x ft wide.
(e) Write a function C(x) for the cost of installing a sidewalk that is x ft wide.
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(f) Use the function C to find the cost of installing a sidewalk 2 ft wide.
(g) If Jayden has $4, 000 to spend, what is the widest he can make the sidewalk to the nearest tenth
of a foot?
2. Maya has a circular garden with a radius of 10m (meters). She would like to add a stone walkway of
constant width around the outside of the garden. The depth of the sidewalk is to be 0.1m.
(a) Draw a picture of the situation.
(b) What is the total area of the garden and a sidewalk that is 2m wide?
(c) What is the area of the garden?
(d) What is the area of a 2m wide sidewalk?
(e) Write an equation for the area A of the sidewalk as a function of the width w of the sidewalk.
(f) Write an equation for the volume V of stone used in constructing the sidewalk as a function of
its width w.
(g) The stone company states that 1m3 of the stone to be used weighs 1 ton. Write an equation for
the number of tons of stone needed as a function of the width w of the sidewalk.
(h) The stone costs $35 per ton. Write an equation for the cost C of the stone to be used as a function
of the width w.
(i) What is the widest sidewalk Maya can make if she has $1, 000 allotted for the stone?
3. The cost C (in dollars) to produce x bracelets is given by the function C(x) = 0.05x2 + 2x + 10. What
is the average rate of change in the cost of producing a bracelet if production is increased from 10 to
20 bracelets?
4. C(x) = 0.1x2 − x + 1500 is the cost to produce x watches. Find the following:
(a) The average rate of change in the cost as x is increased from x = 100 to x = 120.
(b) The average rate of change in cost as x is increased by 40 from the current production level of
100.
(c) The marginal cost at x = 100.
5. The temperature at time t (in minutes) of a plate that is being heated is given by the function T (t) =
9t2 + 10t + 50. You will use a calculator to estimate how rapidly the temperature is changing at t = 4
minutes. First, calculate the average rate of change of temperature per minute between t and t + h
for each of the following increments of h added to t = 4.
(a) h = 1 min
(b) h = 0.1 min
(c) h = 0.01 min
(d) h = 0.001 min
(e) h = 0.0001 min
Look for a pattern in this sequence of average rates of change. The number that the average rates
of change seem to be converging to is called the instantaneous rate of change or, simply, the rate of
change of the temperature at t = 4.
6. For each function below, find the average rate of change of the function between x and x + h and
simplify.
(a) f (x) = 5 − 3x
(b) g(x) = x2 + 4
√
x−2
4
(d) f (x) =
x+1
(c) h(x) =
7. For each of the following functions, find the average rate of change of the function from t = t1 to t = t2
and simplify.
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(d) f (t) =
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(c) h(t) =
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Chapter 3
Preparation for Related Rates
3.1
Problem Solving - Related Rates
“Oh, yes these problems can be nasty. Lots of students fear related rates problems. Why? Maybe because
they are word problems, and students just don’t like word problems. Having to change English into mathematics intimidates many people. It’s as if when they hear it in words, the mathematical sides of their
brain shut down”
Colin Adams, Mathematician, author of How to Ace Calculus: The Streetwise Guide
A related rates problem is a problem that has an equation relating two or more things which change over
time.
For example, consider pumping water into a cylindrical water tank - both the height of the water in the
tank and the volume of the water in the tank are changing. It may be easy for us to determine one of
these rates of change (for example the rate of change of the volume could be determined by measuring the
inflow). The volume of water in the cylinder is related to the height of the water - via the equation for
the volume of a cylinder. Make sure you know the equations in chapter 2 for the volumes and areas of
the basic geometric shapes. This equation, and some calculus, can then be used to find the rate of change
of the height of the water given we know the rate of change of the volume. The main point is that if we
can determine the relationship between (the equation which relates) two variables than we also know the
relationship between the rate of change of those two variables.
Here we look at how to set up these problems, as this is usually the part students find to be the most
difficult, even though it involves no calculus.
3.2
Related Rates - Setting up the Problem
Here are the basic steps to follow to set up for solving a related rates problem:
Step 1: Model the problem using a diagram:
Many related rates problem arise from physical situations that can be modeled with a simple diagram. If
it is possible to do so, always draw a diagram.
Step 2: Identify the independent variable:
The independent variable is the one that all of the other variables depend on. This will nearly always
be time. In our water filling cylinder above, note that the volume and height of the water in the cylinder
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increase as time increases. Time is the independent variable and volume and height are the dependent
variables (there value depends on the time they are observed at).
Step 3: Assign names to all quantities and label your diagram:
On your diagram, label all of the quantities that change over time with a lower case variable name. Use
names that suggest what the quantity is (e.g. h for height, w for width, r for radius etc). If there is any
chance of mixing up which variable name is being used for which quantity, write that information beside
your diagram (e.g. Let r be the radius of the cylinder). Then, label on the diagram, all of the quantities
that do not change with respect to time. Label these with the actual numbers if you know them, or with
capital letters if the quantity is fixed but unknown.
Step 4: Find an equation that relates the dependent variables:
In order to find an equation that relates the dependent variables, use the geometry of the situation as
made evident by your diagram.
Some possibles for determining this equation are:
(a) Use the appropriate volume or area formula.
(b) Identify right angle triangles in your drawing that will allow you to use the Pythagorean Theorem.
(c) Identify similar triangles in your drawing that will allow you to use proportion between the sides.
(d) For problems involving circular motion, use the trigometric functions.
If the relationship between the dependent variable is not readily apparent to you, add lines to your diagram
that will create triangles or squares. These will sometimes allow you to see the relationship between the
dependent variables.
The following several examples will highlight the methods above:
3.2.1
Example - similar triangles
A 2m tall man walks with a speed of 3 m/s away from a street light that is atop an 6-m pole. What is the
relationship between the distance from lamp post to the man and the distance from the lamppost to the
tip of the man’s shadow?
Let the distance from the lamppost to the man be d.
And the distance from the lamppost to the tip of the man’s shadow be s.
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Note that there are two triangles in the above diagram, the large triangle with sides 6m and s and the
small triangle with sides 2m and s − d. These triangles are similar (make sure you know what this means)
because all three of the angles in both triangles are the same. This means that the ratios of the lengths of
the sides of these two triangles are equal. We exploit this fact to draw the following conclusion:
6
2
=
s−d
s
And this can be simplified as follows:
2s = 6(s − d)
s = 3(s − d)
s = 3s − 3d
−2s = −3d
s = 32 d
3.2.2
Example - Pythagorean Theorem
A particle is moving along the parabola y = x2 starting from the origin. What is the relationship between
the x coordinate of the particle and the distance of the particle from the origin?
The particle is at some point on the curve, labelled P . This point has coordinates x and y and because the
particle is on the parabola y = x2 .
The distance to the origin is given by the Pythagorean theorem:
d2 = x2 + y 2 and since y = x2
d2 = x2 + (x2 )2 or
d2 = x2 + x4 .
3.2.3
Example - Appropriate Volume or Surface Areas formula
Sawdust is being poured into a cone-shaped pile. At all times, the height of the pile is equal to the radius
of the base. What is the relationship between the volume of the pile and the height of the pile? Give a
relationship that does not involve the radius of the base of the pile.
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The formula for calculating the volume of a cone is:
V = 31 πr2 h where r is the radius of the base of the cone and h is the height of the cone.
In this particular program, we are told that r = h at all times. Therefore we can substitute into the volume
to obtain:
V = 13 πh2 h = 13 πh3
3.2.4
Example - When Adding lines to the diagram helps
A man starts running north at 4 m/s from a point P . Five minutes later a woman starts running south at
5 m/s from a point 500 m due east of P . What is the relationship between the postion of the man and the
woman?
We will draw the diagram at some time t which is greater than 5 minutes because for the first 5 minutes
the woman is standing still. The problem becomes interesting after she starts moving. We want to know
how to relate z to x and w and this is not immediately apparent from the diagram.
Consider the diagram below once two additional lines have been added. It is much more apparent now
that we can use the Pythagorean theorem to establish the following:
z 2 = 5002 + (x + w)2 where x and w can easily be calculated once we are told how much time has passed.
x and w are calculated using the speed of the man and the woman. Consider a time of 10 minutes measured
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from the time the man starts walking and calculate both x and w. You should find that x = 2400m and
w = 1500m.
3.2.5
Example - Using a Trigonometric Function
This example is a little more complicated. Consider a vertical post with a square cross-section of 1m by 1m
set squarely on a cubical block 3m on a side. Supports from the floor to the sides of the post are desired.
Find the relationship between the length of support and the angle, θ, the support makes with the floor?
First, carefully observe the diagram and reread the words of the problem. Make sure that you that believe
that this diagram depicts the verbal description of the situation.
We want a relationship between L and θ. Why? Perhaps so that we can decide what angle to mount that
supports at the minimize the length of the required supports.
L consists of two pieces, one contribution from the larger triangle, L1 , and one contribution from the
smaller triangle, L2 .
3
3
or L1 = 3 csc θ
or L1 =
L1
sin θ
1
1
For L2 , note that cos θ =
or L2 = sec θ
or L2 =
L2
cos θ
For L1 , note that sin θ =
Therefore L = L1 + L2 = 3 csc θ + sec θ.
3.3
Exercises
1. An airplane is flying parallel to the ground at an altitude of 3 miles. It flies directly over an observer
on the ground and continues flying parallel to the ground. Let d be the distance from the observer to
the airplane.
(a) Draw a picture that represents the situation.
(b) Identify and label all constants and variable quantities.
(c) Find a relationship between d and the other variable quantities; express this relationship as an
equation.
2. Oil is pumped out of an inverted cone-shaped tank. The radius of the top of the tank is 3m, and the
height of the tank is 10m.
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(a) Draw a picture of the tank. Identify and label all constants and variables.
(b) Find an equation that relates the volume of oil in the tank and just one other variable.
3. Two cars leave an intersection at the same time. One car travels north and the other travels west.
Let d be the distance between the two cars.
(a) Draw a picture that represents the situation.
(b) Identify and label all constants and variable quantities.
(c) Find a relationship between d and the other variable quantities and express this relationship as
an equation.
Bibliography
[1] C. Adams, A. Thompson, and J. Hass. How to Ace Calculus: The Streetwise Guide. W.H. Freeman and
Company, 1998.
[2] J. Stewart. Calculus: Early Transcendentals. Brooks/Cole Sengage Learning, 7th edition, 2011.
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Index
Absolute Value, 7
Adding and Subtracting Fractions, 8
Average Rate of Change, 16
Compound Fractions, 9
Difference of Two Squares, 1
Factor Theorem, 3
Factoring a polynomial of degree 2:, 2
long division, polynomial, 3
Rationalizing a radical, 6
Sum/Difference of Two Cubes, 2
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