content stringlengths 86 994k | meta stringlengths 288 619 |
|---|---|
Go to the source code of this file.
subroutine slapmt (FORWRD, M, N, X, LDX, K)
SLAPMT performs a forward or backward permutation of the columns of a matrix.
Function/Subroutine Documentation
subroutine slapmt ( logical FORWRD,
integer M,
integer N,
real, dimension( ldx, * ) X,
integer LDX,
integer, dimension( * ) K
SLAPMT performs a forward or backward permutation of the columns of a matrix.
Download SLAPMT + dependencies
[TGZ] [ZIP] [TXT]
SLAPMT rearranges the columns of the M by N matrix X as specified
by the permutation K(1),K(2),...,K(N) of the integers 1,...,N.
If FORWRD = .TRUE., forward permutation:
X(*,K(J)) is moved X(*,J) for J = 1,2,...,N.
If FORWRD = .FALSE., backward permutation:
X(*,J) is moved to X(*,K(J)) for J = 1,2,...,N.
FORWRD is LOGICAL
[in] FORWRD = .TRUE., forward permutation
= .FALSE., backward permutation
M is INTEGER
[in] M The number of rows of the matrix X. M >= 0.
N is INTEGER
[in] N The number of columns of the matrix X. N >= 0.
X is REAL array, dimension (LDX,N)
[in,out] X On entry, the M by N matrix X.
On exit, X contains the permuted matrix X.
LDX is INTEGER
[in] LDX The leading dimension of the array X, LDX >= MAX(1,M).
K is INTEGER array, dimension (N)
[in,out] K On entry, K contains the permutation vector. K is used as
internal workspace, but reset to its original value on
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
September 2012
Definition at line 105 of file slapmt.f. | {"url":"https://netlib.org/lapack/explore-html-3.4.2/d4/dd7/slapmt_8f.html","timestamp":"2024-11-06T06:13:16Z","content_type":"application/xhtml+xml","content_length":"11142","record_id":"<urn:uuid:a808797d-1c01-4c85-84ec-384273de71b1>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00044.warc.gz"} |
Rich association between the arts and mathematics - Free Essay Example | WritingUniverse
The arts and mathematics have always had a close relationship. Although many people regard the arts and mathematics as distinct fields, there are several ties between them. Art is regarded as a
method of expression and admiration for beauty, as well as an emotional connection. The American Mathematical Society's (AMS) Joseph Malkevitch identified various parallels between art and math.
Throughout the Renaissance, painters and mathematicians recognized one another as truth seekers. Malkevitch claims that painters like Luca Pacioli, Leonardo da Vinci, Albrecht Dürer, and M.C. Escher
all employed mathematical thought to achieve their aesthetic vision. Pacioli, who taught mathematics to Leonardo da Vinci, is said to be the initiator of the double entry accounting system, among
many other accomplishments. However, it is in his first textbook Summa de Arithmetica, Geometrica, proportioni et propotionalita¸ which was printed in 1494, that he explored the connection of
mathematics and proportion, particularly in works of art. He also wrote the book De Divina Proportione, with beautiful illustrations of three-dimensional geometric solids and templates created by his
pupil Leonardo da Vinci (Meisner). Pacioli’s main goal in writing De Divina Proportione is to reveal to artists the secret of harmonic forms. Here is a translated quote from the book:
A work necessary for all the clear-sighted and inquiring human minds, in which everyone who loves to study philosophy, perspective, painting, sculpture, architecture, music and other mathematical
disciplines will find a very delicate, subtle and admirable teaching and will delight in diverse questions touching on a very secret science. (Meisner)
Da Vinci incorporated the principle of Divine proportion in his works of art, particularly in his paintings “The Annunciation”, “The Last Supper”, and “Mona Lisa.” To Da Vinci and other Renaissance
masters like Botticelli and Raphael, creating works of art is synonymous to accurate proportionality, a formula first recorded by Euclid in 300 B.C. As we can see, the importance of mathematics and
art is very evident in Da Vinci’s work. Not only did he use the Divine Proportion extensively in his work, he also used geometry (“The Vitruvian Man.”)
Another Renaissance man, Albrecht Dürer created a series of “Master Prints”, one of which, “Melencolia I”, has influenced generations of artists who are interested in applying math to their works of
art. Dürer used 3D solids in a way that was never used before, an influence from scientist Johannes Kepler, who was seeking a method to pack spheres in the densest way possible. Dürer was a
practicing mathematician and artist, and the polyhedron in the center of “Melancolia I” was studied by generations of mathematicians, and became relevant in applied sciences in the late 20th century,
being used in information and communication theory, and molecular biology (Chudnovksy and Chudnovsky).
However great the Renaissance artists are and their brilliant application of mathematics in their work, this paper will specifically discuss one particular 20th century artist, M.C. Escher. Escher
studied decorative arts, and in 1922 while travelling through Spain and Italy, he experienced a breakthrough in his work. He was fascinated with the decorative interlocking symmetrical designs of the
Moorish-influenced Alhambra, which led him to study tessellation. Tessellation became one of the basic elements of his work, which then led to other mathematical concepts.
M.C. Escher created a lot of work based on mathematics, and to guide us along that path and to better explain the mathematical concepts involved in his work, we have put together some of the
terminologies related to the discipline and his works of art.
An elliptic curve can be used in artwork and architecture. Its particular used is in M.C. Escher’s lithograph entitled Print Gallery. An elliptic curve is a kind of cubic curve. In equation form, an
elliptic curve is y2 z = x3 + axz2 +bz3.
A cyclic number is an (n – 1) digit integer that, when multiplied by 1, 2, 3 …., n- 1, produces the digits in a different order. It has been conjectured, but not yet proven, that an infinite number
of cyclic numbers exist.
Symmetry is another concept that Escher worked with, and his works contains numerous symmetrical images. Symmetry involves more than just repeated images. Symmetry of an object in the plane is a
rigid motion of the plane that leaves the object apparently unchanged. It means that if a viewer closes his or her eyes during the repositioning of a certain object, the object would look exactly the
same when she opened her eyes (Tapp 3).
Symmetry is also defined as looking at the collection of all rigid plane motions, which compose of compositions of rotations, mirror reflections, and sliding translations. According to Silver, any
two symmetries can be “multiplied” to produce another, possibly different symmetry.
The description of all possible symmetry groups of bounded objects is usually attributed to Leonardo da Vinci. Here is Da Vinci’s theorem on bounded objects: The symmetry group of any bounded object
in the place is either infinite or isomorphic to a dihedral or cyclic group (Tapp 51). A stronger rigid version of the theorem is also true: Any bounded object in the place with a finite symmetry
group is rigidly equivalent to one of the objects pictured. The corollary of Da Vinci’s symmetry theorem is: “If a bound object has exactly n rotations and zero flips, then its symmetry group is
isomorphic to Cn. If a bounded object has exactly n rotations, and n flips, then its symmetric group is isomorphic to Dn.
When it comes to the symmetry of wallpaper patterns, there is also a theorem related to this. The symmetry group of any wallpaper pattern is isomorphic to the symmetry group of one of the 17 model
patterns (Tapp 58). However, this theorem is not quite optimal because among the 17model patterns, certain pairs have isomorphic symmetry groups. The symmetry groups of 17 model patterns are often
called “wallpaper groups”. M.C. Escher incorporated many of these patterns into his paintings. It also occurs in nature, such as in honeycombs.
In a way, symmetries behave just like numbers because they can be added or subtracted. When two rotations of symmetrical shapes or two reflections of symmetrical shapes are added or subtracted, a
rotation is the result. However, if one reflection and one rotation is added or subtracted, we get a reflection.
In symmetry, addition of shapes can be repeated. Various figures such as triangles, pentagons, and three-dimensional objects such as cubes can be used. In mathematics, an overall collection of all
symmetries all at one is called a Group. A Group is a collection of objects with the following properties:
If A and B are elements of the group then A ∗ B is also an element of the group.
The operation ∗ is associative, meaning that A ∗ (B ∗ C) = (A ∗ B) ∗ C for any elements A, B and C in the group.
There is an identity (a 0-element) which does nothing.
Every element has an opposite or inverse, which, when combined, gives the identity. (Mathigon)
One way of looking at symmetry is through the design of an equilateral triangle. Equilateral triangles can be rotated through 60 degrees, 120 degrees, and 360 degrees. The reflections across each of
the three lines joining a vertex to the midpoint of the opposite edge can also be rotated. All of these six symmetries comprise a dihedral group (Silver).
A Tessellation is a pattern that can cover a flat surface without any gaps or overlaps. M.C. Escher created a lot of tessellations in his work. Most tessellations have translational, rotational, or
reflectional symmetries. A combination of reflection and translation is called glide reflection (Mathigon). The transformations that we can do to a tessellation are called the Isometries of the
Plane. The symmetry groups in tessellations that tell how a pattern is repeated.
Main Content
The Works of M.C. Escher
Prentententoonstelling (Print Gallery)
M.C. is a world-renowned Dutch graphic artist who made lithographs, woodcuts, and wood engravings. His art-work is mathematically inspired, and he explored mathematical concepts such as symmetry,
perspective, hyperbolic geometry, tessellations, and many more. He was interested in curvilinear perspectives, platonic solids like spheres and cubes, and cylinders and stellated polyhedra. He was
also inspired by the Möbius Strip with his work Möbius Strip I and II (1963 and 1961, respectively).
His 1948 Wood Engraving Stars are made up of cubes formed into a cage with chameleons inside. His interest in spheres and concentric circles inspired his workds Sphere Surface with Fish (1958),
Sphere Spirals (1958), and Concentric Rinds (1953) wood engraving (M.C. Escher).
Dutch graphic artist Maurits Cornelis Escher created a lithograph in 1956 which he entitled Prentententoonstelling (Print Gallery in English). It is an unusual lithograph showing a man viewing a
Mediterranean seaport print inside an art gallery, and among the quayside buildings, as the young man’s eyes travel from left to right, and then down, he sees that one of the buildings is the very
gallery where he is standing. The Droste effect is a famous visual effect in which the image contains itself on a smaller scale, and the image realistically appears on the picture. The Droste effect
is used on the Print Gallery.
It is this work of Escher that we will take a special interest on, as we delve into the mathematics behind this interesting lithograph. The lithograph is drawn on a certain elliptic curve over the
field of complex numbers. There is a deduction that an idealized version of the picture repeats itself in the middle (de Smit and Lenstra, Jr., 446). In precise terms, the lithograph in question has
a copy of itself which is rotated 157.63 degrees and scaled won by a factor of 22 (de Smit and Lenstra, Jr., 446).
In creating Print Gallery, Escher said that he began with the idea of the possibility of creating an annular bulge, a cyclic expansion…without beginning or end. What is a cyclic expansion? The number
142, 857 have long been recognized by those interested in number oddities as one of the most remarkable of integers. When multiplied by any number from 1 to 6 the result always consists of the same
digits and, more remarkably still, these digits always appear in the same cyclic order, but with each number commencing at a different point (Lines, 141). This striking property is called a cyclic
Escher used curved lines to create the lithograph, which he duplicated a number of times. In the beginning, he used straight lines and attempted to expand it, however, the individual small square
only retained its shape with the use of a cyclical expansion of curved lines. After several attempts, Escher created a grid that expands by a factor of 4 in each direction. With this grid, when you
go clockwise going to the center, it folds unto itself, expanded by a factor of 44 = 256.
The grid rotates and scales, which could lead to many different images. Escher then needed an undistorted representation of the scene, but reduced to a factor of 256. The painstaking detail in which
Escher created his work is evidenced by making four different studies, one for each corner of the lithograph. He then fitted the straight square onto the curved grid that he made.
Escher created a complex multiplicative method, a very accurate way of going back and forth between the straight and curved lines of the space he created.
Symmetry in M.C. Escher’s Work
What distinguishes M.C. Escher’s work with symmetry and pattern is that every print or drawing is not abstract. They are easily recognizable, and the viewer can identify each separate object as
composed of everyday objects that we see (MacGillavry, 123).
M.C. Escher used straightedge and compass constructions to create his hyperbolic patterns. These days, symmetrical patterns can be used using computer graphics. Mathematician H.SM. Coxeter hugely
influenced Escher’s work on symmetrical patterns. When Coxeter sent him a copy of his paper entitled “Crystal symmetry and its generalizations, he was shocked because he had found the long-desired
solution on the problem of designing repeated patterns (Dunham 2003).
As Durnham (2003) quotes Goodman Strauss, the center of an orthogonal circular arc is external to the disk and is called its pole. The locus of all poles of arcs through a point in the disk is a line
called the polar of that point.
In creating his first ever work on tessellation, Circle I, Escher used the external web of poles and polar segments, which is called the scaffolding of the tessellation.
Escher’s work Another World depicts the inside of a cube-shaped building. It shows symmetry because it contains a clever three directions of view, at 90 degree angles, which shows an apparent
fourfold rotational symmetry. Another World shows the faces of the cube building with large windows separated by pillars. The axes of the pillars converge towards one point, exactly in the center of
the print (MacGillavry, 124).
Escher also loves plane-filling repeating patterns, which are tessellations. Wallpaper or block printed material did not satisfy him, and instead he sought to work in constructing repeating patterns
from motifs with varying sizes. Circle Limits I to IV are examples of Escher putting small motifs in the middle of the picture, which grew larger and larger towards the edges, and then largest motifs
at the center, becoming smaller and smaller towards the circumference of the circle (MacGillavry 131).
Mathematicians have always assisted artists by providing them tools, which consists of theorems and formulas in order for them to represent their vision in various representations. Although he does
not consider himself a mathematician, M.C. Escher’s work has apparent mathematical qualities because he approached them in a mathematical way.
In Escher’s Print Gallery, we learned about cyclic numbers and the multiplicative method in curved and straight lines. We also saw the possibility of placing undistorted images in curved lines, which
creates a dramatic visual effect.
In Escher’s work on symmetry and tessellations appeal to our intellect rather than emotions, and through there we learn various types of symmetries.
Escher’s work is a landmark foundation in the study of how art and mathematics are harmoniously melded together. These days, a computer program can replicate and even manipulate Escher’s images, but
the main contribution of Escher cannot be undermined. He laid this foundation for other artists and even scientists and mathematicians, not only for printed art, but also for architecture, sculpture,
and graphic design.
Works Cited
Chudnovsky, David and Chudnovsky, Gregory. “After 500 Years, Dürer’s Art Still
Engraved on Mathematician’s Minds (Op-Ed).” Live Science, May 13, 2014. www.livescience.com/45557-durer-engraving-shaped-science-and-math.html. Accessed 11 April 2017.
De Smit, B. and Lenstra, Jr., H.W. “The Mathematical Structure of Escher’s Print
Gallery.” Notices of the AMA, April 2003, 50 (4), pp. 446 – 457. www.ams.org/notices/200304/fea-escher.pdf
Accessed 10 April 2017.
MacGillavry, Caroline H. “The Symmetry of M.C. Escher’s “Impossible” Images. Comp
& Maths, with Appls. 1986, 12B (/2), pp.123-138 ac.els-cdn.com/089812218690146X/1-s2.0-089812218690146Xmain.pdf?_tid=a084e74c-23c2-11e7-92d2 00000aab0f26&acdnat=
1492470628_fee1325e312942bf317dfe388ec447c9. Accessed 10 April 2017.
Malkevitch, Joseph. “Mathematics and Art.” American Mathematical Society.
www.ams.org/samplings/feature-column/fcarc-art1. Accessed 10 April 2017.
Mathigon. “Symmetry and Groups.” Mathigon, n.d.
world.mathigon.org/Symmetry_and_Groups. Accessed 10 April 2017.
Meisner, Gary. “Da Vinci and the Divine Proportion in Art Composition.” The Golden
Number, July 7, 2014. www.goldennumber.net/leonardo-da-vinci-golden-ratio-art/ Accessed 11 April 2017.
M.C. Escher. “M.C. Escher – Biography”. www.mcescher.com/about/biography/
Accessed 10 April 2017.
Lines, M.E. A Number for Your Thoughts. Taylor and Francis, 1986. books.google.com.ph/books?id=Am9og6q_ny4C&pg=PA141&dq=cyclic+number&hl=en&sa=X&ved=0ahUKEwi4zNuWzqvTAhUJv7wKHYsADv8Q6AEIIzAA#v=
onepage&q=cyclic%20number&f=false. Accessed 10 April 2017.
Silver, Daniel S. “Fearless Symmetry.” American Scientist, 2016.
www.americanscientist.org/bookshelf/pub/fearless-symmetry. Accessed 10 April
Tapp, Kristopher. Symmetry: A Mathematical Exploration. Springer, 2012. | {"url":"https://writinguniverse.com/rich-association-between-the-arts-and-mathematics/","timestamp":"2024-11-06T11:08:59Z","content_type":"text/html","content_length":"99693","record_id":"<urn:uuid:ad9b054c-9541-4ea3-904f-6dda1358fa1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00437.warc.gz"} |
Code and documentation for a updated release to the Omic Kriging R package. Focused primarily on improving efficiency, improving ease of use, and reducing dependencies.
This package provides functions to generate a correlation matrix from a genetic dataset and to use this matrix to predict the phenotype of an individual by using the phenotypes of the remaining
individuals through kriging. Kriging is a geostatistical method for optimal prediction or best unbiased linear prediction. It consists of predicting the value of a variable at an unobserved location
as a weighted sum of the variable at observed locations. Intuitively, it works as a reverse linear regression: instead of computing correlation (univariate regression coefficients are simply scaled
correlation) between a dependent variable Y and independent variables X, it uses known correlation between X and Y to predict Y. More updated versions can be found here
Authors and Contributors:
• Hae Kyung Im
• Heather E. Wheeler
• Keston Aquino Michaels
• Vassily Trubetskoy | {"url":"https://cran.rstudio.com/web/packages/OmicKriging/readme/README.html","timestamp":"2024-11-03T22:24:36Z","content_type":"application/xhtml+xml","content_length":"2710","record_id":"<urn:uuid:21f8a3b1-d768-4b28-84f2-6b28bbf4253e>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00737.warc.gz"} |
Help Pls!
Find a linear inequality with the following solution set. Each grid line represents one unit.
(Give your answer in "standard form" ax + by + c > 0 or ax + by + c >= 0 where a b and c are integers with no common factor other than 1.)
Cutiecupcake Nov 28, 2023
To find a linear inequality with the given solution set, we need to consider the shaded region. Let's denote the coordinates of the shaded region's vertices.
Looking at the image, we can see that the shaded region is below and to the left of the line passing through the points (4, 1) and (5, 3). We can use these two points to find the equation of the line
in slope-intercept form (y = mx + b):
First, find the slope (m): m=5−43−1=2
Now, we can use the slope and one of the points (let's use (4, 1)) to find the y-intercept (b): 2⋅47b=−7
So, the equation of the line is y=2x−7.
Now, since we want the shaded region to be below and to the left of this line, we want the region where y<2x−7.
We can rewrite this inequality in standard form (ax + by + c > 0) by moving all terms to one side
So, the linear inequality that represents the shaded region in standard form is −2x+y+7<0.
bingboy Nov 29, 2023 | {"url":"https://web2.0calc.com/questions/help-pls_71","timestamp":"2024-11-06T17:52:14Z","content_type":"text/html","content_length":"21379","record_id":"<urn:uuid:68c486ba-9a49-42e7-915c-c20b3d224a2c>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00609.warc.gz"} |
Square & Square Root of 60 - Methods, Calculation, Formula, How to find
Square & Square Root of 60
Square of 60
60² (60×60) =3600
The square number of 60, denoted as 60², results in 3600. It’s the product of multiplying 60 by itself. This operation is common in mathematics and geometry.
Square Root of 60
√60 = 7.74596669241
√60 = 7.745 up to three places of decimal
The square root of 60, denoted as √60, It is an irrational number, approximately equal to 7.746, and falls between the square roots of 49 and 64.
Square Root of 60: 7.74596669241
Exponential Form: 60^½
Radical Form: √60
Is the Square Root of 60 Rational or Irrational?
The square root of 60 is irrational
The square root of 60, denoted as √60, does not result in a neat, simple fraction. To understand why, let’s break down the factors of 60. The prime factorization of 60 is2²×3×5.
For a square root to be rational, the exponents in its prime factorization need to be even, because a square root function essentially divides these exponents by 2. In the case of
√60, the prime factors 3 and 5 are raised to an odd power (1), which, when halved, results in non-integer exponents. Since a rational number is defined as one that can be expressed as the quotient of
two integers, and
√60 cannot be expressed in such a manner due to its prime factorization, it is deemed irrational. This property highlights a fundamental aspect of irrational numbers: their inability to be neatly
expressed as fractions, making
√60 an endless, non-repeating decimal.
Rational numbers Rational numbers are fractions representing the quotient of two integers, where the denominator is not zero. They can be written as a/b, with both a and b being integers and b ≠ 0.
Rational numbers include integers, proper fractions, improper fractions, and finite decimals.
Examples: 1/2, 3/4, and 7/8.
Irrational numbers Irrational numbers are real numbers that cannot be expressed as a simple fraction, meaning their decimal representation is infinite and non-repeating.
Examples: 2, π (pi), and e
• Square of 60:The square of 60, which means multiplying 60 by itself, results in 3600. This is a basic arithmetic operation in mathematics 60²=60×60=7.746.
• Square root of 60: The square root of 60, denoted as √60 It’s an irrational number, It’s approximately 7.746, a non-integer value.
• Representation: The number 60 is a composite, even number, significant in time measurement as seconds per minute and minutes per hour, and forms the basis of the sexagesimal system.
Methods to Find Value of Root 60
1. Estimation
Since √60 lies between the square roots of perfect squares 49 (7²) and 64 (8²), we can estimate that √60 is slightly less than 8.
2. Prime Factorization
Break down 60 into its prime factors: 60=2×2×3×5=2×3×5² .
Pair the prime factors into squares if possible: 2² can be paired, while 3 and 5 cannot.
Take the square root of each pair: The square root of 2²is 2.
The square root of 60 is then 2√15 because we bring out the square (2) and leave the remaining factors under the square root.
3. Long Division Method
Group the digits of 60 into pairs (from right to left). Since 60 is a two-digit number, it’s already a pair.
Find a number (a) such that a×a is less than or equal to 60. In this case, a is 7 because
Subtract 49 from 60 to get 11.
Bring down two zeros beside 11, making it 1100.
Now, find a number (b) such that 2a×b×b is less than or equal to This involves trial and error, and the number you get will be part of the decimal places of √60
4. Using a Calculator
For most purposes, using a calculator to find the square root of 60 will give you a quick and precise answer. On most calculators, you would enter 60 followed by the square root function.
Square Root of 60 by Long Division Method
Step 1: Find the Largest Square
Identify the largest square number less than 60. Here,
7×7=49 is the largest square less than 60.
Step 2: Subtract and Find Remainder
Subtract 49 from 60 to get the remainder, 60−49=11.
Our initial quotient (which also serves as the divisor for this step) is 7.
Step 3: Update the Divisor
Double the current quotient (which is 7) to get 14. This will be the starting part of our new divisor.
Step 4: Extend the Dividend
Append two zeros to the remainder to get
1100 (as if we’re bringing down two zeros in traditional division).
Step 5: Find the New Divisor and Quotient Digit
Look for a digit X to place at the unit’s place of 14X such that 14X×X is less than 1100.
Here, 147×7=1029, fitting the condition. So, the new digit in the quotient is 7.
Step 6: Update Dividend and Divisor for the Next Step
Subtract 1029 from 1100 to get a new remainder, and then bring down two more zeros to form a new dividend.
Update the divisor by adding the last digit of the quotient to it, making 147+7=154.
Step 7: Repeat the Process
Repeat the process with the new dividend and divisor. Find a digit X for the divisor 154X such that 154X×X is less than the new dividend.
For example, 1544×4=6176, which is less than 7100, the new dividend after bringing down two zeros.
Add 4 to the quotient, following 7.7, to get 7.74.
60 is Perfect Square root or Not
60 is Perfect Square root or Not
The number 60 is not a perfect square. A perfect square is a number that can be expressed as the product of an integer with itself. For example, 1 (1×1), 4 (2×2), 9 (3×3), 16 (4×4), and so on are
perfect squares. To determine if 60 is a perfect square, one would typically look for an integer that, when multiplied by itself, equals 60. However, there is no integer that satisfies this condition
for 60, making it not a perfect square.
FAQs Short Question & Answers
Where is the square root of 60 on the number line?
The square root of 60 is not an integer, but it falls between the square roots of two perfect squares: 49 (7×7) and 64 (8×8). Since the square root of 49 is 7 and the square root of 64 is 8, the
square root of 60 will be a decimal number between 7 and 8.
What are factors of 60?
The factors of 60 are the integers that can be evenly divided into 60. These include:
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
What is prime factor 60?
The prime factors of 60 are 2, 3, and 5, with 2 appearing twice, because 60 equals 2 x 2 x 3 x 5. These are the smallest prime numbers that multiply together to give 60.
Why is 2 a factor of 60?
2 is a factor of 60 because 60 can be evenly divided by 2 with no remainder. This means 2 is a prime number that multiplies with others to result in 60. | {"url":"https://www.examples.com/maths/square-and-square-root-of-60.html","timestamp":"2024-11-09T07:02:19Z","content_type":"text/html","content_length":"107150","record_id":"<urn:uuid:343b7623-009d-449c-bdfc-6ca6c37fb84b>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00730.warc.gz"} |
Mr. Nohner
Resources Targets & Tutorials
Target 8a: I can calculate the are of rectangles, parallelograms, triangles, trapezoids and kites
Target 8b: I can calculate the area of regular polygons
Target 8c: I can calculate an approximate area for irregular shapes
Unit 8 Packet Target 8d: I can calculate the area of circles
Target 8e: I can calculate the area of portions of circles
Target 8f: I can use area to determine geometric probability
Target 8g: I can calculate surface areas of 3D shapes. | {"url":"http://mrnohner.com/geight.html","timestamp":"2024-11-09T12:20:21Z","content_type":"text/html","content_length":"2800","record_id":"<urn:uuid:f18ad928-eeea-479f-9c00-fd051678fd8c>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00179.warc.gz"} |
Most Common Data Science Interview Questions | .cult by Honeypot
We analysed 900+ data science interview questions from companies over the past few years. Here’s a guide to the most common data science interview question categories. Every category explained with
an example.
Becoming a data scientist is considered a prestigious trait. Back in 2012, Harvard Business Review called 'data scientist' the sexiest job of the 21st century, and the growing trend of roles in the
industry seems to be confirming that statement. To confirm this sexiness is still ongoing, the info from Glassdoor shows being a data scientist is the second-best job in America in 2021.
(Source: Glassdoor)
To get such a prestigious job, you have to go through rigorous job interviews. Data science questions asked can be very broad and complex. This is expected, considering the role of a data scientist
usually incorporates so many areas. To help you prepare for the data science job interviews, I have reviewed all the applicable questions and separated them into different question categories.
Here’s how I did that.
Description and Methodology of the Analysis
I have gathered data from various job search boards and websites and company review platforms such as Glassdoor, Indeed, Reddit, and Blind App. To be more precise, there are 903 questions collected
over the past four years.
The questions are sectioned into pre-determined categories. These categories are the result of an expert analysis of the interview experience description taken from our sources. The categories are:
1. Algorithms 2. Business case 3. Coding 4. Modelling 5. Probability 6. Product 7. Statistics 8. System design 9. Technical
What types of interview questions should you expect?
This chart shows you the question type per category according to the collected data.
Translated to percentages, the chart looks like this:
As you can see, the coding and modelling questions are most dominant. More than half of all questions come from that area. It’s not surprising when you think about it. Coding and modelling are
probably the two most important skills for a data scientist. Coding-type questions are widespread, comprising more than one-third of all questions. Other question types, such as algorithms and
statistics, are also fairly significant; 24% of all questions come from these two categories. Other categories are not as represented. I find that reasonable, considering the nature of a data
scientist role.
Now I want to guide you through every question category and show you some examples of the questions being asked.
Are you looking to progress in your career? Companies on Honeypot are looking for Data Scientists! Visit honeypot.io to sign up!
The most tested concepts on data science interview questions
As you already saw, coding questions are the single most important topic in data science. Such questions will require some sort of data manipulation using the code to identify insights. The questions
are designed to test coding ability, problem-solving skills, and creativity. You’ll usually do that on a computer or a whiteboard.
Coding interview question Example
One example from Microsoft is this one:
QUESTION: “Calculate the share of new and existing users. Output the month, share of new users, and share of existing users as a ratio. New users are defined as users who started using services in
the current month. Existing users are users who started using services in the current month and used services in any previous month. Assume that the dates are all from the year 2020.”
You’ll be using the table fact_events, with the sample data looking like this:
To get the desired output, you should write this code:
with all_users as ( SELECT date_part('month', time_id) AS month, count(DISTINCT user_id) as all_users FROM fact_events GROUP BY month), new_users as ( SELECT date_part
('month', new_user_start_date) AS month, count(DISTINCT user_id) as new_users FROM (SELECT user_id, min(time_id) as new_user_start_date FROM fact_events
GROUP BY user_id) sq GROUP BY month ) SELECT au.month, new_users / all_users::decimal as share_new_users, 1- (new_users / all_users::decimal) as share_existing_users FROM
all_users au JOIN new_users nu ON nu.month = au.month
Writing a code in SQL is the most often tested concept when it comes to coding. It’s no surprise since SQL has been the most used tool in data science. One of the concepts you almost can’t avoid in
the interviews is the joins. So make sure you know the difference between different joins and how to use them to get the required result.
Also, you can expect to group data using the GROUP BY clause very often. Some other concepts that are usually asked are filtering data using the WHERE and/or HAVING clause. You’ll also be asked to
select distinct data. And also make sure that you know the aggregate functions, such as SUM(), AVG(), COUNT(), MIN(), MAX().
Some concepts don’t occur that much often, but it’s worth mentioning them and being prepared for such questions. For example, Common Table Expressions or CTEs is one such topic. The other one is the
CASE() clause. Also, don’t forget to refresh your memory on handling the string data types and dates.
1. Modeling
Modelling was the second-largest category in our research data, with 20% of all questions coming from here. These questions are designed to test your knowledge on building statistical models and
implementing machine learning models.
Modelling interview question Examples
Regression, the most common technical data science concept asked in interviews. It’s not surprising, considering the nature of the statistical modelling.
One example from Galvanise would be the following:
QUESTION: “What is regularisation in regression?”
Here is how you could answer this question:
ANSWER: “A regularisation is a special type of regression where the coefficient estimates are constrained (or regularised) to zero. By doing this, it is possible to reduce the variance of the model
while at the same time decreasing the sampling error. Regularisation is used to avoid or reduce overfitting. Overfitting happens when the model learns training data so well it undermines the model’s
performance on new data. To avoid overfitting, Ridge or Lasso regularisations are usually used.”
Some of the concepts tested regularly are, again, other regression analysis concepts, such as logistic regression, Bayesian logistic regression, and naive Bayes classifiers. You can also be asked
about the random forests, as well as testing and evaluating models.
2. Algorithms
Questions on algorithms are all questions that require solving a mathematical problem, mainly through code by using one of the programming languages. These questions involve a step-by-step process,
usually requiring adjustment or computation to produce an answer. These questions test the basic knowledge of problem-solving and data manipulation, which can be implemented for complex problems at
Algorithm interview question examples
The technical concept tested most under algorithms is solving a mathematical or syntax problem with a programming language.
Here is one example you can find on Leetcode:
QUESTION: “You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two
numbers and return the sum as a linked list. “
The example of the data could be something like this:
(Source: Leetcode)
ANSWER (code written in Java should be):
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode dummyHead = new ListNode(0); ListNode p = l1, q = l2, curr = dummyHead; int carry = 0; while (p != null || q !=
null) { int x = (p != null) ? p.val : 0; int y = (q != null) ? q.val : 0; int sum = carry + x + y; carry = sum / 10; curr.next = new ListNode(sum % 10);
curr = curr.next; if (p != null) p = p.next; if (q != null) q = q.next; } if (carry > 0) { curr.next = new ListNode(carry); } return dummyHead.next; }
The other general concepts often tested by this type of question are arrays, dynamic programming, strings, greedy algorithm, depth-first search, tree, hash table, and binary search.
3. Statistics
The statistics interview questions are questions testing the knowledge of statistical theory and associated principles. These questions intend to try how familiar you are with the founding
theoretical principles in data science. Being able to understand the theoretical and mathematical background of analyses being done is important. Answer those questions well, and every interviewer
will appreciate you.
Statistics interview question examples
The most mentioned technical concept is sampling and distribution. For a data scientist, this is one of the most commonly used statistics principles the data scientist implements daily.
For example, an interview question from IBM asks:
QUESTION: “What is an example of a data type with a non-Gaussian distribution?”
To answer the question, you could first define a Gaussian distribution. Then you could follow this by giving examples of the non-Gaussian distribution. Something like this:
ANSWER: “A Gaussian distribution is a distribution where a certain known percentage of the data can be found when examining standard deviations from the mean, otherwise known as a normal
distribution. Some of the examples of the non-Gaussian distribution can be exponential distribution or binomial distribution.”
When preparing for the job interview, make sure you also cover the following topics: variance and standard deviation, covariance and correlation, the p-value, mean and median, hypothesis testing, and
Bayesian statistics. These are all concepts you’ll need as a data scientist, so expect them in the job interviews too.
4. Probability
These questions require theoretical knowledge only on probability concepts. Interviewers ask these questions to get a deep understanding of your knowledge on the methods and uses of probability to
complete the complex data studies usually performed in the workplace.
Probability interview question example
It’s highly probable, pun intended, that the question you’ll get is to calculate the probability of getting a certain card/number from a set of dice/cards. This seems to be the most common element of
questioning for most companies in our research, as many of them have asked these types of questions.
An example of such a probability question from Facebook:
QUESTION: “What is the probability of getting a pair by drawing two cards separately in a 52-card deck?”
Here is how you can answer this:
ANSWER: “This first card you draw can be whatever, so it does not impact the result other than that there is one card less left in the deck. Once the first card is drawn, there are three remaining
cards in the deck that can be drawn to get a pair. So, the chance of matching your first card with a pair is 3 out of 51 (remaining cards). This means that the probability of this event occurring is
3/51 or 5.89%.”
Since this is a kind of “specialised” question that deals only with probability, no other concepts are asked. The only difference is how imaginative the question is. But basically, you’ll always have
to calculate the probability of some event and show your thinking.
5. Product
Product interview questions will ask you to evaluate the performance of a product/service through data. These questions test your knowledge of adapting and using data science principles in any
environment, as is the case with daily work.
Product interview question example
The most prominent technical concept in this category is identifying a company’s product and proposing improvements from a data scientist’s perspective. The high variance in technical concepts tested
on the product side can be explained by the nature of product questions and the higher level of creativity required to answer these.
An example of a product question from Facebook would be:
QUESTION: “What is your favourite Facebook product, and how would you improve it?” ANSWER: Due to the nature of the question, we will let you answer this one yourself.
The general concepts tested heavily depend on the company that’s interviewing you. Just make sure you are familiar with the company’s business and their products (ideally, you’re their user, as
well), and you’ll be fine.
6. Business Case
This category includes case studies and generic questions related to the business that would test a data science skill. The significance of knowing how to answer these questions can be enormous as
some interviewers would like the candidates to know how to apply data science principles to solve a company’s specific problems before hiring them.
Business case question example
Due to the nature of the question type, I could not identify a single technical concept that stands out. Since most of the questions categorised here are case studies, they are unique in a certain
However, here is an example of a business case question from Uber:
QUESTION: “There is a pool of people who took Uber rides from two cities that were close in proximity, for example, Menlo Park and Palo Alto, and any data you could think of could be collected. What
data would you collect so that the city the passenger took a ride from could be determined?”
ANSWER: “To determine the city, we need to have access to the location/geographical data. The data collected could be GPS coordinates, longitude/latitude, and ZIP code.”
7. System Design
System design questions are all questions related to designing technology systems. They are asked to analyse the candidate’s process in solving problems, creating, and designing systems to help
customers/clients. Knowing system design can be quite important for a data scientist; even if your role is not to design a system, you will most likely play a role in an established system and need
to know how it works in order to do your work.
System design interview question example
These questions cover different topics and tasks. But the one that stands out is building a database. Data scientists deal heavily with databases daily, so it makes sense to ask this question to see
whether you can build a database from scratch.
Here is one question example from Audible uncovered in our research:
QUESTION: “Can you walk us through how you would build a recommendation system?”
ANSWER: Since there is such a variety of approaches to answer this question, we will leave you to come up with your own way of building one.
Again, to answer these questions, it’s essential to know the company’s business. Think a little about databases that the company most probably needs, and try to elaborate your approach a little
before the interview.
8. Technical
Technical questions are all questions that are asking about the explanation of various data science technical concepts. The technical questions are theoretical and require knowledge of the technology
you will be using at the company. Due to nature, they can seem similar to coding questions. Knowing the theory behind what you are doing is quite important, so technical questions can often be asked
in interviews.
Technical interview question example
The most tested area is theoretical knowledge of Python and SQL. Not surprising, since these two languages are dominant in data science, along with R to complement Python.
An example of a real-world technical question from Walmart would be:
QUESTION: “What are the data structures in Python?”
ANSWER: “The data structures are used for storing data. There are four data structures in Python: List, Dictionary, Tuple, and Set. Those are the built-in data structures. Lists are used for creating
lists that can contain different types of data. Dictionary is basically a set of keys; they are used to store a value with a key and getting the data using the same key. Tuples are the same as lists.
The difference is that in a tuple, the data can’t be changed. Set contains the unordered elements with no duplicates. Along with the built-in data structures, there are also the user-defined data
These are catch-all types of questions. It’s a category for all the questions that can’t cleanly fit into other categories. Due to that, there are no specific concepts that occur more or less often.
This data science interview guide has been written to support the research undertaken to understand the types of questions being asked at a data science interview. The interview questions’ data are
taken from dozens of companies over a four-year period and analysed. The questions have been categorised under nine different question types (algorithms, business case, coding, modelling,
probability, product, statistics, system design, and technical questions).
As part of the analysis, I talked about some of the most common technical concepts from each question type category. For example, the most asked statistics questions have to do with sampling and
distribution. Every question category is supported by one practical example of the real question.
The article is intended to serve you as an important guide for interview preparation or simply learning more about data science. I hope I have helped you to feel more comfortable about the data
science interview process. Good luck with the interviews! | {"url":"https://cult.honeypot.io/reads/most-common-data-science-interview-questions/","timestamp":"2024-11-13T22:31:59Z","content_type":"text/html","content_length":"246742","record_id":"<urn:uuid:0694c173-9831-468c-a411-5b1e1267a296>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00119.warc.gz"} |
1.Calculus, Metric Edition by James Stewart, Saleem Watson, Daniel K. Clegg (z-lib.org)
Study Smarter.
Ever wonder if you studied enough? WebAssign from
Cengage can help.
WebAssign is an online learning platform for your math,
statistics, physical sciences and engineering courses.
It helps you practice, focus your study time and absorb
what you learn. When class comes—you’re way
more confident.
With WebAssign you will:
Get instant feedback
and grading
Know how well you
understand concepts
Watch videos and tutorials
when you’re stuck
Perform better on
in-class assignments
Ask your instructor today how you can get
access to WebAssign!
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
REFERENCE page 1
Cut here and keep for reference
Arithmetic Operations
Geometric Formulas
ad 1 bc
1 −
− 3 −
asb 1 cd − ab 1 ac
− 1
Formulas for area A, circumference C, and volume V:
Sector of Circle
A − 12 bh
A − r 2
A − 12 r 2
C − 2r
s − r s in radiansd
− 12 ab sin Exponents and Radicals
− x m2n
x2n − n
x m x n − x m1n
sx mdn − x m n
sxydn − x n y n
n m
x myn − s
x − (s
x 1yn − s
s xy − s x s y
− n
3 r
V − 13 r 2h
V − r h
A − 4r 2
A − rsr 2 1 h 2
Factoring Special Polynomials
x 2 2 y 2 − sx 1 ydsx 2 yd
x 3 1 y 3 − sx 1 ydsx 2 2 xy 1 y 2d
x 3 2 y 3 − sx 2 ydsx 2 1 xy 1 y 2d
Binomial Theorem
Distance and Midpoint Formulas
sx 1 yd2 − x 2 1 2xy 1 y 2sx 2 yd2 − x 2 2 2xy 1 y 2
Distance between P1sx1, y1d and P2sx 2, y2d:
sx 1 yd3 − x 3 1 3x 2 y 1 3xy 2 1 y 3
d − ssx 2 2 x1d2 1 s y2 2 y1d2
sx 2 yd3 − x 3 2 3x 2 y 1 3xy 2 2 y 3
sx 1 ydn − x n 1 nx n21y 1
nsn 2 1d n22 2
x y
n n2k k …
x y 1
1 nxy n21 1 y n
nsn 2 1d … sn 2 k 1 1d
1 … 1
Midpoint of P1 P2:
y 2 y1 − msx 2 x1d
Slope-intercept equation of line with slope m and y-intercept b:
If a , b and c . 0, then ca , cb.
y − mx 1 b
If a , b and c , 0, then ca . cb.
| |
| |
| x | . ameansx . aorx , 2a
x − ameansx − aorx − 2a
y2 2 y1
x 2 2 x1
Point-slope equation of line through P1sx1, y1d with slope m:
If a , b and b , c, then a , c.
If a . 0, then
Slope of line through P1sx1, y1d and P2sx 2, y2d:
Inequalities and Absolute Value
If a , b, then a 1 c , b 1 c.
x1 1 x 2 y1 1 y2
Quadratic Formula
2b 6 sb 2 2 4ac
If ax 2 1 bx 1 c − 0, then x −
Equation of the circle with center sh, kd and radius r:
x , ameans2a , x , a
sx 2 hd2 1 s y 2 kd2 − r 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
REFERENCE page 2
Angle Measurement
Fundamental Identities
csc −
sin sec −
cos tan −
sin cos cot −
cos sin s in radiansd
cot −
tan sin2 1 cos2 − 1
Right Angle Trigonometry
1 1 tan2 − sec 2 1 1 cot 2 − csc 2 sins2d − 2sin coss2d − cos tans2d − 2tan sin
radians − 1808
18 −
1 rad −
s − r
csc −
cos −
sec −
tan −
cot −
sin −
Trigonometric Functions
sin −
csc −
cos −
tan −
S D
2 − sin 2
S D
S D
2 − cos 2
2 − cot 2
The Law of Sines
sec −
cot −
sin A
sin B
sin C
(x, y)
The Law of Cosines
a 2 − b 2 1 c 2 2 2bc cos A
Graphs of Trigonometric Functions
b 2 − a 2 1 c 2 2 2ac cos B
y=sin x
y=tan x
c 2 − a 2 1 b 2 2 2ab cos C
y=cos x
Addition and Subtraction Formulas
2π x
sinsx 1 yd − sin x cos y 1 cos x sin y
sinsx 2 yd − sin x cos y 2 cos x sin y
cossx 1 yd − cos x cos y 2 sin x sin y
y=csc x
y=cot x
cossx 2 yd − cos x cos y 1 sin x sin y
y=sec x
2π x
2π x
2π x
tansx 1 yd −
tan x 1 tan y
1 2 tan x tan y
tansx 2 yd −
tan x 2 tan y
1 1 tan x tan y
Double-Angle Formulas
sin 2x − 2 sin x cos x
Trigonometric Functions of Important Angles
sin cos tan 0
cos 2x − cos 2x 2 sin 2x − 2 cos 2x 2 1 − 1 2 2 sin 2x
tan 2x −
2 tan x
1 2 tan2x
Half-Angle Formulas
sin 2x −
1 2 cos 2x
1 1 cos 2x
cos 2x −
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Me t r i c Ver s io n
Australia • Brazil • Mexico • Singapore • United Kingdom • United States
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
This is an electronic version of the print textbook. Due to electronic rights restrictions,
some third party content may be suppressed. Editorial review has deemed that any suppressed
content does not materially affect the overall learning experience. The publisher reserves the right
to remove content from this title at any time if subsequent rights restrictions require it. For
valuable information on pricing, previous editions, changes to current editions, and alternate
formats, please visit www.cengage.com/highered to search by ISBN#, author, title, or keyword for
materials in your areas of interest.
Important Notice: Media content referenced within the product description or the product
text may not be available in the eBook version.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Calculus, Ninth Edition, Metric Version
© 2021, 2016 Cengage Learning, Inc.
James Stewart, Daniel Clegg, Saleem Watson
WCN: 02-300
ALL RIGHTS RESERVED. No part of this work covered by the copyright herein
Metric Version Prepared by Anthony Tan and
Michael Verwer both at McMaster University
may be reproduced or distributed in any form or by any means, except as
permitted by U.S. copyright law, without the prior written permission of the
copyright owner.
International Product Director, Global Editions:
For product information and technology assistance, contact us at
Timothy L. Anderson
Cengage Customer & Sales Support, 1-800-354-9706
or support.cengage.com.
Product Assistant: Andrew Reddish
Content Manager: Emma Collins
For permission to use material from this text or product, submit all
requests online at www.cengage.com/permissions.
Production Service: Kathi Townes, TECHart
Compositor: Graphic World
Art Director: Angela Sheehan, Vernon Boes
IP Analyst: Ashley Maynard
ISBN: 978-0-357-11346-2
Cengage International Offices
Australia/New Zealand
tel: (65) 6410 1200
tel: (61) 3 9685 4111
Cover Designer: Nadine Ballard
tel: (55) 11 3665 9900
tel: (91) 11 4364 1111
Cover Image: WichitS/ShutterStock.com
Latin America
UK/Europe/Middle East/Africa
tel: (52) 55 1500 6000
tel: (44) 0 1264 332 424
IP Project Manager: Carly Belcher
Manager, Global IP Integration: Eleanor Rummer
Text Designer: Diane Beasley
Represented in Canada by
Nelson Education, Ltd.
tel: (416) 752 9100 / (800) 668 0671
Cengage Learning is a leading provider of customized learning solutions
with office locations around the globe, including Singapore, the United
Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at:
For product information: www.cengage.com/international
Visit your local office: www.cengage.com/global
Visit our corporate website: www.cengage.com
Printed in China
Print Number: 01
Print Year: 2020
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A Tribute to James Stewart
About the Authors xxiii
Technology in the Ninth Edition
To the Student
Diagnostic Tests
A Preview of Calculus 1
1 Functions and Limits
Four Ways to Represent a Function 8
Mathematical Models: A Catalog of Essential Functions
New Functions from Old Functions 36
The Tangent and Velocity Problems 45
The Limit of a Function 51
Calculating Limits Using the Limit Laws 62
The Precise Definition of a Limit 73
Continuity 83
Review 95
Principles of Problem Solving 99
2 Derivatives
2.1 Derivatives and Rates of Change 108
wr i t in g proj ec t
• Early Methods for Finding Tangents
2.2 The Derivative as a Function 120
2.3 Differentiation Formulas 133
applied proj ec t
• Building a Better Roller Coaster
2.4 Derivatives of Trigonometric Functions 148
2.5 The Chain Rule 156
applied proj ec t
• Where Should a Pilot Start Descent?
2.6 Implicit Differentiation 164
d is cov ery proj ec t
• Families of Implicit Curves
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2.7 Rates of Change in the Natural and Social Sciences 172
2.8 Related Rates 185
2.9 Linear Approximations and Differentials 192
d is cov ery proj ec t
• Polynomial Approximations
Review 199
Problems Plus 204
Applications of Differentiation
3.1 Maximum and Minimum Values 210
applied proj ec t
• The Calculus of Rainbows 219
The Mean Value Theorem 220
What Derivatives Tell Us about the Shape of a Graph
Limits at Infinity; Horizontal Asymptotes 237
Summary of Curve Sketching 250
Graphing with Calculus and Technology 258
Optimization Problems 265
applied proj ec t
• The Shape of a Can 278
applied proj ec t
• Planes and Birds: Minimizing Energy
3.8 Newton’s Method 280
3.9 Antiderivatives 285
Review 292
Problems Plus 297
4.1 The Area and Distance Problems 302
4.2 The Definite Integral 314
d is cov ery proj ec t
• Area Functions 328
4.3 The Fundamental Theorem of Calculus 329
4.4 Indefinite Integrals and the Net Change Theorem 339
wr i t in g proj ec t
• Newton, Leibniz, and the Invention of Calculus 348
4.5 The Substitution Rule 349
Review 357
Problems Plus 361
Applications of Integration
5.1 Areas Between Curves 364
applied proj ec t
• The Gini Index 373
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Volumes 374
Volumes by Cylindrical Shells 388
Work 395
Average Value of a Function 401
applied proj ec t
• Calculus and Baseball 404
Review 405
Problems Plus 408
Inverse Functions:
Exponential, Logarithmic, and Inverse Trigonometric Functions
6.1 Inverse Functions and Their Derivatives 412
Instructors may cover either Sections 6.2–6.4 or Sections 6.2*–6.4*. See the Preface.
Exponential Functions and
Their Derivatives 420
6.2* The Natural Logarithmic
Functions 433
6.3* The Natural Exponential
Derivatives of Logarithmic
Functions 440
6.4* General Logarithmic and
Exponential Functions
6.5 Exponential Growth and Decay 478
applied proj ec t
• Controlling Red Blood Cell Loss During Surgery 486
6.6 Inverse Trigonometric Functions 486
applied proj ec t
• Where to Sit at the Movies 495
6.7 Hyperbolic Functions 495
6.8 Indeterminate Forms and l’Hospital’s Rule 503
wr i t in g proj ec t
• The Origins of l’Hospital’s Rule 515
Review 516
Problems Plus 520
Techniques of Integration
Integration by Parts 524
Trigonometric Integrals 531
Trigonometric Substitution 538
Integration of Rational Functions by Partial Fractions
Strategy for Integration 555
Integration Using Tables and Technology 561
d is cov ery proj ec t
• Patterns in Integrals 566
7.7 Approximate Integration 567
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
7.8 Improper Integrals 580
Review 590
Problems Plus 594
Further Applications of Integration
8.1 Arc Length 598
d is cov ery proj ec t
• Arc Length Contest 605
8.2 Area of a Surface of Revolution 605
d is cov ery proj ec t
• Rotating on a Slant 613
8.3 Applications to Physics and Engineering 614
d is cov ery proj ec t
• Complementary Coffee Cups 625
8.4 Applications to Economics and Biology 625
8.5 Probability 630
Review 638
Problems Plus 640
Differential Equations
9.1 Modeling with Differential Equations 644
9.2 Direction Fields and Euler’s Method 650
9.3 Separable Equations 659
applied proj ec t
• How Fast Does a Tank Drain? 668
9.4 Models for Population Growth 669
9.5 Linear Equations 679
applied proj ec t
• Which Is Faster, Going Up or Coming Down? 686
9.6 Predator-Prey Systems 687
Review 694
Problems Plus 697
10 Parametric Equations and Polar Coordinates
Curves Defined by Parametric Equations
d is cov ery proj ec t
Polar Coordinates
• Running Circles Around Circles 710
Calculus with Parametric Curves
d is cov ery proj ec t
• Bézier Curves 722
d is cov ery proj ec t
• Families of Polar Curves 732
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Calculus in Polar Coordinates
Conic Sections 740
Conic Sections in Polar Coordinates
Review 757
Problems Plus 760
11 Sequences, Series, and Power Series
Sequences 762
d is cov ery proj ec t
• Logistic Sequences 776
Series 776
The Integral Test and Estimates of Sums 789
The Comparison Tests 798
Alternating Series and Absolute Convergence 803
The Ratio and Root Tests 812
Strategy for Testing Series 817
Power Series 819
Representations of Functions as Power Series 825
Taylor and Maclaurin Series 833
d is cov ery proj ec t
wr i t in g proj ec t
• An Elusive Limit 848
• How Newton Discovered the Binomial Series 849
11.11 Applications of Taylor Polynomials 849
applied proj ec t
• Radiation from the Stars 858
Review 859
Problems Plus 863
12 Vectors and the Geometry of Space
Three-Dimensional Coordinate Systems
Vectors 874
The Dot Product 885
The Cross Product 893
d is cov ery proj ec t
d is cov ery proj ec t
• The Shape of a Hanging Chain 884
• The Geometry of a Tetrahedron 902
Equations of Lines and Planes
d is cov ery proj ec t
• Putting 3D in Perspective 912
Cylinders and Quadric Surfaces
Review 921
Problems Plus 925
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13 Vector Functions
Vector Functions and Space Curves 928
Derivatives and Integrals of Vector Functions 936
Arc Length and Curvature 942
Motion in Space: Velocity and Acceleration 954
applied proj ec t
• Kepler’s Laws 963
Review 965
Problems Plus 968
14 Partial Derivatives
Functions of Several Variables
Limits and Continuity 989
Partial Derivatives 999
d is cov ery proj ec t
• Deriving the Cobb-Douglas Production Function 1011
• The Speedo LZR Racer 1022
The Chain Rule 1023
Directional Derivatives and the Gradient Vector
Maximum and Minimum Values 1046
d is cov ery proj ec t
Tangent Planes and Linear Approximations
applied proj ec t
• Quadratic Approximations and Critical Points 1057
Lagrange Multipliers 1058
applied proj ec t
• Rocket Science 1066
applied proj ec t
• Hydro-Turbine Optimization 1068
Review 1069
Problems Plus 1073
15 Multiple Integrals
Double Integrals over Rectangles 1076
Double Integrals over General Regions 1089
Double Integrals in Polar Coordinates 1100
Applications of Double Integrals 1107
Surface Area 1117
Triple Integrals 1120
d is cov ery proj ec t
• Volumes of Hyperspheres 1133
Triple Integrals in Cylindrical Coordinates
d is cov ery proj ec t
• The Intersection of Three Cylinders
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Triple Integrals in Spherical Coordinates
Change of Variables in Multiple Integrals
Review 1155
applied proj ec t
• Roller Derby 1146
Problems Plus 1159
16 Vector Calculus
Vector Fields 1162
Line Integrals 1169
The Fundamental Theorem for Line Integrals
Green’s Theorem 1192
Curl and Divergence 1199
Parametric Surfaces and Their Areas 1208
Surface Integrals 1220
Stokes’ Theorem 1233
The Divergence Theorem 1239
Summary 1246
Review 1247
Problems Plus 1251
Numbers, Inequalities, and Absolute Values A2
Coordinate Geometry and Lines A10
Graphs of Second-Degree Equations A16
Trigonometry A24
Sigma Notation A36
Proofs of Theorems A41
Answers to Odd-Numbered Exercises A51
Index A135
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A great discovery solves a great problem but there is a grain of discovery in the solution
of any problem. Your problem may be modest; but if it challenges your curiosity and
brings into play your inventive faculties, and if you solve it by your own means, you may
experience the tension and enjoy the triumph of discovery.
george polya
The art of teaching, Mark Van Doren said, is the art of assisting discovery. In this Ninth
Edition, Metric Version, as in all of the preceding editions, we continue the tradition
of writing a book that, we hope, assists students in discovering calculus — both for its
practical power and its surprising beauty. We aim to convey to the student a sense of the
utility of calculus as well as to promote development of technical ability. At the same
time, we strive to give some appreciation for the intrinsic beauty of the subject. Newton
undoubtedly experienced a sense of triumph when he made his great discoveries. We
want students to share some of that excitement.
The emphasis is on understanding concepts. Nearly all calculus instructors agree that
conceptual understanding should be the ultimate goal of calculus instruction; to implement this goal we present fundamental topics graphically, numerically, algebraically,
and verbally, with an emphasis on the relationships between these different representations. Visualization, numerical and graphical experimentation, and verbal descriptions
can greatly facilitate conceptual understanding. Moreover, conceptual understanding
and technical skill can go hand in hand, each reinforcing the other.
We are keenly aware that good teaching comes in different forms and that there
are different approaches to teaching and learning calculus, so the exposition and exercises are designed to accommodate different teaching and learning styles. The features
(including projects, extended exercises, principles of problem solving, and historical
insights) provide a variety of enhancements to a central core of fundamental concepts
and skills. Our aim is to provide instructors and their students with the tools they need
to chart their own paths to discovering calculus.
Alternate Versions
The Stewart Calculus series includes several other calculus textbooks that might be
preferable for some instructors. Most of them also come in single variable and multivariable versions.
• Calculus: Early Transcendentals, Ninth Edition, Metric Version, is similar to the
present textbook except that the exponential, logarithmic, and inverse trigonometric
functions are covered in the first semester.
• Essential Calculus, Second Edition, is a much briefer book (840 pages), though it
contains almost all of the topics in Calculus, Ninth Edition. The relative brevity is
achieved through briefer exposition of some topics and putting some features on the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
• Essential Calculus: Early Transcendentals, Second Edition, resembles Essential
Calculus, but the exponential, logarithmic, and inverse trigonometric functions are
covered in Chapter 3.
• Calculus: Concepts and Contexts, Fourth Edition, emphasizes conceptual understanding even more strongly than this book. The coverage of topics is not encyclopedic and the material on transcendental functions and on parametric equations is
woven throughout the book instead of being treated in separate chapters.
• Brief Applied Calculus is intended for students in business, the social sciences, and
the life sciences.
• Biocalculus: Calculus for the Life Sciences is intended to show students in the life
sciences how calculus relates to biology.
• Biocalculus: Calculus, Probability, and Statistics for the Life Sciences contains all
the content of Biocalculus: Calculus for the Life Sciences as well as three additional chapters covering probability and statistics.
What’s New in the Ninth Edition, Metric Version?
The overall structure of the text remains largely the same, but we have made many
improvements that are intended to make the Ninth Edition, Metric Version even more
usable as a teaching tool for instructors and as a learning tool for students. The changes
are a result of conversations with our colleagues and students, suggestions from users
and reviewers, insights gained from our own experiences teaching from the book, and
from the copious notes that James Stewart entrusted to us about changes that he wanted
us to consider for the new edition. In all the changes, both small and large, we have
retained the features and tone that have contributed to the success of this book.
• More than 20% of the exercises are new:
Basic exercises have been added, where appropriate, near the beginning of exercise sets. These exercises are intended to build student confidence and reinforce
understanding of the fundamental concepts of a section. (See, for instance, Exercises 7.3.1 –4, 9.1.1 – 5, 11.4.3 – 6.)
Some new exercises include graphs intended to encourage students to understand
how a graph facilitates the solution of a problem; these exercises complement
subsequent exercises in which students need to supply their own graph. (See
Exercises 5.2.1–4, Exercises 10.4.43 –46 as well as 53 – 54, 15.5.1 – 2, 15.6.9 – 12,
16.7.15 and 24, 16.8.9 and 13.)
Some exercises have been structured in two stages, where part (a) asks for the
setup and part (b) is the evaluation. This allows students to check their answer
to part (a) before completing the problem. (See Exercises 5.1.1 –4, 5.3.3 –4,
15.2.7 – 10.)
Some challenging and extended exercises have been added toward the end of
selected exercise sets (such as Exercises 5.2.87, 9.3.56, 11.2.79 – 81, and 11.9.47).
Titles have been added to selected exercises when the exercise extends a concept
discussed in the section. (See, for example, Exercises 3.4.64, 10.1.55 – 57, and
Some of our favorite new exercises are 1.3.71, 2.6.63, 3.5.41 – 44, 5.2.79, 5.5.18,
6.4.99 (also 6.4*.67), 10.5.69, 15.1.38, and 15.4.3 – 4. In addition, Problem 14 in
the Problems Plus following Chapter 5 and Problem 4 in the Problems Plus following Chapter 15 are interesting and challenging.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
• New examples have been added, and additional steps have been added to the solutions of some existing examples. (See, for instance, Example 2.1.5, Example 5.3.5,
Example 10.1.5, Examples 14.8.1 and 14.8.4, and Example 16.3.4.)
• Several sections have been restructured and new subheads added to focus the
organization around key concepts. (Good illustrations of this are Sections 1.6, 11.1,
11.2, and 14.2.)
• Many new graphs and illustrations have been added, and existing ones updated, to
provide additional graphical insights into key concepts.
• A few new topics have been added and others expanded (within a section or
in extended exercises) that were requested by reviewers. (Examples include a
sub­section on torsion in Section 13.3, symmetric difference quotients in Exercise 2.1.60, and improper integrals of more than one type in Exercises 7.8.65 – 68.)
• New projects have been added and some existing projects have been updated.
(For instance, see the Discovery Project following Section 12.2, The Shape of a
Hanging Chain.)
• Alternating series and absolute convergence are now covered in one section (11.5).
• The chapter on Second-Order Differential Equations, as well as the associated
appendix section on complex numbers, has been moved to the website.
Each feature is designed to complement different teaching and learning practices.
Throughout the text there are historical insights, extended exercises, projects, problemsolving principles, and many opportunities to experiment with concepts by using technology. We are mindful that there is rarely enough time in a semester to utilize all of
these features, but their availability in the book gives the instructor the option to assign
some and perhaps simply draw attention to others in order to emphasize the rich ideas
of calculus and its crucial importance in the real world.
n Conceptual Exercises
The most important way to foster conceptual understanding is through the problems that
the instructor assigns. To that end we have included various types of problems. Some
exercise sets begin with requests to explain the meanings of the basic concepts of the
section (see, for instance, the first few exercises in Sections 1.5, 1.8, 11.2, 14.2, and
14.3) and most exercise sets contain exercises designed to reinforce basic understanding
(such as Exercises 1.8.3 – 10, 4.5.1 – 8, 5.1.1 – 4, 7.3.1 – 4, 9.1.1 – 5, and 11.4.3 – 6). Other
exercises test conceptual understanding through graphs or tables (see Exercises 2.1.17,
2.2.34 – 36, 2.2.45 – 50, 9.1.23 – 25, 10.1.30 – 33, 13.2.1 – 2, 13.3.37 –43, 14.1.41 –44,
14.3.2, 14.3.4 – 6, 14.6.1 – 2, 14.7.3 –4, 15.1.6 – 8, 16.1.13 – 22, 16.2.19 – 20, and 16.3.1 – 2).
Many exercises provide a graph to aid in visualization (see for instance Exercises 5.2.1 –4, 10.4.43 –46, 15.5.1 – 2, 15.6.9 – 12, and 16.7.24). Another type of exercise
uses verbal descriptions to gauge conceptual understanding (see Exercises 1.8.12,
2.2.64, 3.3.65 – 66, and 7.8.79). In addition, all the review sections begin with a Concept
Check and a True-False Quiz.
We particularly value problems that combine and compare graphical, numerical, and
algebraic approaches (see Exercises 2.7.27, 3.4.33 –34, and 9.4.4).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
n Graded Exercise Sets
Each exercise set is carefully graded, progressing from basic conceptual exercises, to
skill-development and graphical exercises, and then to more challenging exercises that
often extend the concepts of the section, draw on concepts from previous sections, or
involve applications or proofs.
n Real-World Data
Real-world data provide a tangible way to introduce, motivate, or illustrate the concepts
of calculus. As a result, many of the examples and exercises deal with functions defined
by such numerical data or graphs. These real-world data have been obtained by contacting companies and government agencies as well as researching on the Internet and in
libraries. See, for instance, Figure 1 in Section 1.1 (seismograms from the Northridge
earthquake), Exercise 2.2.34 (number of cosmetic surgeries), Exercise 4.1.12 (velocity
of the space shuttle Endeavour), Exercise 4.4.73 (power consumption in the New England states), Example 3 in Section 14.4 (the heat index), Figure 1 in Section 14.6 (temperature contour map), Example 9 in Section 15.1 (snowfall in Colorado), and Figure 1
in Section 16.1 (velocity vector fields of wind in San Francisco Bay).
n Projects
One way of involving students and making them active learners is to have them work
(perhaps in groups) on extended projects that give a feeling of substantial accomplishment when completed. There are three kinds of projects in the text.
Applied Projects involve applications that are designed to appeal to the imagination of students. The project after Section 9.5 asks whether a ball thrown upward takes
longer to reach its maximum height or to fall back to its original height (the answer
might surprise you). The project after Section 14.8 uses Lagrange multipliers to determine the masses of the three stages of a rocket so as to minimize the total mass while
enabling the rocket to reach a desired velocity.
Discovery Projects anticipate results to be discussed later or encourage discovery
through pattern recognition (see the project following Section 7.6, which explores patterns in integrals). Other discovery projects explore aspects of geometry: tetrahedra
(after Section 12.4), hyperspheres (after Section 15.6), and intersections of three cylinders (after Section 15.7). Additionally, the project following Section 12.2 uses the
geometric definition of the derivative to find a formula for the shape of a hanging chain.
Some projects make substantial use of technology; the one following Section 10.2
shows how to use Bézier curves to design shapes that represent letters for a laser printer.
Writing Projects ask students to compare present-day methods with those of the
founders of calculus — Fermat’s method for finding tangents, for instance, following
Section 2.1. Suggested references are supplied.
More projects can be found in the Instructor’s Guide. There are also extended exercises that can serve as smaller projects. (See Exercise 3.7.53 on the geometry of beehive
cells, Exercise 5.2.87 on scaling solids of revolution, or Exercise 9.3.56 on the formation of sea ice.)
n Problem Solving
Students usually have difficulties with problems that have no single well-defined
procedure for obtaining the answer. As a student of George Polya, James Stewart
experienced first-hand Polya’s delightful and penetrating insights into the process
of problem solving. Accordingly, a modified version of Polya’s four-stage problemsolving strategy is presented following Chapter 1 in Principles of Problem Solving.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
These principles are applied, both explicitly and implicitly, throughout the book. Each of
the other chapters is followed by a section called Problems Plus, which features examples
of how to tackle challenging calculus problems. In selecting the Problems Plus problems we have kept in mind the following advice from David Hilbert: “A mathematical
problem should be difficult in order to entice us, yet not inaccessible lest it mock our
efforts.” We have used these problems to great effect in our own calculus classes; it is
gratifying to see how students respond to a challenge. James Stewart said, “When I put
these challenging problems on assignments and tests I grade them in a different way . . .
I reward a student significantly for ideas toward a solution and for recognizing which
problem-solving principles are relevant.”
n Dual Treatment of Exponential and Logarithmic Functions
There are two possible ways of treating the exponential and logarithmic functions and
each method has its passionate advocates. Because one often finds advocates of both
approaches teaching the same course, we include full treatments of both methods. In
Sections 6.2, 6.3, and 6.4 the exponential function is defined first, followed by the logarithmic function as its inverse. (Students have seen these functions introduced this way
in previous courses.) In the alternative approach, presented in Sections 6.2*, 6.3*, and
6.4*, the logarithm is defined as an integral and the exponential function is its inverse.
This latter method is, of course, less intuitive but more elegant. You can use whichever
treatment you prefer.
If the first approach is taken, then much of Chapter 6 can be covered before Chapters 4 and 5, if desired. To accommodate this choice of presentation there are specially
identified exercises involving integrals of exponential and logarithmic functions at the
end of the appropriate sections of Chapters 4 and 5. This order of presentation allows a
faster-paced course to teach the transcendental functions and the definite integral in the
first semester of the course.
For instructors who would like to go even further in this direction an alternate edition
of this book, called Calculus: Early Transcendentals, Ninth Edition, Metric Version, is
available. In this version the exponential and logarithmic functions are introduced in the
first chapter. Their limits and derivatives are found in the second and third chapters at
the same time as polynomials and the other elementary functions.
n Technology
When using technology, it is particularly important to clearly understand the concepts that underlie the images on the screen or the results of a calculation. When
properly used, graphing calculators and computers are powerful tools for discovering
and understanding those concepts. This textbook can be used either with or without
technology — we use two special symbols to indicate clearly when a particular type of
assistance from technology is required. The icon ; indicates an exercise that definitely
requires the use of graphing software or a graphing calculator to aid in sketching a
graph. (That is not to say that the technology can’t be used on the other exercises as
well.) The symbol
means that the assistance of software or a graphing calculator is
needed beyond just graphing to complete the exercise. Freely available websites such
as WolframAlpha.com or Symbolab.com are often suitable. In cases where the full
resources of a computer algebra system, such as Maple or Mathematica, are needed, we
state this in the exercise. Of course, technology doesn’t make pencil and paper obsolete.
Hand calculation and sketches are often preferable to technology for illustrating and
reinforcing some concepts. Both instructors and students need to develop the ability
to decide where using technology is appropriate and where more insight is gained by
working out an exercise by hand.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
n WebAssign: webassign.net
This Ninth Edition is available with WebAssign, a fully customizable online solution
for STEM disciplines from Cengage. WebAssign includes homework, an interactive
mobile eBook, videos, tutorials and Explore It interactive learning modules. Instructors
can decide what type of help students can access, and when, while working on assignments. The patented grading engine provides unparalleled answer evaluation, giving
students instant feedback, and insightful analytics highlight exactly where students are
struggling. For more information, visit cengage.com /WebAssign.
n Stewart Website
Visit StewartCalculus.com for these additional materials:
• Homework Hints
• Solutions to the Concept Checks (from the review section of each chapter)
• Algebra and Analytic Geometry Review
• Lies My Calculator and Computer Told Me
• History of Mathematics, with links to recommended historical websites
• Additional Topics (complete with exercise sets): Fourier Series, Rotation of Axes,
Formulas for the Remainder Theorem in Taylor Series
• Additional chapter on second-order differential equations, including the method of
series solutions, and an appendix section reviewing complex numbers and complex
exponential functions
• Instructor Area that includes archived problems (drill exercises that appeared in
previous editions, together with their solutions)
• Challenge Problems (some from the Problems Plus sections from prior editions)
• Links, for particular topics, to outside Web resources
Diagnostic Tests
The book begins with four diagnostic tests, in Basic Algebra, Analytic Geometry, Functions, and Trigonometry.
A Preview of Calculus
This is an overview of the subject and includes a list of questions to motivate the study
of calculus.
Functions and Limits
From the beginning, multiple representations of functions are stressed: verbal, numerical, visual, and algebraic. A discussion of mathematical models leads to a review of the
standard functions from these four points of view. The material on limits is motivated
by a prior discussion of the tangent and velocity problems. Limits are treated from
descriptive, graphical, numerical, and algebraic points of view. Section 1.7, on the
precise epsilon–delta defintion of a limit, is an optional section.
The material on derivatives is covered in two sections in order to give students more time
to get used to the idea of a derivative as a function. The examples and exercises explore
the meanings of derivatives in various contexts. Higher derivatives are introduced in
Section 2.2.
Applications of Differentiation
The basic facts concerning extreme values and shapes of curves are deduced from the
Mean Value Theorem. Graphing with technology emphasizes the interaction between
calculus and machines and the analysis of families of curves. Some substantial optimization problems are provided, including an explanation of why you need to raise your
head 42° to see the top of a rainbow.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
The area problem and the distance problem serve to motivate the definite integral, with
sigma notation introduced as needed. (Full coverage of sigma notation is provided in
Appendix E.) Emphasis is placed on explaining the meanings of integrals in various
contexts and on estimating their values from graphs and tables.
5 Applications of Integration
This chapter presents the applications of integration — area, volume, work, average
value — that can reasonably be done without specialized techniques of integration.
General methods are emphasized. The goal is for students to be able to divide a quantity
into small pieces, estimate with Riemann sums, and recognize the limit as an integral.
6 Inverse Functions:
As discussed more fully on page xiv, only one of the two treatments of these functions
need be covered. Exponential growth and decay are covered in this chapter.
7 Techniques of Integration
All the standard methods are covered but, of course, the real challenge is to be able to
recognize which technique is best used in a given situation. Accordingly, a strategy for
evaluating integrals is explained in Section 7.5. The use of mathematical software is
discussed in Section 7.6.
Exponential, Logarithmic, and
Inverse Trigonometric Functions
Further Applications of Integration This chapter contains the applications of integration — arc length and surface area — for
which it is useful to have available all the techniques of integration, as well as applications to biology, economics, and physics (hydrostatic force and centers of mass). A section on probability is included. There are more applications here than can realistically be
covered in a given course. Instructors may select applications suitable for their students
and for which they themselves have enthusiasm.
Differential Equations
Modeling is the theme that unifies this introductory treatment of differential equations.
Direction fields and Euler’s method are studied before separable and linear equations
are solved explicitly, so that qualitative, numerical, and analytic approaches are given
equal consideration. These methods are applied to the exponential, logistic, and other
models for population growth. The first four or five sections of this chapter serve as a
good introduction to first-order differential equations. An optional final section uses
predator-prey models to illustrate systems of differential equations.
Parametric Equations and
Polar Coordinates
This chapter introduces parametric and polar curves and applies the methods of calculus to them. Parametric curves are well suited to projects that require graphing with
technology; the two presented here involve families of curves and Bézier curves. A brief
treatment of conic sections in polar coordinates prepares the way for Kepler’s Laws in
Chapter 13.
Sequences, Series, and
Power Series
The convergence tests have intuitive justifications (see Section 11.3) as well as formal
proofs. Numerical estimates of sums of series are based on which test was used to prove
convergence. The emphasis is on Taylor series and polynomials and their applications
to physics.
12 Vectors and the Geometry
of Space
The material on three-dimensional analytic geometry and vectors is covered in this and
the next chapter. Here we deal with vectors, the dot and cross products, lines, planes,
and surfaces.
13 Vector Functions
This chapter covers vector-valued functions, their derivatives and integrals, the length
and curvature of space curves, and velocity and acceleration along space curves, culminating in Kepler’s laws.
14 Partial Derivatives
Functions of two or more variables are studied from verbal, numerical, visual, and algebraic points of view. In particular, partial derivatives are introduced by looking at a
specific column in a table of values of the heat index (perceived air temperature) as a
function of the actual temperature and the relative humidity.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Multiple Integrals
Contour maps and the Midpoint Rule are used to estimate the average snowfall and
average temperature in given regions. Double and triple integrals are used to compute
volumes, surface areas, and (in projects) volumes of hyperspheres and volumes of intersections of three cylinders. Cylindrical and spherical coordinates are introduced in the
context of evaluating triple integrals. Several applications are considered, including
computing mass, charge, and probabilities.
16 Vector Calculus
Vector fields are introduced through pictures of velocity fields showing San Francisco
Bay wind patterns. The similarities among the Fundamental Theorem for line integrals,
Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem are emphasized.
17 Second-Order Differential
Since first-order differential equations are covered in Chapter 9, this online chapter deals
with second-order linear differential equations, their application to vibrating springs and
electric circuits, and series solutions.
Calculus, Ninth Edition, Metric Version, is supported by a complete set of ancillaries.
Each piece has been designed to enhance student understanding and to facilitate creative
n Ancillaries for Instructors
Instructor’s Guide
by Douglas Shaw
Each section of the text is discussed from several viewpoints. Available online at the Instructor’s Companion Site, the Instructor’s Guide contains suggested time to allot, points to stress,
text discussion topics, core materials for lecture, workshop / discussion suggestions, group work
exercises in a form suitable for handout, and suggested homework assignments.
Complete Solutions Manual
Single Variable Calculus, Ninth Edition, Metric Version
Chapters 1–11
By Joshua Babbin, Scott Barnett, and Jeffery A. Cole, with metrication by Anthony Tan and
Michael Verwer, both from McMaster University.
Multivariable Calculus, Ninth Edition, Metric Version
Chapters 10 –16
By Joshua Babbin and Gina Sanders, with metrication by Anthony Tan and Michael Verwer,
both from McMaster University.
Includes worked-out solutions to all exercises in the text. Both volumes of the Complete Solutions Manual are available online at the Instructor’s Companion Site.
Test BankContains text-specific multiple-choice and free response test items and is available online at the
Instructor’s Companion Site.
Cengage Learning Testing
Powered by Cognero
This flexible online system allows you to author, edit, and manage test bank content; create
multiple test versions in an instant; and deliver tests from your LMS, your class­room, or
wherever you want.
n Ancillaries for Instructors and Students
Stewart Website
Homework Hints n Algebra Review
Challenge Problems n Web links n
Access to WebAssign
Printed Access Code: ISBN 978-0-357-43944-9
Instant Access Code: ISBN 978-0-357-43943-2
Additional Topics n
History of Mathematics
Drill exercises
Prepare for class with confidence using WebAssign from Cengage. This online learning
platform—which includes an interactive ebook—fuels practice, so you absorb what you learn
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
and prepare better for tests. Videos and tutorials walk you through concepts and deliver instant
feedback and grading, so you always know where you stand in class. Focus your study time and
get extra practice where you need it most. Study smarter! Ask your instructor today how you can
get access to WebAssign, or learn about self-study options at Cengage.com / WebAssign.
One of the main factors aiding in the preparation of this edition is the cogent advice
from a large number of reviewers, all of whom have extensive experience teaching calculus. We greatly appreciate their suggestions and the time they spent to understand the
approach taken in this book. We have learned something from each of them.
n Ninth Edition Reviewers
Malcolm Adams, University of Georgia
Ulrich Albrecht, Auburn University
Bonnie Amende, Saint Martin’s University
Champike Attanayake, Miami University Middletown
Amy Austin, Texas A&M University
Elizabeth Bowman, University of Alabama
Joe Brandell, West Bloomfield High School / Oakland University
Lorraine Braselton, Georgia Southern University
Mark Brittenham, University of Nebraska – Lincoln
Michael Ching, Amherst College
Kwai-Lee Chui, University of Florida
Arman Darbinyan, Vanderbilt University
Roger Day, Illinois State University
Toka Diagana, Howard University
Karamatu Djima, Amherst College
Mark Dunster, San Diego State University
Eric Erdmann, University of Minnesota – Duluth
Debra Etheridge, The University of North Carolina at Chapel Hill
Mahmoud Fathelden, South Texas College
Jerome Giles, San Diego State University
Mark Grinshpon, Georgia State University
Katie Gurski, Howard University
John Hall, Yale University
David Hemmer, University at Buffalo – SUNY, N. Campus
Kathlene Hockaday, University of Buffalo
Frederick Hoffman, Florida Atlantic University
Keith Howard, Mercer University
Iztok Hozo, Indiana University Northwest
Shu-Jen Huang, University of Florida
Matthew Isom, Arizona State University – Polytechnic
James Kimball, University of Louisiana at Lafayette
Thomas Kinzel, Boise State University
Denise LeGrand, University of Arkansas at Little Rock
Anastasios Liakos, United States Naval Academy
Chris Lim, Rutgers University – Camden
Jia Liu, University of West Florida
Joseph Londino, University of Memphis
Colton Magnant, Georgia Southern University
Mark Marino, University at Buffalo – SUNY, N. Campus
Kodie Paul McNamara, Georgetown University
Mariana Montiel, Georgia State University
Russell Murray, Saint Louis Community College
Ashley Nicoloff, Glendale Community College
Daniella Nokolova-Popova, Florida Atlantic University
Giray Okten, Florida State University – Tallahassee
George Pelekanos, Southern Illinois University
Aaron Peterson, Northwestern University
Alice Petillo, Marymount University
Mihaela Poplicher, University of Cincinnati
Cindy Pulley, Illinois State University
Russell Richins, Thiel College
Jim Rolf, University of Virginia
Lorenzo Sadun, University of Texas at Austin
Michael Santilli, Mesa Community College
Christopher Shaw, Columbia College
Brian Shay, Canyon Crest Academy
Mike Shirazi, Germanna Community College – Fredericksburg
Pavel Sikorskii, Michigan State University
Mary Smeal, University of Alabama
Edwin Smith, Jacksonville State University
Sandra Spiroff, University of Mississippi
Stan Stascinsky, Tarrant County College
Jinyuan Tao, Loyola University of Maryland
Ilham Tayahi, University of Memphis
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Michael Tom, Louisiana State University – Baton Rouge
Josaphat Uvah, University of West Florida
Michael Westmoreland, Denison University
Scott Wilde, Baylor University
Larissa Williamson, University of Florida
Jie Wu, Thiel College
Michael Yatauro, Penn State Brandywine
Gang Yu, Kent State University
Loris Zucca, Lone Star College – Kingwood
n Previous Edition Reviewers
Jay Abramson, Arizona State University
B. D. Aggarwala, University of Calgary
John Alberghini, Manchester Community College
Michael Albert, Carnegie-Mellon University
Daniel Anderson, University of Iowa
Maria Andersen, Muskegon Community College
Eric Aurand, Eastfield College
Amy Austin, Texas A&M University
Donna J. Bailey, Northeast Missouri State University
Wayne Barber, Chemeketa Community College
Joy Becker, University of Wisconsin – Stout
Marilyn Belkin, Villanova University
Neil Berger, University of Illinois, Chicago
David Berman, University of New Orleans
Anthony J. Bevelacqua, University of North Dakota
Richard Biggs, University of Western Ontario
Robert Blumenthal, Oglethorpe University
Martina Bode, Northwestern University
Przemyslaw Bogacki, Old Dominion University
Barbara Bohannon, Hofstra University
Jay Bourland, Colorado State University
Adam Bowers, University of California San Diego
Philip L. Bowers, Florida State University
Amy Elizabeth Bowman, University of Alabama in Huntsville
Stephen W. Brady, Wichita State University
Michael Breen, Tennessee Technological University
Monica Brown, University of Missouri – St. Louis
Robert N. Bryan, University of Western Ontario
David Buchthal, University of Akron
Roxanne Byrne, University of Colorado at Denver and
Health Sciences Center
Jenna Carpenter, Louisiana Tech University
Jorge Cassio, Miami-Dade Community College
Jack Ceder, University of California, Santa Barbara
Scott Chapman, Trinity University
Zhen-Qing Chen, University of Washington – Seattle
James Choike, Oklahoma State University
Neena Chopra, The Pennsylvania State University
Teri Christiansen, University of Missouri – Columbia
Barbara Cortzen, DePaul University
Carl Cowen, Purdue University
Philip S. Crooke, Vanderbilt University
Charles N. Curtis, Missouri Southern State College
Daniel Cyphert, Armstrong State College
Robert Dahlin
Bobby Dale Daniel, Lamar University
Jennifer Daniel, Lamar University
M. Hilary Davies, University of Alaska Anchorage
Gregory J. Davis, University of Wisconsin – Green Bay
Elias Deeba, University of Houston – Downtown
Daniel DiMaria, Suffolk Community College
Seymour Ditor, University of Western Ontario
Edward Dobson, Mississippi State University
Andras Domokos, California State University, Sacramento
Greg Dresden, Washington and Lee University
Daniel Drucker, Wayne State University
Kenn Dunn, Dalhousie University
Dennis Dunninger, Michigan State University
Bruce Edwards, University of Florida
David Ellis, San Francisco State University
John Ellison, Grove City College
Martin Erickson, Truman State University
Garret Etgen, University of Houston
Theodore G. Faticoni, Fordham University
Laurene V. Fausett, Georgia Southern University
Norman Feldman, Sonoma State University
Le Baron O. Ferguson, University of California – Riverside
Newman Fisher, San Francisco State University
Timothy Flaherty, Carnegie Mellon University
José D. Flores, The University of South Dakota
William Francis, Michigan Technological University
James T. Franklin, Valencia Community College, East
Stanley Friedlander, Bronx Community College
Patrick Gallagher, Columbia University – New York
Paul Garrett, University of Minnesota – Minneapolis
Frederick Gass, Miami University of Ohio
Lee Gibson, University of Louisville
Bruce Gilligan, University of Regina
Matthias K. Gobbert, University of Maryland, Baltimore County
Gerald Goff, Oklahoma State University
Isaac Goldbring, University of Illinois at Chicago
Jane Golden, Hillsborough Community College
Stuart Goldenberg, California Polytechnic State University
John A. Graham, Buckingham Browne & Nichols School
Richard Grassl, University of New Mexico
Michael Gregory, University of North Dakota
Charles Groetsch, University of Cincinnati
Semion Gutman, University of Oklahoma
Paul Triantafilos Hadavas, Armstrong Atlantic State University
Salim M. Haïdar, Grand Valley State University
D. W. Hall, Michigan State University
Robert L. Hall, University of Wisconsin – Milwaukee
Howard B. Hamilton, California State University, Sacramento
Darel Hardy, Colorado State University
Shari Harris, John Wood Community College
Gary W. Harrison, College of Charleston
Melvin Hausner, New York University/Courant Institute
Curtis Herink, Mercer University
Russell Herman, University of North Carolina at Wilmington
Allen Hesse, Rochester Community College
Diane Hoffoss, University of San Diego
Randall R. Holmes, Auburn University
Lorraine Hughes, Mississippi State University
James F. Hurley, University of Connecticut
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Amer Iqbal, University of Washington – Seattle
Matthew A. Isom, Arizona State University
Jay Jahangiri, Kent State University
Gerald Janusz, University of Illinois at Urbana-Champaign
John H. Jenkins, Embry-Riddle Aeronautical University, Prescott Campus
Lea Jenkins, Clemson University
John Jernigan, Community College of Philadelphia
Clement Jeske, University of Wisconsin, Platteville
Carl Jockusch, University of Illinois at Urbana-Champaign
Jan E. H. Johansson, University of Vermont
Jerry Johnson, Oklahoma State University
Zsuzsanna M. Kadas, St. Michael’s College
Brian Karasek, South Mountain Community College
Nets Katz, Indiana University Bloomington
Matt Kaufman
Matthias Kawski, Arizona State University
Frederick W. Keene, Pasadena City College
Robert L. Kelley, University of Miami
Akhtar Khan, Rochester Institute of Technology
Marianne Korten, Kansas State University
Virgil Kowalik, Texas A&I University
Jason Kozinski, University of Florida
Kevin Kreider, University of Akron
Leonard Krop, DePaul University
Carole Krueger, The University of Texas at Arlington
Mark Krusemeyer, Carleton College
Ken Kubota, University of Kentucky
John C. Lawlor, University of Vermont
Christopher C. Leary, State University of New York at Geneseo
David Leeming, University of Victoria
Sam Lesseig, Northeast Missouri State University
Phil Locke, University of Maine
Joyce Longman, Villanova University
Joan McCarter, Arizona State University
Phil McCartney, Northern Kentucky University
Igor Malyshev, San Jose State University
Larry Mansfield, Queens College
Mary Martin, Colgate University
Nathaniel F. G. Martin, University of Virginia
Gerald Y. Matsumoto, American River College
James McKinney, California State Polytechnic University, Pomona
Tom Metzger, University of Pittsburgh
Richard Millspaugh, University of North Dakota
John Mitchell, Clark College
Lon H. Mitchell, Virginia Commonwealth University
Michael Montaño, Riverside Community College
Teri Jo Murphy, University of Oklahoma
Martin Nakashima, California State Polytechnic University,
Ho Kuen Ng, San Jose State University
Richard Nowakowski, Dalhousie University
Hussain S. Nur, California State University, Fresno
Norma Ortiz-Robinson, Virginia Commonwealth University
Wayne N. Palmer, Utica College
Vincent Panico, University of the Pacific
F. J. Papp, University of Michigan – Dearborn
Donald Paul, Tulsa Community College
Mike Penna, Indiana University – Purdue University Indianapolis
Chad Pierson, University of Minnesota, Duluth
Mark Pinsky, Northwestern University
Lanita Presson, University of Alabama in Huntsville
Lothar Redlin, The Pennsylvania State University
Karin Reinhold, State University of New York at Albany
Thomas Riedel, University of Louisville
Joel W. Robbin, University of Wisconsin – Madison
Lila Roberts, Georgia College and State University
E. Arthur Robinson, Jr., The George Washington University
Richard Rockwell, Pacific Union College
Rob Root, Lafayette College
Richard Ruedemann, Arizona State University
David Ryeburn, Simon Fraser University
Richard St. Andre, Central Michigan University
Ricardo Salinas, San Antonio College
Robert Schmidt, South Dakota State University
Eric Schreiner, Western Michigan University
Christopher Schroeder, Morehead State University
Mihr J. Shah, Kent State University – Trumbull
Angela Sharp, University of Minnesota, Duluth
Patricia Shaw, Mississippi State University
Qin Sheng, Baylor University
Theodore Shifrin, University of Georgia
Wayne Skrapek, University of Saskatchewan
Larry Small, Los Angeles Pierce College
Teresa Morgan Smith, Blinn College
William Smith, University of North Carolina
Donald W. Solomon, University of Wisconsin – Milwaukee
Carl Spitznagel, John Carroll University
Edward Spitznagel, Washington University
Joseph Stampfli, Indiana University
Kristin Stoley, Blinn College
Mohammad Tabanjeh, Virginia State University
Capt. Koichi Takagi, United States Naval Academy
M. B. Tavakoli, Chaffey College
Lorna TenEyck, Chemeketa Community College
Magdalena Toda, Texas Tech University
Ruth Trygstad, Salt Lake Community College
Paul Xavier Uhlig, St. Mary’s University, San Antonio
Stan Ver Nooy, University of Oregon
Andrei Verona, California State University – Los Angeles
Klaus Volpert, Villanova University
Rebecca Wahl, Butler University
Russell C. Walker, Carnegie-Mellon University
William L. Walton, McCallie School
Peiyong Wang, Wayne State University
Jack Weiner, University of Guelph
Alan Weinstein, University of California, Berkeley
Roger Werbylo, Pima Community College
Theodore W. Wilcox, Rochester Institute of Technology
Steven Willard, University of Alberta
David Williams, Clayton State University
Robert Wilson, University of Wisconsin – Madison
Jerome Wolbert, University of Michigan – Ann Arbor
Dennis H. Wortman, University of Massachusetts, Boston
Mary Wright, Southern Illinois University – Carbondale
Paul M. Wright, Austin Community College
Xian Wu, University of South Carolina
Zhuan Ye, Northern Illinois University
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
We thank all those who have contributed to this edition—and there are many—as
well as those whose input in previous editions lives on in this new edition. We thank
Marigold Ardren, David Behrman, George Bergman, R. B. Burckel, Bruce Colletti,
John Dersch, Gove Effinger, Bill Emerson, Alfonso Gracia-Saz, Jeffery Hayen, Dan
Kalman, Quyan Khan, John Khoury, Allan MacIsaac, Tami Martin, Monica Nitsche,
Aaron Peterson, Lamia Raffo, Norton Starr, Jim Trefzger, Aaron Watson, and Weihua
Zeng for their suggestions; Joseph Bennish, Craig Chamberlin, Kent Merryfield, and
Gina Sanders for insightful conversations on calculus; Al Shenk and Dennis Zill for permission to use exercises from their calculus texts; COMAP for permission to use project
material; David Bleecker, Victor Kaftal, Anthony Lam, Jamie Lawson, Ira Rosenholtz,
Paul Sally, Lowell Smylie, Larry Wallen, and Jonathan Watson for ideas for exercises;
Dan Drucker for the roller derby project; Thomas Banchoff, Tom Farmer, Fred Gass,
John Ramsay, Larry Riddle, Philip Straffin, and Klaus Volpert for ideas for projects;
Josh Babbin, Scott Barnett, and Gina Sanders for solving the new exercises and suggesting ways to improve them; Jeff Cole for overseeing all the solutions to the exercises
and ensuring their correctness; Mary Johnson and Marv Riedesel for accuracy in proofreading, and Doug Shaw for accuracy checking. In addition, we thank Dan Anderson,
Ed Barbeau, Fred Brauer, Andy Bulman-Fleming, Bob Burton, David Cusick, Tom
DiCiccio, Garret Etgen, Chris Fisher, Barbara Frank, Leon Gerber, Stuart Goldenberg,
Arnold Good, Gene Hecht, Harvey Keynes, E. L. Koh, Zdislav Kovarik, Kevin Kreider, Emile LeBlanc, David Leep, Gerald Leibowitz, Larry Peterson, Mary Pugh, Carl
Riehm, John Ringland, Peter Rosenthal, Dusty Sabo, Dan Silver, Simon Smith, Alan
Weinstein, and Gail Wolkowicz.
We are grateful to Phyllis Panman for assisting us in preparing the manuscript, solving the exercises and suggesting new ones, and for critically proofreading the entire
We are deeply indebted to our friend and colleague Lothar Redlin who began working with us on this revision shortly before his untimely death in 2018. Lothar’s deep
insights into mathematics and its pedagogy, and his lightning fast problem-solving
skills, were invaluable assets.
We especially thank Kathi Townes of TECHarts, our production service and copyeditor (for this as well as the past several editions). Her extraordinary ability to recall
any detail of the manuscript as needed, her facility in simultaneously handling different editing tasks, and her comprehensive familiarity with the book were key factors in
its accuracy and timely production. We also thank Lori Heckelman for the elegant and
precise rendering of the new illustrations.
At Cengage Learning we thank Timothy Bailey, Teni Baroian, Diane Beasley, Carly
Belcher, Vernon Boes, Laura Gallus, Stacy Green, Nikkita Kendrick, Mark Linton, Samantha Lugtu, Ashley Maynard, Irene Morris, Lynh Pham, Jennifer Risden, Tim Rogers, Mark
Santee, Angela Sheehan, and Tom Ziolkowski. They have all done an outstanding job.
This textbook has benefited greatly over the past three decades from the advice and
guidance of some of the best mathematics editors: Ron Munro, Harry Campbell, Craig
Barth, Jeremy Hayhurst, Gary Ostedt, Bob Pirtle, Richard Stratton, Liz Covello, Neha
Taleja, and now Gary Whalen. They have all contributed significantly to the success
of this book. Prominently, Gary Whalen’s broad knowledge of current issues in the
teaching of mathematics and his continual research into creating better ways of using
technology as a teaching and learning tool were invaluable resources in the creation of
this edition.
JA M E S S T E WA RT
DA N I E L C L E G G
S A L E E M WAT S O N
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A Tribute to James Stewart
james stewart had a singular gift for teaching mathematics. The large lecture
halls where he taught his calculus classes were always packed to capacity with students,
whom he held engaged with interest and anticipation as he led them to discover a new
concept or the solution to a stimulating problem. Stewart presented calculus the way
he viewed it — as a rich subject with intuitive concepts, wonderful problems, powerful applications, and a fascinating history. As a testament to his success in teaching
and lecturing, many of his students went on to become mathematicians, scientists, and
engineers — and more than a few are now university professors themselves. It was his
students who first suggested that he write a calculus textbook of his own. Over the years,
former students, by then working scientists and engineers, would call him to discuss
mathematical problems that they encountered in their work; some of these discussions
resulted in new exercises or projects in the book.
We each met James Stewart—or Jim as he liked us to call him—through his teaching
and lecturing, resulting in his inviting us to coauthor mathematics textbooks with him.
In the years we have known him, he was in turn our teacher, mentor, and friend.
Jim had several special talents whose combination perhaps uniquely qualified him
to write such a beautiful calculus textbook — a textbook with a narrative that speaks to
students and that combines the fundamentals of calculus with conceptual insights on
how to think about them. Jim always listened carefully to his students in order to find
out precisely where they may have had difficulty with a concept. Crucially, Jim really
enjoyed hard work — a necessary trait for completing the immense task of writing a
calculus book. As his coauthors, we enjoyed his contagious enthusiasm and optimism,
making the time we spent with him always fun and productive, never stressful.
Most would agree that writing a calculus textbook is a major enough feat for one
lifetime, but amazingly, Jim had many other interests and accomplishments: he played
violin professionally in the Hamilton and McMaster Philharmonic Orchestras for many
years, he had an enduring passion for architecture, he was a patron of the arts and cared
deeply about many social and humanitarian causes. He was also a world traveler, an
eclectic art collector, and even a gourmet cook.
James Stewart was an extraordinary person, mathematician, and teacher. It has been
our honor and privilege to be his coauthors and friends.
DA N I E L C L E G G
S A L E E M WAT S O N
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
About the Authors
For more than two decades, Daniel Clegg and Saleem Watson have worked with James
Stewart on writing mathematics textbooks. The close working relationship between
them was particularly productive because they shared a common viewpoint on teaching
mathematics and on writing mathematics. In a 2014 interview James Stewart remarked
on their collaborations: “We discovered that we could think in the same way . . . we
agreed on almost everything, which is kind of rare.”
Daniel Clegg and Saleem Watson met James Stewart in different ways, yet in each
case their initial encounter turned out to be the beginning of a long association. Stewart
spotted Daniel’s talent for teaching during a chance meeting at a mathematics conference and asked him to review the manuscript for an upcoming edition of Calculus and
to author the multivariable solutions manual. Since that time Daniel has played an everincreasing role in the making of several editions of the Stewart calculus books. He and
Stewart have also coauthored an applied calculus textbook. Stewart first met Saleem
when Saleem was a student in his graduate mathematics class. Later Stewart spent a
sabbatical leave doing research with Saleem at Penn State University, where Saleem
was an instructor at the time. Stewart asked Saleem and Lothar Redlin (also a student of
Stewart’s) to join him in writing a series of precalculus textbooks; their many years of
collaboration resulted in several editions of these books.
james stewart was professor of mathematics at McMaster University and the
University of Toronto for many years. James did graduate studies at Stanford University and the University of Toronto, and subsequently did research at the University of
London. His research field was Harmonic Analysis and he also studied the connections
between mathematics and music.
daniel clegg is professor of mathematics at Palomar College in Southern California. He did undergraduate studies at California State University, Fullerton and
graduate studies at the University of California, Los Angeles (UCLA). Daniel is a
consummate teacher; he has been teaching mathematics ever since he was a graduate
student at UCLA.
saleem watson is professor emeritus of mathematics at California State University,
Long Beach. He did undergraduate studies at Andrews University in Michigan and
graduate studies at Dalhousie University and McMaster University. After completing
a research fellowship at the University of Warsaw, he taught for several years at Penn
State before joining the mathematics department at California State University, Long
Stewart and Clegg have published Brief Applied Calculus.
Stewart, Redlin, and Watson have published Precalculus: Mathematics for
Calculus, College Algebra, Trigonometry, Algebra and Trigonometry, and (with Phyllis
Panman) College Algebra: Concepts and Contexts.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Technology in the Ninth Edition
Graphing and computing devices are valuable tools for learning and exploring calculus,
and some have become well established in calculus instruction. Graphing calculators are
useful for drawing graphs and performing some numerical calculations, like approximating solutions to equations or numerically evaluating derivatives (Chapter 2) or definite integrals (Chapter 4). Mathematical software packages called computer algebra
systems (CAS, for short) are more powerful tools. Despite the name, algebra represents
only a small subset of the capabilities of a CAS. In particular, a CAS can do mathematics symbolically rather than just numerically. It can find exact solutions to equations and
exact formulas for derivatives and integrals.
We now have access to a wider variety of tools of varying capabilities than ever
before. These include Web-based resources (some of which are free of charge) and
apps for smartphones and tablets. Many of these resources include at least some CAS
functionality, so some exercises that may have typically required a CAS can now be
completed using these alternate tools.
In this edition, rather than refer to a specific type of device (a graphing calculator, for
instance) or software package (such as a CAS), we indicate the type of capability that is
needed to work an exercise.
Graphing Icon
The appearance of this icon beside an exercise indicates that you are expected to use a
machine or software to help you draw the graph. In many cases, a graphing calculator
will suffice. Websites such as Desmos.com provide similar capability. If the graph is in
3D (see Chapters 12 – 16), WolframAlpha.com is a good resource. There are also many
graphing software applications for computers, smartphones, and tablets. If an exercise
asks for a graph but no graphing icon is shown, then you are expected to draw the graph
by hand. In Chapter 1 we review graphs of basic functions and discuss how to use transformations to graph modified versions of these basic functions.
Technology Icon
This icon is used to indicate that software or a device with abilities beyond just graphing
is needed to complete the exercise. Many graphing calculators and software resources
can provide numerical approximations when needed. For working with mathematics
symbolically, websites like WolframAlpha.com or Symbolab.com are helpful, as are
more advanced graphing calculators such as the Texas Instrument TI-89 or TI-Nspire
CAS. If the full power of a CAS is needed, this will be stated in the exercise, and access
to software packages such as Mathematica, Maple, MATLAB, or SageMath may be
required. If an exercise does not include a technology icon, then you are expected to
evaluate limits, derivatives, and integrals, or solve equations by hand, arriving at exact
answers. No technology is needed for these exercises beyond perhaps a basic scientific
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
To the Student
Reading a calculus textbook is different from reading a story or a news article. Don’t
be discouraged if you have to read a passage more than once in order to understand it.
You should have pencil and paper and calculator at hand to sketch a diagram or make a
Some students start by trying their homework problems and read the text only if they
get stuck on an exercise. We suggest that a far better plan is to read and understand a
section of the text before attempting the exercises. In particular, you should look at the
definitions to see the exact meanings of the terms. And before you read each example,
we suggest that you cover up the solution and try solving the problem yourself.
Part of the aim of this course is to train you to think logically. Learn to write the
solutions of the exercises in a connected, step-by-step fashion with explanatory
sentences — not just a string of disconnected equations or formulas.
The answers to the odd-numbered exercises appear at the back of the book, in Appendix G. Some exercises ask for a verbal explanation or interpretation or description. In
such cases there is no single correct way of expressing the answer, so don’t worry that
you haven’t found the definitive answer. In addition, there are often several different
forms in which to express a numerical or algebraic answer, so if your answer differs
from the given one, don’t immediately assume you’re wrong. For example, if the answer
given in the back of the book is s2 2 1 and you obtain 1y(1 1 s2 ), then you’re correct and rationalizing the denominator will show that the answers are equivalent.
The icon ; indicates an exercise that definitely requires the use of either a graphing calculator or a computer with graphing software to help you sketch the graph. But
that doesn’t mean that graphing devices can’t be used to check your work on the other
exercises as well. The symbol
indicates that technological assistance beyond just
graphing is needed to complete the exercise. (See Technology in the Ninth Edition for
more details.)
You will also encounter the symbol , which warns you against committing an error.
This symbol is placed in the margin in situations where many students tend to make the
same mistake.
Homework Hints are available for many exercises. These hints can be found on
Stewart­Calculus.com as well as in WebAssign. The homework hints ask you questions that allow you to make progress toward a solution without actually giving you the
answer. If a particular hint doesn’t enable you to solve the problem, you can click to
reveal the next hint.
We recommend that you keep this book for reference purposes after you finish the
course. Because you will likely forget some of the specific details of calculus, the book
will serve as a useful reminder when you need to use calculus in subsequent courses.
And, because this book contains more material than can be covered in any one course, it
can also serve as a valuable resource for a working scientist or engineer.
Calculus is an exciting subject, justly considered to be one of the greatest achievements of the human intellect. We hope you will discover that it is not only useful but
also intrinsically beautiful.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Diagnostic Tests
Success in calculus depends to a large extent on knowledge of the mathematics that
precedes calculus: algebra, analytic geometry, functions, and trigonometry. The following tests are intended to diagnose weaknesses that you might have in these areas. After
taking each test you can check your answers against the given answers and, if necessary,
refresh your skills by referring to the review materials that are provided.
A Diagnostic Test: Algebra
Evaluate each expression without using a calculator.
(a) s23d4
(b) 234
(c) 324
(f) 16 23y4
2.Simplify each expression. Write your answer without negative exponents.
(a) s200 2 s32
(b) s3a 3b 3 ds4ab 2 d 2
3x 3y2 y 3
x 2 y21y2
Expand and simplify.
(a) 3sx 1 6d 1 4s2x 2 5d
(b) sx 1 3ds4x 2 5d
(c) (sa 1 sb )(sa 2 sb )
(d) s2x 1 3d2
(e) sx 1 2d3
Factor each expression.
(a) 4x 2 2 25
(c) x 3 2 3x 2 2 4x 1 12
(e) 3x
2 9x
1 6x
(b) 2x 2 1 5x 2 12
(d) x 4 1 27x
(f) x 3 y 2 4xy
5. Simplify the rational expression.
x 2 1 3x 1 2
x2 2 x 2 2
x2 2 4
2x 2 2 x 2 1
x2 2 9
2x 1 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Rationalize the expression and simplify.
s5 2 2
Rewrite by completing the square.
(a) x 2 1 x 1 1
s4 1 h 2 2
(b) 2x 2 2 12x 1 11
Solve the equation. (Find only the real solutions.)
2x 2 1
(a) x 1 5 − 14 2 12 x
(c) x 2 2 x 2 12 − 0
(d) 2x 2 1 4x 1 1 − 0
(e) x 4 2 3x 2 1 2 − 0
(f) 3 x 2 4 − 10
(g) 2xs4 2 xd21y2 2 3 s4 2 x − 0
9.Solve each inequality. Write your answer using interval notation.
(a) 24 , 5 2 3x < 17
(b) x 2 , 2x 1 8
(c) xsx 2 1dsx 1 2d . 0
(d) x 2 4 , 3
2x 2 3
10.State whether each equation is true or false.
(a) s p 1 qd2 − p 2 1 q 2
(b) sab − sa sb
(c) sa 2 1 b 2 − a 1 b
1 1 TC
ayx 2 byx
− 2
1.(a) 81
(d) 25
2.(a) 6s2
(b) 281
(b) 48a 5b7
3.(a) 11x 2 2
(b) 4x 2 1 7x 2 15
(c) a 2 b
(d) 4x 2 1 12x 1 9
(e) x 1 6x 1 12x 1 8
4.(a) s2x 2 5ds2x 1 5d
(c) sx 2 3dsx 2 2dsx 1 2d
(e) 3x21y2sx 2 1dsx 2 2d
(b) s2x 2 3dsx 1 4d
(d) xsx 1 3dsx 2 2 3x 1 9d
(f) xysx 2 2dsx 1 2d
(d) 2sx 1 yd
6.(a) 5s2 1 2s10
7.(a) ( x 1 2 ) 1 34
2sx 2 3d2 2 7
8.(a) 6
(d) 21 6
2 s2
(b) 1
(c) 23, 4
(e) 61, 6s2
(f) 23 , 22
9.(a) f24, 3d
(c) s22, 0d ø s1, `d
(e) s21, 4g
10.(a) False
(d) False
s4 1 h 1 2
(b) s22, 4d
(d) s1, 7d
(b) True
(e) False
(c) False
(f) True
If you had difficulty with these problems, you may wish to consult the
Review of Algebra on the website StewartCalculus.com.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B Diagnostic Test: Analytic Geometry
Find an equation for the line that passes through the point s2, 25d and
(a) has slope 23
(b) is parallel to the x-axis
(c) is parallel to the y-axis
(d) is parallel to the line 2x 2 4y − 3
2.Find an equation for the circle that has center s21, 4d and passes through the point s3, 22d.
3.Find the center and radius of the circle with equation x 2 1 y 2 2 6x 1 10y 1 9 − 0.
4.Let As27, 4d and Bs5, 212d be points in the plane.
(a)Find the slope of the line that contains A and B.
(b)Find an equation of the line that passes through A and B. What are the intercepts?
(c) Find the midpoint of the segment AB.
(d) Find the length of the segment AB.
(e) Find an equation of the perpendicular bisector of AB.
(f ) Find an equation of the circle for which AB is a diameter.
5.Sketch the region in the xy-plane defined by the equation or inequalities.
| |
(a) 21 < y < 3
(c) y , 1 2
| |
(b) x , 4 and y , 2
(d) y > x 2 1
(f) 9x 2 1 16y 2 − 144
(e) x 1 y , 4
1.(a) y − 23x 1 1
(c) x − 2
(b) y − 25
(d) y − 12 x 2 6
2. sx 1 1d2 1 s y 2 4d2 − 52
4x 1 3y 1 16 − 0; x-intercept 24, y-intercept 2 16
s21, 24d
3x 2 4y − 13
sx 1 1d2 1 s y 1 4d2 − 100
y=1- 2 x
3.Center s3, 25d, radius 5
(f )
4 x
If you had difficulty with these problems, you may wish to consult
the review of analytic geometry in Appendixes B and C.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
C Diagnostic Test: Functions
The graph of a function f is given at the left.
(a) State the value of f s21d.
(b) Estimate the value of f s2d.
(c) For what values of x is f sxd − 2?
(d) Estimate the values of x such that f sxd − 0.
(e) State the domain and range of f .
2. If f sxd − x 3, evaluate the difference quotient
3. Find the domain of the function.
(a) f sxd −
2x 1 1
x2 1 x 2 2
(b) tsxd −
f s2 1 hd 2 f s2d
and simplify your answer.
x 11
(c) hsxd − s4 2 x 1 sx 2 2 1
4. How are graphs of the functions obtained from the graph of f ?
(a) y − 2f sxd
(b) y − 2 f sxd 2 1
(c) y − f sx 2 3d 1 2
5. Without using a calculator, make a rough sketch of the graph.
(a) y − x 3
(b) y − sx 1 1d3
(c) y − sx 2 2d3 1 3
(d) y − 4 2 x
(e) y − sx
(f) y − 2 sx
(g) y − 22
(h) y − 1 1 x
1 2 x 2 if x < 0
6. Let f sxd −
2x 1 1 if x . 0
(a) Evaluate f s22d and f s1d.
(b) Sketch the graph of f .
7.If f sxd − x 2 1 2x 2 1 and tsxd − 2x 2 3, find each of the following functions.
(a) f 8 t
(b) t 8 f
(c) t 8 t 8 t
1.(a) 22
(c) 23, 1
(e) f23, 3g, f22, 3g
(b) 2.8
(d) 22.5, 0.3
5. (a)
4.(a) Reflect about the x-axis
(b)Stretch vertically by a factor of 2, then shift 1 unit
(c) Shift 3 units to the right and 2 units upward
(2, 3)
2. 12 1 6h 1 h 2
3.(a) s2`, 22d ø s22, 1d ø s1, `d
(b) s2`, `d
(c) s2`, 21g ø f1, 4g
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
6.(a) 23, 3(b)
7. (a) s f 8 tdsxd − 4x 2 2 8x 1 2
(b) s t 8 f dsxd − 2x 2 1 4x 2 5
(c) s t 8 t 8 tdsxd − 8x 2 21
If you had difficulty with these problems, you should look at sections 1.1–1.3 of this book.
D Diagnostic Test: Trigonometry
Convert from degrees to radians.
(a) 3008
(b) 2188
2. Convert from radians to degrees.
(a) 5y6
(b) 2
3.Find the length of an arc of a circle with radius 12 cm if the arc subtends a central angle
of 308.
4. Find the exact values.
(a) tansy3d
(b) sins7y6d
(c) secs5y3d
5.Express the lengths a and b in the figure in terms of .
6.If sin x − 13 and sec y − 54, where x and y lie between 0 and y2, evaluate sinsx 1 yd.
7. Prove the identities.
(a) tan sin 1 cos − sec (b)
2 tan x
− sin 2x
1 1 tan 2x
8.Find all values of x such that sin 2x − sin x and 0 < x < 2.
9.Sketch the graph of the function y − 1 1 sin 2x without using a calculator.
1.(a) 5y3
(b) 2y10
2.(a) 1508
(b) 3608y < 114.68
8. 0, y3, , 5y3, 2
3. 2 cm
4.(a) s3
(4 1 6 s2 )
(c) 2
5. a − 24 sin , b − 24 cos _π
If you had difficulty with these problems, you should look at Appendix D of this book.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
By the time you finish this course, you will be able to determine where a pilot should start descent for a smooth landing,
find the length of the curve used to design the Gateway Arch in St. Louis, compute the force on a baseball bat when it
strikes the ball, predict the population sizes for competing predator-prey species, show that bees form the cells of a
beehive in a way that uses the least amount of wax, and estimate the amount of fuel needed to propel a rocket into orbit.
Top row: Who is Danny / Shutterstock.com; iStock.com/gnagel; Richard Paul Kane / Shutterstock
Bottom row: Bruce Ellis / Shutterstock.com; Kostiantyn Kravchenko / Shutterstock.com; Ben Cooper / Science Faction / Getty Images
A Preview of Calculus
CALCULUS IS FUNDAMENTALLY DIFFERENT from the mathematics that you have studied previously:
calculus is less static and more dynamic. It is concerned with change and motion; it deals with quantities that approach other quantities. For that reason it may be useful to have an overview of calculus
before beginning your study of the subject. Here we give a preview of some of the main ideas of
calculus and show how their foundations are built upon the concept of a limit.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A PREVIEW OF CALCULUSWhat Is Calculus?
The world around us is continually changing — populations increase, a cup of coffee
cools, a stone falls, chemicals react with one another, currency values fluctuate, and
so on. We would like to be able to analyze quantities or processes that are undergoing
continuous change. For example, if a stone falls 10 feet each second we could easily
tell how fast it is falling at any time, but this is not what happens — the stone falls faster
and faster, its speed changing at each instant. In studying calculus, we will learn how
to model (or describe) such instantaneously changing processes and how to find the
cumulative effect of these changes.
Calculus builds on what you have learned in algebra and analytic geometry but
advances these ideas spectacularly. Its uses extend to nearly every field of human activity. You will encounter numerous applications of calculus throughout this book.
At its core, calculus revolves around two key problems involving the graphs of functions — the area problem and the tangent problem — and an unexpected relationship
between them. Solving these problems is useful because the area under the graph of a
function and the tangent to the graph of a function have many important interpretations
in a variety of contexts.
The Area Problem
The origins of calculus go back at least 2500 years to the ancient Greeks, who found
areas using the “method of exhaustion.” They knew how to find the area A of any polygon by dividing it into triangles, as in Figure 1, and adding the areas of these triangles.
It is a much more difficult problem to find the area of a curved figure. The Greek
method of exhaustion was to inscribe polygons in the figure and circumscribe polygons
about the figure, and then let the number of sides of the polygons increase. Figure 2
illustrates this process for the special case of a circle with inscribed regular polygons.
FIGURE 1
FIGURE 2
Let An be the area of the inscribed regular polygon of n sides. As n increases, it appears
that An gets closer and closer to the area of the circle. We say that the area A of the circle
is the limit of the areas of the inscribed polygons, and we write
A − lim An
n l`
FIGURE 3
The area A of the region under
the graph of f
The Greeks themselves did not use limits explicitly. However, by indirect reasoning,
Eudoxus (fifth century bc) used exhaustion to prove the familiar formula for the area of
a circle: A − r 2.
We will use a similar idea in Chapter 4 to find areas of regions of the type shown
in Figure 3. We approximate such an area by areas of rectangles as shown in Figure 4.
If we approximate the area A of the region under the graph of f by using n rectangles
R1 , R2 , . . . , Rn , then the approximate area is
An − R1 1 R2 1 c 1 Rn
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
FIGURE 4
Approximating the area A using rectangles
Now imagine that we increase the number of rectangles (as the width of each one
decreases) and calculate A as the limit of these sums of areas of rectangles:
A − lim An
n l`
In Chapter 4 we will learn how to calculate such limits.
The area problem is the central problem in the branch of calculus called integral
calculus; it is important because the area under the graph of a function has different
interpretations depending on what the function represents. In fact, the techniques that
we develop for finding areas will also enable us to compute the volume of a solid, the
length of a curve, the force of water against a dam, the mass and center of mass of a rod,
the work done in pumping water out of a tank, and the amount of fuel needed to send a
rocket into orbit.
The Tangent Problem
FIGURE 5
The tangent line at P
Consider the problem of trying to find an equation of the tangent line L to a curve with
equation y − f sxd at a given point P. (We will give a precise definition of a tangent line
in Chapter 1; for now you can think of it as the line that touches the curve at P and follows the direction of the curve at P, as in Figure 5.) Because the point P lies on the
tangent line, we can find the equation of L if we know its slope m. The problem is that
we need two points to compute the slope and we know only one point, P, on L. To get
around the problem we first find an approximation to m by taking a nearby point Q on
the curve and computing the slope m PQ of the secant line PQ.
Now imagine that Q moves along the curve toward P as in Figure 6. You can see that
the secant line PQ rotates and approaches the tangent line L as its limiting position. This
FIGURE 6
The secant lines approach the tangent line as Q approaches P.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
means that the slope m PQ of the secant line becomes closer and closer to the slope m of
the tangent line. We write
Q { x, ƒ}
P { a, f(a)}
m − lim mPQ
and say that m is the limit of m PQ as Q approaches P along the curve.
Notice from Figure 7 that if P is the point sa, f sadd and Q is the point sx, f sxdd, then
FIGURE 7
The secant line PQ
mPQ −
f sxd 2 f sad
Because x approaches a as Q approaches P, an equivalent expression for the slope of
the tangent line is
m − lim
f sxd 2 f sad
In Chapter 1 we will learn rules for calculating such limits.
The tangent problem has given rise to the branch of calculus called differential
calculus; it is important because the slope of a tangent to the graph of a function can
have different interpretations depending on the context. For instance, solving the tangent problem allows us to find the instantaneous speed of a falling stone, the rate of
change of a chemical reaction, or the direction of the forces on a hanging chain.
A Relationship between the Area and Tangent Problems
The area and tangent problems seem to be very different problems but, surprisingly,
the problems are closely related — in fact, they are so closely related that solving one
of them leads to a solution of the other. The relationship between these two problems
is introduced in Chapter 4; it is the central discovery in calculus and is appropriately
named the Fundamental Theorem of Calculus. Perhaps most importantly, the Fundamental Theorem vastly simplifies the solution of the area problem, making it possible
to find areas without having to approximate by rectangles and evaluate the associated
Isaac Newton (1642 –1727) and Gottfried Leibniz (1646 –1716) are credited with
the invention of calculus because they were the first to recognize the importance of the
Fundamental Theorem of Calculus and to utilize it as a tool for solving real-world problems. In studying calculus you will discover these powerful results for yourself.
We have seen that the concept of a limit arises in finding the area of a region and in finding the slope of a tangent line to a curve. It is this basic idea of a limit that sets calculus
apart from other areas of mathematics. In fact, we could define calculus as the part of
mathematics that deals with limits. We have mentioned that areas under curves and
slopes of tangent lines to curves have many different interpretations in a variety of contexts. Finally, we have discussed that the area and tangent problems are closely related.
After Isaac Newton invented his version of calculus, he used it to explain the motion
of the planets around the sun, giving a definitive answer to a centuries-long quest for a
description of our solar system. Today calculus is applied in a great variety of contexts,
such as determining the orbits of satellites and spacecraft, predicting population sizes,
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
forecasting weather, measuring cardiac output, and gauging the efficiency of an economic market.
In order to convey a sense of the power and versatility of calculus, we conclude with
a list of some of the questions that you will be able to answer using calculus.
1. How can we design a roller coaster for a safe and smooth ride?
(See the Applied Project following Section 2.3.)
2. How far away from an airport should a pilot start descent?
(See the Applied Project following Section 2.5.)
3. How can we explain the fact that the angle of elevation from an observer up to the
highest point in a rainbow is always 42°?
(See the Applied Project following Section 3.1.)
4. How can we estimate the amount of work that was required to build the Pyramid
of Khufu in ancient Egypt?
(See Exercise 36 in Section 5.4.)
5. With what speed must a projectile be launched so that it escapes the earth’s gravitation pull?
(See Exercise 77 in Section 7.8.)
6. How can we explain the changes in the thickness of sea ice over time and why
cracks in the ice tend to “heal”?
(See Exercise 56 in Section 9.3.)
7. Does a ball thrown upward take longer to reach its maximum height or to fall back
down to its original height?
(See the Applied Project following Section 9.5.)
8. How can we fit curves together to design shapes to represent letters on a laser
(See the Applied Project following Section 10.2.)
9. How can we explain the fact that planets and satellites move in elliptical orbits?
(See the Applied Project following Section 13.4.)
10. How can we distribute water flow among turbines at a hydroelectric station so as
to maximize the total energy production?
(See the Applied Project following Section 14.8.)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
The electrical power produced by a wind turbine can be estimated by a mathematical function that incorporates several factors. We will
explore this function in Exercise 1.2.25 and determine the expected power output of a particular turbine for various wind speeds.
chaiviewfinder / Shutterstock.com
Functions and Limits
THE FUNDAMENTAL OBJECTS THAT WE deal with in calculus are functions. We stress that a function can be represented in different ways: by an equation, in a table, by a graph, or in words. We
look at the main types of functions that occur in calculus and describe the process of using these
functions as mathematical models of real-world phenomena.
In A Preview of Calculus (immediately preceding this chapter) we saw how the idea of a limit
underlies the various branches of calculus. It is therefore appropriate to begin our study of calculus
by investigating limits of functions and their properties.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
1.1 Four Ways to Represent a Function
■ Functions
Functions arise whenever one quantity depends on another. Consider the following four
A. T
he area A of a circle depends on the radius r of the circle. The rule that connects r
and A is given by the equation A − r 2. With each positive number r there is associated one value of A, and we say that A is a function of r.
Table 1 World Population
B. T
he human population of the world P depends on the time t. Table 1 gives estimates
of the world population P at time t, for certain years. For instance,
P < 2,560,000,000
when t − 1950
For each value of the time t there is a corresponding value of P, and we say that P is
a function of t.
C. T
he cost C of mailing an envelope depends on its weight w. Although there is no
simple formula that connects w and C, the post office has a rule for determining C
when w is known.
D. T
he vertical acceleration a of the ground as measured by a seismograph during an
earthquake is a function of the elapsed time t. Figure 1 shows a graph generated by
seismic activity during the Northridge earthquake that shook Los Angeles in 1994.
For a given value of t, the graph provides a corresponding value of a.
FIGURE 1
Vertical ground acceleration
during the Northridge earthquake
t (seconds)
Calif. Dept. of Mines and Geology
Each of these examples describes a rule whereby, given a number (r in Example A),
another number (A) is assigned. In each case we say that the second number is a function
of the first number. If f represents the rule that connects A to r in Example A, then we
express this in function notation as A − f srd.
A function f is a rule that assigns to each element x in a set D exactly one element, called f sxd, in a set E.
We usually consider functions for which the sets D and E are sets of real numbers.
The set D is called the domain of the function. The number f sxd is the value of f at x
and is read “ f of x.” The range of f is the set of all possible values of f sxd as x varies
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.1Four Ways to Represent a Function
FIGURE 2
Machine diagram for a function f
throughout the domain. A symbol that represents an arbitrary number in the domain of a
function f is called an independent variable. A symbol that represents a number in the
range of f is called a dependent variable. In Example A, for instance, r is the independent variable and A is the dependent variable.
It’s helpful to think of a function as a machine (see Figure 2). If x is in the domain of
the function f, then when x enters the machine, it’s accepted as an input and the machine
produces an output f sxd according to the rule of the function. So we can think of the
domain as the set of all possible inputs and the range as the set of all possible outputs.
The preprogrammed functions in a calculator are good examples of a function as a
machine. For example, if you input a number and press the squaring key, the calculator
displays the output, the square of the input.
Another way to picture a function is by an arrow diagram as in Figure 3. Each arrow
connects an element of D to an element of E. The arrow indicates that f sxd is associated
with x, f sad is associated with a, and so on.
Perhaps the most useful method for visualizing a function is its graph. If f is a function with domain D, then its graph is the set of ordered pairs
hsx, f sxdd x [ Dj
(Notice that these are input-output pairs.) In other words, the graph of f consists of all
points sx, yd in the coordinate plane such that y − f sxd and x is in the domain of f.
The graph of a function f gives us a useful picture of the behavior or “life history” of
a function. Since the y-coordinate of any point sx, yd on the graph is y − f sxd, we can
read the value of f sxd from the graph as being the height of the graph above the point x.
(See Figure 4.) The graph of f also allows us to picture the domain of f on the x-axis and
its range on the y-axis as in Figure 5.
FIGURE 3
Arrow diagram for f
y y
f (1)f (1)
y y
{ x, ƒ
{ x, ƒ
} }
f (2)f (2)
x x
x x
FIGURE 4
y y ƒ(x)
x x
FIGURE 5
EXAMPLE 1 The graph of a function f is shown in Figure 6.
(a) Find the values of f s1d and f s5d.
(b) What are the domain and range of f ?
(a) We see from Figure 6 that the point s1, 3d lies on the graph of f, so the value of f
at 1 is f s1d − 3. (In other words, the point on the graph that lies above x − 1 is 3 units
above the x-axis.)
When x − 5, the graph lies about 0.7 units below the x-axis, so we estimate that
f s5d < 20.7.
(b) We see that f sxd is defined when 0 < x < 7, so the domain of f is the closed
interval f0, 7g. Notice that f takes on all values from 22 to 4, so the range of f is
FIGURE 6
The notation for intervals is given in
Appendix A.
h y 22 < y < 4j − f22, 4g
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
In calculus, the most common method of defining a function is by an algebraic equation. For example, the equation y − 2x 2 1 defines y as a function of x. We can express
this in function notation as fsxd − 2x 2 1.
EXAMPLE 2 Sketch the graph and find the domain and range of each function.
(a) fsxd − 2x 2 1
(b) tsxd − x 2
FIGURE 7
(2, 4)
(_1, 1)
(a) The equation of the graph is y − 2x 2 1, and we recognize this as being the
equation of a line with slope 2 and y-intercept 21. (Recall the slope-intercept form of
the equation of a line: y − mx 1 b. See Appendix B.) This enables us to sketch a
portion of the graph of f in Figure 7. The expression 2x 2 1 is defined for all real
numbers, so the domain of f is the set of all real numbers, which we denote by R. The
graph shows that the range is also R.
(b) Since ts2d − 2 2 − 4 and ts21d − s21d2 − 1, we could plot the points s2, 4d and
s21, 1d, together with a few other points on the graph, and join them to produce the
graph (Figure 8). The equation of the graph is y − x 2, which represents a parabola (see
Appendix C). The domain of t is R. The range of t consists of all values of tsxd, that is,
all numbers of the form x 2. But x 2 > 0 for all numbers x and any positive number y is a
square. So the range of t is hy y > 0j − f0, `d. This can also be seen from Figure 8.
f sa 1 hd 2 f sad
EXAMPLE 3 If f sxd − 2x 2 2 5x 1 1 and h ± 0, evaluate
SOLUTION We first evaluate f sa 1 hd by replacing x by a 1 h in the expression for f sxd:
FIGURE 8
f sa 1 hd − 2sa 1 hd2 2 5sa 1 hd 1 1
− 2sa 2 1 2ah 1 h 2 d 2 5sa 1 hd 1 1
− 2a 2 1 4ah 1 2h 2 2 5a 2 5h 1 1
The expression
Then we substitute into the given expression and simplify:
f sa 1 hd 2 f sad
s2a 2 1 4ah 1 2h 2 2 5a 2 5h 1 1d 2 s2a 2 2 5a 1 1d
2a 2 1 4ah 1 2h 2 2 5a 2 5h 1 1 2 2a 2 1 5a 2 1
4ah 1 2h 2 5h
− 4a 1 2h 2 5
f sa 1 hd 2 f sad
in Example 3 is called a difference
quotient and occurs frequently in
calculus. As we will see in Chapter 2,
it represents the average rate of
change of f sxd between x − a and
x − a 1 h.
■ Representations of Functions
We consider four different ways to represent a function:
(by a description in words)
(by a table of values)
(by a graph)
(by an explicit formula)
If a single function can be represented in all four ways, it’s often useful to go from one
representation to another to gain additional insight into the function. (In Example 2, for
instance, we started with algebraic formulas and then obtained graphs.) But certain functions are described more naturally by one method than by another. With this in mind, let’s
reexamine the four situations that we considered at the beginning of this section.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.1Four Ways to Represent a Function
Table 2 World Population
since 1900)
A. T
he most useful representation of the area of a circle as a function of its radius is
probably the algebraic formula A − r 2 or, in function notation, Asrd − r 2. It is
also possible to compile a table of values or sketch a graph (half a parabola). Because
a circle has to have a positive radius, the domain is hr r . 0j − s0, `d and the range
is also s0, `d.
B. We are given a description of the function in words: Pstd is the human population of
the world at time t. Let’s measure t so that t − 0 corresponds to the year 1900. Table 2
provides a convenient representation of this function. If we plot the ordered pairs in
the table, we get the graph (called a scatter plot) in Figure 9. It too is a useful representation; the graph allows us to absorb all the data at once. What about a formula?
Of course, it’s impossible to devise an explicit formula that gives the exact human
population Pstd at any time t. But it is possible to find an expression for a function that
approximates Pstd. In fact, using methods explained in Section 1.2, we obtain an
approximation for the population P:
Pstd < f std − s1.43653 3 10 9 d ∙ s1.01395d t
Figure 10 shows that it is a reasonably good “fit.” The function f is called a mathematical model for population growth. In other words, it is a function with an explicit
formula that approximates the behavior of our given function. We will see, however,
that the ideas of calculus can be applied to a table of values; an explicit formula is not
Years since 1900
FIGURE 9
Years since 1900
FIGURE 10
A function defined by a table of
values is called a tabular function.
Table 3
w (grams)
Cswd (dollars)
0 , w , 25
25 , w , 50
50 , w , 75
75 , w , 100
100 , w , 125
The function P is typical of the functions that arise whenever we attempt to apply
calculus to the real world. We start with a verbal description of a function. Then we
may be able to construct a table of values of the function, perhaps from instrument
readings in a scientific experiment. Even though we don’t have complete knowledge
of the values of the function, we will see throughout the book that it is still possible
to perform the operations of calculus on such a function.
C. Again, the function is described in words: Let Cswd be the cost of mailing a large
envelope with weight w. The rule that the US Postal Service used as of 2019 is as
follows: The cost is 1 dollar for up to 25 g plus 15 cents for each additional gram (or
less) up to 350 g. A table of values is the most convenient representation for this function (see Table 3), though it is possible to sketch a graph (see Example 10).
D. The graph shown in Figure 1 is the most natural representation of the vertical acceleration function astd. It’s true that a table of values could be compiled, and it is
even possible to devise an approximate formula. But everything a geologist needs to
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
know— amplitudes and patterns — can be seen easily from the graph. (The same is
true for the patterns seen in electrocardiograms of heart patients and polygraphs for
In the next example we sketch the graph of a function that is defined verbally.
EXAMPLE 4 When you turn on a hot-water faucet that is connected to a hot-water
tank, the temperature T of the water depends on how long the water has been running.
Draw a rough graph of T as a function of the time t that has elapsed since the faucet
was turned on.
FIGURE 11
SOLUTION The initial temperature of the running water is close to room temperature
because the water has been sitting in the pipes. When the water from the hot-water tank
starts flowing from the faucet, T increases quickly. In the next phase, T is constant at
the tempera­ture of the heated water in the tank. When the tank is drained, T decreases
to the temperature of the water supply. This enables us to make the rough sketch of T
as a function of t shown in Figure 11.
In the following example we start with a verbal description of a function in a physical
situation and obtain an explicit algebraic formula. The ability to do this is a useful skill
in solving calculus problems that ask for the maximum or minimum values of quantities.
EXAMPLE 5 A rectangular storage container with an open top has a volume of 10 m3.
The length of its base is twice its width. Material for the base costs $10 per square
meter; material for the sides costs $6 per square meter. Express the cost of materials as
a function of the width of the base.
FIGURE 12
SOLUTION We draw a diagram as in Figure 12 and introduce notation by letting w and
2w be the width and length of the base, respectively, and h be the height.
The area of the base is s2wdw − 2w 2, so the cost, in dollars, of the material for the
base is 10s2w 2 d. Two of the sides have area wh and the other two have area 2wh, so the
cost of the material for the sides is 6f2swhd 1 2s2whdg. The total cost is therefore
C − 10s2w 2 d 1 6f2swhd 1 2s2whdg − 20 w 2 1 36 wh
To express C as a function of w alone, we need to eliminate h and we do so by using
the fact that the volume is 10 m3. Thus
w s2wdh − 10
which gives
PS In setting up applied functions as in
Example 5, it may be useful to review the
principles of problem solving at the end of
this chapter, particularly Step 1: Understand the Problem.
− 2
2w 2
Substituting this into the expression for C, we have
S D
C − 20w 2 1 36w
− 20w 2 1
Therefore the equation
Cswd − 20w 2 1
expresses C as a function of w.
w . 0
In the next example we find the domain of a function that is defined algebraically.
If a function is given by a formula and the domain is not stated explicitly, we use the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.1Four Ways to Represent a Function
following domain convention: the domain of the function is the set of all inputs for
which the formula makes sense and gives a real-number output.
EXAMPLE 6 Find the domain of each function.
(a) f sxd − sx 1 2 (b) tsxd −
x 2x
(a) Because the square root of a negative number is not defined (as a real number),
the domain of f consists of all values of x such that x 1 2 > 0. This is equivalent to
x > 22, so the domain is the interval f22, `d.
(b) Since
tsxd − 2
x 2x
xsx 2 1d
and division by 0 is not allowed, we see that tsxd is not defined when x − 0 or x − 1.
So the domain of t is
hx x ± 0, x ± 1j
which could also be written in interval notation as
s2`, 0d ø s0, 1d ø s1, `d
■ Which Rules Define Functions?
Not every equation defines a function. The equation y − x 2 defines y as a function of x
because the equation determines exactly one value of y for each value of x. However, the
equation y 2 − x does not define y as a function of x because some input values x correspond to more than one output y; for instance, for the input x − 4 the equation gives the
outputs y − 2 and y − 22.
Similarly, not every table defines a function. Table 3 defined C as a function of w — each
package weight w corresponds to exactly one mailing cost. On the other hand, Table 4 does
not define y as a function of x because some input values x in the table correspond to more
than one output y; for instance, the input x − 5 gives the outputs y − 7 and y − 8.
(a, b)
Table 4
(a) This curve represents a function.
(a, c)
(b) This curve doesn’t represent
a function.
FIGURE 13
What about curves drawn in the xy-plane? Which curves are graphs of functions? The
following test gives an answer.
The Vertical Line Test A curve in the xy-plane is the graph of a function of x if
and only if no vertical line intersects the curve more than once.
(a, b)
The reason for the truth of the Vertical Line Test can be seen in Figure 13. If each
vertical line x − a intersects a curve only once, at sa, bd, then exactly one function value
is defined by f sad − b. But if a line x − a intersects the curve twice, at sa, bd and sa, cd,
then the curve can’t represent a function because a function can’t assign two different
values to a.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
For example, the parabola x − y 2 2 2 shown in Figure 14(a) is not the graph of a
function of x because, as you can see, there are vertical lines that intersect the parabola
twice. The parabola, however, does contain the graphs of two functions of x. Notice
that the equation x − y 2 2 2 implies y 2 − x 1 2, so y − 6sx 1 2 . Thus the upper
and lower halves of the parabola are the graphs of the functions f sxd − s x 1 2 [from
Example 6(a)] and tsxd − 2s x 1 2 . [See Figures 14(b) and (c).]
(_2, 0)
FIGURE 14
_2 0
(b) y=œ„„„„
(a) x=¥-2
(c) y=_œ„„„„
We observe that if we reverse the roles of x and y, then the equation x − hs yd − y 2 2 2
does define x as a function of y (with y as the independent variable and x as the dependent variable). The graph of the function h is the parabola in Figure 14(a).
■ Piecewise Defined Functions
The functions in the following four examples are defined by different formulas in dif­
ferent parts of their domains. Such functions are called piecewise defined functions.
EXAMPLE 7 A function f is defined by
f sxd −
1 2 x if x < 21
if x . 21
Evaluate f s22d, f s21d, and f s0d and sketch the graph.
SOLUTION Remember that a function is a rule. For this particular function the rule is
the following: First look at the value of the input x. If it happens that x < 21, then the
value of f sxd is 1 2 x. On the other hand, if x . 21, then the value of f sxd is x 2. Note
that even though two different formulas are used, f is one function, not two.
Since 22 < 21, we have f s22d − 1 2 s22d − 3.
Since 21 < 21, we have f s21d − 1 2 s21d − 2.
Since 0 . 21, we have f s0d − 0 2 − 0.
FIGURE 15
How do we draw the graph of f ? We observe that if x < 21, then f sxd − 1 2 x,
so the part of the graph of f that lies to the left of the vertical line x − 21 must
coincide with the line y − 1 2 x, which has slope 21 and y-intercept 1. If x . 21,
then f sxd − x 2, so the part of the graph of f that lies to the right of the line x − 21
must coincide with the graph of y − x 2, which is a parabola. This enables us to sketch
the graph in Figure 15. The solid dot indicates that the point s21, 2d is included on the
graph; the open dot indicates that the point s21, 1d is excluded from the graph.
The next example of a piecewise defined function is the absolute value function.
Recall that the absolute value of a number a, denoted by a , is the distance from a to 0
on the real number line. Distances are always positive or 0, so we have
| |
For a more extensive review of
absolute values, see Appendix A.
| a | > 0for every number a
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.1Four Ways to Represent a Function
For example,
| 3 | − 3| 23 | − 3| 0 | − 0| s2 2 1 | − s2 2 1| 3 2 | − 2 3
In general, we have
| a | − aif
| a | − 2a if
(Remember that if a is negative, then 2a is positive.)
EXAMPLE 8 Sketch the graph of the absolute value function f sxd − | x |.
y=| x |
SOLUTION From the preceding discussion we know that
|x| −
if x > 0
2x if x , 0
Using the same method as in Example 7, we see that the graph of f coincides with the
line y − x to the right of the y-axis and coincides with the line y − 2x to the left of the
y-axis (see Figure 16).
FIGURE 16
EXAMPLE 9 Find a formula for the function f graphed in Figure 17.
SOLUTION The line through s0, 0d and s1, 1d has slope m − 1 and y-intercept b − 0,
so its equation is y − x. Thus, for the part of the graph of f that joins s0, 0d to s1, 1d,
we have
f sxd − xif 0 < x < 1
The line through s1, 1d and s2, 0d has slope m − 21, so its point-slope form is
FIGURE 17
y 2 0 − s21dsx 2 2dory − 2 2 x
The point-slope form of the equation
of a line is y 2 y1 − msx 2 x 1 d . See
Appendix B.
So we have
f sxd − 2 2 xif 1 , x < 2
We also see that the graph of f coincides with the x-axis for x . 2. Putting this information together, we have the following three-piece formula for f :
if 0 < x < 1
f sxd − 2 2 x if 1 , x < 2
if x . 2
EXAMPLE 10 In Example C at the beginning of this section we considered the cost
Cswd of mailing a large envelope with weight w. In effect, this is a piecewise defined
function because, from Table 3, we have
Cswd −
FIGURE 18
125 w
0 , w , 25
25 , w , 50
50 , w , 75
75 , w , 100
The graph is shown in Figure 18.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Looking at Figure 18, you can see why a function like the one in Example 10 is called
a step function.
■ Even and Odd Functions
If a function f satisfies f s2xd − f sxd for every number x in its domain, then f is called
an even function. For instance, the function f sxd − x 2 is even because
f s2xd − s2xd2 − x 2 − f sxd
The geometric significance of an even function is that its graph is symmetric with respect
to the y-axis (see Figure 19). This means that if we have plotted the graph of f for x > 0,
we obtain the entire graph simply by reflecting this portion about the y-axis.
If f satisfies f s2xd − 2f sxd for every number x in its domain, then f is called an odd
function. For example, the function f sxd − x 3 is odd because
FIGURE 19
An even function
f s2xd − s2xd3 − 2x 3 − 2f sxd
The graph of an odd function is symmetric about the origin (see Figure 20). If we already
have the graph of f for x > 0, we can obtain the entire graph by rotating this portion
through 1808 about the origin.
EXAMPLE 11 Determine whether each of the following functions is even, odd, or
neither even nor odd.
(a) f sxd − x 5 1 x(b) tsxd − 1 2 x 4(c) hsxd − 2x 2 x 2
FIGURE 20
An odd function
f s2xd − s2xd5 1 s2xd − s21d5x 5 1 s2xd
− 2x 5 2 x − 2sx 5 1 xd
− 2f sxd
Therefore f is an odd function.
ts2xd − 1 2 s2xd4 − 1 2 x 4 − tsxd
So t is even.
hs2xd − 2s2xd 2 s2xd2 − 22x 2 x 2
Since hs2xd ± hsxd and hs2xd ± 2hsxd, we conclude that h is neither even nor odd. ■
The graphs of the functions in Example 11 are shown in Figure 21. Notice that the
graph of h is symmetric neither about the y-axis nor about the origin.
FIGURE 21
■ Increasing and Decreasing Functions
The graph shown in Figure 22 rises from A to B, falls from B to C, and rises again from C
to D. The function f is said to be increasing on the interval fa, bg, decreasing on fb, cg,
and increasing again on fc, dg. Notice that if x 1 and x 2 are any two numbers between
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.1Four Ways to Represent a Function
a and b with x 1 , x 2, then f sx 1 d , f sx 2 d. We use this as the defining property of an
increasing function.
FIGURE 22
a x¡
A function f is called increasing on an interval I if
f sx 1 d , f sx 2 dwhenever x 1 , x 2 in I
It is called decreasing on I if
f sx 1 d . f sx 2 dwhenever x 1 , x 2 in I
In the definition of an increasing function it is important to realize that the inequality
f sx 1 d , f sx 2 d must be satisfied for every pair of numbers x 1 and x 2 in I with x 1 , x 2.
You can see from Figure 23 that the function f sxd − x 2 is decreasing on the interval
s2`, 0g and increasing on the interval f0, `d.
FIGURE 23
1. I f f sxd − x 1 s2 2 x and tsud − u 1 s2 2 u , is it true
that f − t ?
2. If
x2 2 x
andtsxd − x
f sxd −
is it true that f − t ?
3. The
graph of a function t is given.
(a) State the values of ts22d, ts0d, ts2d, and ts3d.
(b) For what value(s) of x is tsxd − 3 ?
(c) For what value(s) of x is tsxd ⩽ 3 ?
(d) State the domain and range of t.
(e) On what interval(s) is t increasing?
4. The graphs of f and t are given.
(a) State the values of f s24d and ts3d.
(b) Which is larger, f s23d or ts23d ?
(c) For what values of x is f sxd − tsxd ?
(d) On what interval(s) is f sxd ⩽ tsxd ?
(e) State the solution of the equation f sxd − 21.
(f ) On what interval(s) is t decreasing?
(g) State the domain and range of f.
(h) State the domain and range of t.
5. Figure
1 was recorded by an instrument operated by the
California Department of Mines and Geology at the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
University Hospital of the University of Southern California
in Los Angeles. Use it to estimate the range of the vertical
ground acceleration function at USC during the Northridge
(c) The years when the temperature was smallest and largest
(d) The range of T
T (•C)
6. In
this section we discussed examples of ordinary, everyday
functions: population is a function of time, postage cost is a
function of package weight, water temperature is a function
of time. Give three other examples of functions from everyday life that are described verbally. What can you say about
the domain and range of each of your functions? If possible,
sketch a rough graph of each function.
7–14 Determine whether the equation or table defines y as a
function of x.
9. x 1 sy 2 3d − 5
10. 2xy 1 5y − 4
11. s y 1 3d3 1 1 − 2x
12. 2x 2 y − 0
Height (cm) Shoe size
| |
Year Tuition cost ($)
15–18 Determine whether the curve is the graph of a function
of x. If it is, state the domain and range of the function.
yy y
yy y
00 0 11 1 xx x
00 0 11 1 xx x
yy y
yy y
00 0 11 1 xx x
0 0 11 1 x x
19. Shown
is a graph of the global average temperature T during
the 20th century. Estimate the following.
(a) The global average temperature in 1950
(b) The year when the average temperature was 14.2°C
2000 t
Source: Adapted from Globe and Mail [Toronto], 5 Dec. 2009. Print.
20. Trees
grow faster and form wider rings in warm years and
grow more slowly and form narrower rings in cooler years.
The figure shows ring widths of a Siberian pine from 1500
to 2000.
(a) What is the range of the ring width function?
(b)What does the graph tend to say about the temperature of
the earth? Does the graph reflect the volcanic eruptions of
the mid-19th century?
Ring width (mm)
8. 3x 2 2 2y − 5
7. 3x 2 5y − 7
2000 t
Source: Adapted from G. Jacoby et al., “Mongolian Tree Rings
and 20th-Century Warming,” Science 273 (1996): 771–73.
21. You
put some ice cubes in a glass, fill the glass with cold
water, and then let the glass sit on a table. Describe how the
temperature of the water changes as time passes. Then sketch
a rough graph of the temperature of the water as a function of
the elapsed time.
22. You
place a frozen pie in an oven and bake it for an hour.
Then you take it out and let it cool. Describe how the temperature of the pie changes as time passes. Then sketch a rough
graph of the temperature of the pie as a function of time.
23. The
graph shows the power consumption for a day in September in San Francisco. (P is measured in megawatts; t is mea­
sured in hours starting at midnight.)
(a) What was the power consumption at 6 am? At 6 pm?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.1Four Ways to Represent a Function
(b)When was the power consumption the lowest? When was
it the highest? Do these times seem reasonable?
(a)Use the readings to sketch a rough graph of T as a
function of t.
(b)Use your graph to estimate the temperature at 9:00 am.
31. T
emperature readings T (in °C) were recorded every two
hours from midnight to 2:00 pm in Atlanta on a day in June.
The time t was measured in hours from midnight.
Pacific Gas & Electric
24. Three
runners compete in a 100-meter race. The graph
depicts the distance run as a function of time for each
runner. Describe in words what the graph tells you about
this race. Who won the race? Did each runner finish the
32. Researchers
measured the blood alcohol concentration (BAC)
of eight adult male subjects after rapid consumption of 30 mL
of ethanol (corresponding to two standard alcoholic drinks).
The table shows the data they obtained by averaging the BAC
(in gydL) of the eight men.
(a)Use the readings to sketch a graph of the BAC as a
function of t.
(b)Use your graph to describe how the effect of alcohol
varies with time.
25. Sketch
a rough graph of the outdoor temperature as a function
of time during a typical spring day.
26. Sketch
a rough graph of the number of hours of daylight as a
function of the time of year.
27. Sketch
a rough graph of the amount of a particular brand of
coffee sold by a store as a function of the price of the coffee.
28. Sketch
a rough graph of the market value of a new car as a
function of time for a period of 20 years. Assume the car is
well maintained.
29. A
homeowner mows the lawn every Wednesday afternoon.
Sketch a rough graph of the height of the grass as a function
of time over the course of a four-week period.
30. An
airplane takes off from an airport and lands an hour later
at another airport, 650 kilometers away. If t represents the
time in minutes since the plane has left the terminal building,
let xstd be the horizontal distance traveled and ystd be the altitude of the plane.
(a) Sketch a possible graph of xstd.
(b) Sketch a possible graph of ystd.
(c) Sketch a possible graph of the ground speed.
(d) Sketch a possible graph of the vertical velocity.
t (hours)
t (hours)
Source: Adapted from P. Wilkinson et al., “Pharmacokinetics of
Ethanol after Oral Administration in the Fasting State,” Journal of
Pharmacokinetics and Biopharmaceutics 5 (1977): 207–24.
33. If f sxd − 3x 2 2 x 1 2, find f s2d, f s22d, f sad, f s2ad,
f sa 1 1d, 2 f sad, f s2ad, f sa 2 d, [ f sad] 2, and f sa 1 hd.
34. If tsxd −
, find ts0d, ts3d, 5tsad, 12 ts4ad, tsa 2 d,
sx 1 1
ftsad 2, tsa 1 hd, and tsx 2 ad.
35–38 Evaluate the difference quotient for the given function.
Simplify your answer.
f s3 1 hd 2 f s3d
35. f sxd − 4 1 3x 2 x 2,h
36. f sxd − x 3,37. f sxd −
f sa 1 hd 2 f sad
f sxd 2 f sad
38. f sxd − sx 1 2,f sxd 2 f s1d
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
39–46 Find the domain of the function.
39. f sxd −
x2 2 9
40. f sxd −
44. f sud −
sx 2 2 5x
45. Fs pd − s2 2 s p
x2 1 1
x 1 4x 2 21
46. hsxd − sx 2 4x 2 5
47. Find
the domain and range and sketch the graph of the
function hsxd − s4 2 x 2 .
48. Find the domain and sketch the graph of the function
x2 2 4
50. f sxd −
51. f sxd −
52. f sxd −
x 2 1 2 if x , 0
if x > 0
65–70 Find a formula for the described function and state its domain.
65. A
rectangle has perimeter 20 m. Express the area of the rectangle as a function of the length of one of its sides.
66. A
rectangle has area 16 m2. Express the perimeter of the rect­
angle as a function of the length of one of its sides.
67. Express
the area of an equilateral triangle as a function of the
length of a side.
69. An
open rectangular box with volume 2 m3 has a square base.
Express the surface area of the box as a function of the length
of a side of the base.
if x < 1
7 2 2x if x . 1
53–58 Sketch the graph of the function.
55. tstd − 1 2 3t
57. f sxd −
58. tsxd −
H| |
f sxd −
54. f sxd − x 1 2
| |
| |
if x < 1
if x . 1
|| x | 2 1|
71. A
box with an open top is to be constructed from a rectan­
gular piece of cardboard with dimensions 30 cm by 50 cm
by cutting out equal squares of side x at each corner
and then folding up the sides as in the figure. Express the
vol­ume V of the box as a function of x.
x 1 1 if x < 21
if x . 21
| |
70. A right circular cylinder has volume 400 cm3. Express the
radius of the cylinder as a function of the height.
if x , 2
53. f sxd − x 1 x
68. A
closed rectangular box with volume 0.25 m3 has length twice
the width. Express the height of the box as a function of the width.
49–52 Evaluate f s23d, f s0d, and f s2d for the piecewise defined
function. Then sketch the graph of the function.
49. f sxd −
f sxd −
42. tstd − s3 2 t 2 s2 1 t
41. f std − s
2t 2 1
43. hsxd −
72. A
Norman window has the shape of a rectangle surmounted
by a semicircle. If the perimeter of the window is 10 m,
express the area A of the window as a function of the width x
of the window.
59–64 Find a formula for the function whose graph is the given
59. The line segment joining the points s1, 23d and s5, 7d
60. The line segment joining the points s25, 10d and s7, 210d
61. The bottom half of the parabola x 1 s y 2 1d 2 − 0
62. The top half of the circle x 2 1 s y 2 2d 2 − 4
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.2Mathematical Models: A Catalog of Essential Functions
73. In
a certain state the maximum speed permitted on freeways
is 100 kmyh and the minimum speed is 60 kmyh. The fine for
violating these limits is $15 for every kilometer per hour
above the maximum speed or below the minimum speed.
Express the amount of the fine F as a function of the driving
speed x and graph Fsxd for 0 < x < 150.
79–80 The graph of a function defined for x > 0 is given.
Complete the graph for x , 0 to make (a) an even function and
(b) an odd function.
79. 74. An
electricity company charges its customers a base rate of $10
a month, plus 6 cents per kilowatt-hour (kWh) for the first
1200 kWh and 7 cents per kWh for all usage over 1200 kWh.
Express the monthly cost E as a function of the amount x of
electricity used. Then graph the function E for 0 < x < 2000.
75. In
a certain country, income tax is assessed as follows. There
is no tax on income up to $10,000. Any income over $10,000
is taxed at a rate of 10%, up to an income of $20,000. Any
income over $20,000 is taxed at 15%.
(a)Sketch the graph of the tax rate R as a function of the
income I.
(b)How much tax is assessed on an income of $14,000?
On $26,000?
(c)Sketch the graph of the total assessed tax T as a function
of the income I.
76. (a)If the point s5, 3d is on the graph of an even function,
what other point must also be on the graph?
(b)If the point s5, 3d is on the graph of an odd function, what
other point must also be on the graph?
77–78 Graphs of f and t are shown. Decide whether each
function is even, odd, or neither. Explain your reasoning.
80. x
81–86 Determine whether f is even, odd, or neither. You may
wish to use a graphing calculator or computer to check your
answer visually.
x 11
81. f sxd −
x 11
82. f sxd −
83. f sxd −
84. f sxd − x x
85. f sxd − 1 1 3x 2 2 x 4
| |
86. f sxd − 1 1 3x 3 2 x 5
87. If f and t are both even functions, is f 1 t even? If f and t
are both odd functions, is f 1 t odd? What if f is even and t
is odd? Justify your answers.
88. If f and t are both even functions, is the product ft even? If f
and t are both odd functions, is ft odd? What if f is even and
t is odd? Justify your answers.
1.2 Mathematical Models: A Catalog of Essential Functions
A mathematical model is a mathematical description (often by means of a function or
an equation) of a real-world phenomenon such as the size of a population, the demand
for a product, the speed of a falling object, the concentration of a product in a chemical
reaction, the life expectancy of a person at birth, or the cost of emissions reductions. The
purpose of the model is to understand the phenomenon and perhaps to make predictions
about future behavior.
Given a real-world problem, our first task in the mathematical modeling process is to
formulate a mathematical model by identifying and naming the independent and dependent variables and making assumptions that simplify the phenomenon enough to make
it mathematically tractable. We use our knowledge of the physical situation and our
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
mathematical skills to obtain equations that relate the variables. In situations where there
is no physical law to guide us, we may need to collect data (either from the Internet or
a library or by conducting our own experiments) and examine the data in the form of a
table in order to discern patterns. From this numeri­cal representation of a function we
may wish to obtain a graphical representation by plotting the data. The graph might even
suggest a suitable algebraic formula in some cases.
The second stage is to apply the mathematics that we know (such as the calculus that
will be developed throughout this book) to the mathematical model that we have formulated in order to derive mathematical conclusions. Then, in the third stage, we take those
mathematical conclusions and interpret them as information about the original realworld phenomenon by way of offering explanations or making predictions. The final step
is to test our predictions by checking against new real data. If the predictions don’t compare well with reality, we need to refine our model or formulate a new model and start
the cycle again. Figure 1 illustrates the process of mathematical modeling.
FIGURE 1
A mathematical model is never a completely accurate representation of a physical
situation—it is an idealization. A good model simplifies reality enough to permit mathematical calculations but is accurate enough to provide valuable conclusions. It is important to realize the limitations of a model.
There are many different types of functions that can be used to model relationships
observed in the real world. In what follows, we discuss the behavior and graphs of some of
these functions and give examples of situations appropriately modeled by such functions.
The modeling process
■ Linear Models
The coordinate geometry of lines is
reviewed in Appendix B.
When we say that y is a linear function of x, we mean that the graph of the function is a
line, so we can use the slope-intercept form of the equation of a line to write a formula
for the function as
y − f sxd − mx 1 b
where m is the slope of the line and b is the y-intercept.
A characteristic feature of linear functions is that they change at a constant rate. For
instance, Figure 2 shows a graph of the linear function f sxd − 3x 2 2 and a table of
sample values. Notice that whenever x increases by 0.1, the value of f sxd increases
by 0.3. So f sxd increases three times as fast as x. This means that the slope of the graph
of y − 3x 2 2, namely 3, can be interpreted as the rate of change of y with respect to x.
FIGURE 2
f sxd − 3x 2 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.2Mathematical Models: A Catalog of Essential Functions
EXAMPLE 1(a) As dry air moves upward, it expands and cools. If the ground temperature is 20°C
and the temperature at a height of 1 km is 10°C, express the temperature T (in °C) as a
function of the height h (in kilometers), assuming that a linear model is appropriate.
(b) Draw the graph of the function in part (a). What does the slope represent?
(c) What is the temperature at a height of 2.5 km?
(a) Because we are assuming that T is a linear function of h, we can write
T − mh 1 b
We are given that T − 20 when h − 0, so
20 − m ? 0 1 b − b
In other words, the y-intercept is b − 20.
We are also given that T − 10 when h − 1, so
10 − m ? 1 1 20
The slope of the line is therefore m − 10 2 20 − 210 and the required linear function is
T − 210h 1 20
( b) The graph is sketched in Figure 3. The slope is m − 210°Cykm, and this represents the rate of change of temperature with respect to height.
(c) At a height of h − 2.5 km, the temperature is
FIGURE 3
T − 210s2.5d 1 20 − 25°C
If there is no physical law or principle to help us formulate a model, we construct an
empirical model, which is based entirely on collected data. We seek a curve that “fits”
the data in the sense that it captures the basic trend of the data points.
Table 1
CO2 level
(in ppm)
CO2 level
(in ppm)
EXAMPLE 2 Table 1 lists the average carbon dioxide level in the atmosphere,
measured in parts per million at Mauna Loa Observatory from 1980 to 2016. Use the
data in Table 1 to find a model for the carbon dioxide level.
SOLUTION We use the data in Table 1 to make the scatter plot in Figure 4, where t
represents time (in years) and C represents the CO2 level (in parts per million, ppm).
C (ppm)
FIGURE 4
Scatter plot for the average CO2 level
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Notice that the data points appear to lie close to a straight line, so it’s natural to
choose a linear model in this case. But there are many possible lines that approximate
these data points, so which one should we use? One possibility is the line that passes
through the first and last data points. The slope of this line is
404.2 2 338.7
< 1.819
We write its equation as
C 2 338.7 − 1.819st 2 1980d
A computer or graphing calculator
finds the regression line by the
method of least squares, which is to
minimize the sum of the squares of
the vertical distances between the
data points and the line. The details
are explained in Exercise 14.7.61.
C − 1.819t 2 3262.92
Equation 1 gives one possible linear model for the carbon dioxide level; it is
graphed in Figure 5. Notice that our model gives values higher than most of the actual
CO2 levels. A better linear model is obtained by a procedure from statistics called
linear regression. Many graphing calculators and computer software applications can
determine the regression line for a set of data. One such calculator gives the slope
and y-intercept of the regression line for the data from Table 1 as
m − 1.78242b − 23192.90
So our least squares model for the CO2 level is
C − 1.78242t 2 3192.90
In Figure 6 we graph the regression line as well as the data points. Comparing with
Figure 5, we see that the regression line gives a better fit.
C (ppm)
C (ppm)
FIGURE 5
FIGURE 6
Linear model through first and last data points
The regression line
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.2Mathematical Models: A Catalog of Essential Functions
EXAMPLE 3 Use the linear model given by Equa­tion 2 to estimate the average CO2
level for 1987 and to predict the level for the year 2025. According to this model, when
will the CO2 level exceed 440 parts per million?
SOLUTION Using Equation 2 with t − 1987, we estimate that the average CO2 level in
1987 was
Cs1987d − 1.78242s1987d 2 3192.90 < 348.77
This is an example of interpolation because we have estimated a value between
observed values. (In fact, the Mauna Loa Observatory reported that the average CO2
level in 1987 was 348.93 ppm, so our estimate is quite accurate.)
With t − 2025, we get
Cs2025d − 1.78242s2025d 2 3192.90 < 416.50
So we predict that the average CO2 level in the year 2025 will be 416.5 ppm. This is
an example of extrapolation because we have predicted a value outside the time
frame of observations. Consequently, we are far less certain about the accuracy of our
Using Equation 2, we see that the CO2 level exceeds 440 ppm when
1.78242t 2 3192.90 . 440
Solving this inequality, we get
< 2038.18
We therefore predict that the CO2 level will exceed 440 ppm by the year 2038. This
pre­diction is risky because it involves a time quite remote from our observations. In
fact, we see from Figure 6 that the trend has been for CO2 levels to increase rather
more rapidly in recent years, so the level might exceed 440 ppm well before 2038.
■ Polynomials
A function P is called a polynomial if
Psxd − a n x n 1 a n21 x n21 1 ∙ ∙ ∙ 1 a 2 x 2 1 a 1 x 1 a 0
(a) y=≈+x+1
where n is a nonnegative integer and the numbers a 0 , a 1, a 2 , . . . , a n are constants called
the coefficients of the polynomial. The domain of any polynomial is R − s2`, `d.
If the leading coefficient a n ± 0, then the degree of the polynomial is n. For example,
the function
Psxd − 2x 6 2 x 4 1 25 x 3 1 s2
(b) y=_2≈+3x+1
FIGURE 7
The graphs of quadratic functions
are parabolas.
is a polynomial of degree 6.
A polynomial of degree 1 is of the form Psxd − mx 1 b and so it is a linear function.
A polynomial of degree 2 is of the form Psxd − ax 2 1 bx 1 c and is called a quadratic
function. Its graph is always a parabola obtained by shifting the parabola y − ax 2, as we
will see in Section 1.3. The parabola opens upward if a . 0 and downward if a , 0.
(See Figure 7.)
A polynomial of degree 3 is of the form
Psxd − ax 3 1 bx 2 1 cx 1 da ± 0
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
and is called a cubic function. Figure 8 shows the graph of a cubic function in part (a)
and graphs of polynomials of degrees 4 and 5 in parts (b) and (c). We will see later why
the graphs have these shapes.
(a) y=˛-x+1
FIGURE 8
(b) y=x$-3≈+x
(c) y=3x%-25˛+60x
Polynomials are commonly used to model various quantities that occur in the natural
and social sciences. For instance, in Section 2.7 we will explain why economists often
use a polynomial Psxd to represent the cost of producing x units of a commodity. In the
following example we use a quadratic function to model the fall of a ball.
Table 2
EXAMPLE 4 A ball is dropped from the upper observation deck of the CN Tower,
450 m above the ground, and its height h above the ground is recorded at 1-second
intervals in Table 2. Find a model to fit the data and use the model to predict the time at
which the ball hits the ground.
SOLUTION We draw a scatter plot of the data in Figure 9 and observe that a linear
model is inappropriate. But it looks as if the data points might lie on a parabola, so we
try a quadratic model instead. Using a graphing calculator or computer algebra system
(which uses the least squares method), we obtain the following quadratic model:
h − 449.36 1 0.96t 2 4.90t 2
h (meters)
FIGURE 9
FIGURE 10
Scatter plot for a falling ball
Quadratic model for a falling ball
In Figure 10 we plot the graph of Equation 3 together with the data points and see
that the quadratic model gives a very good fit.
The ball hits the ground when h − 0, so we solve the quadratic equation
24.90t 2 1 0.96t 1 449.36 − 0
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.2Mathematical Models: A Catalog of Essential Functions
The quadratic formula gives
20.96 6 ss0.96d2 2 4s24.90ds449.36d
The positive root is t < 9.67, so we predict that the ball will hit the ground after falling
about 9.7 seconds.
■ Power Functions
A function of the form f sxd − x a, where a is a constant, is called a power function. We
consider several cases.
(i ) a − n, where n is a positive integer
The graphs of f sxd − x n for n − 1, 2, 3, 4, and 5 are shown in Figure 11. (These are
polynomials with only one term.) We already know the shape of the graphs of y − x
(a line through the origin with slope 1) and y − x 2 [a parabola, see Example 1.1.2(b)].
FIGURE 11 Graphs of f sxd − x n for n − 1, 2, 3, 4, 5
The general shape of the graph of f sxd − x n depends on whether n is even or odd.
If n is even, then f sxd − x n is an even function and its graph is similar to the parabola
y − x 2. If n is odd, then f sxd − x n is an odd function and its graph is similar to that
of y − x 3. Notice from Figure 12, however, that as n increases, the graph of y − x n
becomes flatter near 0 and steeper when x > 1. (If x is small, then x 2 is smaller, x 3
is even smaller, x 4 is smaller still, and so on.)
| |
A family of functions is a collection
of functions whose equations are
related. Figure 12 shows two families
of power functions, one with even
powers and one with odd powers.
y=x ^
(_1, 1)
FIGURE 12y
y=x $
(1, 1)
y=x #
(1, 1)
y=x %
(_1, _1)
(ii ) a − 1yn, where n is a positive integer
The function f sxd − x 1yn − s
x is a root function. For n − 2 it is the square root
function f sxd − sx , whose domain is f0, `d and whose graph is the upper half of the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
parabola x − y 2. [See Figure 13(a).] For other even values of n, the graph of y − s
x is
similar to that of y − sx . For n − 3 we have the cube root function f sxd − sx whose
domain is R (recall that every real number has a cube root) and whose graph is shown
in Figure 13(b). The graph of y − s
x for n odd sn . 3d is similar to that of y − s
(1, 1)
FIGURE 13
(1, 1)
(a) ƒ=œ„
Graphs of root functions
(b) ƒ=Œ„
(iii ) a 5 21
The graph of the reciprocal function f sxd − x 21 − 1yx is shown in Figure 14. Its
graph has the equation y − 1yx, or xy − 1, and is a hyperbola with the coordinate axes
as its asymptotes. This function arises in physics and chemistry in connection with
Boyle’s Law, which says that when the temperature is constant, the volume V of a gas
is inversely proportional to the pressure P:
where C is a constant. Thus the graph of V as a function of P (see Figure 15) has the
same general shape as the right half of Figure 14.
FIGURE 1
FIGURE 1
The reciprocal function
Volume as a function of pressure
at constant temperature
(iv ) a 5 22
Among the remaining negative powers for the power function f sxd − x a, by far
the most important is that of a − 22. Many natural laws state that one quantity is
inversely proportional to the square of another quantity. In other words, the first
quantity is modeled by a function of the form f sxd − Cyx 2 and we refer to this as an
inverse square law. For instance, the illumination I of an object by a light source
is inversely proportional to the square of the distance x from the source:
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.2Mathematical Models: A Catalog of Essential Functions
where C is a constant. Thus the graph of I as a function of x (see Figure 17) has the
same general shape as the right half of Figure 16.
FIGURE 16
FIGURE 17
The reciprocal of the squaring function
Illumination from a light source as a
function of distance from the source
Inverse square laws model gravitational force, loudness of sound, and electrostatic
force between two charged particles. See Exercise 37 for a geometric reason why
inverse square laws often occur in nature.
Power functions are also used to model species-area relationships (Exercises 35–36)
and the period of revolution of a planet as a function of its distance from the sun (see
Exercise 34).
■ Rational Functions
A rational function f is a ratio of two polynomials:
f sxd −
where P and Q are polynomials. The domain consists of all values of x such that
Qsxd ± 0. A simple example of a rational function is the function f sxd − 1yx, whose
domain is hx x ± 0j; this is the reciprocal function graphed in Figure 14. The function
f sxd −
FIGURE 18
f sxd −
2x 4 2 x 2 1 1
x2 2 4
2x 4 2 x 2 1 1
x2 2 4
is a rational function with domain hx x ± 62j. Its graph is shown in Figure 18.
■ Algebraic Functions
A function f is called an algebraic function if it can be constructed using algebraic
operations (such as addition, subtraction, multiplication, division, and taking roots) starting with polynomials. Any rational function is automatically an algebraic function. Here
are two more examples:
f sxd − sx 2 1 1tsxd −
x 4 2 16x 2
x 1 sx
1 sx 2 2d s
In Chapter 3 we will sketch a variety of algebraic functions, and we will see that their
graphs can assume many different shapes.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
An example of an algebraic function occurs in the theory of relativity. The mass of a
particle with velocity v is
m − f svd −
s1 2 v 2yc 2
where m 0 is the rest mass of the particle and c − 3.0 3 10 5 kmys is the speed of light in
a vacuum.
Functions that are not algebraic are called transcendental; these include the trigonometric, exponential, and logarithmic functions.
■ Trigonometric Functions
Trigonometry and the trigonometric functions are reviewed on Reference Page 2 and
also in Appendix D. In calculus the convention is that radian measure is always used
(except when otherwise indicated). For example, when we use the function f sxd − sin x,
it is understood that sin x means the sine of the angle whose radian measure is x. Thus
the graphs of the sine and cosine functions are as shown in Figure 19.
The Reference Pages are located at
the front and back of the book.
(a) ƒ=sin x
FIGURE 19
(b) ©=cos x
Notice that for both the sine and cosine functions the domain is s2`, `d and the range
is the closed interval f21, 1g. Thus, for all values of x, we have
21 < sin x < 121 < cos x < 1
or, in terms of absolute values,
| sin x | < 1| cos x | < 1
An important property of the sine and cosine functions is that they are periodic functions and have period 2. This means that, for all values of x,
sinsx 1 2d − sin xcossx 1 2d − cos x
The periodic nature of these functions makes them suitable for modeling repetitive phenomena such as tides, vibrating springs, and sound waves. For instance, in Example 1.3.4
we will see that a reasonable model for the number of hours of daylight in Philadelphia
t days after January 1 is given by the function
Lstd − 12 1 2.8 sin
st 2 80d
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.2Mathematical Models: A Catalog of Essential Functions
1 2 2 cos x
SOLUTION This function is defined for all values of x except for those that make the
denominator 0. But
1 2n
1 2 2 cos x − 0 &? cos x −
&? x −
1 2norx −
EXAMPLE 5 Find the domain of the function f sxd −
where n is any integer (because the cosine function has period 2). So the domain of f
is the set of all real numbers except for the ones noted above.
_π _ π
The tangent function is related to the sine and cosine functions by the equation
tan x −
sin x
cos x
and its graph is shown in Figure 20. It is undefined whenever cos x − 0, that is, when
x − 6y2, 63y2, . . . . Its range is s2`, `d. Notice that the tangent function has per­iod :
FIGURE 20
tansx 1 d − tan xfor all x
y − tanxx
The remaining three trigonometric functions (cosecant, secant, and cotangent) are
the reciprocals of the sine, cosine, and tangent functions. Their graphs are shown in
Appendix D.
(a) y=2®
■ Exponential Functions
(b) y=(0.5)®
FIGURE 21
■ Logarithmic Functions
The logarithmic functions f sxd − log b x, where the base b is a positive constant, are the
inverse functions of the exponential functions. They will be studied in Chapter 6. Figure 22 shows the graphs of four logarithmic functions with various bases. In each case
the domain is s0, `d, the range is s2`, `d, and the function increases slowly when x . 1.
y=log™ x
y=log£ x
FIGURE 22
The exponential functions are the functions of the form f sxd − b x, where the base b is
a positive constant. The graphs of y − 2 x and y − s0.5d x are shown in Figure 21. In both
cases the domain is s2`, `d and the range is s0, `d.
Exponential functions will be studied in detail in Chapter 6, and we will see that
they are useful for modeling many natural phenomena, such as when populations grow
(if b . 1) or decline (if b , 1d.
y=log¡¸ x
y=log∞ x
EXAMPLE 6 Classify the following functions as one of the types of functions that we
have discussed.
(a) f sxd − 5 x(b) tsxd − x 5(c) hsxd −
(d) ustd − 1 2 t 1 5t 4
1 2 sx
(a) f sxd − 5 x is an exponential function. (The variable x is the exponent.)
(b) tsxd − x 5 is a power function. (The variable x is the base.) We could also consider
it to be a polynomial of degree 5.
(c) hsxd −
is an algebraic function. (It is not a rational function because the
1 2 sx
denominator is not a polynomial.)
(d) ustd − 1 2 t 1 5t 4 is a polynomial of degree 4.
Table 3 (on the following page) shows a summary of graphs of some families of
essential functions that will be used frequently throughout the book.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Table 3 Families of Essential Functions and Their Graphs
f sxd − mx 1 b
f sxd − x n
f sxd − s
ƒ= œ„
f sxd −
ƒ=b® (b>1)
f sxd − b x
f sxd − log b x
and Logarithmic
ƒ=b® (b<1)
ƒ=log b x (b>1)
f sxd − sin x
f sxd − cos x
f sxd − tan x
ƒ=sin x
2π x
ƒ=cos x
ƒ=tan x
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.2Mathematical Models: A Catalog of Essential Functions
1–2 Classify each function as a power function, root function,
polynomial (state its degree), rational function, algebraic function,
trigonometric function, exponential function, or logarithmic
(c) r std − t s3
(d) v std − 8 t
x2 1 1
2. (a) f std −
(f ) tsud − log10 u
3t 2 1 2
(b) hsrd − 2.3 r
(c) sstd − st 1 4
(d) y − x 4 1 5
(e) tsxd − sx
(f ) y − 2
3–4 Match each equation with its graph. Explain your choices.
(Don’t use a computer or graphing calculator.)
(c) y − x
8. What
do all members of the family of linear functions
f sxd − 1 1 msx 1 3d have in common? Sketch several
members of the family.
9. What
do all members of the family of linear functions
f sxd − c 2 x have in common? Sketch several members of
the family.
y y
(d) y − sx
(_2, (_2,
2) 2)
(0, 1)(0, 1)
(4, 2)(4, 2)
(1, _2.5)
(1, _2.5)
14. Recent
studies indicate that the average surface temperature of the earth has been rising steadily. Some scientists
have modeled the temperature by the linear function
T − 0.02t 1 8.50, where T is temperature in °C and t
represents years since 1900.
(a)What do the slope and T-intercept represent?
(b)Use the equation to predict the earth’s average surface
temperature in 2100.
13. Find
a formula for a cubic function f if f s1d − 6 and
f s21d − f s0d − f s2d − 0.
(b) y − 3 x
1 2 tan x
11–12 Find a formula for the quadratic function whose graph is
6. tsxd −
7. (a)Find an equation for the family of linear functions
with slope 2 and sketch several members of the
(b)Find an equation for the family of linear functions
such that f s2d − 1. Sketch several members of the
(c)Which function belongs to both families?
4. (a) y − 3x
cos x
1 2 sin x
; 10. Sketch several members of the family of polynomials
Psxd − x 3 2 cx 2. How does the graph change when c
3. (a) y − x 2(b) y − x 5(c) y − x 8
5. f sxd −
(b) tstd − cos2 t 2 sin t
1. (a) f sxd − x 3 1 3x 2
(e) y −
5–6 Find the domain of the function.
15. If
the recommended adult dosage for a drug is D (in mg),
then to determine the appropriate dosage c for a child of
age a, pharmacists use the equation c − 0.0417Dsa 1 1d.
Suppose the dosage for an adult is 200 mg.
(a)Find the slope of the graph of c. What does it represent?
(b)What is the dosage for a newborn?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
16. The
manager of a weekend flea market knows from past
experience that if he charges x dollars for a rental space at
the market, then the number y of spaces that will be rented is
given by the equation y − 200 2 4x.
(a)Sketch a graph of this linear function. (Remember that
the rental charge per space and the number of spaces
rented can’t be negative quantities.)
(b)What do the slope, the y-intercept, and the x-intercept of
the graph represent?
17. The
relationship between the Fahrenheit sFd and Celsius sCd
temperature scales is given by the linear function F − 95 C 1 32.
(a) Sketch a graph of this function.
(b)What is the slope of the graph and what does it represent?
What is the F-intercept and what does it represent?
18. Jade
and her roommate Jari commute to work each morning,
traveling west on I-10. One morning Jade left for work at
6:50 am, but Jari left 10 minutes later. Both drove at a constant speed. The graphs show the distance (in kilometers)
each of them has traveled on I-10, t minutes after 7:00 am.
(a) Use the graph to decide which driver is traveling faster.
(b)Find the speed (in kmyh) at which each of them is driving.
(c)Find linear functions f and t that model the distances
traveled by Jade and Jari as functions of t (in minutes).
Distance traveled
(6, 25)
(6, 11)
Time since 7:00 AM (min)
19. The
manager of a furniture factory finds that it costs $2200 to
manufacture 100 chairs in one day and $4800 to produce
300 chairs in one day.
(a)Express the cost as a function of the number of chairs
produced, assuming that it is linear. Then sketch the graph.
(b)What is the slope of the graph and what does it represent?
(c)What is the y-intercept of the graph and what does it
20. The
monthly cost of driving a car depends on the number of
kilometers driven. Lynn found that in May it cost her $380 to
drive 770 km and in June it cost her $460 to drive 1290 km.
(a)Express the monthly cost C as a function of the distance
driven d, assuming that a linear relationship gives a
suitable model.
(b)Use part (a) to predict the cost of driving 2400 km per
(c)Draw the graph of the linear function. What does the
slope represent?
(d)What does the C-intercept represent?
(e)Why does a linear function give a suitable model in this
21. At
the surface of the ocean, the water pressure is the same
as the air pressure above the water, 1.05 kgycm2. Below the
surface, the water pressure increases by 0.3 kgycm2 for every
3 m of descent.
(a)Express the water pressure as a function of the depth
below the ocean surface.
(b)At what depth is the pressure 7 kgycm2 ?
22. The resistance R of a wire of fixed length is related to its
diameter x by an inverse square law, that is, by a function of
the form Rsxd − kx 22.
(a) A wire of fixed length and 0.005 meters in diameter has a
resistance of 140 ohms. Find the value of k.
(b) Find the resistance of a wire made of the same material
and of the same length as the wire in part (a) but with a
diameter of 0.008 meters.
23. The illumination of an object by a light source is related to
the distance from the source by an inverse square law. Suppose that after dark you are sitting in a room with just one
lamp, trying to read a book. The light is too dim, so you move
your chair halfway to the lamp. How much brighter is the
24. The pressure P of a sample of oxygen gas that is compressed
at a constant temperature is related to the volume V of gas by
a reciprocal function of the form P − kyV.
(a) A sample of oxygen gas that occupies 0.671 m3 exerts a
pressure of 39 kPa at a temperature of 293 K (absolute
temperature measured on the Kelvin scale). Find the
value of k in the given model.
(b) If the sample expands to a volume of 0.916 m3, find the
new pressure.
25. The power output of a wind turbine depends on many factors.
It can be shown using physical principles that the power P
generated by a wind turbine is modeled by
P − kAv 3
where v is the wind speed, A is the area swept out by the
blades, and k is a constant that depends on air density,
efficiency of the turbine, and the design of the wind turbine
(a) If only wind speed is doubled, by what factor is the
power output increased?
(b) If only the length of the blades is doubled, by what factor
is the power output increased?
(c) For a particular wind turbine, the length of the blades is
30 m and k − 0.214 kgym3. Find the power output (in
watts, W − m2 ? kgys3 ) when the wind speed is 10 mys,
15 mys, and 25 mys.
26. Astronomers infer the radiant exitance (radiant flux
emitted per unit area) of stars using the Stefan Boltzmann
EsT d − s5.67 3 1028 dT 4
where E is the energy radiated per unit of surface area
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.2Mathematical Models: A Catalog of Essential Functions
measured in watts (W) and T is the absolute temperature
measured in kelvins (K).
(a) Graph the function E for temperatures T between
100 K and 300 K.
(b) Use the graph to describe the change in energy E as
the temperature T increases.
30. When
laboratory rats are exposed to asbestos fibers, some of
them develop lung tumors. The table lists the results of several experiments by different scientists.
(a)Find the regression line for the data.
(b)Make a scatter plot and graph the regression line.
Does the regression line appear to be a suitable model
for the data?
(c)What does the y-intercept of the regression line represent?
27–28 For each scatter plot, decide what type of function you
might choose as a model for the data. Explain your choices.
27. ((a)
Asbestos Percent of mice
that develop
lung tumors
28. (a)
(a) y
29. The
table shows (lifetime) peptic ulcer rates (per 100 population) for various family incomes as reported by the National
Health Interview Survey.
(a)Make a scatter plot of these data and decide whether a
linear model is appropriate.
(b)Find and graph a linear model using the first and last
data points.
(c) Find and graph the regression line.
(d)Use the linear model in part (c) to estimate the ulcer rate
for people with an income of $25,000.
(e)According to the model, how likely is someone with
an income of $80,000 to suffer from peptic ulcers?
(f )Do you think it would be reasonable to apply the
model to someone with an income of $200,000?
Ulcer rate
(per 100 population)
Percent of mice
that develop
lung tumors
31.Anthropologists use a linear model that relates human femur
(thighbone) length to height. The model allows an anthropologist to determine the height of an individual when only a
partial skeleton (including the femur) is found. Here we find
the model by analyzing the data on femur length and height
for the eight males given in the table.
(a)Make a scatter plot of the data.
(b) Find and graph the regression line that models the data.
(c)An anthropologist finds a human femur of length
53 cm. How tall was the person?
Femur length
Femur length
32. The
table shows average US retail residential prices of
electricity from 2000 to 2016, measured in cents per kilowatt hour.
(a)Make a scatter plot. Is a linear model appropriate?
(b)Find and graph the regression line.
(c)Use your linear model from part (b) to estimate the
average retail price of electricity in 2005 and 2017.
Years since 2000 CentsykWh Years since 2000 CentsykWh
Source: US Energy Information Administration
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
33. The
table shows world average daily oil consumption from
1985 to 2015, measured in thousands of barrels per day.
(a)Make a scatter plot and decide whether a linear model is
(b)Find and graph the regression line.
(c)Use the linear model to estimate the oil consumption in
2002 and 2017.
since 1985
Thousands of barrels
of oil per day
Source: US Energy Information Administration
34. The
table shows the mean (average) distances d of the planets from the sun (taking the unit of measurement to be the
distance from the earth to the sun) and their periods T (time
of revolution in years).
(a) Fit a power model to the data.
(b)Kepler’s Third Law of Planetary Motion states that
“The square of the period of revolution of a planet is
propor­tional to the cube of its mean distance from the sun.”
Does your model corroborate Kepler’s Third Law?
35. It
makes sense that the larger the area of a region, the
larger the number of species that inhabit the region.
Many ecologists have modeled the species-area relation
with a power function. In particular, the number of
species S of bats living in caves in central Mexico has
been related to the surface area A of the caves by the
equation S − 0.7A0.3.
(a)The cave called Misión Imposible near Puebla,
Mexico, has a surface area of A − 60 m2. How
many species of bats would you expect to find in
that cave?
(b)If you discover that four species of bats live in a cave,
estimate the area of the cave.
36. The
table shows the number N of species of reptiles and
amphibians inhabiting Caribbean islands and the area A of
the island in square kilometers.
(a)Use a power function to model N as a function
of A.
(b)The Caribbean island of Dominica has area 753 km 2.
How many species of reptiles and amphibians would
you expect to find on Dominica?
Puerto Rico
37. Suppose that a force or energy originates from a point
source and spreads its influence equally in all directions,
such as the light from a lightbulb or the gravitational force
of a planet. So at a distance r from the source, the intensity
I of the force or energy is equal to the source strength S
divided by the surface area of a sphere of radius r. Show
that I satisfies the inverse square law I − kyr 2, where k is
a positive constant.
1.3 New Functions from Old Functions
In this section we start with the basic functions we discussed in Section 1.2 and obtain
new functions by shifting, stretching, and reflecting their graphs. We also show how to
combine pairs of functions by the standard arithmetic operations and by composition.
■ Transformations of Functions
By applying certain transformations to the graph of a given function we can obtain
the graphs of related functions. This will give us the ability to sketch the graphs of
many functions quickly by hand. It will also enable us to write equations for given graphs.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.3New Functions from Old Functions
Let’s first consider translations of graphs. If c is a positive number, then the graph of
y − f sxd 1 c is just the graph of y − f sxd shifted upward a distance of c units (because
each y-coordinate is increased by the same number c). Likewise, if tsxd − f sx 2 cd,
where c . 0, then the value of t at x is the same as the value of f at x 2 c (c units to the
left of x). Therefore the graph of y − f sx 2 cd is just the graph of y − f sxd shifted c units
to the right (see Figure 1).
Vertical and Horizontal ShiftsSuppose c . 0. To obtain the graph of
y − f sxd 1 c, shift the graph of y − f sxd a distance c units upward
y − f sxd 2 c, shift the graph of y − f sxd a distance c units downward
y − f sx 2 cd, shift the graph of y − f sxd a distance c units to the right
y − f sx 1 cd, shift the graph of y − f sxd a distance c units to the left
y =ƒ
y= 1c ƒ
FIGURE 1 Translating the graph of f
FIGURE 2 Stretching and reflecting the graph of f
Now let’s consider the stretching and reflecting transformations. If c . 1, then the
graph of y − cf sxd is the graph of y − f sxd stretched by a factor of c in the vertical
direction (because each y-coordinate is multiplied by the same number c). The graph of
y − 2f sxd is the graph of y − f sxd reflected about the x-axis because the point sx, yd is
replaced by the point sx, 2yd. (See Figure 2 and the following chart, where the results of
other stretching, shrinking, and reflecting transformations are also given.)
Vertical and Horizontal Stretching and ReflectingSuppose c . 1. To obtain the
graph of
y − cf sxd, stretch the graph of y − f sxd vertically by a factor of c
y − s1ycd f sxd, shrink the graph of y − f sxd vertically by a factor of c
y − f scxd, shrink the graph of y − f sxd horizontally by a factor of c
y − f sxycd, stretch the graph of y − f sxd horizontally by a factor of c
y − 2f sxd, reflect the graph of y − f sxd about the x-axis
y − f s2xd, reflect the graph of y − f sxd about the y-axis
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Figure 3 illustrates these stretching transformations when applied to the cosine function with c − 2. For instance, in order to get the graph of y − 2 cos x we multiply the
y-coordinate of each point on the graph of y − cos x by 2. This means that the graph of
y − cos x gets stretched vertically by a factor of 2.
y=2 cos x
y=cos x
cos x
y=cos 1 x
y=cos 2x
y=cos x
FIGURE 3
EXAMPLE 1 Given the graph of y − sx , use transformations to graph y − sx 2 2,
y − sx 2 2 , y − 2sx , y − 2sx , and y − s2x .
SOLUTION The graph of the square root function y − sx , obtained from Figure 1.2.13(a), is shown in Figure 4(a). In the other parts of the figure we sketch
y − sx 2 2 by shifting 2 units downward, y − sx 2 2 by shifting 2 units to the
right, y − 2sx by reflecting about the x-axis, y − 2sx by stretching vertically by a
factor of 2, and y − s2x by reflecting about the y-axis.
(a) y=œ„x
(b) y=œ„-2
(d) y=_ œ„x
(c) y=œ„„„„
(f ) y=œ„„
(e) y=2 œ„x
FIGURE 4
EXAMPLE 2 Sketch the graph of the function f sxd − x 2 1 6x 1 10.
SOLUTION Completing the square, we write the equation of the graph as
y − x 2 1 6x 1 10 − sx 1 3d2 1 1
This means we obtain the desired graph by starting with the parabola y − x 2 and
shifting 3 units to the left and then 1 unit upward (see Figure 5).
(_3, 1)
FIGURE 5
(a) y=≈
(b) y=(x+3)@+1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.3New Functions from Old Functions
EXAMPLE 3 Sketch the graph of each function.
(a) y − sin 2x
(b) y − 1 2 sin x
(a) We obtain the graph of y − sin 2x from that of y − sin x by compressing horizontally by a factor of 2. (See Figures 6 and 7.) Because the period of y − sin x is 2,
the period of y − sin 2x is 2y2 − .
y=sin x
y=sin 2x
0 π π
FIGURE 6
FIGURE 7
(b) To obtain the graph of y − 1 2 sin x, we again start with y − sin x. We reflect
about the x-axis to get the graph of y − 2sin x and then we shift 1 unit upward to get
y − 1 2 sin x. (See Figure 8.)
y=1-sin x
FIGURE 8
EXAMPLE 4 Figure 9 shows graphs of the number of hours of daylight as functions of
the time of the year at several latitudes. Given that Philadelphia is located at approximately 408N latitude, find a function that models the length of daylight at Philadelphia.
20° N
30° N
40° N
50° N
Hours 10
FIGURE 9
Graph of the length of daylight from
March 21 through December 21
at various latitudes
Source: Adapted from L. Harrison,
Daylight, Twilight, Darkness and Time
(New York: Silver, Burdett, 1935), 40.
60° N
Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
SOLUTION Notice that each curve resembles a shifted and stretched sine function.
By looking at the blue curve we see that, at the latitude of Philadelphia, daylight
lasts about 14.8 hours on June 21 and 9.2 hours on December 21, so the amplitude
of the curve (the factor by which we have to stretch the sine curve vertically) is
2 s14.8 2 9.2d − 2.8.
By what factor do we need to stretch the sine curve horizontally if we measure the
time t in days? Because there are about 365 days in a year, the period of our model
should be 365. But the period of y − sin t is 2, so the horizontal stretching factor
is 2y365.
We also notice that the curve begins its cycle on March 21, the 80th day of the year,
so we have to shift the curve 80 units to the right. In addition, we shift it 12 units
upward. Therefore we model the length of daylight in Philadelphia on the t th day of the
year by the function
Lstd − 12 1 2.8 sin
st 2 80d 365
(a) y=≈-1
Another transformation of some interest is taking the absolute value of a function. If
y − f sxd , then according to the definition of absolute value, y − f sxd when f sxd > 0
and y − 2f sxd when f sxd , 0. This tells us how to get the graph of y − f sxd from the
graph of y − f sxd: the part of the graph that lies above the x-axis remains the same, and
the part that lies below the x-axis is reflected about the x-axis.
EXAMPLE 5 Sketch the graph of the function y − | x 2 2 1 |.
SOLUTION We first graph the parabola y − x 2 2 1 in Figure 10(a) by shifting the
parabola y − x 2 downward 1 unit. We see that the graph lies below the x-axis when
21 , x , 1, so we reflect that part of the graph about the x-axis to obtain the graph of
y − x 2 2 1 in Figure 10(b).
(b) y=| ≈-1 |
FIGURE 10
■ Combinations of Functions
Two functions f and t can be combined to form new functions f 1 t, f 2 t, ft, and fyt
in a manner similar to the way we add, subtract, multiply, and divide real numbers.
Definition Given two functions f and t, the sum, difference, product, and
quotient functions are defined by
s f 1 tdsxd − f sxd 1 tsxds f 2 tdsxd − f sxd 2 tsxd
s ftdsxd − f sxd tsxd
f sxd
sxd −
If the domain of f is A and the domain of t is B, then the domain of f 1 t (and the
domain of f 2 t) is the intersection A > B because both f sxd and tsxd have to be defined.
For example, the domain of f sxd − sx is A − f0, `d and the domain of tsxd − s2 2 x
is B − s2`, 2g, so the domain of s f 1 tdsxd − sx 1 s2 2 x is A > B − f0, 2g.
The domain of ft is also A > B. Because we can’t divide by 0, the domain of fyt is
hx [ A > B tsxd ± 0j. For instance, if f sxd − x 2 and tsxd − x 2 1, then the domain
of the rational function s fytdsxd − x 2ysx 2 1d is hx x ± 1j, or s2`, 1d ø s1, `d.
There is another way of combining two functions to obtain a new function. For
example, suppose that y − f sud − su and u − tsxd − x 2 1 1. Since y is a function
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.3New Functions from Old Functions
of u and u is, in turn, a function of x, it follows that y is ultimately a function of x.
We compute this by substitution:
y − f sud − f s tsxdd − f sx 2 1 1d − sx 2 1 1
x (input)
The procedure is called composition because the new function is composed of the two
given functions f and t.
In general, given any two functions f and t, we start with a number x in the domain
of t and calculate tsxd. If this number tsxd is in the domain of f, then we can calculate
the value of f stsxdd. Notice that the output of one function is used as the input to the next
function. The result is a new function hsxd − f s tsxdd obtained by substituting t into f. It
is called the composition (or composite) of f and t and is denoted by f 8 t (“ f circle t”).
Definition Given two functions f and t, the composite function f 8 t (also
called the composition of f and t) is defined by
s f 8 tdsxd − f s tsxdd
f { ©} (output)
FIGURE 11
The f 8 t machine is composed of
the t machine (first) and then the
f machine.
The domain of f 8 t is the set of all x in the domain of t such that tsxd is in the domain
of f. In other words, s f 8 tdsxd is defined whenever both tsxd and f stsxdd are defined.
Figure 11 shows how to picture f 8 t in terms of machines.
EXAMPLE 6 If f sxd − x 2 and tsxd − x 2 3, find the composite functions f 8 t and t 8 f.
SOLUTION We have
s f 8 tdsxd − f s tsxdd − f sx 2 3d − sx 2 3d2
st 8 f dsxd − ts f sxdd − tsx 2 d − x 2 2 3
NOTE You can see from Example 6 that, in general, f 8 t ± t 8 f. Remember, the
notation f 8 t means that the function t is applied first and then f is applied second. In
Example 6, f 8 t is the function that first subtracts 3 and then squares; t 8 f is the function
that first squares and then subtracts 3.
EXAMPLE 7 If f sxd − sx and tsxd − s2 2 x , find each function and its domain.
(a) f 8 t(b) t 8 f (c) f 8 f (d) t 8 t
s f 8 tdsxd − f stsxdd − f (s2 2 x ) − ss2 2 x − s
The domain of f 8 t is hx 2 2 x > 0j − hx x < 2j − s2`, 2g.
If 0 < a < b, then a 2 < b 2.
s t 8 f dsxd − ts f sxdd − t (sx ) − s2 2 sx
For sx to be defined we must have x > 0. For s2 2 sx to be defined we must have
2 2 sx > 0, that is, sx < 2, or x < 4. Thus we have 0 < x < 4, so the domain of
t 8 f is the closed interval f0, 4g.
s f 8 f dsxd − f s f sxdd − f (sx ) − ssx − s
The domain of f 8 f is f0, `d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
s t 8 tdsxd − ts tsxdd − t (s2 2 x ) − s2 2 s2 2 x
This expression is defined when both 2 2 x > 0 and 2 2 s2 2 x > 0. The first
inequality means x < 2, and the second is equivalent to s2 2 x < 2, or 2 2 x < 4, or
x > 22. Thus 22 < x < 2, so the domain of t 8 t is the closed interval f22, 2g.
It is possible to take the composition of three or more functions. For instance, the
composite function f 8 t 8 h is found by first applying h, then t, and then f as follows:
s f 8 t 8 hdsxd − f stshsxddd
EXAMPLE 8 Find f 8 t 8 h if f sxd − xysx 1 1d, tsxd − x 10, and hsxd − x 1 3.
s f 8 t 8 hdsxd − f stshsxddd − f stsx 1 3dd
sx 1 3d10
− f ssx 1 3d10 d −
sx 1 3d10 1 1
So far we have used composition to build complicated functions from simpler ones.
But in calculus it is often useful to be able to decompose a complicated function into
simpler ones, as in the following example.
EXAMPLE 9 Given Fsxd − cos2sx 1 9d, find functions f , t, and h such that F − f 8 t 8 h.
SOLUTION Since Fsxd − fcossx 1 9dg 2, the formula for F says: first add 9, then take
the cosine of the result, and finally square. So we let
hsxd − x 1 9tsxd − cos xf sxd − x 2
s f 8 t 8 hdsxd − f s tshsxddd − f s tsx 1 9dd − f scossx 1 9dd
− fcossx 1 9dg 2 − Fsxd
(a) Shift 3 units upward.
3. The graph of y − f sxd is given. Match each equation with its
graph and give reasons for your choices.
(a) y − f sx 2 4d
(b) y − f sxd 1 3
(b) Shift 3 units downward.
(c) y − 13 f sxd
(c) Shift 3 units to the right.
(e) y − 2 f sx 1 6d
1. Suppose
the graph of f is given. Write equations for the
graphs that are obtained from the graph of f as follows.
(d) Shift 3 units to the left.
(d) y − 2f sx 1 4d
(e) Reflect about the x-axis.
(f ) Reflect about the y-axis.
(g) Stretch vertically by a factor of 3.
(h) Shrink vertically by a factor of 3.
2. Explain
how each graph is obtained from the graph
of y − f sxd.
(a) y − f sxd 1 8
(b) y − f sx 1 8d
(c) y − 8 f sxd
(d) y − f s8xd
(e) y − 2f sxd 2 1
(f ) y − 8 f ( 18 x)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.3New Functions from Old Functions
4. T
he graph of f is given. Draw the graphs of the following
(a) y − f sxd 2 3
(b) y − f sx 1 1d
(c) y − f sxd
(d) y − 2f sxd
16. y − 1 1
17. y − 2 1 sx 1 1
18. y − 2sx 2 1d2 1 3
19. y − x 2 2 2x 1 5
20. y − sx 1 1d3 1 2
21. y − 2 2 x
22. y − 2 2 2 cos x
23. y − 3 sin 12 x 1 1
24. y −
5. T
he graph of f is given. Use it to graph the following
(a) y − f s2xd
(b) y − f ( 12 x)
(c) y − f s2xd
–7 The graph of y − s3x 2 x 2 is given. Use transformations
to create a function whose graph is as shown.
_1 0
8. (a)How is the graph of y − 1 1 sx related to the graph of
y − sx ? Use your answer and Figure 4(a) to sketch the
graph of y − 1 1 sx .
(b)How is the graph of y − 5 sin x related to the graph of
y − sin x ? Use your answer and Figure 6 to sketch the
graph of y − 5 sin x .
9–26 Graph the function by hand, not by plotting points, but by
starting with the graph of one of the standard functions given in
Table 1.2.3, and then applying the appropriate transformations.
9. y − 1 1 x
10. y − sx 1 1d
11. y − x 1 2
S D
tan x 2
26. y − sx 2 1
27. The
city of New Orleans is located at latitude 30°N. Use Figure 9 to find a function that models the number of hours of daylight at New Orleans as a function of the time of year. To check
the accuracy of your model, use the fact that on March 31 the
sun rises at 5:51 am and sets at 6:18 pm in New Orleans.
(d) y − 2f s2xd
15. y − sin 4x
25. y − cos x
14. y − 2sx 2 1
| |
13. y −
28. A
variable star is one whose brightness alternately increases
and decreases. For the most visible variable star, Delta Cephei,
the time between periods of maximum brightness is 5.4 days,
the average brightness (or magnitude) of the star is 4.0, and its
brightness varies by 60.35 magnitude. Find a function that
models the brightness of Delta Cephei as a function of time.
29. Some
of the highest tides in the world occur in the Bay of
Fundy on the Atlantic Coast of Canada. At Hopewell Cape
the water depth at low tide is about 2.0 m and at high tide
it is about 12.0 m. The natural period of oscillation is
about 12 hours and on a particular day, high tide occurred
at 6:45 am. Find a function involving the cosine function
that models the water depth Dstd (in meters) as a function
of time t (in hours after midnight) on that day.
30. In
a normal respiratory cycle the volume of air that moves
into and out of the lungs is about 500 mL. The reserve and
residue volumes of air that remain in the lungs occupy about
2000 mL and a single respiratory cycle for an average human
takes about 4 seconds. Find a model for the total volume of
air Vstd in the lungs as a function of time.
| |
| |
| |
31. ( a)How is the graph of y − f ( x ) related to the graph of f ?
(b) Sketch the graph of y − sin x .
(c) Sketch the graph of y − s x .
32. Use
the given graph of f to sketch the graph of y − 1yf sxd.
Which features of f are the most important in sketching
y − 1yf sxd? Explain how they are used.
12. y − 1 2 x 3
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
33–34 Find (a) f 1 t, (b) f 2 t, (c) f t, and (d) fyt and state their
33. f sxd − s25 2 x 2,
34. f sxd −
tsxd − sx 1 1
tsxd −
35– 40 Find the functions (a) f 8 t, (b) t 8 f , (c) f 8 f , and (d) t 8 t
and their domains.
57. Use
the given graphs of f and t to evaluate each expression,
or explain why it is undefined.
(a) f s ts2dd
(b) ts f s0dd
(c) s f 8 tds0d
(d) s t 8 f ds6d
(e) s t 8 tds22d
(f) s f 8 f ds4d
tsxd − 2x 1 1
37. f sxd −
(b) s f + f + f ds1d
(d) s t + f + tds3d
tsxd − s
35. f sxd − x 3 1 5,
36. f sxd −
56. (a) ts ts ts2ddd
(c) s f + f + tds1d
tsxd − x 1 1
38. f sxd −
39. f sxd −
40. f sxd − s5 2 x ,
58. Use
the given graphs of f and t to estimate the value of
f s tsxdd for x − 25, 24, 23, . . . , 5. Use these estimates to
sketch a rough graph of f 8 t.
tsxd − sx 2 1
41– 44 Find f 8 t 8 h.
41. f sxd − 3x 2 2, tsxd − sin x,hsxd − x 2
42. f sxd − x 2 4 ,
tsxd − 2 x,hsxd − sx
tsxd − x 2,hsxd − x 3 1 2
tsxd −
,hsxd − s
43. f sxd − sx 2 3 ,
44. f sxd − tan x,
45. Fsxd − s2 x 1 x 2 d 4
46. Fsxd − cos2 x
48. Gsxd −
49. v std − secst 2 d tanst 2 d
51. Rsxd − ssx 2 1
50. Hsxd − s1 1 sx
51–54 Express the function in the form f 8 t 8 h.
| |
52. Hsxd − s
21 x
54. Hstd − cos (s tan t 1 1)
53. Sstd − sin scos td
55–56 Use the table to evaluate each expression.
f sxd
55. (a)f s ts3dd
(c) s f 8 tds5d
(b) ts f s2dd
(d) s t 8 f ds5d
45–50 Express the function in the form f 8 t.
47. Fsxd −
tsxd − 2x 2 1
tsxd − sin x
59. A
stone is dropped into a lake, creating a circular ripple that
travels outward at a speed of 60 cmys.
(a)Express the radius r of this circle as a function of the
time t (in seconds).
(b)If A is the area of this circle as a function of the radius,
find A 8 r and interpret it.
60. A
spherical balloon is being inflated and the radius of the
balloon is increasing at a rate of 2 cmys.
(a)Express the radius r of the balloon as a function of the
time t (in seconds).
(b)If V is the volume of the balloon as a function of the
radius, find V 8 r and interpret it.
61. A
ship is moving at a speed of 30 kmyh parallel to a straight
shoreline. The ship is 6 km from shore and it passes a lighthouse at noon.
(a)Express the distance s between the lighthouse and the
ship as a function of d, the distance the ship has traveled
since noon; that is, find f so that s − f sd d.
(b)Express d as a function of t, the time elapsed since noon;
that is, find t so that d − tstd.
(c)Find f 8 t. What does this function represent?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.4The Tangent and Velocity Problems
62. An
airplane is flying at a speed of 560 kmyh at an altitude
of two kilometers and passes directly over a radar station at
time t − 0.
(a)Express the horizontal distance d (in kilometers) that the
plane has flown as a function of t.
(b)Express the distance s between the plane and the radar
station as a function of d.
(c) Use composition to express s as a function of t.
63. The Heaviside Function The Heaviside function H is defined by
0 if t , 0
Hstd −
1 if t > 0
It is used in the study of electric circuits to represent the sudden surge of electric current, or voltage, when a switch is
instantaneously turned on.
(a) Sketch the graph of the Heaviside function.
(b)Sketch the graph of the voltage Vstd in a circuit if the
switch is turned on at time t − 0 and 120 volts are
applied instantaneously to the circuit. Write a formula
for Vstd in terms of Hstd.
(c)Sketch the graph of the voltage Vstd in a circuit if the
switch is turned on at time t − 5 seconds and 240 volts
are applied instantaneously to the circuit. Write a formula
for Vstd in terms of Hstd. (Note that starting at t − 5
corre­sponds to a translation.)
64. The Ramp Function The Heaviside function defined in
Exercise 63 can also be used to define the ramp function
y − ctHstd, which represents a gradual increase in voltage or
current in a circuit.
(a) Sketch the graph of the ramp function y − tHstd.
(b)Sketch the graph of the voltage Vstd in a circuit if the
switch is turned on at time t − 0 and the voltage is gradually increased to 120 volts over a 60-second time interval.
Write a formula for Vstd in terms of Hstd for t < 60.
(c)Sketch the graph of the voltage Vstd in a circuit if the
switch is turned on at time t − 7 seconds and the
voltage is gradually increased to 100 volts over a period
of 25 seconds. Write a formula for Vstd in terms of Hstd
for t < 32.
65. Let f and t be linear functions with equations
f sxd − m1 x 1 b1 and tsxd − m 2 x 1 b 2. Is f 8 t also a
linear function? If so, what is the slope of its graph?
66. I f you invest x dollars at 4% interest compounded annually,
then the amount Asxd of the investment after one year is
Asxd − 1.04x. Find A 8 A, A 8 A 8 A, and A 8 A 8 A 8 A. What
do these compositions represent? Find a formula for the composition of n copies of A.
67. (a)If tsxd − 2x 1 1 and hsxd − 4x 2 1 4x 1 7, find a function f such that f 8 t − h. (Think about what operations
you would have to perform on the formula for t to end up
with the formula for h.)
(b)If f sxd − 3x 1 5 and hsxd − 3x 2 1 3x 1 2, find a
function t such that f 8 t − h.
68. If f sxd − x 1 4 and hsxd − 4x 2 1, find a function t such
that t 8 f − h.
69. Suppose t is an even function and let h − f 8 t. Is h always an
even function?
70. Suppose t is an odd function and let h − f 8 t. Is h always an
odd function? What if f is odd? What if f is even?
f sxd be a function with domain R .
Show that Esxd − f sxd 1 f s2xd is an even function.
Show that Osxd − f sxd 2 f s2xd is an odd function.
Prove that every function f sxd can be written as a sum of
an even function and an odd function.
(d) Express the function f sxd − 2 x 1 sx 2 3d2 as a sum of
an even function and an odd function.
71. Let
1.4 The Tangent and Velocity Problems
In this section we see how limits arise when we attempt to find the tangent to a curve or
the velocity of an object.
■ The Tangent Problem
FIGURE 1
The word tangent is derived from the Latin word tangens, which means “touching.” We
can think of a tangent to a curve as a line that touches the curve and follows the same
direction as the curve at the point of contact. How can this idea be made precise?
For a circle we could simply follow Euclid and say that a tangent is a line that intersects the circle once and only once, as in Figure 1(a). For more complicated curves this
definition is inadequate. Figure l(b) shows a line that appears to be a tangent to the
curve C at point P, but it intersects C twice.
To be specific, let’s look at the problem of trying to find a tangent line to the parabola y − x 2 in the following example.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
EXAMPLE 1 Find an equation of the tangent line to the parabola y − x 2 at the
point Ps1, 1d.
Q {x, ≈}
P (1, 1)
SOLUTION We will be able to find an equation of the tangent line as soon as we
know its slope m. The difficulty is that we know only one point, P, on , whereas
we need two points to compute the slope. But observe that we can compute an
approximation to m by choosing a nearby point Qsx, x 2 d on the parabola (as in
Figure 2) and computing the slope mPQ of the secant line PQ. (A secant line, from
the Latin word secans, meaning cutting, is a line that cuts [intersects] a curve more
than once.)
We choose x ± 1 so that Q ± P. Then
mPQ −
FIGURE 2
x2 2 1
For instance, for the point Qs1.5, 2.25d we have
mPQ −
2.25 2 1
− 2.5
1.5 2 1
The following tables show the values of mPQ for several values of x close to 1. The
closer Q is to P, the closer x is to 1 and, it appears from the tables, the closer mPQ is
to 2. This suggests that the slope of the tangent line should be m − 2.
We say that the slope of the tangent line is the limit of the slopes of the secant lines,
and we express this symbolically by writing
lim mPQ − mandlim
Q lP
x2 2 1
Assuming that the slope of the tangent line is indeed 2, we use the point-slope form
of the equation of a line [ y 2 y1 − msx 2 x 1d, see Appendix B] to write the equation
of the tangent line through s1, 1d as
y 2 1 − 2sx 2 1dory − 2x 2 1
Figure 3 illustrates the limiting process that occurs in Example 1. As Q approaches P
along the parabola, the corresponding secant lines rotate about P and approach the tangent line .
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.4The Tangent and Velocity Problems
Q approaches P from the right
Q approachesy P from the right
Q approaches P from the left
Q approaches P from the left
FIGURE 3
Many functions that occur in the sciences are not described by explicit equations; they
are defined by experimental data. The next example shows how to estimate the slope of
the tangent line to the graph of such a function.
EXAMPLE 2 A pulse laser operates by storing charge on a capacitor and releasing it
suddenly when the laser is fired. The data in the table describe the charge Q remaining
on the capacitor (measured in coulombs) at time t (measured in seconds after the laser
is fired). Use the data to draw the graph of this function and estimate the slope of the
tangent line at the point where t − 0.04. (Note: The slope of the tangent line represents
the electric current flowing from the capacitor to the laser [measured in amperes].)
SOLUTION In Figure 4 we plot the given data and use these points to sketch a curve
that approximates the graph of the function.
Q (coulombs)
FIGURE 4
t (seconds)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Given the points Ps0.04, 6.703d and Rs0, 10d on the graph, we find that the slope of
the secant line PR is
mPR −
(0, 10)
(0.02, 8.187)
(0.06, 5.488)
(0.08, 4.493)
(0.1, 3.676)
10 2 6.703
− 282.425
0 2 0.04
The table at the left shows the results of similar calculations for the slopes of other
secant lines. From this table we would expect the slope of the tangent line at t − 0.04
to lie somewhere between 274.20 and 260.75. In fact, the average of the slopes of the
two closest secant lines is
2 s274.20
2 60.75d − 267.475
So, by this method, we estimate the slope of the tangent line to be about 267.5.
Another method is to draw an approximation to the tangent line at P and measure
the sides of the triangle ABC, as in Figure 5.
Q (coulombs)
FIGURE 5
The physical meaning of the answer
in Example 2 is that the electric
current flowing from the capacitor to
the laser after 0.04 seconds is about
265 amperes.
t (seconds)
This gives an estimate of the slope of the tangent line as
| AB | < 2 8.0 2 5.4 − 265.0
0.06 2 0.02
| BC |
■ The Velocity Problem
If you watch the speedometer of a car as you drive in city traffic, you see that the speed
doesn’t stay the same for very long; that is, the velocity of the car is not constant. We
assume from watching the speedometer that the car has a definite velocity at each
moment, but how is the “instantaneous” velocity defined?
Let’s consider the velocity problem: Find the instantaneous velocity of an object moving along a straight path at a specific time if the position of the object at any time is
known. In the next example, we investigate the velocity of a falling ball. Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any
freely falling body is proportional to the square of the time it has been falling. (This
model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted
by sstd and measured in meters, then (at the earth’s surface) Galileo’s observation is
expressed by the equation
sstd − 4.9t 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.4The Tangent and Velocity Problems
EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of
the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after
5 seconds.
SOLUTION The difficulty in finding the instantaneous velocity at 5 seconds is that
we are dealing with a single instant of time st − 5d, so no time interval is involved.
However, we can approximate the desired quantity by computing the average velocity
over the brief time interval of a tenth of a second from t − 5 to t − 5.1:
Steve Allen / Stockbyte / Getty Images
average velocity −
change in position
time elapsed
ss5.1d 2 ss5d
4.9s5.1d2 2 4.9s5d2
− 49.49 mys
The following table shows the results of similar calculations of the average velocity
over successively smaller time periods.
CN Tower in Toronto
Time interval
s=4.9t @
5 < t < 5.1
5 < t < 5.05
5 < t < 5.01
5 < t < 5.001
It appears that as we shorten the time period, the average velocity is becoming closer to
49 mys. The instantaneous velocity when t − 5 is defined to be the limiting value of
these average velocities over shorter and shorter time periods that start at t − 5. Thus it
appears that the (instantaneous) velocity after 5 seconds is 49 mys.
slope of secant line
average velocity
s=4.9t @
You may have the feeling that the calculations used in solving this problem are very
similar to those used earlier in this section to find tangents. In fact, there is a close connection between the tangent problem and the velocity problem. If we draw the graph of
the distance function of the ball (as in Figure 6) and we consider the points Ps5, 4.9s5d 2 d
and Qs5 1 h, 4.9s5 1 hd2 d on the graph, then the slope of the secant line PQ is
mPQ −
slope of tangent line
instantaneous velocity
FIGURE 6
Average velocity smysd
4.9s5 1 hd2 2 4.9s5d 2
s5 1 hd 2 5
which is the same as the average velocity over the time interval f5, 5 1 hg. Therefore the
velocity at time t − 5 (the limit of these average velocities as h approaches 0) must be
equal to the slope of the tangent line at P (the limit of the slopes of the secant lines).
Examples 1 and 3 show that in order to solve tangent and velocity problems we must
be able to find limits. After studying methods for computing limits in the next four sections, we will return to the problems of finding tangents and velocities in Chapter 2.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
1. A
tank holds 1000 liters of water, which drains from the
bottom of the tank in half an hour. The values in the table
show the volume V of water remaining in the tank (in liters)
after t minutes.
t smind
V sLd
(a)If P is the point s15, 250d on the graph of V, find the
slopes of the secant lines PQ when Q is the point on the
graph with t − 5, 10, 20, 25, and 30.
(b)Estimate the slope of the tangent line at P by averaging
the slopes of two secant lines.
(c)Use a graph of V to estimate the slope of the tangent line
at P. (This slope represents the rate at which the water is
flowing from the tank after 15 minutes.)
2. A
student bought a smartwatch that tracks the number of
steps she walks throughout the day. The table shows the number of steps recorded t minutes after 3:00 pm on the first day
she wore the watch.
t smind
(a) Find the slopes of the secant lines corresponding to the
given intervals of t. What do these slopes represent?
(i) [0, 40]
(ii) [10, 20]
(iii) [20, 30]
(b) Estimate the student’s walking pace, in steps per minute,
at 3:20 pm by averaging the slopes of two secant lines.
3. T
he point Ps2, 21d lies on the curve y − 1ys1 2 xd.
(a)If Q is the point sx, 1ys1 2 xdd, find the slope of the
secant line PQ (correct to six decimal places) for the
following values of x :
(i) 1.5
(ii) 1.9
(iii) 1.99
(iv) 1.999
(v) 2.5
(vi) 2.1
(vii) 2.01
(viii) 2.001
(b)Using the results of part (a), guess the value of the slope
of the tangent line to the curve at Ps2, 21d.
(c)Using the slope from part (b), find an equation of the
tangent line to the curve at Ps2, 21d.
4. T
he point Ps0.5, 0d lies on the curve y − cos x .
(a)If Q is the point s x, cos xd, find the slope of the secant
line PQ (correct to six decimal places) for the following
values of x :
(i) 0
(ii) 0.4
(iii) 0.49
(iv) 0.499
(v) 1
(vi) 0.6
(vii) 0.51
(viii) 0.501
(b)Using the results of part (a), guess the value of the slope
of the tangent line to the curve at Ps0.5, 0d.
(c)Using the slope from part (b), find an equation of the
tangent line to the curve at Ps0.5, 0d.
(d)Sketch the curve, two of the secant lines, and the tangent
5. T
he deck of a bridge is suspended 80 meters above a river. If
a pebble falls off the side of the bridge, the height, in meters,
of the pebble above the water surface after t seconds is given
by y − 80 2 4.9t 2.
(a)Find the average velocity of the pebble for the time
period beginning when t − 4 and lasting
(i) 0.1 seconds (ii) 0.05 seconds (iii) 0.01 seconds
(b) Estimate the instantaneous velocity of the pebble after
4 seconds.
6. I f a rock is thrown upward on the planet Mars with a velocity
of 10 mys, its height in meters t seconds later is given by
y − 10 t 2 1.86t 2.
(a) Find the average velocity over the given time intervals:
(i) [1, 2]
(ii) [1, 1.5]
(iii) [1, 1.1]
(iv) [1, 1.01]
(v) [1, 1.001]
(b) Estimate the instantaneous velocity when t − 1.
7. T
he table shows the position of a motorcyclist after acceler­
ating from rest.
t ssecondsd
s (meters)
6.3 14.2 24.1 38.0
(a)Find the average velocity for each time period:
(i) f2, 4g
(ii) f3, 4g
(iii) f4, 5g
(iv) f4, 6g
(b)Use the graph of s as a function of t to estimate the
instantaneous velocity when t − 3.
8. T
he displacement (in centimeters) of a particle moving back
and forth along a straight line is given by the equation of
motion s − 2 sin t 1 3 cos t, where t is measured in
(a)Find the average velocity during each time period:
(i) f1, 2g
(ii) f1, 1.1g
(iii) f1, 1.01g
(iv) f1, 1.001g
(b)Estimate the instantaneous velocity of the particle
when t − 1.
9. The point Ps1, 0d lies on the curve y − sins10yxd.
(a)If Q is the point sx, sins10yxdd, find the slope of the
secant line PQ (correct to four decimal places) for x − 2,
1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9. Do the
slopes appear to be approaching a limit?
(b)Use a graph of the curve to explain why the slopes of the
secant lines in part (a) are not close to the slope of the
tangent line at P.
(c)By choosing appropriate secant lines, estimate the slope
of the tangent line at P.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.5The Limit of a Function
1.5 The Limit of a Function
Having seen in the preceding section how limits arise when we want to find the tangent
to a curve or the velocity of an object, we now turn our attention to limits in general and
numerical and graphical methods for computing them.
■ Finding Limits Numerically and Graphically
Let’s investigate the behavior of the function f defined by f sxd − sx 2 1dysx 2 2 1d for
values of x near 1. The following table gives values of f sxd for values of x close to 1 but
not equal to 1.
f sxd
f sxd
approaches 0.5
From the table and the graph of f shown in Figure 1 we see that the closer x is to 1
(on either side of 1), the closer f sxd is to 0.5. In fact, it appears that we can make the
values of f sxd as close as we like to 0.5 by taking x sufficiently close to 1. We express
this by saying “the limit of the function f sxd − sx 2 1dysx 2 2 1d as x approaches 1 is
equal to 0.5.” The notation for this is
as x approaches 1
FIGURE 1
x l1
− 0.5
x2 2 1
In general, we use the following notation.
1 Intuitive Definition of a Limit Suppose f sxd is defined when x is near the
number a. (This means that f is defined on some open interval that contains a,
except possibly at a itself.) Then we write
lim f sxd − L
and say
“the limit of f sxd, as x approaches a, equals L”
if we can make the values of f sxd arbitrarily close to L (as close to L as we like)
by restricting x to be sufficiently close to a (on either side of a) but not equal to a.
Roughly speaking, this says that the values of f sxd approach L as x approaches a. In
other words, the values of f sxd tend to get closer and closer to the number L as x gets
closer and closer to the number a (from either side of a) but x ± a. (A more precise
definition will be given in Section 1.7.)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
An alternative notation for
lim f sxd − L
f sxd l Lasx l a
which is usually read “ f sxd approaches L as x approaches a.”
Notice the phrase “but x not equal to a” in the definition of limit. This means that in
find­ing the limit of f sxd as x approaches a, we never consider x − a. In fact, f sxd need
not even be defined when x − a. The only thing that matters is how f is defined near a.
Figure 2 shows the graphs of three functions. Note that in part (b), f sad is not defined
and in part (c), f sad ± L. But in each case, regardless of what happens at a, it is true
that lim x l a f sxd − L.
FIGURE 2 lim f sxd − L in all three cases
EXAMPLE 1 Estimate the value of lim
st 2 1 9 2 3
SOLUTION The table lists values of the function for several values of t near 0.
st 2 1 9 2 3
st 2 1 9 2 3
0.162277 . . .
0.165525 . . .
0.166620 . . .
0.166655 . . .
0.166666 . . .
As t approaches 0, the values of the function seem to approach 0.1666666 . . . and so
we guess that
st 2 1 9 2 3
− ■
In Example 1 what would have happened if we had taken even smaller values of t?
The table in the margin shows the results from one calculator; you can see that something
strange seems to be happening.
If you try these calculations on your own calculator you might get different values,
but eventually you will get the value 0 if you make t sufficiently small. Does this
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.5The Limit of a Function
For a further explanation of why
calculators sometimes give false
values, click on Lies My Calculator
and Computer Told Me. In particular, see the section called The Perils
of Subtraction.
mean that the answer is really 0 instead of 16 ? No, the value of the limit is 16 , as we will
show in the next section. The problem is that the calculator gave false values because
st 2 1 9 is very close to 3 when t is small. (In fact, when t is sufficiently small, a
calculator’s value for st 2 1 9 is 3.000 . . . to as many digits as the calculator is
capable of carrying.)
Something similar happens when we try to graph the function
f std −
st 2 1 9 2 3
of Example 1 on a graphing calculator or computer. Parts (a) and (b) of Figure 3
show quite accurate graphs of f , and if we trace along the curve, we can estimate
easily that the limit is about 16 . But if we zoom in too much, as in parts (c) and (d), then
we get inaccurate graphs, again due to rounding errors within the calculations.
(a) _5¯t¯5
(b) _0.1¯t¯0.1
(c) _10 _6¯t¯10 _6
(d) _10 _7¯t¯10 _7
FIGURE 3
EXAMPLE 2 Guess the value of lim
sin x
sin x
SOLUTION The function f sxd − ssin xdyx is not defined when x − 0. Using a calculator (and remembering that, if x [ R, sin x means the sine of the angle whose radian
measure is x), we construct a table of values correct to eight decimal places. From the
table at the left and the graph in Figure 4 we guess that
sin x
This guess is in fact correct, as will be proved in Chapter 2 using a geometric
FIGURE 4
sin x
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
EXAMPLE 3 Find lim x 3 1
cos 5x
SOLUTION As before, we construct a table of values. From the first table it appears that
the limit might be zero.
x3 1
cos 5x
x3 1
cos 5x
But if we persevere with smaller values of x, the second table suggests that the limit is
more likely to be 0.0001. In Section 1.8 we will be able to show that lim x l 0 cos 5x − 1
and that it follows that
cos 5x
lim x 3 1
− 0.0001
■ One-Sided Limits
The Heaviside function H is defined by
FIGURE 5
The Heaviside function
Hstd −
0 if t , 0
1 if t > 0
(This function is named after the electrical engineer Oliver Heaviside [1850 –1925] and
can be used to describe an electric current that is switched on at time t − 0.) Its graph is
shown in Figure 5.
There is no single number that Hstd approaches as t approaches 0, so lim t l 0 Hstd does
not exist. However, as t approaches 0 from the left, Hstd approaches 0. As t approaches 0
from the right, Hstd approaches 1. We indicate this situation symbolically by writing
lim Hstd − 0andlim1 Hstd − 1
t l 02
and we call these one-sided limits. The notation t l 0 2 indicates that we consider only
values of t that are less than 0. Likewise, t l 0 1 indicates that we consider only values
of t that are greater than 0.
2 Intuitive Definition of One-Sided Limits We write
lim f sxd − L
x la2
and say that the left-hand limit of f sxd as x approaches a [or the limit of f sxd as x
approaches a from the left] is equal to L if we can make the values of f sxd arbitrarily close to L by restricting x to be sufficiently close to a with x less than a.
We write
lim1 f sxd − L
x la
and say that the right-hand limit of f sxd as x approaches a [or the limit of f sxd as x
approaches a from the right] is equal to L if we can make the values of f sxd arbitrarily
close to L by restricting x to be sufficiently close to a with x greater than a.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.5The Limit of a Function
For instance, the notation x l 52 means that we consider only x , 5, and x l 51
means that we consider only x . 5. Definition 2 is illustrated in Figure 6.
FIGURE 6
(b) lim ƒ=L
(a) lim ƒ=L
x a+
x a_
Notice that Definition 2 differs from Definition 1 only in that we require x to be less
than (or greater than) a. By comparing these definitions, we see that the following is true.
lim f sxd − L if and only iflim2 f sxd − Landlim1 f sxd − L
x la
x la
EXAMPLE 4 The graph of a function t is shown in Figure 7.
FIGURE 7
Use the graph to state the values (if they exist) of the following:
(a) lim2 tsxd(b) lim1 tsxd(c) lim tsxd
(d) lim2 tsxd(e) lim1 tsxd(f ) lim tsxd
SOLUTION Looking at the graph we see that the values of tsxd approach 3 as x
approaches 2 from the left, but they approach 1 as x approaches 2 from the right.
(a) lim2 tsxd − 3and(b) lim1 tsxd − 1
(c) Since the left and right limits are different, we conclude from (3) that lim x l 2 tsxd
does not exist.
The graph also shows that
(d) lim2 tsxd − 2and(e) lim1 tsxd − 2
(f ) This time the left and right limits are the same and so, by (3), we have
lim tsxd − 2
Despite this fact, notice that ts5d ± 2.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
■ How Can a Limit Fail to Exist?
We have seen that a limit fails to exist at a number a if the left- and right-hand limits are
not equal (as in Example 4). The next two examples illustrate additional ways that a limit
can fail to exist.
EXAMPLE 5 Investigate lim sin
Limits and Technology
Some software applications, including computer algebra systems
(CAS), can compute limits. In order
to avoid the types of pitfalls demonstrated in Examples 1, 3, and 5, such
applications don’t find limits by
numerical experimentation. Instead,
they use more sophisticated techniques such as computing infinite
series. You are encouraged to use
one of these resources to compute
the limits in the examples of this
section and check your answers to
the exercises in this chapter.
SOLUTION Notice that the function f sxd − sinsyxd is undefined at 0. Evaluating the
function for some small values of x, we get
f ( 12 ) − sin 2 − 0
f ( 31) − sin 3 − 0 f ( 14) − sin 4 − 0
f s1d − sin − 0
f s0.1d − sin 10 − 0 f s0.01d − sin 100 − 0
Similarly, f s0.001d − f s0.0001d − 0. On the basis of this information we might
be tempted to guess that the limit is 0, but this time our guess is wrong. Note that
although f s1ynd − sin n − 0 for any integer n, it is also true that f sxd − 1 for
infinitely many values of x (such as 2y5 or 2y101) that approach 0. You can see this
from the graph of f shown in Figure 8.
FIGURE 8
The dashed lines near the y-axis indicate that the values of sinsyxd oscillate
between 1 and 21 infinitely often as x approaches 0.
Since the values of f sxd do not approach a fixed number as x approaches 0,
lim sin
does not exist
Examples 3 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is
easy to guess the wrong value if we use inappropriate values of x, but it is difficult to
know when to stop calculating values. And, as the discussion after Example 1 shows,
sometimes calculators and computers give the wrong values. In the next section, however, we will develop foolproof methods for calculating limits.
Another way a limit at a number a can fail to exist is when the function values grow
arbitrarily large (in absolute value) as x approaches a.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.5The Limit of a Function
EXAMPLE 6 Find lim
if it exists.
SOLUTION As x becomes close to 0, x 2 also becomes close to 0, and 1yx 2 becomes
very large. (See the following table.) In fact, it appears from the graph of the function
f sxd − 1yx 2 shown in Figure 9 that the values of f sxd can be made arbitrarily large by
taking x close enough to 0. Thus the values of f sxd do not approach a number, so
lim x l 0 s1yx 2 d does not exist.
y= 1
FIGURE 9
■ Infinite Limits; Vertical Asymptotes
To indicate the kind of behavior exhibited in Example 6, we use the notation
This does not mean that we are regarding ` as a number. Nor does it mean that the limit
exists. It simply expresses the particular way in which the limit does not exist: 1yx 2 can
be made as large as we like by taking x close enough to 0.
In general, we write symbolically
lim f sxd − `
to indicate that the values of f sxd tend to become larger and larger (or “increase without
bound”) as x becomes closer and closer to a.
4 Intuitive Definition of an Infinite Limit Let f be a function defined on
both sides of a, except possibly at a itself. Then
lim f sxd − `
means that the values of f sxd can be made arbitrarily large (as large as we please)
by taking x sufficiently close to a, but not equal to a.
Another notation for lim x l a f sxd − ` is
f sxd l `asx l a
FIGURE 10
lim f sxd − `
Again, the symbol ` is not a number, but the expression lim x l a f sxd − ` is often read as
“the limit of f sxd, as x approaches a, is infinity”
“ f sxd becomes infinite as x approaches a”
“ f sxd increases without bound as x approaches a”
This definition is illustrated graphically in Figure 10.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
A similar sort of limit, for functions that become large negative as x gets close to a, is
defined in Definition 5 and is illustrated in Figure 11.
When we say a number is “large
negative,” we mean that it is negative but its magnitude (absolute
value) is large.
5 Definition Let f be a function defined on both sides of a, except possibly at
a itself. Then
lim f sxd − 2`
means that the values of f sxd can be made arbitrarily large negative by taking x
sufficiently close to a, but not equal to a.
The symbol lim x l a f sxd − 2` can be read as “the limit of f sxd, as x approaches a, is
negative infinity” or “ f sxd decreases without bound as x approaches a.” As an example
we have
FIGURE 11
lim f sxd − 2`
S D
lim 2
− 2`
Similar definitions can be given for the one-sided infinite limits
lim f sxd − `
lim f sxd − `
x l a2
x l a1
lim f sxd − 2`
lim f sxd − 2`
x l a2
x l a1
remembering that x l a2 means that we consider only values of x that are less than a,
and similarly x l a1 means that we consider only x . a. Illustrations of these four
cases are given in Figure 12.
(a) lim ƒ=`
(b) lim ƒ=`
(c) lim ƒ=_`
(d) lim ƒ=_`
FIGURE 12
6 Definition The vertical line x − a is called a vertical asymptote of the
curve y − f sxd if at least one of the following statements is true:
lim f sxd − `
x la
lim f sxd − 2`
x la
lim f sxd − `
x l a2
lim f sxd − 2`
x l a2
lim f sxd − `
x l a1
lim f sxd − 2`
x l a1
For instance, the y-axis is a vertical asymptote of the curve y − 1yx 2 because
lim x l 0 s1yx 2 d − `. In Figure 12 the line x − a is a vertical asymptote in each of the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.5The Limit of a Function
four cases shown. In general, knowledge of vertical asymptotes is very useful in sketching graphs.
EXAMPLE 7 Does the curve y −
SOLUTION There is a potential vertical asymptote where the denominator is 0, that
is, at x − 3, so we investigate the one-sided limits there.
If x is close to 3 but larger than 3, then the denominator x 2 3 is a small positive
number and 2x is close to 6. So the quotient 2xysx 2 3d is a large positive number.
[For instance, if x − 3.01 then 2xysx 2 3d − 6.02y0.01 − 602.] Thus, intuitively, we
see that
x l3 x 2 3
Likewise, if x is close to 3 but smaller than 3, then x 2 3 is a small negative number
but 2x is still a positive number (close to 6). So 2xysx 2 3d is a numerically large
negative number. Thus
− 2`
x l 32 x 2 3
have a vertical asymptote?
The graph of the curve y − 2xysx 2 3d is given in Figure 13. According to Definition 6, the line x − 3 is a vertical asymptote.
FIGURE 13
NOTE Neither of the limits in Examples 6 and 7 exist, but in Example 6 we can write
lim x l 0 s1yx 2 d − ` because f sxd l ` as x approaches 0 from either the left or the
right. In Example 7, f sxd l ` as x approaches 3 from the right but f sxd l 2` as x
approaches 3 from the left, so we simply say that lim x l 3 f sxd does not exist.
EXAMPLE 8 Find the vertical asymptotes of f sxd − tan x.
SOLUTION Because
tan x −
there are potential vertical asymptotes where cos x − 0. In fact, since cos x l 01 as
x l sy2d2 and cos x l 02 as x l sy2d1, whereas sin x is positive (near 1) when x
is near y2, we have
3π _π
_ 2
y − tan x
tan x − `lim
tan x − 2`
x l sy2d2
FIGURE 14
sin x
cos x
x l sy2d1
This shows that the line x − y2 is a vertical asymptote. Similar reasoning shows
that the lines x − y2 1 n, where n is an integer, are all vertical asymptotes of
f sxd − tan x. The graph in Figure 14 confirms this.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
1. Explain in your own words what is meant by the equation
lim f sxd − 5
(g) lim hsxd
(h) hs0d
( i) lim hsxd
( j) hs2d
(k) lim1 hsxd
(l) lim2 hsxd
x l5
Is it possible for this statement to be true and yet f s2d − 3?
x l5
2. Explain what it means to say that
lim f sxd − 3andlim1 f sxd − 7
x l 12
x l1
In this situation is it possible that lim x l 1 f sxd exists?
3. Explain the meaning of each of the following.
(a) lim f sxd − `
4. U
se the given graph of f to state the value of each quantity,
if it exists. If it does not exist, explain why.
(a) lim2 f sxd
(b) lim1 f sxd
(c) lim f sxd
x l2
(d) f s2d
(e) lim f sxd
(f ) f s4d
7. F
or the function t whose graph is shown, find a number a that
satisfies the given description.
(a) lim tsxd does not exist but tsad is defined.
(b) lim tsxd exists but tsad is not defined.
(c) lim2 tsxd and lim1 tsxd both exist but lim tsxd does
not exist.
5. F
or the function f whose graph is given, state the value of
each quantity, if it exists. If it does not exist, explain why.
(a) lim f sxd
(b) lim2 f sxd
(c) lim1 f sxd
(d) lim f sxd
6 x
8. For the function A whose graph is shown, state the following.
(a) lim Asxd
(b) lim2 Asxd
(c) lim1 Asxd
(d) lim Asxd
x l23
x l2
x l2
x l21
(e) The equations of the vertical asymptotes
(e) f s3d
(d) lim1 tsxd − tsad but lim2 tsxd Þ tsad .
6. F
or the function h whose graph is given, state the value of
each quantity, if it exists. If it does not exist, explain why.
(a) lim 2 hsxd
(b) lim 1 hsxd
(c) lim hsxd
(d) hs23d
(b) lim1 f sxd − 2`
x l 23
x l 23
x l 23
x l 23
(e) lim2 hsxd
(f ) lim1 hsxd
xl 0
x l0
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.5The Limit of a Function
9. F
or the function f whose graph is shown, state the
(a) lim f sxd
(b) lim f sxd
(d) lim2 f sxd
(e) lim1 f sxd
x l27
(c) lim f sxd
x l23
15–18 Sketch the graph of an example of a function f that
satisfies all of the given conditions.
15. lim2 f sxd − 3,lim1 f sxd − 0,f s1d − 2
16. lim f sxd − 4,lim2 f sxd − 1,lim1 f sxd − 23,
(f ) The equations of the vertical asymptotes
f s0d − 6,f s8d − 21
x l 212
f sxd − 0,lim 1 f sxd − 1,lim f sxd − 3,
x l 21
f s21d − 2,f s2d − 1
x l 232
f sxd − 3,lim 1 f sxd − 2,lim2 f sxd − 21,
x l 23
lim1 f sxd − 2,f s23d − 2,f s3d − 0
10. A
patient receives a 150-mg injection of a drug every
4 hours. The graph shows the amount f std of the drug in the
blood­stream after t hours. Find
lim f stdandlim1 f std
t l 122
t l 12
19–22 Guess the value of the limit (if it exists) by evaluating the
function at the given numbers (correct to six decimal places).
x 2 2 3x
,x l3 x 2 2 9
x − 3.1, 3.05, 3.01, 3.001, 3.0001,
2.9, 2.95, 2.99, 2.999, 2.9999
19. lim
and explain the significance of these one-sided limits.
x 2 2 3x
x l 23 x 2 2 9
x − 22.5, 22.9, 22.95, 22.99, 22.999, 22.9999,
23.5, 23.1, 23.05, 23.01, 23.001, 23.0001
20. lim
sin x
x 1 tan x
x − 61, 60.5, 60.2, 60.1, 60.05, 60.01
21. lim
s2 1 hd5 2 32
hl 0
h − 60.5, 60.1, 60.01, 60.001, 60.0001
22. lim
11–12 Sketch the graph of the function and use it to determine
the values of a for which lim x l a f sxd exists.
cos x if x < 0
11. f sxd − 1 2 x if 0 , x , 1
if x > 1
25. lim1 x x
x l0
p l 21
26. lim
t l0
1 1 p9
1 1 p 15
5t 2 1
27–36 Determine the infinite limit.
; 13–14 Use the graph of the function f to state the value of each
limit, if it exists. If it does not exist, explain why.
(a) lim2 f sxd(b) lim1 f sxd(c) lim f sxd
13. f sxd − x s1 1 x 22
sin 3
tan 2
23. lim
if x < 21
12. f sxd − x
if 21 , x < 2
sx 2 1d if x . 2
23–26 Use a table of values to estimate the value of the limit.
If you have a graphing device, use it to confirm your result
14. f sxd −
31yx 2 2
31yx 1 1
29. lim
x l2
28. lim2
sx 2 2d2
30. lim2
sx 2 3d 5
x l51
x l 221
x 2sx 1 2d
x l5
x l3
32. lim
x 2sx 1 2d
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
sec x
x 2 1 2x
35. lim 2
x l 1 x 2 2x 1 1
42. (a)Evaluate the function
lim x cot x
x l 2
hsxd −
x 2 1 4x
36. lim2 2
x l 3 x 2 2x 2 3
for x − 1, 0.5, 0.1, 0.05, 0.01, and 0.005.
37. Find the vertical asymptote of the function
f sxd −
2x 1 4
x2 1 1
3x 2 2x 2
x l1
(b)Confirm your answer to part (a) by graphing the
39. Determine lim2
and lim1 3
x l1 x 2 1
x3 2 1
(a)by evaluating f sxd − 1ysx 3 2 1d for values of x that
approach 1 from the left and from the right,
(b) by reasoning as in Example 7, and
(c) from a graph of f.
se a graph to estimate the equations of all the vertical
; 43. U
asymptotes of the curve
y − tans2 sin xd
2 < x < Then find the exact equations of these asymptotes.
; 40. (a)By graphing the function
f sxd −
tan x 2 x
(c)Evaluate hsxd for successively smaller values of x until
you finally reach a value of 0 for hsxd. Are you still
confident that your guess in part (b) is correct? Explain
why you eventually obtained 0 values. (In Section 6.8 a
method for evaluating this limit will be explained.)
(d)Graph the function h in the viewing rectangle f21, 1g
by f0, 1g. Then zoom in toward the point where the
graph crosses the y-axis to estimate the limit of hsxd as
x approaches 0. Continue to zoom in until you observe
distortions in the graph of h. Compare with the results
of part (c).
(b) Guess the value of lim
38. (a) Find the vertical asymptotes of the function
tan x 2 x
cos 2x 2 cos x
and zooming in toward the point where the graph
crosses the y-axis, estimate the value of lim x l 0 f sxd.
(b)Check your answer in part (a) by evaluating f sxd for
values of x that approach 0.
44. Consider the function f sxd − tan
(a)Show that f sxd − 0 for x −
(b)Show that f sxd − 1 for x −
41. (a)Evaluate the function
f sxd − x 2
for x − 1, 0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the
value of
x2 2
(b)Evaluate f sxd for x − 0.04, 0.02, 0.01, 0.005, 0.003,
and 0.001. Guess again.
(c) What can you conclude about lim1 tan
45. I n the theory of relativity, the mass of a particle with
velocity v is
s1 2 v 2yc 2
where m 0 is the mass of the particle at rest and c is the
speed of light. What happens as v l c2?
1.6 Calculating Limits Using the Limit Laws
■ Properties of Limits
In Section 1.5 we used calculators and graphs to guess the values of limits, but we saw
that such methods don’t always lead to the correct answers. In this section we use the
following properties of limits, called the Limit Laws, to calculate limits.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.6Calculating Limits Using the Limit Laws
Limit Laws Suppose that c is a constant and the limits
lim f sxdandlim tsxd
exist. Then
1. lim f f sxd 1 tsxdg − lim f sxd 1 lim tsxd
2. lim f f sxd 2 tsxdg − lim f sxd 2 lim tsxd
3. lim fcf sxdg − c lim f sxd
4. lim f f sxd tsxdg − lim f sxd ? lim tsxd
lim f sxd
5. lim
f sxd
lim tsxd
if lim tsxd ± 0
These five laws can be stated verbally as follows:
Sum Law
Difference Law
Constant Multiple Law
Product Law
Quotient Law
1. The limit of a sum is the sum of the limits.
2. The limit of a difference is the difference of the limits.
3. The limit of a constant times a function is the constant times the limit of the
4. The limit of a product is the product of the limits.
5. The limit of a quotient is the quotient of the limits (provided that the limit of
the denominator is not 0).
It is easy to believe that these properties are true. For instance, if f sxd is close to L and
tsxd is close to M, it is reasonable to conclude that f sxd 1 tsxd is close to L 1 M. This
gives us an intuitive basis for believing that Law 1 is true. In Section 1.7 we give a precise definition of a limit and use it to prove this law. The proofs of the remaining laws are
given in Appendix F.
EXAMPLE 1 Use the Limit Laws and the graphs of f and t in Figure 1 to evaluate the
following limits, if they exist.
(a) lim f f sxd 1 5tsxdg(b) lim f f sxdtsxdg(c) lim
x l 22
f sxd
(a) From the graphs of f and t we see that
lim f sxd − 1andlim tsxd − 21
x l 22
FIGURE 1
x l 22
Therefore we have
lim f f sxd 1 5tsxdg − lim f sxd 1 lim f5tsxdg(by Limit Law 1)
x l 22
x l 22
x l 22
− lim f sxd 1 5 lim tsxd (by Limit Law 3)
x l 22
x l 22
− 1 1 5s21d − 24
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
(b) We see that lim x l 1 f sxd − 2. But lim x l 1 tsxd does not exist because the left and
right limits are different:
lim tsxd − 22lim1 tsxd − 21
x l 12
So we can’t use Law 4 for the desired limit. But we can use Law 4 for the one-sided
lim f f sxd tsxdg − lim2 f sxd lim2 tsxd − 2 s22d − 24
x l 12
x l1
x l1
lim f f sxd tsxdg − lim1 f sxd lim1 tsxd − 2 s21d − 22
x l 11
x l1
x l1
The left and right limits aren’t equal, so lim x l 1 f f sxd tsxdg does not exist.
(c) The graphs show that
lim f sxd < 1.4andlim tsxd − 0
Because the limit of the denominator is 0, we can’t use Law 5. The given limit does
not exist because the denominator approaches 0 while the numerator approaches a
nonzero number.
If we use the Product Law repeatedly with tsxd − f sxd, we obtain the following law.
Power Law
f x la
6. lim f f sxdg n − lim f sxd nwhere n is a positive integer
x la
A similar property, which you are asked to prove in Exercise 1.8.69, holds for roots:
Root Law
n lim f sxd
f sxd − s
7. lim s
where n is a positive integer
x la
x la
f sxd . 0.g
fIf n is even, we assume that xlim
In applying these seven limit laws, we need to use two special limits:
8. lim c − c
9. lim x − a
These limits are obvious from an intuitive point of view (state them in words or draw
graphs of y − c and y − x), but proofs based on the precise definition are requested in
Exercises 1.7.23–24.
If we now put f sxd − x in Law 6 and use Law 9, we get a useful special limit for
power functions.
10. lim x n − a nwhere n is a positive integer
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.6Calculating Limits Using the Limit Laws
Newton and Limits
Isaac Newton was born on Christmas
Day in 1642, the year of Galileo’s
death. When he entered Cambridge
University in 1661 Newton didn’t
know much mathematics, but he
learned quickly by reading Euclid and
Descartes and by attending the lec­
tures of Isaac Barrow. Cam­bridge was
closed because of the plague from
1665 to 1666, and Newton returned
home to reflect on what he had
learned. Those two years were amaz­
ingly productive for at that time he
made four of his major discoveries:
(1) his repre­senta­tion of functions as
sums of infinite series, including the
binomial theorem; (2) his work on dif­
ferential and integral calculus; (3) his
laws of motion and law of universal
gravitation; and (4) his prism experi­
ments on the nature of light and color.
Because of a fear of controversy and
criticism, he was reluctant to publish
his discoveries and it wasn’t until
1687, at the urging of the astronomer
Halley, that Newton published Principia Mathematica. In this work, the
great­est scientific treatise ever writ­
ten, Newton set forth his version of
calculus and used it to investigate
mechanics, fluid dynamics, and wave
motion, and to explain the motion
of planets and comets.
The beginnings of calculus are
found in the calculations of areas and
volumes by ancient Greek scholars
such as Eudoxus and Archimedes.
Although aspects of the idea of a
limit are implicit in their “method of
exhaustion,” Eudoxus and Archimedes
never explicitly formulated the con­
cept of a limit. Like­wise, mathemati­
cians such as Cavalieri, Fermat, and
Barrow, the immediate precursors of
Newton in the development of calcu­
lus, did not actually use limits. It was
Isaac Newton who was the first to talk
explicitly about limits. He explained
that the main idea behind limits is
that quantities “approach nearer than
by any given difference.” Newton
stated that the limit was the basic
concept in calculus, but it was left to
later mathe­maticians like Cauchy to
clarify his ideas about limits.
If we put f sxd − x in Law 7 and use Law 9, we get a similar special limit for roots.
(For square roots the proof is outlined in Exercise 1.7.37.)
x −s
a where n is a positive integer
11. lim s
(If n is even, we assume that a . 0.)
EXAMPLE 2 Evaluate the following limits and justify each step.
(b) lim
(a) lim s2x 2 2 3x 1 4d
x l5
x l 22
x 3 1 2x 2 2 1
5 2 3x
(a) lim s2x 2 2 3x 1 4d − lim s2x 2 d 2 lim s3xd 1 lim 4(by Laws 2 and 1)
x l5
x l5
x l5
x l5
− 2 lim x 2 2 3 lim x 1 lim 4 (by 3)
x l5
x l5
x l5
− 2s5 2 d 2 3s5d 1 4
(by 10, 9, and 8)
− 39
(b) We start by using Law 5, but its use is fully justified only at the final stage when we
see that the limits of the numerator and denominator exist and the limit of the denominator is not 0.
lim sx 3 1 2x 2 2 1d
x 3 1 2x 2 2 1
x l 22
5 2 3x
x l 22
x l 22
lim x 3 1 2 lim x 2 2 lim 1
x l 22
x l 22
lim 5 2 3 lim x
x l 22
(by Law 5)
lim s5 2 3xd
x l 22
x l 22
s22d3 1 2s22d2 2 1
5 2 3s22d
(by 1, 2, and 3)
(by 10, 9, and 8)
■ Evaluating Limits by Direct Substitution
In Example 2(a) we determined that lim x l 5 f sxd − 39, where f sxd − 2x 2 2 3x 1 4.
Notice that f s5d − 39; in other words, we would have gotten the correct result simply by
substituting 5 for x. Similarly, direct substitution provides the correct answer in part (b).
The functions in Example 2 are a polynomial and a rational function, respectively, and
similar use of the Limit Laws proves that direct substitution always works for such functions (see Exercises 59 and 60). We state this fact as follows.
Direct Substitution Property If f is a polynomial or a rational function and a is
in the domain of f , then
lim f sxd − f sad
x la
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Functions that have the Direct Substitution Property are called continuous at a and
will be studied in Section 1.8. However, not all limits can be evaluated initially by direct
substitution, as the following examples show.
EXAMPLE 3 Find lim
x2 2 1
SOLUTION Let f sxd − sx 2 2 1dysx 2 1d. We can’t find the limit by substituting x − 1
because f s1d isn’t defined. Nor can we apply the Quotient Law, because the limit of
the denominator is 0. Instead, we need to do some preliminary algebra. We factor the
numerator as a difference of squares:
x2 2 1
sx 2 1dsx 1 1d
Notice that in Example 3 we do not
have an infinite limit even though the
denominator approaches 0 as x l 1.
When both numerator and denominator approach 0, the limit may be infinite or it may be some finite value.
The numerator and denominator have a common factor of x 2 1. When we take the
limit as x approaches 1, we have x ± 1 and so x 2 1 ± 0. Therefore we can cancel the
common factor, x 2 1, and then compute the limit by direct substitution as follows:
x2 2 1
sx 2 1dsx 1 1d
− lim
− lim sx 1 1d − 1 1 1 − 2
The limit in this example arose in Example 1.4.1 in finding the tangent to the parabola
y − x 2 at the point s1, 1d.
NOTE In Example 3 we were able to compute the limit by replacing the given function
f sxd − sx 2 2 1dysx 2 1d by a simpler function, tsxd − x 1 1, that has the same limit.
This is valid because f sxd − tsxd except when x − 1, and in computing a limit as x
approaches 1 we don’t consider what happens when x is actually equal to 1. In general,
we have the following useful fact.
If f sxd − tsxd when x ± a, then lim f sxd − lim tsxd, provided the limits exist.
x la
x l1
EXAMPLE 4 Find lim tsxd where
FIGURE 2
The graphs of the functions f (from
Example 3) and t (from Example 4)
tsxd −
x 1 1 if x ± 1
if x − 1
SOLUTION Here t is defined at x − 1 and ts1d − , but the value of a limit as x
approaches 1 does not depend on the value of the function at 1. Since tsxd − x 1 1
for x ± 1, we have
lim tsxd − lim sx 1 1d − 2
Note that the values of the functions in Examples 3 and 4 are identical except when
x − 1 (see Figure 2) and so they have the same limit as x approaches 1.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.6Calculating Limits Using the Limit Laws
EXAMPLE 5 Evaluate lim
s3 1 hd2 2 9
SOLUTION If we define
Fshd −
s3 1 hd2 2 9
then, as in Example 3, we can’t compute lim h l 0 Fshd by letting h − 0 because Fs0d is
undefined. But if we simplify Fshd algebraically, we find that
Fshd −
s9 1 6h 1 h 2 d 2 9
6h 1 h 2
hs6 1 hd
(Recall that we consider only h ± 0 when letting h approach 0.) Thus
EXAMPLE 6 Find lim
s3 1 hd2 2 9
− lim s6 1 hd − 6
st 2 1 9 2 3
SOLUTION We can’t apply the Quotient Law immediately because the limit of the
denominator is 0. Here the preliminary algebra consists of rationalizing the numerator:
st 2 1 9 2 3
st 2 1 9 2 3 st 2 1 9 1 3
− lim
st 2 1 9 1 3
− lim
st 2 1 9d 2 9
t 2 (st 2 1 9 1 3)
− lim
t 2 (st 2 1 9 1 3)
− lim
st 2 1 9 1 3
(Here we use several properties
slim st 1 9d 1 3
of limits: 5, 1, 7, 8, 10.)
t l0
This calculation confirms the guess that we made in Example 1.5.1.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
■ Using One-Sided Limits
Some limits are best calculated by first finding the left- and right-hand limits. The following theorem is a reminder of what we discovered in Section 1.5. It says that a twosided limit exists if and only if both of the one-sided limits exist and are equal.
1 Theorem lim f sxd − Lif and only iflim2 f sxd − L − lim1 f sxd
x la
x la
When computing one-sided limits, we use the fact that the Limit Laws also hold for
one-sided limits.
EXAMPLE 7 Show that lim | x | − 0.
SOLUTION Recall that
The result of Example 7 looks plausible
from Figure 3.
|x| −
| |
if x > 0
2x if x , 0
Since x − x for x . 0, we have
| |
lim x − lim1 x − 0
x l 01
y=| x|
| |
x l0
For x , 0 we have x − 2x and so
| |
lim x − lim2 s2xd − 0
x l 02
x l0
Therefore, by Theorem 1,
| |
lim x − 0
FIGURE 3
EXAMPLE 8 Prove that lim
does not exist.
| |
| |
SOLUTION Using the facts that x − x when x . 0 and x − 2x when x , 0,
we have
|x |
FIGURE 4
|x| −
|x| −
x l 01
x l 02
− lim1 1 − 1
x l0
− lim2 s21d − 21
x l0
x l 01
x l 02
Since the right- and left-hand limits are different, it follows from Theorem 1 that
lim x l 0 x yx does not exist. The graph of the function f sxd − x yx is shown in
Figure 4 and supports the one-sided limits that we found.
| |
| |
EXAMPLE 9 If
f sxd −
sx 2 4
8 2 2x
if x . 4
if x , 4
determine whether lim x l 4 f sxd exists.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.6Calculating Limits Using the Limit Laws
SOLUTION Since f sxd − sx 2 4 for x . 4, we have
It is shown in Example 1.7.4 that
lim x l 01 sx − 0.
lim f sxd − lim1 s x 2 4 − s4 2 4 − 0
x l 41
x l4
Since f sxd − 8 2 2x for x , 4, we have
lim f sxd − lim2 s8 2 2xd − 8 2 2 ? 4 − 0
x l 42
x l4
The right- and left-hand limits are equal. Thus the limit exists and
lim f sxd − 0
FIGURE 5
The graph of f is shown in Figure 5.
Other notations for v x b are fxg and
EXAMPLE 10 The greatest integer function is defined by v x b − the largest integer
that is less than or equal to x. (For instance, v 4 b − 4, v4.8b − 4, v b − 3, vs2 b − 1,
v212 b − 21.) Show that lim x l3 v x b does not exist.
:x;. The greatest integer function is
sometimes called the floor function.
SOLUTION The graph of the greatest integer function is shown in Figure 6. Since
v x b − 3 for 3 < x , 4, we have
lim v x b − lim1 3 − 3
x l 31
y=[ x]
x l3
Since v x b − 2 for 2 < x , 3, we have
lim v x b − lim2 2 − 2
x l 32
x l3
Because these one-sided limits are not equal, lim xl3 v x b does not exist by Theorem 1. ■
■ The Squeeze Theorem
FIGURE 6
The following two theorems describe how the limits of functions are related when the
values of one function are greater than (or equal to) those of another. Their proofs can be
found in Appendix F.
Greatest integer function
2 Theorem If f sxd < tsxd when x is near a (except possibly at a) and the limits
of f and t both exist as x approaches a, then
lim f sxd < lim tsxd
FIGURE 7
3 The Squeeze Theorem If f sxd < tsxd < hsxd when x is near a (except
possibly at a) and
lim f sxd − lim hsxd − L
lim tsxd − L
The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the
Pinching Theorem, is illustrated by Figure 7. It says that if tsxd is squeezed between f sxd
and hsxd near a, and if f and h have the same limit L at a, then t is forced to have the same
limit L at a.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
EXAMPLE 11 Show that lim x 2 sin
− 0.
SOLUTION First note that we cannot rewrite the limit as the product of the
limits lim x l 0 x 2 and lim x l 0 sins1yxd because lim x l 0 sins1yxd does not exist (see
Example 1.5.5).
We can find the limit by using the Squeeze Theorem. To apply the Squeeze Theorem
we need to find a function f smaller than tsxd − x 2 sins1yxd and a function h bigger
than t such that both f sxd and hsxd approach 0 as x l 0. To do this we use our knowledge of the sine function. Because the sine of any number lies between 21 and 1, we
can write
21 < sin
Any inequality remains true when multiplied by a positive number. We know that
x 2 > 0 for all x and so, multiplying each side of the inequalities in (4) by x 2, we get
2x 2 < x 2 sin
< x2
as illustrated by Figure 8. We know that
lim x 2 − 0andlim s2x 2 d − 0
Taking f sxd − 2x 2, tsxd − x 2 sins1yxd, and hsxd − x 2 in the Squeeze Theorem, we
FIGURE 8
y − x sins1yxd
x l2
xl 2
(a) lim f f sxd 1 5tsxdg
(c) lim sf sxd
3f sxd
(f ) lim
tsxd hsxd
f sxd
2. T
he graphs of f and t are given. Use them to evaluate each
limit, if it exists. If the limit does not exist, explain why.
(a) lim f f sxd 1 tsxdg
x l2
(c) lim f f sxd tsxdg
x l 21
(b) lim f f sxd 2 tsxdg
x l3
f sxd
3–9 Evaluate the limit and justify each step by indicating the
appropriate Limit Law(s).
3. lim s4x 2 2 5xd
4. lim s2x 3 1 6x 2 2 9d
5. lim sv 2 1 2vds2v 3 2 5d
6. lim
7. lim s9 2 u 3 1 2u 2
8. lim s
x 1 5 s2x 2 2 3xd
x l5
xl 23
v l2
u l 22
(d) lim
(b) lim f tsxdg 3
(d) lim
x l 21
find the limits that exist. If the limit does not exist, explain
x l2
(f ) f s21d 1 lim tsxd
(e) lim fx 2 f sxdg
lim f sxd − 4lim tsxd − 22lim hsxd − 0
1. Given that
(e) lim
− 0
lim x 2 sin
9. lim
tl 21
2t 5 2 t 4
5t 2 1 4
3t 2 1 1
t 2 2 5t 1 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.6Calculating Limits Using the Limit Laws
; 36. (a) Use a graph of
10. (a) What is wrong with the following equation?
x2 1 x 2 6
f sxd −
to estimate the value of lim x l 0 f sxd to two decimal
(b)Use a table of values of f sxd to estimate the limit to
four decimal places.
(c)Use the Limit Laws to find the exact value of the limit.
(b) In view of part (a), explain why the equation
x2 1 x 2 6
− lim sx 1 3d
is correct.
; 37. Use the Squeeze Theorem to show that
11–34 Evaluate the limit, if it exists.
13. lim
t 2 2 2t 2 8
14. lim
15. lim
x 2 1 5x 1 4
16. lim
x l2
17. lim
x l 22
x l 23
x2 2 x 2 6
3x 2 1 5x 2 2
x l 25
21. lim
sh 2 3d 2 9
22. lim
s9 1 h 2 3
24. lim
u l 21
hl 0
23. lim
25. lim
x l3 x 2 3
26. lim
; 38. Use the Squeeze Theorem to show that
x 2 1 3x
x 2 x 2 12
18. lim
t 3 2 27
t2 2 9
Illustrate by graphing the functions f sxd − 2x 2,
tsxd − x 2 cos 20x, and hsxd − x 2 on the same screen.
x 2 1 3x
x 2 x 2 12
19. lim
xl 6
xl 22
lim x 2 cos 20x − 0
12. lim (8 2 12 x)
11. lim s3x 2 7d
s3 1 x 2 s3
lim sx 3 1 x 2 sin
2x 2 1 9x 2 5
x 2 2 25
Illustrate by graphing the functions f, t, and h (in the notation of the Squeeze Theorem) on the same screen.
u3 1 1
39. If 4x 2 9 < f sxd < x 2 2 4x 1 7 for x > 0, find lim f sxd.
40. If 2x < tsxd < x 4 2 x 2 1 2 for all x, evaluate lim tsxd.
3 2 sx
41. Prove that lim x cos − 0.
xl 0
sx 1 2 2 2
42. Prove that lim1 sx f1 1 sin 2s2yxdg − 0.
x l0
s22 1 hd21 1 2 21
43–48 Find the limit, if it exists. If the limit does not exist,
explain why.
43. lim
(| x 1 4 | 2 2x)
2x 2 1
2x 3 2 x 2
| x 1 4|
s1 1 t 2 s1 2 t
27. lim
28. lim
4 2 sx
29. lim
x l 16 16x 2 x 2
x 2 2 4x 1 4
30. lim 4
x l 2 x 2 3x 2 2 4
sx 2 1 9 2 5
32. lim
x l24
47. lim2
sx 1 hd2
34. lim
49. The Signum Function The signum (or sign) function,
denoted by sgn, is defined by
31. lim
t s1 1 t
sx 1 hd3 2 x 3
33. lim
; 35. (a) Estimate the value of
x l0
t 1t
x l 24
x l 0.52
xl 0
| |D
sgn x −
s1 1 3x 2 1
by graphing the function f sxd − xyss1 1 3x 2 1d.
(b)Make a table of values of f sxd for x close to 0 and guess
the value of the limit.
(c)Use the Limit Laws to prove that your guess is correct.
x l 24
x l 22
48. lim1
xl 0
2x 1 8
| |
22 x
| |
21 if x , 0
20 if x − 0
21 if x . 0
(a) Sketch the graph of this function.
(b)Find each of the following limits or explain why it does
not exist.
(i) lim1 sgn x
(ii) lim2 sgn x
(iii) lim sgn x
xl 0
(iv) lim sgn x
xl 0
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
50. Let tsxd − sgnssin xd.
(a)Find each of the following limits or explain why it does
not exist.
(i) lim1 tsxd
(ii) lim2 tsxd
(iii) lim tsxd
(iv) lim1 tsxd
(v) lim2 tsxd
(vi) lim tsxd
(b)For which values of a does lim x l a tsxd not exist?
(c) Sketch a graph of t.
(ii) lim2 tsxd
x l2
61. If lim
x l2
(b) Does lim x l 2 tsxd exist?
(c) Sketch the graph of t.
52. Let
f sxd −
Bstd −
f sxd
− 5, find the following limits.
f sxd
(a) lim f sxd
(b) lim
x2 1 1
if x , 1
sx 2 2d 2 if x > 1
63. If
4 2 12 t
if t , 2
st 1 c
if t > 2
f sxd −
tsxd −
2 2 x2
65. S
how by means of an example that lim x l a f f sxd tsxdg may
exist even though neither lim x l a f sxd nor lim x l a tsxd exists.
66. Evaluate lim
s6 2 x 2 2
s3 2 x 2 1
67. Is there a number a such that
(iv) lim2 tsxd
x l2
(b) Sketch the graph of t.
(v) lim1 tsxd
(vi) lim tsxd
55. (a)If the symbol v b denotes the greatest integer function
defined in Example 10, evaluate
(i) lim 1 v x b
(ii) lim v x b
(iii) lim v x b
x l 22
x l 22
x l 22.4
(b) If n is an integer, evaluate
(i) lim2 v x b
(ii) lim1 v x b
x ln
x l 22
68. T
he figure shows a fixed circle C1 with equation
sx 2 1d2 1 y 2 − 1 and a shrinking circle C2 with radius r
and center the origin. P is the point s0, rd, Q is the upper point
of intersection of the two circles, and R is the point of intersection of the line PQ and the x-axis. What happens to R
as C2 shrinks, that is, as r l 0 1?
x l sy2d1
f sxd
x l sy2d2
f sxd
(iv) lim f sxd
x l y2
(c) For what values of a does lim x l a f sxd exist?
3x 2 1 ax 1 a 1 3
x2 1 x 2 2
exists? If so, find the value of a and the value of the limit.
(c) For what values of a does lim x l a v x b exist?
56. Let f sxd − v cos x b , 2 < x < .
(a) Sketch the graph of f.
(b) Evaluate each limit, if it exists.
(i) lim f sxd
x 2 if x is rational
0 if x is irrational
64. S
how by means of an example that lim x l a f f sxd 1 tsxdg may
exist even though neither lim x l a f sxd nor lim x l a tsxd exists.
(a) Evaluate each of the following, if it exists.
(i) lim2 tsxd
(ii) lim tsxd
(iii) ts1d
x l1
prove that lim x l 0 f sxd − 0.
Find the value of c so that lim t l2 Bstd exists.
54. Let
f sxd 2 8
− 10, find lim f sxd.
62. If lim
(a) Find lim x l12 f sxd and lim x l11 f sxd.
(b) Does lim x l1 f sxd exist?
(c) Sketch the graph of f.
53. Let
expresses the length L of an object as a function of its velocity v with respect to an observer, where L 0 is the length of the
object at rest and c is the speed of light. Find lim v l c2 L and
interpret the result. Why is a left-hand limit necessary?
60. If r is a rational function, use Exercise 59 to show that
lim x l a rsxd − rsad for every number a in the domain of r.
(a) Find
(i) lim1 tsxd
L − L 0 s1 2 v 2yc 2
59. If p is a polynomial, show that lim xl a psxd − psad.
x2 1 x 2 6
51. Let tsxd −
58. In the theory of relativity, the Lorentz contraction formula
57. If f sxd − v x b 1 v 2x b , show that lim x l 2 f sxd exists but is
not equal to f s2d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.7The Precise Definition of a Limit
1.7 The Precise Definition of a Limit
The intuitive definition of a limit given in Section 1.5 is inadequate for some purposes
because such phrases as “x is close to 2” and “ f sxd gets closer and closer to L” are vague.
In order to be able to prove conclusively that
x3 1
cos 5x
− 0.0001orlim
sin x
we must make the definition of a limit precise.
■ The Precise Definition of a Limit
To motivate the precise definition of a limit, let’s consider the function
f sxd −
2x 2 1 if x ± 3
if x − 3
Intuitively, it is clear that when x is close to 3 but x ± 3, then f sxd is close to 5, and so
lim x l3 f sxd − 5.
To obtain more detailed information about how f sxd varies when x is close to 3, we
ask the following question:
How close to 3 does x have to be so that f sxd differs from 5 by less than 0.l ?
The distance from x to 3 is x 2 3 and the distance from f sxd to 5 is f sxd 2 5 , so our
problem is to find a number (the Greek letter delta) such that
| f sxd 2 5 | , 0.1if| x 2 3 | , but x ± 3
If x 2 3 . 0, then x ± 3, so an equivalent formulation of our problem is to find a
number such that
| f sxd 2 5 | , 0.1if0 , | x 2 3 | , Notice that if 0 , | x 2 3 | , s0.1dy2 − 0.05, then
| f sxd 2 5 | − | s2x 2 1d 2 5 | − | 2x 2 6 | − 2| x 2 3 | , 2s0.05d − 0.1
that is,
| f sxd 2 5 | , 0.1if0 , | x 2 3 | , 0.05
Thus an answer to the problem is given by − 0.05; that is, if x is within a distance of
0.05 from 3, then f sxd will be within a distance of 0.1 from 5.
If we change the number 0.l in our problem to the smaller number 0.01, then by using
the same method we find that f sxd will differ from 5 by less than 0.01 provided that x
differs from 3 by less than (0.01)y2 − 0.005:
| f sxd 2 5 | , 0.01if0 , | x 2 3 | , 0.005
| f sxd 2 5 | , 0.001if0 , | x 2 3 | , 0.0005
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
The numbers 0.1, 0.01, and 0.001 that we have considered are error tolerances that we
might allow. For 5 to be the precise limit of f sxd as x approaches 3, we must not only be
able to bring the difference between f sxd and 5 below each of these three numbers; we
must be able to bring it below any positive number. And, by the same reasoning, we can!
If we write « (the Greek letter epsilon) for an arbitrary positive number, then we find as
before that
| f sxd 2 5 | , «if0 , | x 2 3 | , − 2
is in
This is a precise way of saying that f sxd is close to 5 when x is close to 3 because (1) says
that we can make the values of f sxd within an arbitrary distance « from 5 by restricting
the val­ues of x to be within a distance «y2 from 3 (but x ± 3).
Note that (1) can be rewritten as follows:
if3 2 , x , 3 1 sx ± 3dthen5 2 « , f sxd , 5 1 «
and this is illustrated in Figure 1. By taking the values of x (± 3) to lie in the interval
s3 2 , 3 1 d we can make the values of f sxd lie in the interval s5 2 «, 5 1 «d.
Using (1) as a model, we give a precise definition of a limit.
when x is in here
FIGURE 1
2 Precise Definition of a Limit Let f be a function defined on some open
interval that contains the number a, except possibly at a itself. Then we say that
the limit of f sxd as x approaches a is L, and we write
lim f sxd − L
It is traditional to use the Greek
letters « and in the precise
definition of a limit.
if for every number « . 0 there is a number . 0 such that
if0 , x 2 a , thenf sxd 2 L , «
Since x 2 a is the distance from x to a and f sxd 2 L is the distance from f sxd to
L, and since « can be arbitrarily small, the definition of a limit can be expressed in words
as follows:
lim x l a f sxd − L means that the distance between f sxd and L can be made arbitrarily small by
requiring that the distance from x to a be sufficiently small (but not 0).
lim x l a f sxd − L means that the values of f sxd can be made as close as we please to L
by requiring x to be close enough to a (but not equal to a).
We can also reformulate Definition 2 in terms of intervals by observing that the inequality x 2 a , is equivalent to 2 , x 2 a , , which in turn can be written as
a 2 , x , a 1 . Also 0 , x 2 a is true if and only if x 2 a ± 0, that is, x ± a.
Similarly, the inequality f sxd 2 L , « is equivalent to the pair of inequalities
L 2 « , f sxd , L 1 «. Therefore, in terms of intervals, Definition 2 can be stated as
lim x l a f sxd − L means that for every « . 0 (no matter how small « is) we can find
. 0 such that if x lies in the open interval sa 2 , a 1 d and x ± a, then f sxd lies in
the open interval sL 2 «, L 1 «d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.7The Precise Definition of a Limit
We interpret this statement geometrically by representing a function by an arrow diagram as in Figure 2, where f maps a subset of R onto another subset of R.
FIGURE 2
The definition of limit says that if any small interval sL 2 «, L 1 «d is given around
L, then we can find an interval sa 2 , a 1 d around a such that f maps all the points
in sa 2 , a 1 d (except possibly a) into the interval sL 2 «, L 1 «d. (See Figure 3.)
FIGURE 3
Another geometric interpretation of limits can be given in terms of the graph of a
function. If « . 0 is given, then we draw the horizontal lines y 5 L 1 « and y 5 L 2 «
and the graph of f. (See Figure 4.) If lim x l a f sxd − L, then we can find a number . 0
such that if we restrict x to lie in the interval sa 2 , a 1 d and take x ± a, then the
curve y 5 f sxd lies between the lines y 5 L 2 « and y − L 1 «. (See Figure 5.) You can
see that if such a has been found, then any smaller will also work.
It is important to realize that the process illustrated in Figures 4 and 5 must work for
every positive number «, no matter how small it is chosen. Figure 6 shows that if a
smaller « is chosen, then a smaller may be required.
is in
FIGURE 4
when x is in here
FIGURE 5
FIGURE 6
EXAMPLE 1 Since f sxd − x 3 2 5x 1 6 is a polynomial, we know from the Direct
Substitution Property that lim x l1 f sxd − f s1d − 13 2 5s1d 1 6 − 2. Use a graph to
find a number such that if x is within of 1, then y is within 0.2 of 2, that is,
ifx 2 1 , thensx 3 2 5x 1 6d 2 2 , 0.2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
In other words, find a number that corresponds to « 5 0.2 in the definition of a limit
for the function f sxd 5 x 3 2 5x 1 6 with a 5 1 and L 5 2.
SOLUTION A graph of f is shown in Figure 7; we are interested in the region near the
point s1, 2d. Notice that we can rewrite the inequality
| sx
FIGURE 7
(1, 2)
1.8 , x 3 2 5x 1 6 , 2.2
if0.92 , x , 1.12then1.8 , x 3 2 5x 1 6 , 2.2
FIGURE 8
20.2 , sx 2 5x 1 6d 2 2 , 0.2
So we need to determine the values of x for which the curve y 5 x 3 2 5x 1 6 lies
between the horizontal lines y 5 1.8 and y 5 2.2. Therefore we graph the curves
y 5 x 3 2 5x 1 6, y 5 1.8, and y 5 2.2 near the point s1, 2d in Figure 8. We estimate
that the x-coordinate of the point of intersection of the line y 5 2.2 and the curve
y 5 x 3 2 5x 1 6 is about 0.911. Similarly, y 5 x 3 2 5x 1 6 intersects the line
y 5 1.8 when x < 1.124. So, rounding toward 1 to be safe, we can say that
2 5x 1 6d 2 2 , 0.2
or equivalently
This interval s0.92, 1.12d is not symmetric about x 5 1. The distance from x 5 1 to the
left endpoint is 1 2 0.92 5 0.08 and the distance to the right endpoint is 0.12. We can
choose to be the smaller of these numbers, that is, − 0.08. Then we can rewrite our
inequalities in terms of distances as follows:
ifx 2 1 , 0.08thensx 3 2 5x 1 6d 2 2 , 0.2
This just says that by keeping x within 0.08 of 1, we are able to keep f sxd within 0.2
of 2.
Although we chose − 0.08, any smaller positive value of would also have
The graphical procedure used in Example 1 gives an illustration of the definition for
« 5 0.2, but it does not prove that the limit is equal to 2. A proof has to provide a for
every «.
In proving limit statements it may be helpful to think of the definition of limit as a
challenge. First it challenges you with a number «. Then you must be able to produce a
suitable . You have to be able to do this for every « . 0, not just a particular «.
Imagine a contest between two people, A and B, and imagine yourself to be B. Person A stipulates that the fixed number L should be approximated by the values of f sxd
to within a degree of accuracy « (say, 0.01). Person B then responds by finding a number such that if 0 , x 2 a , , then f sxd 2 L , «. Then A may become more
exacting and challenge B with a smaller value of « (say, 0.0001). Again B has to
respond by finding a corresponding . Usually the smaller the value of «, the smaller
the corresponding value of must be. If B always wins, no matter how small A makes
«, then lim x l a f sxd − L.
EXAMPLE 2 Prove that lim s4x 2 5d − 7.
x l3
1. Preliminary analysis of the problem (guessing a value for ). Let « be a given
positive number. We want to find a number such that
if0 , x 2 3 , thens4x 2 5d 2 7 , «
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.7The Precise Definition of a Limit
Cauchy and Limits
After the invention of calculus in the
17th century, there followed a period
of free development of the subject
in the 18th century. Mathematicians
like the Bernoulli brothers and Euler
were eager to exploit the power of
calculus and boldly explored the con­
sequences of this new and wonderful
mathematical theory without worry­
ing too much about whether their
proofs were completely correct.
The 19th century, by contrast,
was the Age of Rigor in mathematics.
There was a movement to go back to
the foundations of the subject—to
provide careful definitions and rigor­
ous proofs. At the forefront of this
movement was the French mathema­
tician Augustin-Louis Cauchy (1789–
1857), who started out as a military
engineer before becoming a math­
ematics professor in Paris. Cauchy
took Newton’s idea of a limit, which
was kept alive in the 18th century
by the French mathematician Jean
d’Alembert, and made it more precise.
His definition of a limit reads as fol­
lows: “When the successive values
attributed to a variable approach
indefinitely a fixed value so as to end
by differing from it by as little as one
wishes, this last is called the limit of all
the others.” But when Cauchy used
this definition in examples and proofs,
he often employed delta-epsilon
inequalities similar to the ones in this
section. A typical Cauchy proof starts
with: “Designate by and « two very
small numbers; . . .” He used « because
of the correspondence between epsilon and the French word erreur and
because delta corresponds to différence. Later, the German mathema­
tician Karl Weierstrass (1815–1897)
stated the definition of a limit exactly
as in our Definition 2.
| |
| |
But s4x 2 5d 2 7 5 4x 2 12 5 4sx 2 3d 5 4 x 2 3 . Therefore we want such that
if0 , x 2 3 , then4 x 2 3 , «
that is,
if0 , x 2 3 , thenx 2 3 ,
This suggests that we should choose 5 «y4.
2. Proof (showing that this works). Given « . 0, choose 5 «y4. If
0 , x 2 3 , , then
| s4x 2 5d 2 7 | − | 4x 2 12 | − 4| x 2 3 | , 4 − 4
if0 , x 2 3 , thens4x 2 5d 2 7 , «
Therefore, by the definition of a limit,
lim s4x 2 5d − 7
x l3
This example is illustrated by Figure 9.
FIGURE 9
Note that in the solution of Example 2 there were two stages—guessing and proving.
We made a preliminary analysis that enabled us to guess a value for . But then in the
second stage we had to go back and prove in a careful, logical fashion that we had made
a correct guess. This procedure is typical of much of mathematics. Sometimes it is necessary to first make an intelligent guess about the answer to a problem and then prove that
the guess is correct.
EXAMPLE 3 Prove that lim x 2 − 9.
1. Guessing a value for . Let « . 0 be given. We have to find a number . 0
such that
if0 , x 2 3 , thenx 2 2 9 , «
To connect | x 2 9 | with | x 2 3 | we write | x 2 9 | 5 | sx 1 3dsx 2 3d |. Then
we want
if0 , x 2 3 , thenx 1 3 x 2 3 , «
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Notice that if we can find a positive constant C such that x 1 3 , C, then
| x 1 3 || x 2 3 | , C | x 2 3 |
and we can make C | x 2 3 | , « by taking | x 2 3 | , «yC, so we could
choose − «yC.
We can find such a number C if we restrict x to lie in some interval centered at 3.
In fact, since we are interested only in values of x that are close to 3, it is reasonable to
assume that x is within a distance l from 3, that is, x 2 3 , 1. Then 2 , x , 4, so
5 , x 1 3 , 7. Thus we have x 1 3 , 7, and so C 5 7 is a suitable choice for the
But now there are two restrictions on x 2 3 , namely
| x 2 3 | , 1and| x 2 3 | , C 5 7
To make sure that both of these inequalities are satisfied, we take to be the smaller of
the two numbers 1 and «y7. The notation for this is − minh1, «y7j.
2. Showing that this works. Given « . 0, let − minh1, «y7j. If
0 , x 2 3 , , then x 2 3 , 1 ? 2 , x , 4 ? x 1 3 , 7 (as in part l).
We also have x 2 3 , «y7, so
29 − x13 x23 ,7?
| |
This shows that lim x l3 x 2 − 9.
■ One-Sided Limits
The intuitive definitions of one-sided limits that were given in Section 1.5 can be pre­
cisely reformulated as follows.
3 Precise Definition of Left-Hand Limit
lim f sxd − L
x l a2
if for every number « . 0 there is a number . 0 such that
ifa 2 , x , athenf sxd 2 L , «
4 Precise Definition of Right-Hand Limit
lim f sxd − L
x la1
if for every number « . 0 there is a number . 0 such that
ifa , x , a 1 thenf sxd 2 L , «
Notice that Definition 3 is the same as Definition 2 except that x is restricted to lie in
the left half sa 2 , ad of the interval sa 2 , a 1 d. In Definition 4, x is restricted to lie
in the right half sa, a 1 d of the interval sa 2 , a 1 d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.7The Precise Definition of a Limit
EXAMPLE 4 Use Definition 4 to prove that lim1 sx − 0.
1. Guessing a value for . Let « be a given positive number. Here a 5 0 and L 5 0,
so we want to find a number such that
if0 , x , thensx 2 0 , «
that is,
if0 , x , thensx , «
or, squaring both sides of the inequality sx , «, we get
if0 , x , thenx , « 2
This suggests that we should choose 5 « 2.
2. Showing that this works. Given « . 0, let 5 « 2. If 0 , x , , then
sx , s 5 s« 2 5 «
| sx 2 0 | , «
According to Definition 4, this shows that lim x l 01 sx − 0.
■ The Limit Laws
As the preceding examples show, it is not always easy to prove that limit statements are
true using the «, definition. In fact, if we had been given a more complicated function
such as f sxd 5 s6x 2 2 8x 1 9dys2x 2 2 1d, a proof would require a great deal of ingenuity. Fortunately this is unnecessary because the Limit Laws stated in Section 1.6 can be
proved using Definition 2, and then the limits of complicated functions can be found
rigorously from the Limit Laws without resorting to the definition directly.
For instance, we prove the Sum Law: If lim x l a f sxd − L and lim x l a tsxd − M both
exist, then
lim f f sxd 1 tsxdg − L 1 M
The remaining laws are proved in the exercises and in Appendix F.
PROOF OF THE SUM LAW Let « . 0 be given. We must find . 0 such that
if0 , x 2 a , thenf sxd 1 tsxd 2 sL 1 Md , «
Triangle Inequality:
|a 1 b| < |a| 1 |b|
(See Appendix A.)
Using the Triangle Inequality we can write
| f sxd 1 tsxd 2 sL 1 Md | 5 | s f sxd 2 Ld 1 s tsxd 2 Md |
< | f sxd 2 L | 1 | tsxd 2 M |
We make | f sxd 1 tsxd 2 sL 1 Md | less than « by making each of the terms
| f sxd 2 L | and | tsxd 2 M | less than «y2.
Since «y2 . 0 and lim x l a f sxd − L, there exists a number 1 . 0 such that
if0 , x 2 a , 1thenf sxd 2 L ,
Similarly, since lim x l a tsxd − M, there exists a number 2 . 0 such that
if0 , x 2 a , 2thentsxd 2 M ,
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Let − minh1, 2j, the smaller of the numbers 1 and 2. Notice that
if0 , x 2 a , then0 , x 2 a , 1and0 , x 2 a , 2
| f sxd 2 L | , 2 and| tsxd 2 M | , 2
and so
Therefore, by (5),
| f sxd 1 tsxd 2 sL 1 Md | < | f sxd 2 L | 1 | tsxd 2 M |
1 5«
To summarize,
if0 , x 2 a , thenf sxd 1 tsxd 2 sL 1 Md , «
Thus, by the definition of a limit,
lim f f sxd 1 tsxdg − L 1 M ■
■ Infinite Limits
Infinite limits can also be defined in a precise way. The following is a precise version of
Definition 1.5.4.
6 Precise Definition of an Infinite Limit Let f be a function defined on some
open interval that contains the number a, except possibly at a itself. Then
lim f sxd − `
means that for every positive number M there is a positive number such that
FIGURE 10
if0 , x 2 a , thenf sxd . M
This says that the values of f sxd can be made arbitrarily large (larger than any given
number M) by requiring x to be close enough to a (within a distance , where depends
on M, but with x ± a). A geometric illustration is shown in Figure 10.
Given any horizontal line y 5 M, we can find a number . 0 such that if we restrict
x to lie in the interval sa 2 , a 1 d but x ± a, then the curve y 5 f sxd lies above the
line y − M. You can see that if a larger M is chosen, then a smaller may be required.
EXAMPLE 5 Use Definition 6 to prove that lim
− `.
SOLUTION Let M be a given positive number. We want to find a number such that
| |
if0 , x , then1yx 2 . M
. M&?x 2 , &?sx 2 ,
&?x ,
| |
| |
So if we choose − 1ysM and 0 , x , − 1ysM , then 1yx 2 . M. This shows
that 1yx 2 l ` as x l 0.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.7The Precise Definition of a Limit
Similarly, the following is a precise version of Definition 1.5.5. It is illustrated by
Figure 11.
7 Definition Let f be a function defined on some open interval that contains
the number a, except possibly at a itself. Then
lim f sxd − 2`
means that for every negative number N there is a positive number such that
FIGURE 11
if0 , x 2 a , thenf sxd , N
1. Use the given graph of f to find a number such that
ifx 2 1 , thenf sxd 2 1 , 0.2
4. Use the given graph of f sxd 5 x 2 to find a number such that
ifx 2 1 , thenx 2 2 1 , 12
1 1.1
2. Use the given graph of f to find a number such that
if0 , x 2 3 , thenf s xd 2 2 , 0.5
; 5. Use a graph to find a number such that
ifx 2 2 , thensx 2 1 5 2 3 , 0.3
; 6. Use a graph to find a number such that
ifx 2
, thencos 2 x 2
, 0.1
; 7. For the limit
lim sx 3 2 3x 1 4d − 6
2.6 3
3. U
se the given graph of f sxd − sx to find a number such
ifx 2 4 , thensx 2 2 , 0.4
illustrate Definition 2 by finding values of that correspond
to « 5 0.2 and « 5 0.1.
; 8. For the limit
x l2
4x 1 1
− 4.5
3x 2 4
illustrate Definition 2 by finding values of that correspond
to « 5 0.5 and « 5 0.1.
; 9. (a) Use a graph to find a number such that
if4 , x , 4 1 then0
x2 1 4
sx 2 4
. 100
(b) What limit does part (a) suggest is true?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
; 10. Given that lim x l csc x − `, illustrate Definition 6 by
finding values of that correspond to (a) M − 500 and
(b) M − 1000.
11. A
machinist is required to manufacture a circular metal disk
with area 1000 cm2.
(a)What radius produces such a disk?
(b)If the machinist is allowed an error tolerance of 65 cm 2
in the area of the disk, how close to the ideal radius in
part (a) must the machinist control the radius?
(c)In terms of the «, definition of lim x l a f sxd − L, what
is x? What is f sxd? What is a? What is L? What value
of « is given? What is the corresponding value of ?
rystal growth furnaces are used in research to determine
; 12. C
how best to manufacture crystals used in electronic components. For proper growth of a crystal, the temperature must
be controlled accurately by adjusting the input power. Suppose the relationship is given by
T swd − 0.1w 2 1 2.155w 1 20
where T is the temperature in degrees Celsius and w is the
power input in watts.
(a)How much power is needed to maintain the temperature
at 200°C?
(b)If the temperature is allowed to vary from 200°C by
up to 61°C, what range of wattage is allowed for the
input power?
(c)In terms of the «, definition of lim x l a f sxd − L, what
is x? What is f sxd? What is a? What is L? What value
of « is given? What is the corresponding value of ?
13. (a)Find a number such that if x 2 2 , , then
4x 2 8 , «, where « 5 0.1.
(b) Repeat part (a) with « 5 0.01.
14. Given that limx l 2 s5x 2 7d − 3, illustrate Definition 2 by
finding values of that correspond to « 5 0.1, « 5 0.05,
and « 5 0.01.
15–18 Prove the statement using the «, definition of a limit
and illustrate with a diagram like Figure 9.
15. lim
( 12 x 2 1) − 1
16. lim s2 2 3xd − 24
17. lim s22x 1 1d − 5
18. lim s2x 2 5d − 23
x l2
19. lim (1 2 13 x) − 22
20. lim
x 2 2 2x 2 8
21. lim
9 2 4x 2
22. lim
x l21.5 3 1 2x
23. lim x − a
24. lim c − c
25. lim x − 0
| |
27. lim x − 0
30. lim sx 2 1 2x 2 7d − 1
31. lim sx 2 1d − 3
32. lim x 3 − 8
x l 22
( 32 x 2 12 ) − 7
26. lim x 3 − 0
lim 1 s
61x −0
x l 26
33. V
erify that another possible choice of for showing that
lim x l3 x 2 − 9 in Example 3 is − min{2, «y8}.
34. V
erify, by a geometric argument, that the largest possible
choice of for showing that lim x l3 x 2 − 9 is
− s 9 1 « 2 3.
35. (a)For the limit lim x l 1 sx 3 1 x 1 1d − 3, use a graph to
find a value of that corresponds to « 5 0.4.
(b)By solving the cubic equation x 3 1 x 1 1 5 3 1 «,
find the largest possible value of that works for any
given « . 0.
(c)Put « 5 0.4 in your answer to part (b) and compare
with your answer to part (a).
36. Prove that lim
x l2
− .
37. Prove that lim sx − sa if a . 0.
Hint: Use sx 2 sa
| − s|x
1 sa
38. If H is the Heaviside function defined in Section 1.5,
prove, using Definition 2, that lim t l 0 Hstd does not exist.
[Hint: Use an indirect proof as follows. Suppose that the
limit is L. Take « 5 12 in the definition of a limit and try to
arrive at a contradiction.]
39. If the function f is defined by
f sxd −
0 if x is rational
1 if x is irrational
prove that lim x l 0 f sxd does not exist.
40. By comparing Definitions 2, 3, and 4, prove Theorem 1.6.1:
lim f sxd − L
if and only if
lim f sxd − L − lim1 f sxd
x l a2
x la
41. How close to 23 do we have to take x so that
. 10,000
sx 1 3d4
x l1
19–32 Prove the statement using the «, definition of a limit.
29. lim sx 2 2 4x 1 5d − 1
42. Prove, using Definition 6, that lim
x l23
43. Prove that lim 2
x l21
− `.
sx 1 3d4
− 2`.
sx 1 1d 3
44. Suppose that lim x l a f sxd − ` and lim x l a tsxd − c, where
c is a real number. Prove each statement.
(a) lim f f sxd 1 tsxdg − `
(b) lim f f sxd tsxdg − `if c . 0
(c) lim f f sxd tsxdg − 2`if c , 0
xl a
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.8Continuity
1.8 Continuity
■ Continuity of a Function
We noticed in Section 1.6 that the limit of a function as x approaches a can often be
found simply by calculating the value of the function at a. Functions having this property
are called continuous at a. We will see that the mathematical definition of continuity corresponds closely with the meaning of the word continuity in everyday language. (A continuous process is one that takes place without interruption.)
1 Definition A function f is continuous at a number a if
lim f sxd − f sad
xl a
As illustrated in Figure 1, if f is continuous, then the points sx, f sxdd on
the graph of f approach the point
sa, f sadd on the graph. So there is no
gap in the curve.
Notice that Definition l implicitly requires three things if f is continuous at a:
1. f sad is defined (that is, a is in the domain of f )
2. lim f sxd exists
x la
3. lim f sxd − f sad
x la
as x approaches a
FIGURE 1
The definition says that f is continuous at a if f sxd approaches f sad as x approaches
a. Thus a continuous function f has the property that a small change in x produces only
a small change in f sxd. In fact, the change in f sxd can be kept as small as we please by
keeping the change in x sufficiently small.
If f is defined near a (in other words, f is defined on an open interval containing a,
except perhaps at a), we say that f is discontinuous at a (or f has a discontinuity at a)
if f is not continuous at a.
Physical phenomena are usually continuous. For instance, the displacement or velocity of a moving vehicle varies continuously with time, as does a person’s height. But
discontinuities do occur in such situations as electric currents. [The Heaviside function,
introduced in Section 1.5, is discontinuous at 0 because lim t l 0 Hstd does not exist.]
Geometrically, you can think of a function that is continuous at every number in an
interval as a function whose graph has no break in it: the graph can be drawn without
removing your pen from the paper.
EXAMPLE 1 Figure 2 shows the graph of a function f. At which numbers is f discontinuous? Why?
FIGURE 2
SOLUTION It looks as if there is a discontinuity when a − 1 because the graph has a
break there. The official reason that f is discontinuous at 1 is that f s1d is not defined.
The graph also has a break when a 5 3, but the reason for the discontinuity is
different. Here, f s3d is defined, but lim x l3 f sxd does not exist (because the left and
right limits are different). So f is discontinuous at 3.
What about a 5 5? Here, f s5d is defined and lim x l5 f sxd exists (because the left and
right limits are the same). But
lim f sxd ± f s5d
So f is discontinuous at 5.
Now let’s see how to detect discontinuities when a function is defined by a formula.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
EXAMPLE 2 Where are each of the following functions discontinuous?
x2 2 x 2 2
(a) f sxd 5
(c) f sxd −
if x ± 0
if x − 0
(b) f sxd −
x2 2 x 2 2
if x ± 2
if x − 2
(d) f sxd − v x b
(a) Notice that f s2d is not defined, so f is discontinuous at 2. Later we’ll see why f is
continuous at all other numbers.
(b) Here f s2d 5 1 is defined and
x2 2 x 2 2
sx 2 2dsx 1 1d
− lim
− lim sx 1 1d − 3
x l2
x l2
x l2
lim f sxd − lim
x l2
exists. But
lim f sxd ± f s2d
x l2
so f is not continuous at 2.
(c) Here f s0d 5 1 is defined but
lim f sxd − lim
does not exist. (See Example 1.5.6.) So f is discontinuous at 0.
(d) The greatest integer function f sxd − v x b has discontinuities at all of the inte­gers because lim x ln v x b does not exist if n is an integer. (See Example 1.6.10 and
Exercise 1.6.55.)
Figure 3 shows the graphs of the functions in Example 2. In each case the graph can’t
be drawn without lifting the pen from the paper because a hole or break or jump occurs
in the graph. The kind of discontinuity illustrated in parts (a) and (b) is called removable
because we could remove the discontinuity by redefining f at just the single number 2.
[If we redefine f to be 3 at x − 2, then f is equivalent to the function tsxd − x 1 1,
which is continuous.] The discontinuity in part (c) is called an infinite discontinuity.
The discontinuities in part (d) are called jump discontinuities because the function
“jumps” from one value to another.
(a) A removable discontinuity
(b) A removable discontinuity
(c) An infinite discontinuity
(d) Jump discontinuities
FIGURE 3 Graphs of the functions in Example 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.8Continuity
2 Definition A function f is continuous from the right at a number a if
lim f sxd − f sad
x l a1
and f is continuous from the left at a if
lim f sxd − f sad
x l a2
EXAMPLE 3 At each integer n, the function f sxd − v x b [see Figure 3(d)] is continuous from the right but discontinuous from the left because
lim f sxd − lim1 v xb − n − f snd
x l n1
x ln
lim f sxd − lim2 v x b − n 2 1 ± f snd
x l n2
x ln
3 Definition A function f is continuous on an interval if it is continuous at
every number in the interval. (If f is defined only on one side of an endpoint of the
interval, we understand continuous at the endpoint to mean continuous from the
right or continuous from the left.)
EXAMPLE 4 Show that the function f sxd − 1 2 s1 2 x 2 is continuous on the
interval f21, 1g.
SOLUTION If 21 , a , 1, then using the Limit Laws from Section 1.6, we have
lim f sxd − lim (1 2 s1 2 x 2 )
− 1 2 lim s1 2 x 2 (by Laws 2 and 8)
− 1 2 s lim s1 2 x 2 d (by 7)
− 1 2 s1 2 a 2
− f sad
Thus, by Definition l, f is continuous at a if 21 , a , 1. Similar calculations show that
lim f sxd − 1 − f s21dandlim2 f sxd − 1 − f s1d
x l 211
FIGURE 4
(by 2, 8, and 10)
x l1
so f is continuous from the right at 21 and continuous from the left at 1. Therefore,
according to Definition 3, f is continuous on f21, 1g.
The graph of f is sketched in Figure 4. It is the lower half of the circle
x 2 1 sy 2 1d2 − 1
■ Properties of Continuous Functions
Instead of always using Definitions 1, 2, and 3 to verify the continuity of a function as
we did in Example 4, it is often convenient to use the next theorem, which shows how to
build up complicated continuous functions from simple ones.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
4 Theorem If f and t are continuous at a and c is a constant, then the following functions are also continuous at a:
1. f 1 t
2. f 2 t
3. cf
4. ft
if tsad ± 0
PROOF Each of the five parts of this theorem follows from the corresponding Limit
Law in Section 1.6. For instance, we give the proof of part 1. Since f and t are continuous at a, we have
lim f sxd − f sadandlim tsxd − tsad
lim s f 1 tdsxd − lim f f sxd 1 tsxdg
− lim f sxd 1 lim tsxd(by Law 1)
− f sad 1 tsad
− s f 1 tdsad
This shows that f 1 t is continuous at a.
It follows from Theorem 4 and Definition 3 that if f and t are continuous on an interval, then so are the functions f 1 t, f 2 t, cf, ft, and (if t is never 0) fyt. The following
theorem was stated in Section 1.6 as the Direct Substitution Property.
5 Theorem
(a)Any polynomial is continuous everywhere; that is, it is continuous on
R − s2`, `d.
(b)Any rational function is continuous wherever it is defined; that is, it is contin­
uous on its domain.
(a) A polynomial is a function of the form
Psxd − cn x n 1 cn21 x n21 1 ∙ ∙ ∙ 1 c1 x 1 c0
where c0 , c1, . . . , cn are constants. We know that
lim c0 − c0(by Law 8)
lim x m − a mm − 1, 2, . . . , n(by 10)
This equation is precisely the statement that the function f sxd − x m is a continuous
function. Thus, by part 3 of Theorem 4, the function tsxd − cx m is continuous. Since P
is a sum of functions of this form and a constant function, it follows from part 1 of
Theorem 4 that P is continuous.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.8Continuity
(b) A rational function is a function of the form
f sxd −
where P and Q are polynomials. The domain of f is D − hx [ R Qsxd ± 0j. We
know from part (a) that P and Q are continuous everywhere. Thus, by part 5 of Theorem 4, f is continuous at every number in D.
As an illustration of Theorem 5, observe that the volume of a sphere varies continuously with its radius because the formula Vsrd − 43 r 3 shows that V is a polynomial
function of r. Likewise, if a ball is thrown vertically into the air with an initial velocity
of 15 mys, then the height of the ball in meters t seconds later is given by the formula
h − 15t 2 4.9t 2. Again this is a polynomial function, so the height is a continuous function of the elapsed time, as we might expect.
Knowledge of which functions are continuous enables us to evaluate some limits very
quickly, as the following example shows. Compare it with Example 1.6.2(b).
EXAMPLE 5 Find lim
x l 22
x 3 1 2x 2 2 1
5 2 3x
SOLUTION The function
f sxd −
x 3 1 2x 2 2 1
5 2 3x
is rational, so by Theorem 5 it is continuous on its domain, which is h x x ± 53 j.
x 3 1 2x 2 2 1
− lim f sxd − f s22d
x l22
x l22
5 2 3x
P(cos ¨, sin ¨)
(1, 0)
FIGURE 5
It turns out that most of the familiar functions are continuous at every number in their
domains. For instance, Limit Law 11 in Section 1.6 is exactly the statement that root
functions are continuous.
From the appearance of the graphs of the sine and cosine functions (Figure 1.2.19),
we would certainly guess that they are continuous. We know from the definitions of sin and cos that the coordinates of the point P in Figure 5 are scos , sin d. As l 0, we
see that P approaches the point s1, 0d and so cos l 1 and sin l 0. Thus
Another way to establish the limits in
(6) is to use the Squeeze Theorem
with the inequality sin , (for
. 0), which is proved in Section 2.4.
s22d3 1 2s22d2 2 1
−2 5 2 3s22d
lim cos − 1
lim sin − 0
Since cos 0 − 1 and sin 0 − 0, the equations in (6) assert that the cosine and sine functions are continuous at 0. The addition formulas for cosine and sine can then be used to
deduce that these functions are continuous everywhere (see Exercises 66 and 67).
It follows from part 5 of Theorem 4 that
tan x −
sin x
cos x
is continuous except where cos x − 0. This happens when x is an odd integer multiple
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
3π _π
_ 2
FIGURE 6
Functions and Limits
of y2, so y − tan x has infinite discontinuities when x − 6y2, 63y2, 65y2, and
so on (see Figure 6).
7 Theorem The following types of functions are continuous at every number in
their domains:
• polynomials
• rational functions
• root functions
• trigonometric functions
EXAMPLE 6 On what intervals is each function continuous?
y − tan x
(b) tsxd −
(a) f sxd − x 100 2 2x 37 1 75
(c) hsxd − sx 1
x 2 1 2x 1 17
x2 2 1
x 11
(a) f is a polynomial, so it is continuous on s2`, `d by Theorem 5(a).
(b) t is a rational function, so by Theorem 5(b), it is continuous on its domain,
which is D − hx x 2 2 1 ± 0j − hx x ± 61j. Thus t is continuous on the intervals
s2`, 21d, s21, 1d, and s1, `d.
(c) We can write hsxd − Fsxd 1 Gsxd 2 Hsxd, where
Fsxd − sx Gsxd −
Hsxd − 2
x 11
F is continuous on [0, `d by Theorem 7. G is a rational function, so it is continuous
everywhere except when x 2 1 − 0, that is, x − 1. H is also a rational function, but its
denominator is never 0, so H is continuous everywhere. Thus, by parts 1 and 2 of
Theorem 4, h is continuous on the intervals [0, 1d and s1, `d.
EXAMPLE 7 Evaluate lim
x l
sin x
2 1 cos x
SOLUTION Theorem 7 tells us that y − sin x is continuous. The function in the
denomi­nator, y − 2 1 cos x, is the sum of two continuous functions and is therefore
continuous. Notice that this function is never 0 because cos x > 21 for all x and so
2 1 cos x . 0 everywhere. Thus the ratio
f sxd −
sin x
2 1 cos x
is continuous everywhere. Hence, by the definition of a continuous function,
x l
sin x
sin 0
− lim f sxd − f sd −
− 0
2 1 cos x
2 1 cos 221
Another way of combining continuous functions f and t to get a new continuous
function is to form the composite function f 8 t. This fact is a consequence of the following theorem.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.8Continuity
This theorem says that a limit symbol
can be moved through a function symbol if the function is continuous and the
limit exists. In other words, the order of
these two symbols can be reversed.
8 Theorem If f is continuous at b and lim tsxd − b, then lim f s tsxdd − f sbd.
x la
x la
In other words,
lim f stsxdd − f lim tsxd
xl a
Intuitively, Theorem 8 is reasonable because if x is close to a, then tsxd is close to b,
and since f is continuous at b, if tsxd is close to b, then fs tsxdd is close to f sbd. A proof
of Theorem 8 is given in Appendix F.
x , with n being a posiLet’s now apply Theorem 8 in the special case where f sxd − s
tive integer. Then
f stsxdd − s
f lim tsxd − n lim tsxd
If we put these expressions into Theorem 8, we get
n lim tsxd
lim s
tsxd − s
and so Limit Law 7 has now been proved. (We assume that the roots exist.)
9 Theorem If t is continuous at a and f is continuous at tsad, then the composite function f 8 t given by s f 8 tds xd − f s tsxdd is continuous at a.
This theorem is often expressed informally by saying “a continuous function of a
continuous function is a continuous function.”
PROOF Since t is continuous at a, we have
lim tsxd − tsad
Since f is continuous at b − tsad, we can apply Theorem 8 to obtain
lim f s tsxdd − f stsadd
which is precisely the statement that the function hsxd − f stsxdd is continuous at a;
that is, f 8 t is continuous at a.
EXAMPLE 8 Where are the following functions continuous?
(a) hsxd − sinsx 2 d
(b) Fsxd −
sx 1 7 2 4
(a) We have hsxd − f stsxdd, where
tsxd − x 2andf sxd − sin x
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
Now t is continuous on R since it is a polynomial, and f is also continuous everywhere. Thus h − f 8 t is continuous on R by Theorem 9.
(b) Notice that F can be broken up as the composition of four continuous functions:
F − f + t + h + korFsxd − f stshsksxdddd
f sxd −
tsxd − x 2 4hsxd − sx ksxd − x 2 1 7
We know that each of these functions is continuous on its domain (by Theorems 5
and 7), so by Theorem 9, F is continuous on its domain, which is
h x [ R | sx 2 1 7
± 4 j − hx x ± 63j − s2`, 23d ø s23, 3d ø s3, `d
■ The Intermediate Value Theorem
An important property of continuous functions is expressed by the following theorem,
whose proof is found in more advanced books on calculus.
10 The Intermediate Value Theorem Suppose that f is continuous on the
closed interval fa, bg and let N be any number between f sad and f sbd, where
f sad ± f sbd. Then there exists a number c in sa, bd such that f scd − N.
The Intermediate Value Theorem states that a continuous function takes on every intermediate value between the function values f sad and f sbd. It is illustrated by Figure 7. Note
that the value N can be taken on once [as in part (a)] or more than once [as in part (b)].
c b
a c¡
FIGURE 7
FIGURE 8
If we think of a continuous function as a function whose graph has no hole or break,
then it is easy to believe that the Intermediate Value Theorem is true. In geometric
terms it says that if any horizontal line y − N is given between y − f sad and y − f sbd
as in Figure 8, then the graph of f can’t jump over the line. It must intersect y − N
It is important that the function f in Theorem 10 be continuous. The Intermediate
Value Theorem is not true in general for discontinuous functions (see Exercise 52).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.8Continuity
One use of the Intermediate Value Theorem is in locating solutions of equations as in
the following example.
EXAMPLE 9 Show that there is a solution of the equation
4x 3 2 6x 2 1 3x 2 2 − 0
between 1 and 2.
SOLUTION Let f sxd − 4x 3 2 6x 2 1 3x 2 2. We are looking for a solution of the
given equation, that is, a number c between 1 and 2 such that f scd − 0. Therefore we
take a − 1, b − 2, and N − 0 in Theorem 10. We have
f s1d − 4 2 6 1 3 2 2 − 21 , 0
f s2d − 32 2 24 1 6 2 2 − 12 . 0
Thus f s1d , 0 , f s2d; that is, N − 0 is a number between f s1d and f s2d. The
function f is continuous since it is a polynomial, so the Intermediate Value Theorem
says there is a number c between 1 and 2 such that f scd − 0. In other words, the
equation 4x 3 2 6x 2 1 3x 2 2 − 0 has at least one solution c in the interval s1, 2d.
In fact, we can locate a solution more precisely by using the Intermediate Value
Theorem again. Since
f s1.2d − 20.128 , 0andf s1.3d − 0.548 . 0
a solution must lie between 1.2 and 1.3. A calculator gives, by trial and error,
f s1.22d − 20.007008 , 0andf s1.23d − 0.056068 . 0
so a solution lies in the interval s1.22, 1.23d.
We can use a graphing calculator or computer to illustrate the use of the Intermediate Value Theorem in Example 9. Figure 9 shows the graph of f in the viewing rectangle f21, 3g by f23, 3g and you can see that the graph crosses the x-axis between
1 and 2. Fig­ure 10 shows the result of zooming in to the viewing rectangle f1.2, 1.3g
by f20.2, 0.2g.
FIGURE 9
FIGURE 10
In fact, the Intermediate Value Theorem plays a role in the very way these graphing
devices work. A computer calculates a finite number of points on the graph and turns on
the pixels that contain these calculated points. It assumes that the function is continuous
and takes on all the intermediate values between two consecutive points. The computer
therefore “connects the dots” by turning on the intermediate pixels.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
1. W
rite an equation that expresses the fact that a function f is
continuous at the number 4.
2. I f f is continuous on s2`, `d, what can you say about its
10. C
ontinuous only from the left at 21, not continuous from the
left or right at 3
3. (a)From the given graph of f , state the numbers at which f
is discontinuous and explain why.
(b)For each of the numbers stated in part (a), determine
whether f is continuous from the right, or from the left,
or neither.
11. The toll T charged for driving on a certain stretch of a toll
road is $5 except during rush hours (between 7 am and 10 am
and between 4 pm and 7 pm) when the toll is $7.
(a)Sketch a graph of T as a function of the time t, measured
in hours past midnight.
(b)Discuss the discontinuities of this function and their
significance to someone who uses the road.
12. Explain why each function is continuous or discontinuous.
(a) The temperature at a specific location as a function of
(b)The temperature at a specific time as a function of the
distance due west from New York City
(c)The altitude above sea level as a function of the distance
due west from New York City
(d)The cost of a taxi ride as a function of the distance
(e)The current in the circuit for the lights in a room as a
function of time
9. Discontinuities at 0 and 3, but continuous from the right at 0
and from the left at 3
4. F
rom the given graph of t, state the numbers at which t is
discontinuous and explain why.
13–16 Use the definition of continuity and the properties of limits
to show that the function is continuous at the given number a.
13. f sxd − 3x 2 1 sx 1 2d5,a − 21
14. t std −
t 2 1 5t
,a − 2
2t 1 1
15. psvd − 2s3v 2 1 1 ,a − 1
4r 2 2 2r 1 7 ,a − 22
16. f srd − s
5– 6 The graph of a function f is given.
(a)At what numbers a does lim x l a f sxd not exist?
(b)At what numbers a is f not continuous?
(c)At what numbers a does lim x l a f sxd exist but f is not
continuous at a ?
17 –18 Use the definition of continuity and the properties of limits
to show that the function is continuous on the given interval.
17. f sxd − x 1 sx 2 4 ,f4, `d
18. tsxd −
7 – 10 Sketch the graph of a function f that is defined on R and
continuous except for the stated discontinuities.
7. Removable discontinuity at 22, infinite discontinuity at 2
8.Jump discontinuity at 23, removable discontinuity at 4
,s2`, 22d
3x 1 6
19– 24 Explain why the function is discontinuous at the given
number a. Sketch the graph of the function.
19. f sxd −
20. f sxd −
a − 22
if x ± 22
a − 22
if x − 22
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 1.8Continuity
21. f sxd −
1 2 x 2 if x , 1
if x > 1
x2 2 x
22. f sxd − x 2 2 1
if x − 1
cos x
if x , 0
23. f sxd − 0
if x − 01 2 x 2 if x . 0
2x 2 2 5x 2 3
24. f sxd −
if x ± 3
if x − 3
25 –26
(a) Show that f has a removable discontinuity at x − 3.
(b) Redefine f s3d so that f is continuous at x − 3 (and thus the
discontinuity is “removed”).
x2 2 9
25. f sxd −
26. f sxd −
x 2 7x 1 12
27 – 34 Explain, using Theorems 4, 5, 7, and 9, why the
function is continuous at every number in its domain. State
the domain.
27. f sxd −
29. hstd −
sx 4 1 2
cosst 2d
1 2 t2
28. tsvd −
3v 2 1
v 1 2v 2 15
30. tsud −
31. Lsvd − v s9 2 v 2
34. Fsxd − sinscosssin xdd
35 – 38 Use continuity to evaluate the limit.
35. lim x s20 2 x 2
37. lim x 2 tan x
38. lim x 3ysx 2 1 x 2 2
x l2
x l y4
s1 2 sin x
1 2 x 2 if x < 1
sx 2 1 if x . 1
sin x if x , y4
cos x if x > y4
43– 45 Find the numbers at which f is discontinuous. At which
of these numbers is f continuous from the right, from the left,
or neither? Sketch the graph of f .
x 2 if x , 21
43.f sxd − x
if 21 < x , 1
1yx if x > 1
x 2 1 1 if x < 1
44. f sxd − 3 2 x if 1 , x < 4
if x . 4
x 1 2 if x , 0
45. f sxd − 2x 2
if 0 < x < 1
2 2 x if x . 1
46. T
he gravitational force exerted by the planet Earth on a unit
mass at a distance r from the center of the planet is
Fsrd −
if r , R
if r > R
where M is the mass of Earth, R is its radius, and G is the
gravitational constant. Is F a continuous function of r ?
f sxd −
lim sinstanscos dd
x l2
40. y − tan sx
cx 2 1 2x if x , 2
x 3 2 cx
if x > 2
48. Find the values of a and b that make f continuous
l y2
; 39 –40 Locate the discontinuities of the function and illustrate
by graphing.
39. f sxd −
42. f sxd −
47. F
or what value of the constant c is the function f continuous on s2`, `d?
32. Bsud − s3u 2 2 1 s
2u 2 3
33. Msxd −
41 – 42 Show that f is continuous on s2`, `d.
41.f sxd −
if x ± 1
f sxd −
x2 2 4
if x , 2
ax 2 2 bx 1 3 if 2 < x , 3
2x 2 a 1 b
if x > 3
49. S
uppose f and t are continuous functions such that ts2d − 6
and lim x l 2 f3 f sxd 1 f sxd tsxdg − 36. Find f s2d.
50. Let f sxd − 1yx and tsxd − 1yx 2.
(a) Find s f + tds xd.
(b) Is f + t continuous everywhere? Explain.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
51. W
hich of the following functions f has a removable discon­
tinuity at a? If the discontinuity is removable, find a function t that agrees with f for x ± a and is continuous at a.
x4 2 1
(a) f sxd −
,a − 1
x 3 2 x 2 2 2x
(b) f sxd −
,a − 2
(c) f sxd − v sin x b ,a − 52. S
uppose that a function f is continuous on [0, 1] except at
0.25 and that f s0d − 1 and f s1d − 3. Let N − 2. Sketch
two pos­sible graphs of f , one showing that f might not satisfy the conclusion of the Intermediate Value Theorem and
one showing that f might still satisfy the conclusion of the
Intermediate Value Theorem (even though it doesn’t satisfy
the hypothesis).
66. T
o prove that sine is continuous, we need to show that
lim x l a sin x − sin a for every real number a. By Exercise 65 an equivalent statement is that
lim sinsa 1 hd − sin a
Use (6) to show that this is true.
67. Prove that cosine is a continuous function.
68. (a) Prove Theorem 4, part 3.
(b) Prove Theorem 4, part 5.
69. Use Theorem 8 to prove Limit Laws 6 and 7 from
Section 1.6.
70. Is there a number that is exactly 1 more than its cube?
71. For what values of x is f continuous?
0 if x is rational
1 if x is irrational
53. If f sxd − x 2 1 10 sin x, show that there is a number c such
that f scd − 1000.
f sxd −
54. Suppose f is continuous on f1, 5g and the only solutions of
the equation f sxd − 6 are x − 1 and x − 4. If f s2d − 8,
explain why f s3d . 6.
72. For what values of x is t continuous?
55– 58 Use the Intermediate Value Theorem to show that there
is a solution of the given equation in the specified interval.
tsxd −
55. 2x 1 4x 1 1 − 0,s21, 0d
57. cos x − x,s0, 1d
f sxd −
58. sin x − x 2 2 x,s1, 2d
59 – 60
(a) Prove that the equation has at least one real solution.
(b) Use a calculator to find an interval of length 0.01 that
contains a solution.
59. cos x − x 3
60. x 5 2 x 2 1 2x 1 3 − 0
; 61– 62
(a) Prove that the equation has at least one real solution.
(b) Find the solution correct to three decimal places, by
61. x 5 2 x 2 2 4 − 0
62. sx 2 5 −
63– 64 Prove, without graphing, that the graph of the function
has at least two x-intercepts in the specified interval.
63. y − sin x 3,s1, 2d
64. y − x 2 2 3 1 1yx,s0, 2d
65. Prove that f is continuous at a if and only if
lim f sa 1 hd − f sad
0 if x is rational
x if x is irrational
73. Show that the function
56. 2yx − x 2 sx ,s2, 3d
x 4 sins1yxd
if x ± 0
if x − 0
is continuous on s2`, `d.
74. If a and b are positive numbers, prove that the equation
x 3 1 2x 2 2 1
x 1x22
has at least one solution in the interval s21, 1d.
75. A
Tibetan monk leaves the monastery at 7:00 am and
takes his usual path to the top of the mountain, arriving at
7:00 pm. The following morning, he starts at 7:00 am at the
top and takes the same path back, arriving at the monastery
at 7:00 pm. Use the Intermediate Value Theorem to show
that there is a point on the path that the monk will cross at
exactly the same time of day on both days.
76. Absolute Value and Continuity
(a)Show that the absolute value function Fsxd − x is
continuous everywhere.
(b)Prove that if f is a continuous function on an interval,
then so is f .
(c)Is the converse of the statement in part (b) also true?
In other words, if f is continuous, does it follow
that f is continuous? If so, prove it. If not, find a
| |
| |
| |
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1Review
Answers to the Concept Check are available at StewartCalculus.com.
1. (a) What is a function? What are its domain and range?
(b) What is the graph of a function?
(c)How can you tell whether a given curve is the graph of
a function?
2. Discuss
four ways of representing a function. Illustrate your
discussion with examples.
3. (a)What is an even function? How can you tell if a function
is even by looking at its graph? Give three examples of an
even function.
(b)What is an odd function? How can you tell if a function
is odd by looking at its graph? Give three examples of
an odd function.
4. What is an increasing function?
5. What is a mathematical model?
6. Give an example of each type of function.
(a) Linear function
(b) Power function
(c) Exponential function
(d) Quadratic function
(e) Polynomial of degree 5
(f ) Rational function
7. Sketch
by hand, on the same axes, the graphs of the following
(a) f sxd − x
(b) tsxd − x 2
(c) hsxd − x
(d) jsxd − x 4
8. Draw,
by hand, a rough sketch of the graph of each function.
(a) y − sin x
(b) y − cos x
(c) y − tan x
(d) y − 1yx
(e) y − x
(f ) y − sx
| |
9. S
uppose that f has domain A and t has domain B.
(a) What is the domain of f 1 t ?
(b) What is the domain of f t ?
(c) What is the domain of fyt ?
10. How
is the composite function f 8 t defined? What is its
11. Suppose
the graph of f is given. Write an equation for each of
the graphs that are obtained from the graph of f as follows.
(a) Shift 2 units upward.
(b) Shift 2 units downward.
(c) Shift 2 units to the right. (d) Shift 2 units to the left.
(e) Reflect about the x-axis. (f ) Reflect about the y-axis.
(g) Stretch vertically by a factor of 2.
(h) Shrink vertically by a factor of 2.
( i ) Stretch horizontally by a factor of 2.
( j ) Shrink horizontally by a factor of 2.
12. Explain
what each of the following means and illustrate with a
(a) lim f sxd − L
(b) lim1 f sxd − L (c) lim2 f sxd − L
x la
(d) lim f sxd − `
x la
x la
x la
(e) lim f sxd − 2`
x la
13. D
escribe several ways in which a limit can fail to exist.
Illustrate with sketches.
14. W
hat does it mean to say that the line x − a is a vertical
asymptote of the curve y − f sxd? Draw curves to illustrate the
various possibilities.
15. State the following Limit Laws.
(a) Sum Law
(b) Difference Law
(c) Constant Multiple Law
(d) Product Law
(e) Quotient Law
(f ) Power Law
(g) Root Law
16. What does the Squeeze Theorem say?
17. (a) What does it mean for f to be continuous at a?
(b)What does it mean for f to be continuous on the
interval s2`, `d? What can you say about the graph
of such a function?
18. (a)Give examples of functions that are continuous on f21, 1g.
(b)Give an example of a function that is not continuous
on f0, 1g.
19. What does the Intermediate Value Theorem say?
Determine whether the statement is true or false. If it is true,
explain why. If it is false, explain why or give an example that
disproves the statement.
1. If f is a function, then f ss 1 td − f ssd 1 f std.
2. If f ssd − f std, then s − t.
3. If f is a function, then f s3xd − 3 f sxd.
4. If x 1 , x 2 and f is a decreasing function, then f sx 1 d . f sx 2 d.
5. A vertical line intersects the graph of a function at most once.
6. If x is any real number, then sx 2 − x.
7. lim
− lim
2 lim
xl4 x 2 4
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
8. lim
x l1
9. lim
Functions and Limits
lim sx 2 1 6x 2 7d
x 2 1 6x 2 7
x l1
lim sx 2 1 5x 2 6d
x 2 1 5x 2 6
17. If p is a polynomial, then lim x l b psxd − psbd.
lim sx 2 3d
lim sx 2 1 2x 2 4d
x 2 1 2x 2 4
19. I f the line x − 1 is a vertical asymptote of y − f sxd, then
f is not defined at 1.
18. If lim x l 0 f sxd − ` and lim x l 0 tsxd − `, then
lim x l 0 f f sxd 2 tsxdg − 0.
x l1
20. If f s1d . 0 and f s3d , 0, then there exists a number c
between 1 and 3 such that f scd − 0.
x2 2 9
21. If f is continuous at 5 and f s5d − 2 and f s4d − 3, then
lim x l 2 f s4x 2 2 11d − 2.
11. lim
x 29
− lim sx 1 3d
12. If lim x l 5 f sxd − 2 and lim x l 5 tsxd − 0, then
limx l 5 f f sxdytsxdg does not exist.
13. If lim x l5 f sxd − 0 and lim x l 5 tsxd − 0, then
lim x l 5 f f sxdytsxdg does not exist.
14. If neither lim x l a f sxd nor lim x l a tsxd exists, then
lim x l a f f sxd 1 tsxdg does not exist.
22. If f is continuous on f21, 1g and f s21d − 4 and f s1d − 3,
then there exists a number r such that r , 1 and f srd − .
| |
23. Let f be a function such that lim x l 0 f sxd − 6. Then there
exists a positive number such that if 0 , x , , then
f sxd 2 6 , 1.
| |
24. I f f sxd . 1 for all x and lim x l 0 f sxd exists, then
lim x l 0 f sxd . 1.
25. The equation x 10 2 10x 2 1 5 − 0 has a solution in the
interval s0, 2d.
15. If lim x l a f sxd exists but lim x l a tsxd does not exist, then
lim x l a f f sxd 1 tsxdg does not exist.
26. If f is continuous at a, so is f .
16. If lim x l 6 f f sxd tsxdg exists, then the limit must be f s6d ts6d.
27. If f is continuous at a, so is f .
| |
| |
1.Let f be the function whose graph is given.
2. Determine
whether each curve is the graph of a function of x.
If it is, state the domain and range of the function.
3. If f sxd − x 2 2 2x 1 3, evaluate the difference quotient
(f )
Estimate the value of f s2d.
Estimate the values of x such that f sxd − 3.
State the domain of f.
State the range of f.
On what interval is f increasing?
Is f even, odd, or neither even nor odd? Explain.
f sa 1 hd 2 f sad
4. Sketch
a rough graph of the yield of a crop as a function of
the amount of fertilizer used.
5–8 Find the domain and range of the function. Write your
answer in interval notation.
5. f sxd − 2ys3x 2 1d
6. tsxd − s16 2 x 4
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1Review
7. y − 1 1 sin x
8. Fstd − 3 1 cos 2t
9. Suppose
that the graph of f is given. Describe how the graphs
of the following functions can be obtained from the graph of f.
(a) y − f sxd 1 5
(b) y − f sx 1 5d
(c) y − 1 1 2 f sxd
(d) y − f sx 2 2d 2 2
(e) y − 2f sxd
(f ) y − 3 2 f sxd
10. T
he graph of f is given. Draw the graphs of the following
(a) y − f sx 2 8d
(b) y − 2f sxd
(c) y − 2 2 f sxd
(d) y − 12 f sxd 2 1
born in the United States. Use a scatter plot to choose an
appropriate type of model. Use your model to predict the life
span of a male born in the year 2030.
Birth year Life expectancy
Birth year Life expectancy
11–16 Use transformations to sketch the graph of the function.
11. f sxd − x 3 1 2
12. f sxd − sx 2 3d2
13. y − sx 1 2
14. y −
22. A
small-appliance manufacturer finds that it costs $9000 to
produce 1000 toaster ovens a week and $12,000 to produce
1500 toaster ovens a week.
(a)Express the cost as a function of the number of toaster
ovens produced, assuming that it is linear. Then sketch
the graph.
(b)What is the slope of the graph and what does it
(c)What is the y-intercept of the graph and what does it
23. The graph of f is given.
15. tsxd − 1 1 cos 2x
16. f sxd −
1 1 x if x , 0
1 1 x 2 if x > 0
17. Determine whether f is even, odd, or neither even nor odd.
(a) f sxd − 2x 2 3x 1 2
(b) f sxd − x 2 x
(c) f sxd − 1 2 cos 2x
(d) f sxd − 1 1 sin x
(e) f sxd − sx 1 1d2
18. Find
an expression for the function whose graph consists of
the line segment from the point s22, 2d to the point s21, 0d
together with the top half of the circle with center the origin
and radius 1.
19. I f f sxd − sx and tsxd − sin x, find the functions
(a) f 8 t, (b) t 8 f , (c) f 8 f , (d) t 8 t, and their domains.
20. Express the function Fsxd − 1ysx 1 sx as a composition of
three functions.
21. Life
expectancy has improved dramatically in recent decades.
The table gives the life expectancy at birth (in years) of males
(a) Find each limit, or explain why it does not exist.
(i) lim1 f sxd
(ii) lim 1 f sxd (iii) lim f sxd
x l2
x l 23
(iv) lim f sxd
(v) lim f sxd
x l 23
(vi) lim2 f sxd
(b) State the equations of the vertical asymptotes.
(c) At what numbers is f discontinuous? Explain.
24. Sketch
the graph of an example of a function f that satisfies
all of the following conditions:
lim f sxd − 22,lim2 f sxd − 1,f s0d − 21,
x l01
x l0
lim2 f sxd − `,lim1 f sxd − 2`
x l2
x l2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Functions and Limits
25–38 Find the limit.
25. lim cossx 3 1 3xd
x2 2 9
x 1 2x 2 3
27. lim
x l 23
45. Let
26. lim
x l3
x2 2 9
x 1 2x 2 3
28. lim1
x l1
29. lim
sh 2 1d3 1 1
30. lim
31. lim
sr 2 9d4
32. lim1
r 2 2 3r 2 4
4r 2 1 r 2 3
33. lim
r l 21
35. lim
4 2 ss
s 2 16
37. lim
1 2 s1 2 x 2
38. lim
t l2
36. lim
x l3
(v) lim1 f sxd
x l3
(iii) lim f sxd
(vi) lim f sxd
x l3
(b) Where is f discontinuous?
(c) Sketch the graph of f.
46. Let
3 2 st 1 4
v2 1 2v 2 8
v l2
x 2 3x 1 2
(a) Evaluate each limit, if it exists.
(ii) lim2 f sxd
(i) lim1 f sxd
(iv) lim2 f sxd
t2 2 4
t3 2 8
34. lim
x2 2 9
x 1 2x 2 3
if x , 0
f sxd − 3 2 x
if 0 < x , 3
sx 2 3d2 if x . 3
v 2 16
2x 2 x 2
tsxd −
(a)For each of the numbers 2, 3, and 4, determine whether t
is continuous from the left, continuous from the right, or
continuous at the number.
(b) Sketch the graph of t.
47–48 Show that the function is continuous on its domain. State
the domain.
sx 2 2 9
47. hsxd − s
x 1 x 3 cos x
48. tsxd − 2
x 22
39. If 2x 2 1 < f sxd < x 2 for 0 , x , 3, find lim x l1 f sxd.
49–50 Use the Intermediate Value Theorem to show that there is
a solution of the equation in the given interval.
40. Prove that lim x l 0 x 2 coss1yx 2 d − 0.
49. x 5 2 x 3 1 3x 2 5 − 0, s1, 2d
41–44 Prove the statement using the precise definition of
a limit.
50. 2 sin x − 3 2 2x,s0, 1d
41. lim s14 2 5xd − 4
42. lim s
x −0
43. lim sx 2 2 3xd − 22
44. lim1
sx 2 4
51. Suppose that f sxd < tsxd for all x, where lim x l a tsxd − 0.
Find lim x l a f sxd.
52. Let f sxd − v x b 1 v 2x b .
(a) For what values of a does lim x l a f sxd exist?
(b) At what numbers is f discontinuous?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Principles of Problem Solving
There are no hard and fast rules that will ensure success in solving problems. However,
it is possible to outline some general steps in the problem-solving process and to give
some principles that may be useful in the solution of certain problems. These steps and
principles are just common sense made explicit. They have been adapted from George
Polya’s book How To Solve It.
The first step is to read the problem and make sure that you understand it clearly. Ask
yourself the following questions:
What is the unknown?
What are the given quantities?
What are the given conditions?
For many problems it is useful to
draw a diagram
and identify the given and required quantities on the diagram.
Usually it is necessary to
introduce suitable notation
In choosing symbols for the unknown quantities we often use letters such as a, b, c, m, n,
x, and y, but in some cases it helps to use initials as suggestive symbols; for instance, V
for volume or t for time.
2 THINK OF A PLAN
Find a connection between the given information and the unknown that will enable you
to calculate the unknown. It often helps to ask yourself explicitly: “How can I relate the
given to the unknown?” If you don’t see a connection immediately, the following ideas
may be helpful in devising a plan.
Try to Recognize Something FamiliarRelate the given situation to previous knowledge.
Look at the unknown and try to recall a more familiar problem that has a similar unknown.
Try to Recognize PatternsSome problems are solved by recognizing that some kind of
pattern is occurring. The pattern could be geometric, or numerical, or algebraic. If you can
see regularity or repetition in a problem, you might be able to guess what the continuing
pattern is and then prove it.
Use AnalogyTry to think of an analogous problem — that is, a similar problem, a related
problem — but one that is easier than the original problem. If you can solve the similar,
simpler problem, then it might give you the clues you need to solve the original, more
difficult problem. For instance, if a problem involves very large numbers, you could first
try a similar problem with smaller numbers. Or if the problem involves three-dimensional
geometry, you could look for a similar problem in two-dimensional geometry. Or if the
problem you start with is a general one, you could first try a special case.
Introduce Something ExtraIt may sometimes be necessary to introduce something
new — an auxiliary aid — to help make the connection between the given and the unknown.
For instance, in a problem where a diagram is useful the auxiliary aid could be a new line
drawn in a diagram. In a more algebraic problem it could be a new unknown that is
related to the original unknown.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Take CasesWe may sometimes have to split a problem into several cases and give a
different argument for each of the cases. For instance, we often have to use this strategy
in dealing with absolute value.
Work BackwardSometimes it is useful to imagine that your problem is solved and then
to work backward, step by step, until you arrive at the given data. At this point you may
be able to reverse your steps and thereby construct a solution to the original problem.
This procedure is commonly used in solving equations. For instance, in solving the equation 3x 2 5 − 7, we suppose that x is a number that satisfies 3x 2 5 − 7 and work
backward. We add 5 to each side of the equation and then divide each side by 3 to get
x − 4. Since each of these steps can be reversed, we have solved the problem.
Establish SubgoalsIn a complex problem it is often useful to set subgoals (in which the
desired situation is only partially fulfilled). If we can first reach these subgoals, then we
may be able to build on them to reach our final goal.
Indirect ReasoningSometimes it is appropriate to attack a problem indirectly. In using
proof by contradiction to prove that P implies Q, we assume that P is true and Q is false
and try to see why this can’t happen. Somehow we have to use this information and arrive
at a contradiction to what we absolutely know is true.
Mathematical InductionIn proving statements that involve a positive integer n, it is
frequently helpful to use the following principle.
Principle of Mathematical Induction Let Sn be a statement about the positive
integer n. Suppose that
1. S1 is true.
2. Sk11 is true whenever Sk is true.
Then Sn is true for all positive integers n.
This is reasonable because, since S1 is true, it follows from condition 2 swith k − 1d
that S2 is true. Then, using condition 2 with k − 2, we see that S3 is true. Again using
condition 2, this time with k − 3, we have that S4 is true. This procedure can be followed
3 CARRY OUT THE PLAN
In Step 2 a plan was devised. In carrying out that plan we have to check each stage of the
plan and write the details that prove that each stage is correct.
4 LOOK BACK
Having completed our solution, it is wise to look back over it, partly to see if we have
made errors in the solution and partly to see if we can think of an easier way to solve the
problem. Another reason for looking back is that it will familiarize us with the method of
solution and this may be useful for solving a future problem. Descartes said, “Every
problem that I solved became a rule which served afterwards to solve other problems.”
These principles of problem solving are illustrated in the following examples. Before
you look at the solutions, try to solve these problems yourself, referring to these prin­
ciples of problem solving if you get stuck. You may find it useful to refer to this section
from time to time as you solve the exercises in the remaining chapters of this book.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
As the first example illustrates, it is often necessary to use the problem-solving prin­
ciple of taking cases when dealing with absolute values.
EXAMPLE 1 Solve the inequality | x 2 3 | 1 | x 1 2 | , 11.
SOLUTION Recall the definition of absolute value:
|x| −
It follows that
|x 2 3| −
|x 1 2| −
PS Take cases.
if x > 0
2x if x , 0
2sx 2 3d
if x 2 3 > 0
if x 2 3 , 0
2x 1 3
if x > 3
if x , 3
2sx 1 2d
if x 1 2 > 0
if x 1 2 , 0
2x 2 2
if x > 22
if x , 22
These expressions show that we must consider three cases:
x , 2222 < x , 3x > 3
CASE I If x , 22, we have
| x 2 3 | 1 | x 1 2 | , 11
2x 1 3 2 x 2 2 , 11
22x , 10
x . 25
CASE II If 22 < x , 3, the given inequality becomes
2x 1 3 1 x 1 2 , 11
5 , 11(always true)
CASE III If x > 3, the inequality becomes
x 2 3 1 x 1 2 , 11
2x , 12
Combining cases I, II, and III, we see that the inequality is satisfied when 25 , x , 6.
So the solution is the interval s25, 6d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
In the following example we first guess the answer by looking at special cases and
recognizing a pattern. Then we prove our conjecture by mathematical induction.
In using the Principle of Mathematical Induction, we follow three steps:
Step 1 Prove that Sn is true when n − 1.
Step 2 Assume that Sn is true when n − k and deduce that Sn is true when n − k 1 1.
Step 3 Conclude that Sn is true for all n by the Principle of Mathematical Induction.
EXAMPLE 2 If f0sxd − xysx 1 1d and fn11 − f0 8 fn for n − 0, 1, 2, . . . , find a formula
for fnsxd.
PS Analogy: Try a similar, simpler
SOLUTION We start by finding formulas for fnsxd for the special cases n − 1, 2, and 3.
S D
f1sxd − s f0 8 f0dsxd − f0( f0sxd) − f0
2x 1 1
2x 1 1
f2sxd − s f0 8 f1 dsxd − f0( f1sxd) − f0
2x 1 1
2x 1 1
2x 1 1
3x 1 1
3x 1 1
2x 1 1
2x 1 1
f3sxd − s f0 8 f2 dsxd − f0( f2sxd) − f0
3x 1 1
3x 1 1
3x 1 1
4x 1 1
4x 1 1
3x 1 1
3x 1 1
PS Look for a pattern.
We notice a pattern: the coefficient of x in the denominator of fnsxd is n 1 1 in the
three cases we have computed. So we make the guess that, in general,
fnsxd −
sn 1 1dx 1 1
To prove this, we use the Principle of Mathematical Induction. We have already verified
that (1) is true for n − 1. Assume that it is true for n − k, that is,
fksxd −
sk 1 1dx 1 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
fk11sxd − s f0 8 fk dsxd − f0( fksxd) − f0
sk 1 1dx 1 1
sk 1 1d x 1 1
sk 1 1d x 1 1
sk 1 2dx 1 1
sk 1 2d x 1 1
sk 1 1dx 1 1
sk 1 1dx 1 1
This expression shows that (1) is true for n − k 1 1. Therefore, by mathematical
induction, it is true for all positive integers n.
In the following example we show how the problem-solving strategy of introducing
something extra is sometimes useful when we evaluate limits. The idea is to change the
variable — to introduce a new variable that is related to the original variable —in such a
way as to make the problem simpler. Later, in Section 4.5, we will make more extensive
use of this general idea.
EXAMPLE 3 Evaluate lim
1 1 cx 2 1
, where c is a constant.
SOLUTION As it stands, this limit looks challenging. In Section 1.6 we evaluated limits
in which both numerator and denominator approached 0. There our strategy was to
perform some sort of algebraic manipulation that led to a simplifying cancellation, but
here it’s not clear what kind of algebra is necessary.
So we introduce a new variable t by the equation
1 1 cx
We also need to express x in terms of t, so we solve this equation:
t 3 − 1 1 cxx −
t3 2 1
if c ± 0
Notice that x l 0 is equivalent to t l 1. This allows us to convert the given limit into
one involving the variable t :
1 1 cx 2 1
− lim 3
t l1 st 2 1dyc
− lim
t l1
cst 2 1d
t3 2 1
The change of variable allowed us to replace a relatively complicated limit by a simpler
one of a type that we have seen before. Factoring the denominator as a difference of
cubes, we get
t l1
cst 2 1d
cst 2 1d
− lim
t l1 st 2 1dst 2 1 t 1 1d
t3 2 1
− lim
t l1
t2 1 t 1 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
In making the change of variable we had to rule out the case c − 0. But if c − 0, the
function is 0 for all nonzero x and so its limit is 0. Therefore, in all cases, the limit
is cy3.
The following problems are meant to test and challenge your problem-solving skills.
Some of them require a considerable amount of time to think through, so don’t be discouraged if you can’t solve them right away. If you get stuck, you might find it helpful to
refer to the discussion of the principles of problem solving.
|| − 3.
Solve the inequality | x 2 1 | 2 | x 2 3 | > 5.
1. Solve the equation 4x 2 x 1 1
| | |
Sketch the graph of the function tsxd − | x 2 1 | 2 | x 2 4 |.
Draw the graph of the equation x 1 | x | − y 1 | y |.
3. Sketch the graph of the function f sxd − x 2 2 4 x 1 3 .
6. Sketch the region in the plane consisting of all points sx, yd such that
|x 2 y| 1 |x| 2 |y| < 2
7. The notation maxha, b, . . .j means the largest of the numbers a, b, . . . . Sketch the graph of
each function.
(a)f sxd − maxhx, 1yxj
(b) f sxd − maxhsin x, cos xj
(c) f sxd − maxhx 2, 2 1 x, 2 2 xj
8. Sketch the region in the plane defined by each of the following equations or inequalities.
(a)maxhx, 2yj − 1
(b) 21 < maxhx, 2yj < 1
(c) maxhx, y 2 j − 1
9. Find the exact value of
1 sin
1 sin
1 ∙ ∙ ∙ 1 sin
10. Find the number of solutions of the equation sin x −
11. A
driver sets out on a journey. For the first half of the distance she drives at the leisurely
pace of 30 miyh; she drives the second half at 60 miyh. What is her average speed on
this trip?
12. Is it true that f 8 s t 1 hd − f 8 t 1 f 8 h ?
13. Prove that if n is a positive integer, then 7 n 2 1 is divisible by 6.
14. Prove that 1 1 3 1 5 1 ∙ ∙ ∙ 1 s2n 2 1d − n 2.
15. If f0sxd − x 2 and fn11sxd − f0s fnsxdd for n − 0, 1, 2, . . . , find a formula for fnsxd.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
and fn11 − f0 8 fn for n − 0, 1, 2, . . . , find an expression for fnsxd and
use mathematical induction to prove it.
16. (a)If f0sxd −
(b)Graph f0 , f1, f2 , f3 on the same screen and describe the effects of repeated composition.
; 17. Evaluate lim
x l1
x 21
sx 2 1
18. Find numbers a and b such that
x l0
19. Evaluate lim
sax 1 b 2 2
| 2x 2 1 | 2 | 2x 1 1 | .
20. T
he figure shows a point P on the parabola y − x 2 and the point Q where the perpendicular
bisector of OP intersects the y-axis. As P approaches the origin along the parabola, what
happens to Q? Does it have a limiting position? If so, find it.
21. Evaluate the following limits, if they exist, where v x b denotes the greatest integer function.
(a) lim
v xb
(b) lim x v 1yx b
22. Sketch the region in the plane defined by each of the following equations.
(a) v xb 2 1 v yb 2 − 1
(b) v xb 2 2 v yb 2 − 3
(c) v x 1 yb 2 − 1
(d) v xb 1 v yb − 1
23. Find all values of a such that f is continuous on R:
f sxd −
x 1 1 if x < a
if x . a
24. A fixed point of a function f is a number c in its domain such that f scd − c. (The function
doesn’t move c; it stays fixed.)
(a)Sketch the graph of a continuous function with domain f0, 1g whose range also lies
in f0, 1g. Locate a fixed point of f .
(b)Try to draw the graph of a continuous function with domain f0, 1g and range in f0, 1g
that does not have a fixed point. What is the obstacle?
(c)Use the Intermediate Value Theorem to prove that any continuous function with domain
f0, 1g and range in f0, 1g must have a fixed point.
25. If lim x l a f f sxd 1 tsxdg − 2 and lim x l a f f sxd 2 tsxdg − 1, find lim x l a f f sxd tsxdg.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
26. (a)The figure shows an isosceles triangle ABC with /B − /C. The bisector of angle B
intersects the side AC at the point P. Suppose that the base BC remains fixed but the
altitude AM of the triangle approaches 0, so A approaches the midpoint M of BC.
What happens to P during this process? Does it have a limiting position? If so, find it.
(b)Try to sketch the path traced out by P during this process. Then find an equation of this
curve and use this equation to sketch the curve.
27. (a)If we start from 0° latitude and proceed in a westerly direction, we can let Tsxd denote
the temperature at the point x at any given time. Assuming that T is a continuous
function of x, show that at any fixed time there are at least two diametrically opposite
points on the equator that have exactly the same temperature.
(b)Does the result in part (a) hold for points lying on any circle on the earth’s surface?
(c)Does the result in part (a) hold for barometric pressure and for altitude above sea level?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
We know that when an object is dropped from a height it falls faster and faster. Galileo discovered that the distance the object has
fallen is proportional to the square of the time elapsed. Calculus enables us to calculate the precise speed of the object at any time.
In Exercise 2.1.11 you are asked to determine the speed at which a cliff diver plunges into the ocean.
Icealex / Shutterstock.com
IN THIS CHAPTER WE BEGIN our study of differential calculus, which is concerned with how one
quantity changes in relation to another quantity. The central concept of differential calculus is the
derivative, which is an outgrowth of the velocities and slopes of tangents that we considered in
Chapter 1. After learning how to calculate derivatives, we use them to solve problems involving
rates of change and the approximation of functions.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
2.1 Derivatives and Rates of Change
In Chapter 1 we defined limits and learned techniques for computing them. We now
revisit the problems of finding tangent lines and velocities from Section 1.4. The special
type of limit that occurs in both of these problems is called a derivative and we will see
that it can be interpreted as a rate of change in any of the natural or social sciences or
■ Tangents
Q{ x, ƒ }
P { a, f(a)}
If a curve C has equation y − f sxd and we want to find the tangent line to C at the point
Psa, f sadd, then we consider (as we did in Section 1.4) a nearby point Qsx, f sxdd, where
x ± a, and compute the slope of the secant line PQ:
mPQ −
Then we let Q approach P along the curve C by letting x approach a. If mPQ approaches
a number m, then we define the tangent line , to be the line through P with slope m.
(This amounts to saying that the tangent line is the limiting position of the secant line PQ
as Q approaches P. See Figure 1.)
1 Definition The tangent line to the curve y − f sxd at the point Psa, f sadd is
the line through P with slope
f sxd 2 f sad
m − lim
f sxd 2 f sad
provided that this limit exists.
FIGURE 1
In our first example we confirm the guess we made in Example 1.4.1.
EXAMPLE 1 Find an equation of the tangent line to the parabola y − x 2 at the
point Ps1, 1d.
SOLUTION Here we have a − 1 and f sxd − x 2, so the slope is
m − lim
x l1
− lim
x l1
f sxd 2 f s1d
x2 2 1
− lim
x l1 x 2 1
sx 2 1dsx 1 1d
− lim sx 1 1d − 1 1 1 − 2
x l1
Point-slope form for a line through
the point sx1 , y1 d with slope m:
y 2 y1 − msx 2 x 1 d
Using the point-slope form of the equation of a line, we find that an equation of the
tangent line at s1, 1d is
y 2 1 − 2sx 2 1dory − 2x 2 1
We sometimes refer to the slope of the tangent line to a curve at a point as the slope
of the curve at the point. The idea is that if we zoom in far enough toward the point,
the curve looks almost like a straight line. Figure 2 illustrates this procedure for the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.1Derivatives and Rates of Change
curve y − x 2 in Example 1. The more we zoom in, the more the parabola looks like a
line. In other words, the curve becomes almost indistinguishable from its tangent line.
(1, 1)
(1, 1)
(1, 1)
FIGURE 2 Zooming in toward the point (1, 1) on the parabola y − x 2
Q { a+h, f(a+h)}
There is another expression for the slope of a tangent line that is sometimes easier to
use. If h − x 2 a, then x − a 1 h and so the slope of the secant line PQ is
mPQ −
P { a, f(a)}
FIGURE 3
f sa 1 hd 2 f sad
(See Figure 3 where the case h . 0 is illustrated and Q is located to the right of P. If it
happened that h , 0, however, Q would be to the left of P.)
Notice that as x approaches a, h approaches 0 (because h − x 2 a) and so the expression for the slope of the tangent line in Definition 1 becomes
m − lim
f sa 1 hd 2 f sad
EXAMPLE 2 Find an equation of the tangent line to the hyperbola y − 3yx at the
point s3, 1d.
SOLUTION Let f sxd − 3yx. Then, by Equation 2, the slope of the tangent at s3, 1d is
m − lim
f s3 1 hd 2 f s3d
3 2 s3 1 hd
− lim
− lim
− lim
Therefore an equation of the tangent at the point s3, 1d is
(3, 1)
y 2 1 − 213 sx 2 3d
which simplifies to
FIGURE 4
− lim 2
hs3 1 hd
x 1 3y 2 6 − 0
The hyperbola and its tangent are shown in Figure 4.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
■ Velocities
position at
time t=a
position at
time t=a+h
FIGURE 5
In Section 1.4 we investigated the motion of a ball dropped from the CN Tower and
defined its velocity to be the limiting value of average velocities over shorter and shorter
time periods.
In general, suppose an object moves along a straight line according to an equation of
motion s − f std, where s is the displacement (directed distance) of the object from the
origin at time t. The function f that describes the motion is called the position function
of the object. In the time interval from t − a to t − a 1 h , the change in position is
f sa 1 hd 2 f sad. (See Figure 5.)
The average velocity over this time interval is
average velocity −
Q { a+h, f(a+h)}
P { a, f(a)}
which is the same as the slope of the secant line PQ in Figure 6.
Now suppose we compute the average velocities over shorter and shorter time intervals fa, a 1 hg. In other words, we let h approach 0. As in the example of the falling ball,
we define the velocity (or instantaneous velocity) vsad at time t − a to be the limit of
these average velocities.
3 Definition The instantaneous velocity of an object with position function
f std at time t − a is
vsad − lim
average velocity
FIGURE 6
f sa 1 hd 2 f sad
f sa 1 hd 2 f sad
provided that this limit exists.
This means that the velocity at time t − a is equal to the slope of the tangent line at P
(compare Equation 2 and the expression in Definition 3).
Now that we know how to compute limits, let’s reconsider the problem of the falling ball.
EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of the
CN Tower, 450 m above the ground.
(a) What is the velocity of the ball after 5 seconds?
(b) How fast is the ball traveling when it hits the ground?
Recall from Section 1.4: The dis­tance
(in meters) fallen after t seconds
is 4.9t 2.
SOLUTION Since two different velocities are requested, it’s efficient to start by finding
the velocity at a general time t − a. Using the equation of motion s − f std − 4.9t 2,
we have
v sad − lim
f sa 1 hd 2 f sad
4.9sa 1 hd2 2 4.9a 2
− lim
− lim
4.9sa 2 1 2ah 1 h 2 2 a 2 d
4.9s2ah 1 h 2 d
− lim
− lim
4.9hs2a 1 hd
− lim 4.9s2a 1 hd − 9.8a
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.1Derivatives and Rates of Change
(a) The velocity after 5 seconds is vs5d − s9.8ds5d − 49 mys.
(b) Since the observation deck is 450 m above the ground, the ball will hit the ground
at the time t when sstd − 450, that is,
4.9t 2 − 450
This gives
t2 −
andt −
< 9.6 s
The velocity of the ball as it hits the ground is therefore
SÎ D Î
− 9.8
< 94 mys
■ Derivatives
We have seen that the same type of limit arises in finding the slope of a tangent line
(Equation 2) or the velocity of an object (Definition 3). In fact, limits of the form
h l0
f sa 1 hd 2 f sad
arise whenever we calculate a rate of change in any of the sciences or engineering, such
as a rate of reaction in chemistry or a marginal cost in economics. Since this type of limit
occurs so widely, it is given a special name and notation.
4 Definition The derivative of a function f at a number a, denoted by f 9sad,
f 9sad − lim
f 9sad is read “ f prime of a.”
h l0
f sa 1 hd 2 f sad
if this limit exists.
If we write x − a 1 h, then we have h − x 2 a and h approaches 0 if and only if
x approaches a. Therefore an equivalent way of stating the definition of the derivative,
as we saw in finding tangent lines (see Definition 1), is
f 9sad − lim
f sxd 2 f sad
EXAMPLE 4 Use Definition 4 to find the derivative of the function f sxd − x 2 2 8x 1 9
at the numbers (a) 2 and (b) a.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
Definitions 4 and 5 are equivalent, so
we can use either one to compute the
derivative. In practice, Definition 4
often leads to simpler computations.
(a) From Definition 4 we have
f 9s2d − lim
h l0
− lim
s2 1 hd2 2 8s2 1 hd 1 9 2 s23d
− lim
4 1 4h 1 h 2 2 16 2 8h 1 9 1 3
− lim
h 2 2 4h
hsh 2 4d
− lim
− lim sh 2 4d − 24
h l0
h l0
h l0
h l0
h l0
f s2 1 hd 2 f s2d
f 9sad − lim
h l0
f sa 1 hd 2 f sad
− lim
fsa 1 hd2 2 8sa 1 hd 1 9g 2 fa 2 2 8a 1 9g
− lim
a 2 1 2ah 1 h 2 2 8a 2 8h 1 9 2 a 2 1 8a 2 9
− lim
2ah 1 h 2 2 8h
− lim s2a 1 h 2 8d − 2a 2 8
h l0
h l0
h l0
h l0
As a check on our work in part (a), notice that if we let a − 2, then
f 9s2d − 2s2d 2 8 − 24.
EXAMPLE 5 Use Equation 5 to find the derivative of the function f sxd − 1ysx at the
number a sa . 0d.
SOLUTION From Equation 5 we get
f 9sad − lim
f sxd 2 f sad
− lim
− lim
− lim
sa sx sa
− lim
sx sa
sa 2 sx
sax sx 2 ad
− lim
sa 2 sx
sax sx 2 ad
2sx 2 ad
sax sx 2 ad(sa 1 sx )
sa 2 (sa 1 sa )
a ? 2 sa
− lim
sa 1 sx
sa 1 sx
sax (sa 1 sx )
You can verify that using Definition 4 gives the same result.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.1Derivatives and Rates of Change
We defined the tangent line to the curve y − f sxd at the point Psa, f sadd to be the line
that passes through P and has slope m given by Equation 1 or 2. Since, by Defini­tion 4
(and Equation 5), this is the same as the derivative f 9sad, we can now say the following.
The tangent line to y − f sxd at sa, f sadd is the line through sa, f sadd whose slope is
equal to f 9sad, the derivative of f at a.
If we use the point-slope form of the equation of a line, we can write an equation of
the tangent line to the curve y − f sxd at the point sa, f sadd:
y 2 f sad − f 9sadsx 2 ad
EXAMPLE 6 Find an equation of the tangent line to the parabola y − x 2 2 8x 1 9 at
the point s3, 26d.
SOLUTION From Example 4(b) we know that the derivative of f sxd − x 2 2 8x 1 9 at
the number a is f 9sad − 2a 2 8. Therefore the slope of the tangent line at s3, 26d is
f 9s3d − 2s3d 2 8 − 22. Thus an equation of the tangent line, shown in Figure 7, is
y 2 s26d − s22dsx 2 3dory − 22x
(3, _6)
■ Rates of Change
Suppose y is a quantity that depends on another quantity x. Thus y is a function of x and
we write y − f sxd. If x changes from x 1 to x 2 , then the change in x (also called the increment of x) is
Dx − x 2 2 x 1
FIGURE 7
and the corresponding change in y is
Dy − f sx 2d 2 f sx 1d
The difference quotient
f sx 2d 2 f sx 1d
x2 2 x1
is called the average rate of change of y with respect to x over the interval fx 1, x 2g and
can be interpreted as the slope of the secant line PQ in Figure 8.
Q { ¤, ‡}
P {⁄, fl}
average rate of change mPQ
FIGURE 8
instantaneous rate of change slope of tangent at P
By analogy with velocity, we consider the average rate of change over smaller and
smaller intervals by letting x 2 approach x 1 and therefore letting Dx approach 0. The limit
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
of these average rates of change is called the (instantaneous) rate of change of y with
respect to x at x − x1, which (as in the case of velocity) is interpreted as the slope of the
tangent to the curve y − f sxd at Psx 1, f sx 1dd:
instantaneous rate of change − lim
Dx l 0
f sx2 d 2 f sx1d
− lim
x2 2 x1
We recognize this limit as being the derivative f 9sx 1d.
We know that one interpretation of the derivative f 9sad is as the slope of the tangent
line to the curve y − f sxd when x − a.We now have a second interpretation:
The derivative f 9sad is the instantaneous rate of change of y − f sxd with respect
to x when x − a.
FIGURE 9
The y-values are changing rapidly
at P and slowly at Q.
The connection with the first interpretation is that if we sketch the curve y − f sxd,
then the instantaneous rate of change is the slope of the tangent to this curve at the point
where x − a. This means that when the derivative is large (and therefore the curve is
steep, as at the point P in Figure 9), the y-values change rapidly. When the derivative is
small, the curve is relatively flat (as at point Q) and the y-values change slowly.
In particular, if s − f std is the position function of a particle that moves along a
straight line, then f 9sad is the rate of change of the displacement s with respect to the
time t. In other words, f 9sad is the velocity of the particle at time t − a. The speed of the
particle is the absolute value of the velocity, that is, f 9sad .
In the next example we discuss the meaning of the derivative of a function that is
defined verbally.
EXAMPLE 7 A manufacturer produces bolts of a fabric with a fixed width. The cost of
producing x meters of this fabric is C − f sxd dollars.
(a) What is the meaning of the derivative f 9sxd? What are its units?
(b) In practical terms, what does it mean to say that f 9s1000d − 9?
(c) Which do you think is greater, f 9s50d or f 9s500d? What about f 9s5000d?
(a) The derivative f 9sxd is the instantaneous rate of change of C with respect to x; that
is, f 9sxd means the rate of change of the production cost with respect to the number of
meters produced. (Economists call this rate of change the marginal cost. This idea is
discussed in more detail in Sections 3.7 and 4.7.)
f 9sxd − lim
Dx l 0 Dx
Here we are assuming that the cost
function is well behaved; in other
words, Csxd doesn’t oscillate rapidly
near x − 1000.
the units for f 9sxd are the same as the units for the difference quotient DCyDx. Since DC is
measured in dollars and Dx in meters, it follows that the units for f 9sxd are dollars per meter.
(b) The statement that f 9s1000d − 9 means that, after 1000 meters of fabric have been
manufactured, the rate at which the production cost is increasing is $9ymeter. (When
x − 1000, C is increasing 9 times as fast as x.)
Since Dx − 1 is small compared with x − 1000, we could use the approximation
f 9s1000d <
− DC
and say that the cost of manufacturing the 1000th meter (or the 1001st) is about $9.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.1Derivatives and Rates of Change
(c) The rate at which the production cost is increasing (per meter) is probably lower
when x − 500 than when x − 50 (the cost of making the 500th meter is less than the
cost of the 50th meter) because of economies of scale. (The manufacturer makes more
efficient use of the fixed costs of production.) So
f 9s50d . f 9s500d
But, as production expands, the resulting large-scale operation might become inefficient and there might be overtime costs. Thus it is possible that the rate of increase of
costs will eventually start to rise. So it may happen that
f 9s5000d . f 9s500d
In the following example we estimate the rate of change of the national debt with
respect to time. Here the function is defined not by a formula but by a table of values.
EXAMPLE 8 Let Dstd be the US national debt at time t. The table in the margin gives
approximate values of this function by providing end of year estimates, in billions of
dollars, from 2000 to 2016. Interpret and estimate the value of D9s2008d.
SOLUTION The derivative D9s2008d means the rate of change of D with respect to t
when t − 2008, that is, the rate of increase of the national debt in 2008.
According to Equation 5,
Source: US Dept. of the Treasury
D9s2008d − lim
t l 2008
Dstd 2 Ds2008d
t 2 2008
One way we can estimate this value is to compare average rates of change over
different time intervals by computing difference quotients, as compiled in the following table.
A Note on Units
The units for the average rate of
change D DyDt are the units for D D
divided by the units for Dt, namely,
billions of dollars per year. The instantaneous rate of change is the limit of
the average rates of change, so it is
measured in the same units: billions
of dollars per year.
Time interval
Average rate of change −
[2000, 2008]
[2004, 2008]
[2008, 2012]
[2008, 2016]
Dstd 2 Ds2008d
t 2 2008
From this table we see that D9s2008d lies somewhere between 775.93 and 1433.23 billion dollars per year. [Here we are making the reasonable assumption that the debt didn’t
fluctuate wildly between 2004 and 2012.] A good estimate for the rate of increase of the
US national debt in 2008 would be the average of these two numbers, namely
D9s2008d < 1105 billion dollars per year
Another method would be to plot the debt function and estimate the slope of the
tangent line when t − 2008.
In Examples 3, 7, and 8 we saw three specific examples of rates of change: the velocity
of an object is the rate of change of displacement with respect to time; marginal cost is the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
rate of change of production cost with respect to the number of items produced; the rate
of change of the debt with respect to time is of interest in economics. Here is a small
sample of other rates of change: In physics, the rate of change of work with respect to time
is called power. Chemists who study a chemical reaction are interested in the rate of
change in the concentration of a reactant with respect to time (called the rate of reaction).
A biologist is interested in the rate of change of the population of a colony of bacteria
with respect to time. In fact, the computation of rates of change is important in all of the
natural sciences, in engineering, and even in the social sciences. Further examples will
be given in Section 2.7.
All these rates of change are derivatives and can therefore be interpreted as slopes of
tangents. This gives added significance to the solution of the tangent problem. Whenever
we solve a problem involving tangent lines, we are not just solving a problem in geometry. We are also implicitly solving a great variety of problems involving rates of change
in science and engineering.
1. A curve has equation y − f sxd.
(a)Write an expression for the slope of the secant line
through the points Ps3, f s3dd and Qsx, f sxdd.
(b)Write an expression for the slope of the tangent line at P.
raph the curve y − sin x in the viewing rectangles
; 2. G
f22, 2g by f22, 2g, f21, 1g by f21, 1g, and f20.5, 0.5g
by f20.5, 0.5g. What do you notice about the curve as you
zoom in toward the origin?
3. (a)Find the slope of the tangent line to the parabola
y − x 2 1 3x at the point s21, 22d
(i) using Definition 1
(ii) using Equation 2
(b) Find an equation of the tangent line in part (a).
(c)Graph the parabola and the tangent line. As a check on
your work, zoom in toward the point s21, 22d until the
parabola and the tangent line are indistinguishable.
4. (a)Find the slope of the tangent line to the curve
y − x 3 1 1 at the point s1, 2d
(i) using Definition 1
(ii) using Equation 2
(b) Find an equation of the tangent line in part (a).
(c)Graph the curve and the tangent line in successively
smaller viewing rectangles centered at s1, 2d until the
curve and the line appear to coincide.
5–8 Find an equation of the tangent line to the curve at the
given point.
5. y − 2x 2 2 5x 1 1, s3, 4d
7. y −
, s2, 24d
6. y − x 2 2 2x 3, s1, 21d
8. y − s1 2 3x , s21, 2d
9. (a)Find the slope of the tangent to the curve
y − 3 1 4x 2 2 2x 3 at the point where x − a.
(b)Find equations of the tangent lines at the points s1, 5d
and s2, 3d.
(c)Graph the curve and both tangents on a common screen.
10. (a)Find the slope of the tangent to the curve y − 2sx at
the point where x − a.
(b)Find equations of the tangent lines at the points s1, 2d
and s9, 6d.
(c)Graph the curve and both tangents on a common screen.
11. A cliff diver plunges from a height of 30 m above the water
surface. The distance the diver falls in t seconds is given by
the function dstd − 4.9t 2 m.
(a) After how many seconds will the diver hit the water?
(b) With what velocity does the diver hit the water?
12. I f a rock is thrown upward on the planet Mars with a velocity of 10 mys, its height (in meters) after t seconds is given
by H − 10t 2 1.86t 2.
(a) Find the velocity of the rock after one second.
(b) Find the velocity of the rock when t − a.
(c) When will the rock hit the surface?
(d) With what velocity will the rock hit the surface?
13. T
he displacement (in meters) of a particle moving in a
straight line is given by the equation of motion s − 1yt 2,
where t is measured in seconds. Find the velocity of the
par­ticle at times t − a, t − 1, t − 2, and t − 3.
14. T
he displacement (in meters) of a particle moving in a
straight line is given by s − 12 t 2 2 6t 1 23, where t is measured in seconds.
(a)Find the average velocity over each time interval:
(i) f4, 8g
(ii) f6, 8g
(iii) f8, 10g
(iv) f8, 12g
(b) Find the instantaneous velocity when t − 8.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.1Derivatives and Rates of Change
(c)Draw the graph of s as a function of t and draw the secant
lines whose slopes are the average velocities in part (a).
Then draw the tangent line whose slope is the instantaneous velocity in part (b).
(d) Estimate the value of f 9s50d.
(e) Is f 9s10d . f 9s30d?
f s80d 2 f s40d
(f ) Is f 9s60d .
? Explain.
15. (a)A particle starts by moving to the right along a horizontal
line; the graph of its position function is shown in the figure. When is the particle moving to the right? Moving to
the left? Standing still?
(b)Draw a graph of the velocity function.
s (meters)
19. f sxd − s4x 1 1 ,
6 t (seconds)
20. f sxd − 5x 4,
16. S
hown are graphs of the position functions of two runners,
A and B, who run a 100-meter race and finish in a tie.
(a)Describe and compare how the runners run the race.
(b)At what time is the distance between the runners the
(c)At what time do they have the same velocity?
a − 21
21–22 Use Equation 5 to find f 9sad at the given number a.
21. f sxd −
22. f sxd −
s2x 1 2
23–26 Find f 9sad.
s (meters)
23. f sxd − 2x 2 2 5x 1 3
25. f std −
t2 1 1
24. f std − t 3 2 3t
26. f sxd −
1 2 4x
t (seconds)
17. F
or the function t whose graph is given, arrange the following numbers in increasing order and explain your reasoning:
_1 0
27. Find an equation of the tangent line to the graph of y − Bsxd
at x − 6 if Bs6d − 0 and B9s6d − 2 12.
28. Find an equation of the tangent line to the graph of y − tsxd
at x − 5 if ts5d − 23 and t9s5d − 4.
29. I f f sxd − 3x 2 2 x 3, find f 9s1d and use it to find an equation of
the tangent line to the curve y − 3x 2 2 x 3 at the point s1, 2d.
30. I f tsxd − x 4 2 2, find t9s1d and use it to find an equation of
the tangent line to the curve y − x 4 2 2 at the point s1, 21d.
18. T
he graph of a function f is shown.
(a)Find the average rate of change of f on the interval f20, 60g.
(b)Identify an interval on which the average rate of change
of f is 0.
f s40d 2 f s10d
19–20 Use Definition 4 to find f 9sad at the given number a.
What does this value represent geometrically?
31. (a)If Fsxd − 5xys1 1 x 2 d, find F9s2d and use it to find an
equation of the tangent line to the curve y − 5xys1 1 x 2 d
at the point s2, 2d.
(b)Illustrate part (a) by graphing the curve and the tangent
line on the same screen.
32. (a)If Gsxd − 4x 2 2 x 3, find G9sad and use it to find equations of the tangent lines to the curve y − 4x 2 2 x 3 at
the points s2, 8d and s3, 9d.
(b)Illustrate part (a) by graphing the curve and the tangent
lines on the same screen.
33. I f an equation of the tangent line to the curve y − f sxd at the
point where a − 2 is y − 4x 2 5, find f s2d and f 9s2d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
34. I f the tangent line to y − f sxd at (4, 3) passes through the
point (0, 2), find f s4d and f 9s4d.
35–36 A particle moves along a straight line with equation of
motion s − f std, where s is measured in meters and t in seconds.
Find the velocity and the speed when t − 4.
35. f std − 80t 2 6t 2
36. f std − 10 1
37. A
warm can of soda is placed in a cold refrigerator. Sketch
the graph of the temperature of the soda as a function of time.
Is the initial rate of change of temperature greater or less than
the rate of change after an hour?
38. A
roast turkey is taken from an oven when its temperature has
reached 85°C and is placed on a table in a room where the
temperature is 24°C. The graph shows how the temperature of
the turkey decreases and eventually approaches room temperature. By measuring the slope of the tangent, estimate the rate
of change of the temperature after an hour.
t (min)
39. S
ketch the graph of a function f for which f s0d − 0,
f 9s0d − 3, f 9s1d − 0, and f 9s2d − 21.
40. S
ketch the graph of a function t for which
ts0d − ts2d − ts4d − 0, t9s1d − t9s3d − 0,
t9s0d − t9s4d − 1, t9s2d − 21, lim x l52 tsxd − `, and
lim x l 211 tsxd − 2`.
41. S
ketch the graph of a function t that is continuous on its
domain s25, 5d and where ts0d − 1, t9s0d − 1, t9s22d − 0,
lim x l 251 tsxd − `, and lim x l52 tsxd − 3.
42. S
ketch the graph of a function f where the domain is s22, 2d,
f 9s0d − 22, lim x l 22 f sxd − `, f is continuous at all numbers in its domain except 61, and f is odd.
43–48 Each limit represents the derivative of some function f at
some number a. State such an f and a in each case.
s9 1 h 2 3
43. lim
x 2 64
45. lim
x l2 x 2 2
47. lim
1h 21
l y6
sin 2 2
49. T
he cost (in dollars) of producing x units of a certain commodity is Csxd − 5000 1 10x 1 0.05x 2.
(a)Find the average rate of change of C with respect to x
when the production level is changed
(i) from x − 100 to x − 105
(ii) from x − 100 to x − 101
(b)Find the instantaneous rate of change of C with respect to
x when x − 100. (This is called the marginal cost. Its
significance will be explained in Section 2.7.)
50. Let H std be the daily cost (in dollars) to heat an office building when the outside temperature is t degrees Celsius.
(a) What is the meaning of H9s14d? What are its units?
(b)Would you expect H9s14d to be positive or negative?
51. T
he cost of producing x kilograms of gold from a new gold
mine is C − f sxd dollars.
(a)What is the meaning of the derivative f 9sxd? What are its
(b) What does the statement f 9s22d − 17 mean?
(c)Do you think the values of f 9sxd will increase or decrease
in the short term? What about the long term? Explain.
T (°C)
S D
2 31h 2 8
44. lim
46. lim
x l 1y4 x 2 1
52. T
he quantity (in kilograms) of a gourmet ground coffee that is
sold by a coffee company at a price of p dollars per kilogram
is Q − f s pd.
(a)What is the meaning of the derivative f 9s8d? What are its
(b) Is f 9s8d positive or negative? Explain.
53. T
he quantity of oxygen that can dissolve in water depends on
the temperature of the water. (So thermal pollution influences
the oxygen content of water.) The graph shows how oxygen
solubility S varies as a function of the water temperature T.
(a)What is the meaning of the derivative S9sT d? What are its
(b)Estimate the value of S9s16d and interpret it.
S (mg / L)
T (°C)
Source: C. Kupchella et al., Environmental Science: Living Within
the System of Nature, 2d ed. (Boston: Allyn and Bacon, 1989).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.1Derivatives and Rates of Change
54. T
he graph shows the influence of the temperature T on the
maximum sustainable swimming speed S of Coho salmon.
(a)What is the meaning of the derivative S9sT d? What are its
(b)Estimate the values of S9s15d and S9s25d and interpret
(b)Estimate the instantaneous rate of growth in 2010 by
taking the average of two average rates of change.
What are its units?
(c)Estimate the instantaneous rate of growth in 2010 by
measuring the slope of a tangent.
57–58 Determine whether f 9s0d exists.
S (cm/s)
57. f sxd −
58. f sxd −
T (°C)
55. R
esearchers measured the average blood alcohol concen­tration Cstd of eight men starting one hour after consumption
of 30 mL of ethanol (corresponding to two alcoholic drinks).
t (hours)
Cstd sgydLd
(a)Find the average rate of change of C with respect to t
over each time interval:
(i) f1.0, 2.0g
(ii) f1.5, 2.0g
(iii) f2.0, 2.5g
(iv) f2.0, 3.0g
In each case, include the units.
(b)Estimate the instantaneous rate of change at t − 2 and
interpret your result. What are the units?
S ource: Adapted from P. Wilkinson et al., “Pharmacokinetics of Ethanol after
Oral Administration in the Fasting State,” Journal of Pharmacokinetics and
Biopharmaceutics 5 (1977): 207–24.
56. The number N of locations of a popular coffeehouse chain is
given in the table. (The numbers of locations as of October 1
are given.)
(a)Find the average rate of growth
(i) from 2008 to 2010
(ii) from 2010 to 2012
In each case, include the units. What can you conclude?
x sin
if x − 0
x 2 sin
if x ± 0
if x ± 0
if x − 0
; 59. (a)Graph the function
f sxd − sin x 2 1000
in the viewing rectangle f22, 2g by f24, 4g. What slope
does the graph appear to have at the origin?
(b)Zoom in to the viewing window f20.4, 0.4g by
f20.25, 0.25g and estimate the value of f 9s0d. Does
this agree with your answer from part (a)?
(c)Now zoom in to the viewing window f20.008, 0.008g by
f20.005, 0.005g. Do you wish to revise your estimate
for f 9s0d?
60. Symmetric Difference Quotients In Example 8 we approximated an instantaneous rate of change by averaging two average rates of change. An alternative method is to use a single
average rate of change over an interval centered at the desired
value. We define the symmetric difference quotient of a function f at x − a on the interval fa 2 d, a 1 d g as
f sa 1 d d 2 f sa 2 d d
f sa 1 d d 2 f sa 2 d d
sa 1 d d 2 sa 2 d d
(a) Compute the symmetric difference quotient for the
function D in Example 8 on the interval f2004, 2012g
and verify that your result agrees with the estimate for
D9s2008d computed in the example.
(b) Show that the symmetric difference quotient of a function
f at x − a is equivalent to averaging the average rates of
change of f over the intervals fa 2 d, ag and fa, a 1 d g.
(c) Use a symmetric difference quotient to estimate
f 9s1d for f sxd − x 3 2 2x 2 1 2 with d − 0.4. Draw a
graph of f along with secant lines corresponding to
average rates of change over the intervals f1 2 d, 1g,
f1, 1 1 d g, and f1 2 d, 1 1 d g. Which of these secant
lines appears to have slope closest to that of the tangent
line at x − 1?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
The first person to explicitly formulate the ideas of limits and derivatives was Sir Isaac
Newton in the 1660s. But Newton acknowledged that “If I have seen further than other men,
it is because I have stood on the shoulders of giants.” Two of those giants were Pierre Fermat
(1601–1665) and Newton’s mentor at Cambridge, Isaac Barrow (1630–1677). Newton was
familiar with the methods that these men used to find tangent lines, and their methods played a
role in Newton’s eventual formulation of calculus.
Learn about these methods by researching on the Internet or reading one of the references
listed here. Write an essay comparing the methods of either Fermat or Barrow to modern
methods. In particular, use the method of Section 2.1 to find an equation of the tangent line to
the curve y − x 3 1 2x at the point (1, 3) and show how either Fermat or Barrow would have
solved the same problem. Although you used derivatives and they did not, point out similarities between the methods.
1. C. H. Edwards, The Historical Development of the Calculus (New York: SpringerVerlag, 1979), pp. 124, 132.
2. Howard Eves, An Introduction to the History of Mathematics, 6th ed. (New York:
Saunders, 1990), pp. 391, 395.
3. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford
University Press, 1972), pp. 344, 346.
4. Uta Merzbach and Carl Boyer, A History of Mathematics, 3rd ed. (Hoboken, NJ: Wiley,
2011), pp. 323, 356.
2.2 The Derivative as a Function
■ The Derivative Function
In Section 2.1 we considered the derivative of a function f at a fixed number a:
f 9sad − lim
f sa 1 hd 2 f sad
Here we change our point of view and let the number a vary. If we replace a in Equation 1 by a variable x, we obtain
f 9sxd − lim
f sx 1 hd 2 f sxd
Given any number x for which this limit exists, we assign to x the number f 9sxd. So we
can regard f 9 as a new function, called the derivative of f and defined by Equation 2.
We know that the value of f 9 at x, f 9sxd, can be interpreted geometrically as the slope of
the tangent line to the graph of f at the point sx, f sxdd.
The function f 9 is called the derivative of f because it has been “derived” from f by
the limiting operation in Equation 2. The domain of f 9 is the set hx f 9sxd existsj and
may be smaller than the domain of f.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.2The Derivative as a Function
EXAMPLE 1 The graph of a function f is given in Figure 1. Use it to sketch the graph
of the derivative f 9.
FIGURE 1
SOLUTION We can estimate the value of the derivative at any value of x by drawing the
tangent at the point sx, f sxdd and estimating its slope. For instance, for x − 3 we draw
a tangent at P in Figure 2 and estimate its slope to be about 2 23. (We have drawn a
triangle to help estimate the slope.) Thus f 9s3d < 2 23 < 20.67 and this allows us to
plot the point P9s3, 20.67d on the graph of f 9 directly beneath P. (The slope of the
graph of f becomes the y-value on the graph of f 9.)
mÅ_ 3
P ª(3, _0.67)
FIGURE 2
The slope of the tangent drawn at A appears to be about 21, so we plot the point A9
with a y-value of 21 on the graph of f 9 (directly beneath A). The tangents at B and D
are horizontal, so the derivative is 0 there and the graph of f 9 crosses the x-axis (where
y − 0) at the points B9 and D9, directly beneath B and D. Between B and D, the graph
of f is steepest at C and the tangent line there appears to have slope 1, so the largest
value of f 9sxd between B9 and D9 is 1 (at C9).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
Notice that between B and D the tangents have positive slope, so f 9sxd is positive
there. (The graph of f 9 is above the x-axis.) But to the right of D the tangents have
negative slope, so f 9sxd is negative there. (The graph of f 9 is below the x-axis.)
EXAMPLE 2
(a) If f sxd − x 3 2 x, find a formula for f 9sxd.
(b) Illustrate this formula by comparing the graphs of f and f 9.
(a) When using Equation 2 to compute a derivative, we must remember that the variable
is h and that x is temporarily regarded as a constant during the calculation of the limit.
f 9sxd − lim
− lim
x 3 1 3x 2h 1 3xh 2 1 h 3 2 x 2 h 2 x 3 1 x
− lim
3x 2h 1 3xh 2 1 h 3 2 h
f sx 1 hd 2 f sxd
fsx 1 hd3 2 sx 1 hdg 2 fx 3 2 xg
− lim
− lim s3x 2 1 3xh 1 h 2 2 1d − 3x 2 2 1
(b) We use a calculator to graph f and f 9 in Figure 3. Notice that f 9sxd − 0 when f
has horizontal tangents and f 9sxd is positive when the tangents have positive slope. So
these graphs serve as a check on our work in part (a).
FIGURE 3
EXAMPLE 3 If f sxd − sx , find the derivative of f. State the domain of f 9.
f 9sxd − lim
h l0
− lim
h l0
sx 1 h 2 sx sx 1 h 1 sx
(Rationalize the numerator.)
sx 1 h 1 sx
sx 1 hd 2 x
− lim
h (sx 1 h 1 sx )
h (sx 1 h 1 sx )
− lim
sx 1 h 1 sx
sx 1 sx
h l0
− lim
h l0
(a) ƒ=œ„
f sx 1 hd 2 f sxd
sx 1 h 2 sx
− lim
h l0
We see that f 9sxd exists if x . 0, so the domain of f 9 is s0, `d. This is slightly smaller
than the domain of f , which is f0, `d.
(b) f ª (x)=
FIGURE 4
Let’s check to see that the result of Example 3 is reasonable by looking at the graphs
of f and f 9 in Figure 4. When x is close to 0, sx is also close to 0, so f 9sxd − 1y(2 sx )
is very large and this corresponds to the steep tangent lines near s0, 0d in Figure 4(a) and
the large values of f 9sxd just to the right of 0 in Figure 4(b). When x is large, f 9sxd is very
small and this corresponds to the flatter tangent lines at the far right of the graph of f and
the horizontal asymptote of the graph of f 9.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.2The Derivative as a Function
EXAMPLE 4 Find f 9 if f sxd −
f sx 1 hd 2 f sxd
1 2 sx 1 hd
2 1 sx 1 hd
− lim
f 9sxd − lim
ad 2 bc 1
− lim
s1 2 x 2 hds2 1 xd 2 s1 2 xds2 1 x 1 hd
hs2 1 x 1 hds2 1 xd
− lim
s2 2 x 2 2h 2 x 2 2 xhd 2 s2 2 x 1 h 2 x 2 2 xhd
hs2 1 x 1 hds2 1 xd
Gottfried Wilhelm Leibniz was born in
Leipzig in 1646 and studied law, theol­
ogy, philosophy, and mathematics at
the university there, graduating with
a bachelor’s degree at age 17. After
earning his doctorate in law at age 20,
Leibniz entered the diplomatic service
and spent most of his life traveling to
the capitals of Europe on political mis­
sions. In particular, he worked to avert
a French military threat against Ger­
many and attempted to reconcile the
Catholic and Protestant churches.
His serious study of mathematics
did not begin until 1672 while he was
on a diplomatic mission in Paris. There
he built a calculating machine and
met scientists, like Huygens, who
directed his attention to the latest
develop­ments in mathematics and
science. Leibniz sought to develop a
symbolic logic and system of notation
that would simplify logical reasoning.
In particular, the version of calculus
that he published in 1684 established
the notation and the rules for finding
derivatives that we use today.
Unfortunately, a dreadful priority
dispute arose in the 1690s between
the followers of Newton and those of
Leibniz as to who had invented calcu­
lus first. Leibniz was even accused of
plagiarism by members of the Royal
Society in England. The truth is that
each man invented calculus indepen­
dently. Newton arrived at his version
of calculus first but, because of his
fear of controversy, did not publish
it immediately. So Leibniz’s 1684
account of calculus was the first to be
hs2 1 x 1 hds2 1 xd
− lim
h l 0 s2 1 x 1 hds2 1 xd
s2 1 xd2
− lim
■ Other Notations
If we use the traditional notation y − f sxd to indicate that the independent variable is x
and the dependent variable is y, then some common alternative notations for the derivative are as follows:
f 9sxd − y9 −
f sxd − D f sxd − Dx f sxd
The symbols D and dydx are called differentiation operators because they indicate the
operation of differentiation, which is the process of calculating a derivative.
The symbol dyydx, which was introduced by Leibniz, should not be regarded as a
ratio (for the time being); it is simply a synonym for f 9sxd. Nonetheless, it is a very useful
and suggestive notation, especially when used in conjunction with increment notation.
Referring to Equation 2.1.6, we can rewrite the definition of derivative in Leibniz notation in the form
− lim
Dx l 0 Dx
If we want to indicate the value of a derivative dyydx in Leibniz notation at a specific
number a, we use the notation
which is a synonym for f 9sad. The vertical bar means “evaluate at.”
3 Definition A function f is differentiable at a if f 9sad exists. It is differentiable on an open interval sa, bd [or sa, `d or s2`, ad or s2`, `d] if it is differentiable at every number in the interval.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
EXAMPLE 5 Where is the function f sxd − | x | differentiable?
| |
SOLUTION If x . 0, then x − x and we can choose h small enough that x 1 h . 0
and hence x 1 h − x 1 h. Therefore, for x . 0, we have
f 9sxd − lim
− lim
| x 1 h | 2 | x | − lim
sx 1 hd 2 x
− lim 1 − 1
and so f is differentiable for any x . 0.
Similarly, for x , 0 we have x − 2x and h can be chosen small enough that
x 1 h , 0 and so x 1 h − 2sx 1 hd. Therefore, for x , 0,
| |
f 9sxd − lim
− lim
| x 1 h | 2 | x | − lim
2sx 1 hd 2 s2xd
− lim s21d − 21
and so f is differentiable for any x , 0.
For x − 0 we have to investigate
f 9s0d − lim
− lim
f s0 1 hd 2 f s0d
| 0 1 h | 2 | 0 | − lim | h | h
(if it exists)
Let’s compute the left and right limits separately:
(a) y=ƒ=| x |
|h| −
|h| −
h l 02
− lim1 1 − 1
h l0
− lim2 s21d − 21
h l0
h l 01
h l 02
Since these limits are different, f 9s0d does not exist. Thus f is differentiable at all x
except 0.
A formula for f 9 is given by
(b) y=fª(x)
FIGURE 5
h l 01
f 9sxd −
if x . 0
21 if x , 0
and its graph is shown in Figure 5(b). The fact that f 9s0d does not exist is reflected
geometrically in the fact that the curve y − x does not have a tangent line at s0, 0d.
[See Figure 5(a).]
| |
Both continuity and differentiability are desirable properties for a function to have.
The following theorem shows how these properties are related.
4 Theorem If f is differentiable at a, then f is continuous at a.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.2The Derivative as a Function
PROOF To prove that f is continuous at a, we have to show that lim x l a f sxd − f sad.
We will do this by showing that the difference f sxd 2 f sad approaches 0.
The given information is that f is differentiable at a, that is,
f 9sad − lim
PS An important aspect of problem solving
is trying to find a connection between the
given and the unknown. See Step 2 (Think
of a Plan) in Principles of Problem Solving
following Chapter 1.
f sxd 2 f sad
exists (see Equation 2.1.5). To connect the given and the unknown, we divide and
multiply f sxd 2 f sad by x 2 a (which we can do when x ± a):
f sxd 2 f sad −
f sxd 2 f sad
sx 2 ad
Thus, using Limit Law 4 (Section 1.6), we can write
lim f f sxd 2 f sadg − lim
− lim
f sxd 2 f sad
sx 2 ad
f sxd 2 f sad
? lim sx 2 ad
− f 9sad ? 0 − 0
To use what we have just proved, we start with f sxd and add and subtract f sad:
lim f sxd − lim f f sad 1 s f sxd 2 f saddg
− lim f sad 1 lim f f sxd 2 f sadg
− f sad 1 0 − f sad
Therefore f is continuous at a.
NOTE The converse of Theorem 4 is false; that is, there are functions that are continuous but not differentiable. For instance, the function f sxd − x is continuous at 0
lim f sxd − lim x − 0 − f s0d
| |
| |
(See Example 1.6.7.) But in Example 5 we showed that f is not differentiable at 0.
■ How Can a Function Fail To Be Differentiable?
| |
We saw that the function y − x in Example 5 is not differentiable at 0 and Figure 5(a)
shows that its graph changes direction abruptly when x − 0. In general, if the graph of a
function f has a “corner” or “kink” in it, then the graph of f has no tangent at this point
and f is not differentiable there. [In trying to compute f 9sad, we find that the left and right
limits are different.]
Theorem 4 gives another way for a function not to have a derivative. It says that if f
is not continuous at a, then f is not differentiable at a. So at any discontinuity (for
instance, a jump discontinuity) f fails to be differentiable.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
A third possibility is that the curve has a vertical tangent line when x − a; that is, f
is continuous at a and
vertical tangent line
lim f 9sxd − `
This means that the tangent lines become steeper and steeper as x l a. Figure 6 shows
one way that this can happen; Figure 7(c) shows another. Figure 7 illustrates the three
possibilities that we have discussed.
FIGURE 6
FIGURE 7
Three ways for f not to be
differentiable at a
(a) A corner
(b) A discontinuity
(c) A vertical tangent
A graphing calculator or computer provides another way of looking at differentiability. If f is differentiable at a, then when we zoom in toward the point sa, f sadd the graph
straightens out and appears more and more like a line. (See Figure 8. We saw a specific
example of this in Figure 2.1.2.) But no matter how much we zoom in toward a point like
the ones in Figures 6 and 7(a), we can’t eliminate the sharp point or corner (see Figure 9).
FIGURE 8
FIGURE 9
f is differentiable at a.
f is not differentiable at a.
■ Higher Derivatives
If f is a differentiable function, then its derivative f 9 is also a function, so f 9 may have a
derivative of its own, denoted by s f 9d9 − f 0. This new function f 0 is called the second
derivative of f because it is the derivative of the derivative of f. Using Leibniz notation,
we write the second derivative of y − f sxd as
S D
d 2y
dx 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.2The Derivative as a Function
EXAMPLE 6 If f sxd − x 3 2 x, find and interpret f 0sxd.
SOLUTION In Example 2 we found that the first derivative is f 9sxd − 3x 2 2 1. So the
second derivative is
f 99sxd − s f 9d9sxd − lim
h l0
f 9sx 1 hd 2 f 9sxd
f3sx 1 hd2 2 1g 2 f3x 2 2 1g
h l0
3x 2 1 6xh 1 3h 2 2 1 2 3x 2 1 1
− lim
h l0
− lim s6x 1 3hd − 6x
− lim
h l0
The graphs of f , f 9, and f 0 are shown in Figure 10.
FIGURE 10
We can interpret f 0sxd as the slope of the curve y − f 9sxd at the point sx, f 9sxdd. In
other words, it is the rate of change of the slope of the original curve y − f sxd.
Notice from Figure 10 that f 0sxd is negative when y − f 9sxd has negative slope
and positive when y − f 9sxd has positive slope. So the graphs serve as a check on our
In general, we can interpret a second derivative as a rate of change of a rate of change.
The most familiar example of this is acceleration, which we define as follows.
If s − sstd is the position function of an object that moves in a straight line, we know
that its first derivative represents the velocity v std of the object as a function of time:
v std − s9std −
The instantaneous rate of change of velocity with respect to time is called the acceleration astd of the object. Thus the acceleration function is the derivative of the velocity
function and is therefore the second derivative of the position function:
astd − v9std − s0std
or, in Leibniz notation,
d 2s
− 2
Acceleration is the change in velocity you feel when speeding up or slowing down in a car.
The third derivative f - is the derivative of the second derivative: f -− s f 0d9. So
f -sxd can be interpreted as the slope of the curve y − f 0sxd or as the rate of change of
f 0sxd. If y − f sxd, then alternative notations for the third derivative are
y- − f -sxd −
S D
dx 2
d 3y
dx 3
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
We can also interpret the third derivative physically in the case where the function
is the position function s − sstd of an object that moves along a straight line. Because
s-− ss0d9 − a9, the third derivative of the position function is the derivative of the acceleration function and is called the jerk:
d 3s
− 3
Thus the jerk j is the rate of change of acceleration. It is aptly named because a large jerk
means a sudden change in acceleration, which causes an abrupt movement.
The differentiation process can be continued. The fourth derivative f + is usually
denoted by f s4d. In general, the nth derivative of f is denoted by f snd and is obtained from
f by differentiating n times. If y − f sxd, we write
y snd − f sndsxd −
dx n
EXAMPLE 7 If f sxd − x 3 2 x, find f -sxd and f s4dsxd.
SOLUTION In Example 6 we found that f 0sxd − 6x. The graph of the second derivative
has equation y − 6x and so it is a straight line with slope 6. Since the derivative f -sxd
is the slope of f 0sxd, we have
f -sxd − 6
for all values of x. So f - is a constant function and its graph is a horizontal line. Therefore, for all values of x,
f s4dsxd − 0
We have seen that one application of second and third derivatives occurs in analyzing
the motion of objects using acceleration and jerk. We will investigate another application of second derivatives in Section 3.3, where we show how knowledge of f 0 gives us
information about the shape of the graph of f. In Chapter 11 we will see how second and
higher derivatives enable us to represent functions as sums of infinite series.
1–2 Use the given graph to estimate the value of each derivative.
Then sketch the graph of f 9.
1. (a) f 9s0d
(e) f 9s4d
(b) f 9s1d
(f ) f 9s5d
(c) f 9s2d
(g) f 9s6d
(b) f 9s22d
(e) f 9s1d
(c) f 9s21d
(f ) f 9s2d
(d) f 9s3d
(h) f 9s7d
2. (a) f 9s23d
(d) f 9s0d
(g) f 9s3d
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.2The Derivative as a Function
3. M
atch the graph of each function in (a) – (d) with the graph of
its derivative in I–IV. Give reasons for your choices.
12. S
hown is the graph of the population function Pstd for yeast
cells in a laboratory culture. Use the method of Example 1 to
graph the derivative P9std. What does the graph of P9 tell us
about the yeast population?
P (yeast cells)
15 t (hours)
13. A
rechargeable battery is plugged into a charger. The graph
shows Cstd, the percentage of full capacity that the battery
reaches as a function of time t elapsed (in hours).
(a)What is the meaning of the derivative C9std?
(b)Sketch the graph of C9std. What does the graph tell you?
4–11 Trace or copy the graph of the given function f. (Assume
that the axes have equal scales.) Then use the method of
Example 1 to sketch the graph of f 9 below it.
of full charge
t (hours)
14. T
he graph (from the US Department of Energy) shows
how driving speed affects gas mileage. Fuel economy F
is measured in liters per 100 km and speed v is measured in
kilometers per hour.
(a)What is the meaning of the derivative F9svd?
(b)Sketch the graph of F9svd.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
(c)At what speed should you drive if you want to save
on gas?
F (L/100 km)
105 √ (km/ h)
15. T
he graph shows how the average surface water temperature
f of Lake Michigan varies over the course of a year (where
t is measured in months with t − 0 corresponding to January 1). The average was calculated from data obtained over
a 20-year period ending in 2011. Sketch the graph of the
derivative function f 9. When is f 9std largest?
f (°C)
t (months)
16. Make a careful sketch of the graph of the sine function and
below it sketch the graph of its derivative in the same manner as in Example 1. Can you guess what the derivative of
the sine function is from its graph?
; 17. Let f sxd − x .
(a)Estimate the values of f 9s0d, f 9( 12 ), f 9s1d, and f 9s2d by
zooming in on the graph of f.
(b)Use symmetry to deduce the values of f 9(221 ), f 9s21d,
and f 9s22d.
(c)Use the results from parts (a) and (b) to guess a formula
for f 9sxd.
(d)Use the definition of derivative to prove that your guess
in part (c) is correct.
; 18. Let f sxd − x .
(a)Estimate the values of f 9s0d, f 9( 12 ), f 9s1d, f 9s2d, and
f 9s3d by zooming in on the graph of f.
(b)Use symmetry to deduce the values of f 9(221 ), f 9s21d,
f 9s22d, and f 9s23d.
(c) Use the values from parts (a) and (b) to graph f 9.
(d) Guess a formula for f 9sxd.
(e)Use the definition of a derivative to prove that your
guess in part (d) is correct.
19–30 Find the derivative of the function using the definition of
derivative. State the domain of the function and the domain of its
19. f sxd − 3x 2 8
20. f sxd − mx 1 b
21. f std − 2.5t 2 1 6t
22. f sxd − 4 1 8x 2 5x 2
23. As pd − 4p 3 1 3p
24. F std − t 3 2 5t 1 1
25. f sxd −
x2 2 4
26. Fsvd −
27. tsud −
4u 2 1
28. f sxd − x 4
29. f sxd −
s1 1 x
30. tsxd −
1 1 sx
31. (a)Sketch the graph of f sxd − 1 1 sx 1 3 by starting
with the graph of y − s x and using the transformations
of Sec­tion 1.3.
(b) Use the graph from part (a) to sketch the graph of f 9.
(c)Use the definition of a derivative to find f 9sxd. What are
the domains of f and f 9?
(d)Graph f 9 and compare with your sketch in part (b).
32. (a) If f sxd − x 1 1yx, find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
; by comparing the graphs of f and f 9.
33. (a) If f sxd − x 4 1 2x, find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
by comparing the graphs of f and f 9.
34. T
he table gives the number Nstd, measured in thousands, of
minimally invasive cosmetic surgery procedures performed
in the United States for various years t.
Nstd (thousands)
Source: American Society of Plastic Surgeons
(a) What is the meaning of N9std? What are its units?
(b) Construct a table of estimated values for N9std.
(c) Graph N and N9.
(d)How would it be possible to get more accurate values
for N9std?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.2The Derivative as a Function
35. T
he table gives the height as time passes of a typical pine tree
grown for lumber at a managed site.
Tree age (years)
Height (meters)
41. y
4 x
Source: Arkansas Forestry Commission
If Hstd is the height of the tree after t years, construct a table
of estimated values for H9 and sketch its graph.
36. W
ater temperature affects the growth rate of brook trout.
The table shows the amount of weight gained by brook trout
after 24 days in various water temperatures.
Temperature (°C)
Weight gained (g)
| |
raph the function f sxd − x 1 s x . Zoom in repeatedly,
; 43. G
first toward the point (21, 0) and then toward the origin.
What is different about the behavior of f in the vicinity of
these two points? What do you conclude about the differentiability of f ?
oom in toward the points (1, 0), (0, 1), and (21, 0) on the
; 44. Z
graph of the function tsxd − sx 2 2 1d2y3. What do you
notice? Account for what you see in terms of the differen­
tiability of t.
45–46 The graphs of a function f and its derivative f 9 are shown.
Which is bigger, f 9s21d or f 99s1d ?
If Wsxd is the weight gain at temperature x, construct a table
of estimated values for W9 and sketch its graph. What are the
units for W9sxd?
Source: Adapted from J. Chadwick Jr., “Temperature Effects on Growth
and Stress Physiology of Brook Trout: Implications for Climate Change
Impacts on an Iconic Cold-Water Fish.” Masters Theses. Paper 897. 2012.
37. Let P represent the percentage of a city’s electrical
power that is produced by solar panels t years after
January 1, 2020.
(a) What does dPydt represent in this context?
(b) Interpret the statement
− 3.5
47. T
he figure shows the graphs of f , f 9, and f 0. Identify each
curve, and explain your choices.
39–42 The graph of f is given. State, with reasons, the numbers
at which f is not differentiable.
t −2
38. Suppose N is the number of people in the United States
who travel by car to another state for a vacation in a
year when the average price of gasoline is p dollars per
liter. Do you expect dNydp to be positive or negative?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
48. T
he figure shows graphs of f, f 9, f 0, and f -. Identify each
curve, and explain your choices.
a b c d
54. (a)The graph of a position function of a car is shown,
where s is measured in meters and t in seconds. Use it
to graph the velocity and acceleration of the car. What
is the acceleration at t − 10 seconds?
49. T
he figure shows the graphs of three functions. One is the
position function of a car, one is the velocity of the car, and
one is its acceleration. Identify each curve, and explain your
(b)Use the acceleration curve from part (a) to estimate the
jerk at t − 10 seconds. What are the units for jerk?
55. Let f sxd − s
(a) If a ± 0, use Equation 2.1.5 to find f 9sad.
(b) Show that f 9s0d does not exist.
(c)Show that y − s
x has a vertical tangent line at s0, 0d.
(Recall the shape of the graph of f . See Figure 1.2.13.)
56. (a)
If tsxd − x 2y3, show that t9s0d does not exist.
If a ± 0, find t9sad.
Show that y − x 2y3 has a vertical tangent line at s0, 0d.
Illustrate part (c) by graphing y − x 2y3.
57. S
how that the function f sxd − x 2 6 is not differentiable
at 6. Find a formula for f 9 and sketch its graph.
50. T
he figure shows the graphs of four functions. One is the
position function of a car, one is the velocity of the car, one
is its acceleration, and one is its jerk. Identify each curve,
and explain your choices.
| |
; 51–52 Use the definition of a derivative to find f 9sxd and f 0sxd.
Then graph f , f 9, and f 0 on a common screen and check to see
if your answers are reasonable.
51. f sxd − 3x 2 1 2x 1 1
| |
59. (a) Sketch the graph of the function f sxd − x x .
(b) For what values of x is f differentiable?
(c) Find a formula for f 9.
60. (a) Sketch the graph of the function tsxd − x 1 x .
(b) For what values of x is t differentiable?
(c) Find a formula for t9.
58. W
here is the greatest integer function f sxd − v x b not differentiable? Find a formula for f 9 and sketch its graph.
61. D
erivatives of Even and Odd Functions Recall that a function f is called even if f s2xd − f sxd for all x in its domain
and odd if f s2xd − 2f sxd for all such x. Prove each of the
(a) The derivative of an even function is an odd function.
(b) The derivative of an odd function is an even function.
62–63 Left- and Right-Hand Derivatives The left-hand and
right-hand derivatives of f at a are defined by
52. f sxd − x 3 2 3x
; 53. If f sxd − 2x 2 x , find f 9sxd, f 0sxd, f -sxd, and f sxd.
Graph f , f 9, f 0, and f - on a common screen. Are the
graphs consistent with the geometric interpretations of
these derivatives?
f 92 sad − lim2
f sa 1 hd 2 f sad
f 91 sad − lim1
f sa 1 hd 2 f sad
if these limits exist. Then f 9sad exists if and only if these onesided derivatives exist and are equal.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.3Differentiation Formulas
62. Find f 92s0d and f 91s0d for the given function f. Is f differentiable at 0?
(a) f sxd −
63. Let
0 if x < 0
x if x . 0
(b) f sxd −
f sxd −
Example 1.1.4 we sketched a possible graph of T as a function
of the time t that has elapsed since the faucet was turned on.
(a)Describe how the rate of change of T with respect to t
varies as t increases.
(b) Sketch a graph of the derivative of T.
0 if x < 0
x 2 if x . 0
if x < 0
if 0 , x , 4
65. N
ick starts jogging and runs faster and faster for 3 mintues,
then he walks for 5 minutes. He stops at an intersection for
2 minutes, runs fairly quickly for 5 minutes, then walks for
4 minutes.
(a)Sketch a possible graph of the distance s Nick has
covered after t minutes.
(b) Sketch a graph of dsydt.
if x > 4
Find f 92s4d and f 91s4d.
Sketch the graph of f.
Where is f discontinuous?
Where is f not differentiable?
66. L
et be the tangent line to the parabola y − x 2 at the
point s1, 1d. The angle of inclination of is the angle that makes with the positive direction of the x-axis.
Calculate correct to the nearest degree.
64. W
hen you turn on a hot-water faucet, the temperature T of the
water depends on how long the water has been running. In
2.3 Differentiation Formulas
If it were always necessary to compute derivatives directly from the definition, as we did
in the preceding section, such computations would be tedious and the evaluation of some
limits would require ingenuity. Fortunately, several rules have been developed for finding
derivatives without having to use the definition directly. These formulas greatly simplify
the task of differentiation.
■ Constant Functions
Let’s start with the simplest of all functions, the constant function f sxd − c. The graph of
this function is the horizontal line y − c, which has slope 0, so we must have f 9sxd − 0.
(See Figure 1.) A formal proof, from the definition of a derivative, is also easy:
f 9sxd − lim
FIGURE 1
The graph of f sxd − c is the line
y − c, so f 9sxd − 0.
f sx 1 hd 2 f sxd
− lim
− lim 0 − 0
In Leibniz notation, we write this rule as follows.
Derivative of a Constant Function
scd − 0
■ Power Functions
FIGURE 2
The graph of f sxd − x is the line
y − x, so f 9sxd − 1.
We next look at the functions f sxd − x n, where n is a positive integer. If n − 1, the graph
of f sxd − x is the line y − x, which has slope 1. (See Figure 2.) So
sxd − 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
(You can also verify Equation 1 from the definition of a derivative.) We have already
investigated the cases n − 2 and n − 3. In fact, in Section 2.2 (Exercises 17 and 18) we
found that
sx 2 d − 2xsx 3 d − 3x 2
For n − 4 we find the derivative of f sxd − x 4 as follows:
f 9sxd − lim
f sx 1 hd 2 f sxd
sx 1 hd4 2 x 4
− lim
x 4 1 4x 3h 1 6x 2h 2 1 4xh 3 1 h 4 2 x 4
4x h 1 6x h 1 4xh 3 1 h 4
− lim
− lim
− lim s4x 3 1 6x 2h 1 4xh 2 1 h 3 d − 4x 3
sx 4 d − 4x 3
Comparing the equations in (1), (2), and (3), we see a pattern emerging. It seems to be a
rea­sonable guess that, when n is a positive integer, sdydxdsx n d − nx n21. This turns out to
be true. We prove it in two ways; the second proof uses the Binomial Theorem.
The Power Rule If n is a positive integer, then
sx n d − nx n21
FIRST PROOF The formula
x n 2 a n − sx 2 adsx n21 1 x n22a 1 ∙ ∙ ∙ 1 xa n22 1 a n21 d
can be verified simply by multiplying out the right-hand side (or by summing the
second factor as a geometric series). If f sxd − x n, we can use Equation 2.1.5 for f 9sad
and the equation above to write
f sxd 2 f sad
xn 2 an
− lim
xla x 2 a
f 9sad − lim
− lim sx n21 1 x n22a 1 ∙ ∙ ∙ 1 xa n22 1 a n21 d
− a n21 1 a n22a 1 ∙ ∙ ∙ 1 aa n22 1 a n21
− na n21
f 9sxd − lim
f sx 1 hd 2 f sxd
sx 1 hd n 2 x n
− lim
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.3Differentiation Formulas
In finding the derivative of x 4 we had to expand sx 1 hd4. Here we need to expand
sx 1 hd n and we use the Binomial Theorem to do so:
The Binomial Theorem is given on
Reference Page 1.
f 9sxd − lim
x n 1 nx n21h 1
nsn 2 1d n22 2
x h 1 ∙ ∙ ∙ 1 nxh n21 1 h n 2 x n
nsn 2 1d n22 2
x h 1 ∙ ∙ ∙ 1 nxh n21 1 h n
nx n21h 1
− lim
− lim
nx n21 1
nsn 2 1d n22
x h 1 ∙ ∙ ∙ 1 nxh n22 1 h n21
− nx n21
because every term except the first has h as a factor and therefore approaches 0.
We illustrate the Power Rule using various notations in Example 1.
EXAMPLE 1
(a) If f sxd − x 6, then f 9sxd − 6x 5.
(c) If y − t 4, then
− 4t 3.
(b) If y − x 1000, then y9 − 1000x 999.
d 3
sr d − 3r 2
■ New Derivatives from Old
When new functions are formed from old functions by addition, subtraction, or multiplica­tion by a constant, their derivatives can be calculated in terms of derivatives of the old
func­tions. In particular, the following formula says that the derivative of a constant times
a function is the constant times the derivative of the function.
Geometric Interpretation
of the Constant Multiple Rule
The Constant Multiple Rule If c is a constant and f is a differentiable function,
fcf sxdg − c
f sxd
PROOF Let tsxd − cf sxd. Then
Multiplying by c − 2 stretches the
graph vertically by a factor of 2. All
the rises have been doubled but the
runs stay the same. So the slopes are
also doubled.
t9sxd − lim
tsx 1 hd 2 tsxd
cf sx 1 hd 2 cf sxd
− lim
− lim c
− c lim
− cf 9sxd
f sx 1 hd 2 f sxd
f sx 1 hd 2 f sxd
(by Limit Law 3)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
EXAMPLE 2
s3x 4 d − 3
sx 4 d − 3s4x 3 d − 12x 3
s2xd −
fs21dxg − s21d
sxd − 21s1d − 21
The next rule tells us that the derivative of a sum (or difference) of functions is the
sum (or difference) of the derivatives.
Using prime notation, we can write
the Sum and Difference Rules as
The Sum and Difference Rules If f and t are both differentiable, then
f f sxd 1 tsxdg −
f sxd 1
s f 1 td9 − f 9 1 t9
s f 2 td9 − f 9 2 t9
f f sxd 2 tsxdg −
f sxd 2
PROOF To prove the Sum Rule, we let Fsxd − f sxd 1 tsxd. Then
F9sxd − lim
− lim
− lim
− lim
Fsx 1 hd 2 Fsxd
f f sx 1 hd 1 tsx 1 hdg 2 f f sxd 1 tsxdg
f sx 1 hd 2 f sxd
tsx 1 hd 2 tsxd
f sx 1 hd 2 f sxd
tsx 1 hd 2 tsxd
1 lim
(by Limit Law 1)
hl 0
− f 9sxd 1 t9sxd
To prove the Difference Rule, we write f 2 t as f 1 s21dt and apply the Sum Rule
and the Constant Multiple Rule.
The Sum Rule can be extended to the sum of any number of functions. For instance,
using this theorem twice, we get
s f 1 t 1 hd9 − fs f 1 td 1 hg9 − s f 1 td9 1 h9 − f 9 1 t9 1 h9
The Constant Multiple Rule, the Sum Rule, and the Difference Rule can be combined
with the Power Rule to differentiate any polynomial, as the following three examples
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.3Differentiation Formulas
EXAMPLE 3
sx 8 1 12x 5 2 4x 4 1 10x 3 2 6x 1 5d
sx 8 d 1 12
sx 5 d 2 4
sx 4 d 1 10
sx 3 d 2 6
sxd 1
− 8x 7 1 12s5x 4 d 2 4s4x 3 d 1 10s3x 2 d 2 6s1d 1 0
− 8x 7 1 60x 4 2 16x 3 1 30x 2 2 6
EXAMPLE 4 Find the points on the curve y − x 4 2 6x 2 1 4 where the tangent line is
(0, 4)
SOLUTION Horizontal tangents occur where the derivative is zero. We have
{_ œ„
3, _5}
sx 4 d 2 6
sx 2 d 1
3, _5}
FIGURE 3
The curve y − x 4 2 6x 2 1 4 and its
horizontal tangents
− 4x 3 2 12x 1 0 − 4xsx 2 2 3d
Thus dyydx − 0 if x − 0 or x 2 2 3 − 0, that is, x − 6s3 . So the given curve has
horizontal tangents when x − 0, s3 , and 2s3 . The corresponding points are s0, 4d,
(s3 , 25), and (2s3 , 25). (See Figure 3.)
EXAMPLE 5 The equation of motion of a particle is s − 2t 3 2 5t 2 1 3t 1 4, where
s is measured in centimeters and t­ in seconds. Find the acceleration as a function of
time. What is the acceleration after 2 seconds?
SOLUTION The velocity and acceleration are
vstd −
− 6t 2 2 10t 1 3
astd −
− 12t 2 10
The acceleration after 2 seconds is as2d − 12s2d 2 10 − 14 cmys2.
Next we need a formula for the derivative of a product of two functions.
■ The Product Rule
By analogy with the Sum and Difference Rules, we might be tempted to guess, as Leibniz did three centuries ago, that the derivative of a product is the product of the derivatives. We can see, however, that this guess is wrong by looking at a particular example.
Let f sxd − x and tsxd − x 2. Then the Power Rule gives f 9sxd − 1 and t9sxd − 2x. But
s ftdsxd − x 3, so s ftd9sxd − 3x 2. Thus s ftd9 ± f 9t9. The correct formula was discovered
by Leibniz (soon after his false start) and is called the Product Rule.
In prime notation the Product Rule is
written as
s ftd9 − ft9 1 t f 9
The Product Rule If f and t are both differentiable, then
f f sxdtsxdg − f sxd
ftsxdg 1 tsxd
f f sxdg
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
PROOF Let Fsxd − f sxdtsxd. Then
Fsx 1 hd 2 Fsxd
f sx 1 hd tsx 1 hd 2 f sxdtsxd
− lim
F9sxd − lim
In order to evaluate this limit, we would like to separate the functions f and t as in
the proof of the Sum Rule. We can achieve this separation by subtracting and adding
the term f sx 1 hd tsxd in the numerator:
F9sxd − lim
− lim
f sx 1 hd tsx 1 hd 2 f sx 1 hd tsxd 1 f sx 1 hdtsxd 2 f sxd tsxd
f sx 1 hd
tsx 1 hd 2 tsxd
f sx 1 hd 2 f sxd
1 tsxd
− lim f sx 1 hd lim
tsx 1 hd 2 tsxd
f sx 1 hd 2 f sxd
1 lim tsxd lim
hl 0
− f sxd t9sxd 1 tsxd f 9sxd
Note that lim h l 0 tsxd − tsxd because tsxd is a constant with respect to the variable h.
Also, since f is differentiable at x, it is continuous at x by Theorem 2.2.4, and so
lim h l 0 f sx 1 hd − f sxd. (See Exercise 1.8.65.)
In words, the Product Rule says that the derivative of a product of two functions is the
first function times the derivative of the second function plus the second function times
the derivative of the first function.
EXAMPLE 6 Find F9sxd if Fsxd − s6x 3 ds7x 4 d.
SOLUTION By the Product Rule, we have
F9sxd − s6x 3 d
s7x 4 d 1 s7x 4 d
s6x 3 d
− s6x 3 ds28x 3 d 1 s7x 4 ds18x 2 d − 168x 6 1 126x 6 − 294x 6
Notice that we could verify the answer to Example 6 directly by first multiplying the
Fsxd − s6x 3 ds7x 4 d − 42x 7?F9sxd − 42s7x 6 d − 294x 6
But later we will meet functions, such as y − x 2 sin x, for which the Product Rule is the
only possible method.
EXAMPLE 7 If hsxd − xtsxd and it is known that ts3d − 5 and t9s3d − 2, find h9s3d.
SOLUTION Applying the Product Rule, we get
fxtsxdg − x
ftsxdg 1 tsxd
− x t9sxd 1 tsxd s1d
h9sxd −
h9s3d − 3t9s3d 1 ts3d − 3 2 1 5 − 11
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.3Differentiation Formulas
■ The Quotient Rule
The following formula tells us how to differentiate the quotient of two functions.
In prime notation the Quotient Rule
is written as
The Quotient Rule If f and t are differentiable, then
f 9
t f 9 2 ft9
F G
f sxd
f f sxdg 2 f sxd
ftsxdg 2
PROOF Let Fsxd − f sxdytsxd. Then
f sx 1 hd
f sxd
Fsx 1 hd 2 Fsxd
tsx 1 hd
F9sxd − lim
− lim
hl 0
− lim
f sx 1 hdtsxd 2 f sxd tsx 1 hd
htsx 1 hdtsxd
We can separate f and t in this expression by subtracting and adding the term f sxdtsxd
in the numerator:
F9sxd − lim
− lim
f sx 1 hd tsxd 2 f sxdtsxd 1 f sxdtsxd 2 f sxdtsx 1 hd
htsx 1 hdtsxd
f sx 1 hd 2 f sxd
tsx 1 hd 2 tsxd
2 f sxd
tsx 1 hdtsxd
lim tsxd lim
f sx 1 hd 2 f sxd
tsx 1 hd 2 tsxd
2 lim f sxd lim
lim tsx 1 hd lim tsxd
tsxd f 9sxd 2 f sxdt9sxd
ftsxdg 2
Again t is continuous by Theorem 2.2.4, so lim h l 0 tsx 1 hd − tsxd.
In words, the Quotient Rule says that the derivative of a quotient is the denominator
times the derivative of the numerator minus the numerator times the derivative of the
denominator, all divided by the square of the denominator.
The theorems of this section show that any polynomial is differentiable on R and any
rational function is differentiable on its domain. Furthermore, the Quotient Rule and the
other differentiation formulas enable us to compute the derivative of any rational function, as the next example illustrates.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
Figure 4 shows the graphs of the
function of Example 8 and its derivative. Notice that when y grows rapidly
(near 2s
6 < 21.8), y9 is large. And
when y grows slowly, y9 is near 0.
EXAMPLE 8 Let y −
sx 3 1 6d
y9 −
FIGURE 4
sx 2 1 x 2 2d 2 sx 2 1 x 2 2d
sx 3 1 6d
sx 3 1 6d2
sx 3 1 6ds2x 1 1d 2 sx 2 1 x 2 2ds3x 2 d
sx 3 1 6d2
s2x 4 1 x 3 1 12x 1 6d 2 s3x 4 1 3x 3 2 6x 2 d
sx 3 1 6d2
2x 4 2 2x 3 1 6x 2 1 12x 1 6
sx 3 1 6d2
x2 1 x 2 2
. Then
x3 1 6
NOTE Don’t use the Quotient Rule every time you see a quotient. Sometimes it’s easier
to rewrite a quotient first to put it in a form that is simpler for the purpose of differentiation. For instance, although it is possible to differentiate the function
Fsxd −
3x 2 1 2sx
using the Quotient Rule, it is much easier to perform the division first and write the function as
Fsxd − 3x 1 2x 21y2
before differentiating.
■ General Power Functions
The Quotient Rule can be used to extend the Power Rule to the case where the exponent
is a negative integer.
If n is a positive integer, then
d 2n
sx d − 2nx 2n21
sx 2n d −
s1d 2 1 sx n d
x n 0 2 1 nx n21
sx n d2
x 2n
2nx n21
− 2nx n2122n − 2nx 2n21
x 2n
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.3Differentiation Formulas
EXAMPLE 9
(a) If y −
, then
sx 21 d − 2x 22 − 2 2
d 23
st d − 6s23dt 24 − 2 4
So far we know that the Power Rule holds if the exponent n is a positive or negative
integer. If n − 0, then x 0 − 1, which we know has a derivative of 0. Thus the Power
Rule holds for any integer n. What if the exponent is a fraction? In Example 2.2.3 we
found that
sx −
2 sx
which can be written as
d 1y2
sx d − 12 x21y2
This shows that the Power Rule is true even when n − 12. In fact, it also holds for any real
number n, as we will prove in Chapter 6. (A proof for rational values of n is indicated in
Exercise 2.6.50.) In the meantime we state the general version and use it in the examples
and exercises.
The Power Rule (General Version) If n is any real number, then
sx n d − nx n21
EXAMPLE 10
(a) If f sxd − x , then f 9sxd − x 21.
(b) Let
sx 2
d 22y3
sx d − 223 x2s2y3d21
− 223 x25y3
In Example 11, a and b are constants.
It is customary in mathematics to
use letters near the beginning of the
alphabet to represent constants and
letters near the end of the alphabet to
represent variables.
EXAMPLE 11 Differentiate the function f std − st sa 1 btd.
SOLUTION 1 Using the Product Rule, we have
f 9std − st
sa 1 btd 1 sa 1 btd
− st b 1 sa 1 btd 21 t 21y2
− bst 1
a 1 bt
a 1 3bt
2 st
2 st
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
SOLUTION 2 If we first use the laws of exponents to rewrite f std, then we can proceed
directly without using the Product Rule.
f std − ast 1 btst − at 1y2 1 bt 3y2
f 9std − 12 at21y2 1 32 bt 1y2
which is equivalent to the answer given in Solution 1.
The differentiation rules enable us to find tangent lines without having to resort to the
definition of a derivative. It also enables us to find normal lines. The normal line to a
curve C at a point P is the line through P that is perpendicular to the tangent line at P. (In
the study of optics, one needs to consider the angle between a light ray and the normal
line to a lens.)
EXAMPLE 12 Find equations of the tangent line and normal line to the curve
y − sx ys1 1 x 2 d at the point s1, 12 d.
SOLUTION According to the Quotient Rule, we have
s1 1 x 2 d
s1 1 x 2 d
sx d 2 sx
s1 1 x 2 d2
2 sx s2xd
2 sx
s1 1 x 2 d2
s1 1 x 2 d 2 4x 2
1 2 3x 2
2 sx s1 1 x 2 d2
2 sx s1 1 x 2 d2
s1 1 x 2 d
So the slope of the tangent line at s1, 12 d is
FIGURE 5
2 s1s1 1 12 d2
y 2 12 − 2 41 sx 2 1dory − 214 x 1 34
We use the point-slope form to write an equation of the tangent line at s1, 12 d:
The slope of the normal line at s1, 12 d is the negative reciprocal of 241, namely 4, so
an equation is
y 2 12 − 4sx 2 1dory − 4x 2 72
The curve and its tangent and normal lines are graphed in Figure 5.
EXAMPLE 13 At what points on the hyperbola xy − 12 is the tangent line parallel to
the line 3x 1 y − 0?
SOLUTION Since xy − 12 can be written as y − 12yx, we have
− 12
sx 21 d − 12s2x 22 d − 2 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.3Differentiation Formulas
(2, 6)
Let the x-coordinate of one of the points in question be a. Then the slope of the tangent
line at that point is 212ya 2. This tangent line will be parallel to the line 3x 1 y − 0,
or y − 23x, if it has the same slope, that is, 23. Equating slopes, we get
(_2, _6)
− 23ora 2 − 4ora − 62
Therefore the required points are s2, 6d and s22, 26d. The hyperbola and the tangents
are shown in Figure 6.
We summarize the differentiation formulas we have learned so far as follows.
FIGURE 6
Table of Differentiation Formulas
scd − 0
d n
sx d − nx n21
scf d9 − cf 9
s f 1 td9 − f 9 1 t9
s ftd9 − ft9 1 tf 9
1. tsxd − 4 x 1 7
2. tstd − 5t 1 4 t 2
3. f sxd − x 75 2 x 1 3
4. tsxd − 74 x 2 2 3x 1 12
5. Wsvd − 1.8v23
6. r szd − z 25 2 z 1y2
7. f sxd − x 3y2 1 x 23
8. V std − t 23y5 1 t 4
10. r std −
11. y − 2x 1 sx
13. tsxd −
12. hswd − s2 w 2 s2
14. SsRd − 4 R 2
15. f sxd − x 3sx 1 3d
3x 2 1 x 3
19. Gsqd − s1 1 q d
21. Gsrd −
f 9
tf 9 2 ft9
23. Pswd −
2w 2 2 w 1 4
3r 3y2 1 r 5y2
16. Fstd − s2t 2 3d2
18. y −
sx 1 x
20. Gstd − s5t 1
22. Fszd −
A 1 Bz 1 Cz 2
24. Dstd −
1 1 16t 2
s4td 3
25–26 Find dyydx and dyydt.
25. y − tx 2 1 t 3x
26. y −
27. F
ind the derivative of f sxd − s1 1 2x 2 dsx 2 x 2 d in two
ways: by using the Product Rule and by performing the multiplication first. Do your answers agree?
28. Find the derivative of the function
Fsxd −
15–24 Differentiate the function after first rewriting the function
in a different form. (Do not use the Product or Quotient Rules.)
17. f sxd −
1–14 Differentiate the function.
9. sstd −
s f 2 td9 − f 9 2 t9
x 4 2 5x 3 1 sx
in two ways: by using the Quotient Rule and by simplifying
first. Show that your answers are equivalent. Which method
do you prefer?
29–32 Use the Product Rule to find the derivative of the function.
29. f sxd − s3x 2 2 5xdx 2
30. y − s10x 2 1 7x 2 2ds2 2 x 2 d
31. y − s4x 2 1 3ds2x 1 5d
32. tsxd − sx ( x 1 2sx )
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
33–36 Use the Quotient Rule to find the derivative of the function.
33. y −
34. y −
3 2 2t
5t 1 1
35. tstd −
36. Gsud −
6u 4 2 5u
37–52 Differentiate.
37. f std − 3
t 2t21
39. y −
38. Fsxd −
2x 2 6x 2 1 5
s 2 ss
40. y −
2x 5 1 x 4 2 6x
41. Fsxd −
42. y −
43. Hsud − (u 2 su )(u 1 su
44. Asvd − v
45. Jsud −
sx 1 1
su 1 2d 2
49. Gs yd −
51. f sxd −
59–60 Find an equation of the tangent line to the curve at the
given point.
59. y −
61–64 Find equations of the tangent line and normal line to the
curve at the specified point.
61. y − x 1 sx ,s1, 2d
62. y − x 3y2,s1, 1d
DS D
63. y −
48. y −
Ay 3 1 B
1 1 cx
50. Fstd −
Bt 2 1 Ct 3
52. f sxd −
ax 1 b
cx 1 d
53. The general polynomial of degree n has the form
Psxd − a n x 1 a n21 x
1 ∙ ∙ ∙ 1 a2 x 1 a1 x 1 a0
where a n ± 0. Find the derivative of P.
; 54–56 Find f 9sxd. Compare the graphs of f and f 9 and use them
to explain why your answer is reasonable.
54. f sxd − x 4 2 2x 3 1 x 2
56. f sxd − x 1
,s1, 1d
60. y − 2x 3 2 x 2 1 2, s1, 3d
s2 v 1 1 2 v d
46. hswd − sw 2 1 3wdsw 21 2 w 24 d
47. f std −
; 58. (a)Graph the function tsxd − x ysx 1 1d in the viewing
rectangle f24, 4g by f21, 1.5g.
(b)Using the graph in part (a) to estimate slopes, make a
rough sketch, by hand, of the graph of t9. (See
Example 2.2.1.)
(c)Calculate t9sxd and use this expression to graph t9.
Compare with your sketch in part (b).
55. f sxd − 3x 15 2 5x 3 1 3
; 57. (a)Graph the function
f sxd − x 4 2 3x 3 2 6x 2 1 7x 1 30
in the viewing rectangle f23, 5g by f210, 50g.
(b)Using the graph in part (a) to estimate slopes, make
a rough sketch, by hand, of the graph of f 9. (See
Example 2.2.1.)
(c)Calculate f 9sxd and use this expression to graph f 9.
Compare with your sketch in part (b).
1 1 5x2
(1, 12 )
64. y −
,s4, 0.4d
65. (a)The curve y − 1ys1 1 x 2 d is called a witch of Maria
Agnesi. Find an equation of the tangent line to this
curve at the point (21, 12 ).
(b)Illustrate part (a) by graphing the curve and the tangent
line on the same screen.
66. (a)The curve y − xys1 1 x 2 d is called a serpentine. Find
an equation of the tangent line to this curve at the
point s3, 0.3d.
(b)Illustrate part (a) by graphing the curve and the tangent
line on the same screen.
67–70 Find the first and second derivatives of the function.
67. f sxd − 0.001x 5 2 0.02x 3
68. G srd − sr 1 s
69. f sxd −
1 1 2x
70. f sxd −
; 71–72 Find the first and second derivatives of the function.
Check to see that your answers are reasonable by comparing the
graphs of f , f 9, and f 99.
x2 2 1
71. f sxd − 2 x 2 5x 3y4
72. f sxd − 2
x 11
73. The equation of motion of a particle is s − t 3 2 3t, where
s is in meters and t is in seconds. Find
(a) the velocity and acceleration as functions of t,
(b) the acceleration after 2 s, and
(c) the acceleration when the velocity is 0.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.3Differentiation Formulas
74. T
he equation of motion of a particle is s − t 4 2 2t 3 1 t 2 2 t,
where s is in meters and t is in seconds.
(a)Find the velocity and acceleration as functions of t.
(b)Find the acceleration after 1 s.
(c)Graph the position, velocity, and acceleration functions
on the same screen.
75. B
iologists have proposed a cubic polynomial to model the
length L of Alaskan rockfish at age A:
L − 0.0390A 3 2 0.945A 2 1 10.03A 1 3.07
81. If f sxd − sx tsxd, where ts4d − 8 and t9s4d − 7, find f 9s4d.
82. If hs2d − 4 and h9s2d − 23, find
SsAd − 0.882 A 0.842
where A is measured in square meters. Find S9s100d and
interpret your answer.
(a)Use a calculator or computer to model tire life with a
quadratic function of the pressure.
(b)Use the model to estimate dLydP when P − 200 and
when P − 300. What is the meaning of the derivative?
What are the units? What is the significance of the signs
of the derivatives?
79. Suppose that f s5d − 1, f 9s5d − 6, ts5d − 23, and
t9s5d − 2. Find the following values.
(a) s ftd9s5d
(b) s fytd9s5d
(c) s tyf d9s5d
80. Suppose that f s4d − 2, ts4d − 5, f 9s4d − 6, and
t9s4d − 23. Find h9s4d.
hsxd − 3 f sxd 1 8tsxd
(b) hsxd − f sxdtsxd
(d) hsxd −
f sxd 1 tsxd
84. Let Psxd − FsxdGsxd and Qsxd − FsxdyGsxd, where F and
G are the functions whose graphs are shown.
(a)Find P9s2d.
(b) Find Q9s7d.
77. B
oyle’s Law states that when a sample of gas is compressed
at a constant temperature, the pressure P of the gas is
inversely proportional to the volume V of the gas.
(a)Suppose that the pressure of a sample of air that occupies
0.106 m 3 at 25°C is 50 kPa. Write V as a function of P.
(b)Calculate dVydP when P − 50 kPa. What is the
meaning of the derivative? What are its units?
78. C
ar tires need to be inflated properly because overinflation
or underinflation can cause premature tread wear. The data
in the table show tire life L (in thousands of kilometers) for
a certain type of tire at various pressures P (in kPa).
A −12
76. T
he number of tree species S in a given area A in a forest
reserve has been modeled by the power function
f sxd
83. If f and t are the functions whose graphs are shown, let
usxd − f sxd tsxd and vsxd − f sxdytsxd.
(a) Find u9s1d.
(b) Find v9s4d.
and interpret your answer.
(c) hsxd −
S DZ
where L is measured in centimeters and A in years. Calculate
85. If t is a differentiable function, find an expression for the
derivative of each of the following functions.
(a) y − x tsxd
(b) y −
(c) y −
86. I f f is a differentiable function, find an expression for the
derivative of each of the following functions.
f sxd
(b) y −
(a) y − x 2 f sxd
(c) y −
f sxd
(d) y −
1 1 x f sxd
87. F
ind the points on the curve y − x 3 1 3x 2 2 9x 1 10
where the tangent is horizontal.
88. For what values of x does the graph of
f sxd − x 3 1 3x 2 1 x 1 3
have a horizontal tangent?
89. Show that the curve y − 6x 3 1 5x 2 3 has no tangent line
with slope 4.
90. F
ind an equation of the line that is both tangent to the curve
y − x 4 1 1 and parallel to the line 32 x 2 y − 15.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
91. F
ind equations for two lines that are both tangent to the
curve y − x 3 2 3x 2 1 3x 2 3 and parallel to the line
3x 2 y − 15.
92. Find equations of the tangent lines to the curve
that are parallel to the line x 2 2y − 2.
93. F
ind an equation of the normal line to the curve y − sx
that is parallel to the line 2x 1 y − 1.
94. W
here does the normal line to the parabola y − x 2 2 1 at
the point s21, 0d intersect the parabola a second time? Illustrate with a sketch.
95. D
raw a diagram to show that there are two tangent lines to
the parabola y − x 2 that pass through the point s0, 24d.
Find the coordinates of the points where these tangent lines
intersect the parabola.
96. (a)Find equations of both lines through the point s2, 23d
that are tangent to the parabola y − x 2 1 x.
(b)Show that there is no line through the point s2, 7d that is
tangent to the parabola. Then draw a diagram to see why.
97. F
or what values of a and b is the line 2x 1 y − b tangent to
the parabola y − ax 2 when x − 2?
98. F
ind the nth derivative of each function by calculating the
first few derivatives and observing the pattern that occurs.
(a)f sxd − x n
(b) f sxd − 1yx
99. F
ind a second-degree polynomial P such that Ps2d − 5,
P9s2d − 3, and P99s2d − 2.
100. T
he equation y99 1 y9 2 2y − x 2 is called a differential
equation because it involves an unknown function y and
its derivatives y9 and y99. Find constants A, B, and C such
that the function y − Ax 2 1 Bx 1 C satisfies this equation. (Differential equations will be studied in detail in
Chapter 9.)
101. F
ind a cubic function y − ax 3 1 bx 2 1 cx 1 d whose graph
has horizontal tangents at the points s22, 6d and s2, 0d.
102. F
ind a parabola with equation y − ax 1 bx 1 c that has
slope 4 at x − 1, slope 28 at x − 21, and passes through
the point s2, 15d.
103. In this exercise we estimate the rate at which the total personal income is rising in Boulder, Colorado. In 2015, the
population of this city was 107,350 and the population was
increasing by roughly 1960 people per year. The average
annual income was $60,220 per capita, and this average
was increasing at about $2250 per year (a little above the
national average of about $1810 yearly). Use the Product
Rule and these figures to estimate the rate at which total
personal income was rising in Boulder in 2015. Explain the
meaning of each term in the Product Rule.
104. A
manufacturer produces bolts of a fabric with a fixed
width. The quantity q of this fabric (measured in meters)
that is sold is a function of the selling price p (in dollars per
meter), so we can write q − f s pd. Then the total revenue
earned with selling price p is Rs pd − pf s pd.
(a)What does it mean to say that f s20d − 10,000 and
f 9s20d − 2350?
(b)Assuming the values in part (a), find R9s20d and interpret your answer.
105. T
he Michaelis-Menten equation for the enzyme chymotrypsin is
0.015 1 fSg
where v is the rate of an enzymatic reaction and [S] is the
concentration of a substrate S. Calculate d vyd fSg and interpret it.
106. T
he biomass Bstd of a fish population is the total mass of the
members of the population at time t. It is the product of the
number of individuals Nstd in the population and the average
mass Mstd of a fish at time t. In the case of guppies, breeding
occurs continually. Suppose that at time t − 4 weeks the
population is 820 guppies and is growing at a rate of 50 guppies per week, while the average mass is 1.2 g and is increasing at a rate of 0.14 gyweek. At what rate is the biomass
increasing when t − 4?
107. Extended Product Rule The Product Rule can be extended
to the product of three functions.
(a)Use the Product Rule twice to prove that if f , t, and h
are differentiable, then s fthd9 − f 9th 1 ft9h 1 fth9.
(b) Taking f − t − h in part (a), show that
f f sxdg 3 − 3f f sxdg 2 f 9sxd
(c)Use part (b) to differentiate
y − sx 4 1 3x 3 1 17x 1 82d3
108. Reciprocal Rule If t is differentiable, the Reciprocal Rule
says that
F G
f tsxdg 2
(a)Use the Quotient Rule to prove the Reciprocal Rule.
(b)Use the Reciprocal Rule to differentiate the function in
Exercise 38.
109. An easy proof of the Quotient Rule can be given if we make
the prior assumption that F9sxd exists, where F − fyt. Write
f − Ft; then differentiate using the Product Rule and solve
the resulting equation for F9.
110. (a)If Fsxd − f sxd tsxd, where f and t have derivatives of
all orders, show that F99 − f 99t 1 2 f 9t9 1 f t99.
(b)Find similar formulas for F999 and F s4d.
(c)Guess a formula for F snd.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPLIED PROJECTBuilding a Better Roller Coaster
111. Let
f sxd −
x 1 1 if x , 1
x 1 1 if x > 1
Is f differentiable at 1? Sketch the graphs of f and f 9.
112. At what numbers is the following function t differentiable?
116. A
tangent line is drawn to the hyperbola xy − c at a
point P as shown in the figure.
(a)Show that the midpoint of the line segment cut
from this tangent line by the coordinate axes is P.
(b)Show that the triangle formed by the tangent line and
the coordinate axes always has the same area, no matter
where P is located on the hyperbola.
if x < 0
tsxd − 2x 2 x 2 if 0 , x , 2
if x > 2
Give a formula for t9 and sketch the graphs of t and t9.
113. (a)For what values of x is the function f sxd − x 2 2 9
differentiable? Find a formula for f 9.
(b) Sketch the graphs of f and f 9.
| |
f sxd −
if x < 2
mx 1 b if x . 2
Find the values of m and b that make f differentiable
Susana Ortega / Shutterstock.com
114. W
here is the function hsxd − x 2 1 1 x 1 2 differenti­
able? Give a formula for h9 and sketch the graphs of h and h9.
115. Let
117. I f c . 12, how many lines through the point s0, cd are normal
lines to the parabola y − x 2 ? What if c < 12 ?
118. S
ketch the parabolas y − x 2 and y − x 2 2 2x 1 2. Do you
think there is a line that is tangent to both curves? If so, find
its equation. If not, why not?
119. Evaluate lim
x 1000 2 1
Suppose you are asked to design the first ascent and drop for a new roller coaster. By studying photographs of your favorite coasters, you decide to make the slope of the ascent 0.8 and
the slope of the drop 21.6. You decide to connect these two straight stretches y − L 1sxd and
y − L 2 sxd with part of a parabola y − f sxd − a x 2 1 bx 1 c, where x and f sxd are measured
in meters. For the track to be smooth there can’t be abrupt changes in direction, so you want
the linear segments L 1 and L 2 to be tangent to the parabola at the transition points P and Q.
(See the figure.) To simplify the equations, you decide to place the origin at P.
1. (a)Suppose the horizontal distance between P and Q is 30 m. Write equations in a, b, and
c that will ensure that the track is smooth at the transition points.
(b) Solve the equations in part (a) for a, b, and c to find a formula for f sxd.
; (c) Plot L 1, f , and L 2 to verify graphically that the transitions are smooth.
(d) Find the difference in elevation between P and Q.
2. T
he solution in Problem 1 might look smooth, but it might not feel smooth because the
piecewise defined function [consisting of L 1sxd for x , 0, f sxd for 0 < x < 30, and L 2sxd
for x . 30] doesn’t have a continuous second derivative. So you decide to improve the
design by using a quadratic function qsxd − ax 2 1 bx 1 c only on the interval
3 < x < 27 and connecting it to the linear functions by means of two cubic functions:
tsxd − kx 3 1 lx 2 1 mx 1 n
hsxd − px 1 qx 1 rx 1 s
27 , x < 30
(a)Write a system of equations in 11 unknowns that ensures that the functions and their
first two derivatives agree at the transition points.
(b)Solve the system of equations in part (a) to find formulas for qsxd, tsxd, and hsxd.
(c)Plot L 1, t, q, h, and L 2, and compare with the plot in Problem 1(c).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
2.4 Derivatives of Trigonometric Functions
A review of the trigonometric functions is given in Appendix D.
Before starting this section, you might need to review the trigonometric functions. In
particular, it is important to remember that when we talk about the function f defined for
all real numbers x by
f sxd − sin x
it is understood that sin x means the sine of the angle whose radian measure is x. A similar convention holds for the other trigonometric functions cos, tan, csc, sec, and cot.
Recall from Section 1.8 that all of the trigonometric functions are continuous at every
number in their domains.
■ Derivatives of the Trigonometric Functions
If we sketch the graph of the function f sxd − sin x and use the interpretation of f 9sxd as the
slope of the tangent to the sine curve in order to sketch the graph of f 9 (see Exercise 2.2.16),
then it looks as if the graph of f 9 may be the same as the cosine curve (see Figure 1).
y=ƒ=sin x
y=fª(x )
FIGURE 1
Let’s try to confirm our guess that if f sxd − sin x, then f 9sxd − cos x. From the definition of a derivative, we have
f 9sxd − lim
− lim
sinsx 1 hd 2 sin x
− lim
sin x cos h 1 cos x sin h 2 sin x
(using the addition formula for
sine; see Appendix D)
− lim
− lim
f sx 1 hd 2 f sxd
F S
sin x
cos h 2 1
− lim sin x lim
S DG
sin x cos h 2 sin x
cos x sin h
1 cos x
sin h
cos h 2 1
sin h
1 lim cos x lim
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.4Derivatives of Trigonometric Functions
Two of these four limits are easy to evaluate. Because we regard x as a constant when
com­puting a limit as h l 0, we have
lim sin x − sin xandlim cos x − cos x
Later in this section we will prove that
sin h
cos h 2 1
− 1andlim
Putting these limits into (1), we get
cos h 2 1
sin h
1 lim cos x lim
− ssin xd 0 1 scos xd 1 − cos x
f 9sxd − lim sin x lim
So we have proved the formula for the derivative of the sine function:
ssin xd − cos x
EXAMPLE 1 Differentiate y − x 2 sin x.
Figure 2 shows the graphs of the
function of Example 1 and its deriva­
tive. Notice that y9 − 0 whenever y
has a horizontal tangent.
SOLUTION Using the Product Rule and Formula 2, we have
− x2
ssin xd 1 sin x
sx 2 d
− x 2 cos x 1 2x sin x
Using the same methods as in the proof of Formula 2, we can prove (see Exercise 26)
scos xd − 2sin x
FIGURE 2
The tangent function can also be differentiated by using the definition of a derivative,
but it is easier to use the Quotient Rule together with Formulas 2 and 3:
stan xd −
S D
cos x
sin x
cos x
ssin xd 2 sin x
scos xd
cos x cos x 2 sin x s2sin xd
cos2x 1 sin2x
− sec2xscos2x 1 sin2x − 1d
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
stan xd − sec2x
The derivatives of the remaining trigonometric functions, csc, sec, and cot, can
also be found easily using the Quotient Rule (see Exercises 23 –25). We collect all the
differentiation formulas for trigonometric functions in the following table. Remember
that they are valid only when x is measured in radians.
Table of Derivatives of Trigonometric Functions
ssin xd − cos x
scos xd − 2sin x
stan xd − sec2x
When you memorize this table, it
is helpful to notice that the minus
signs go with the derivatives of
the “cofunctions,” that is, cosine,
cosecant, and cotangent.
scsc xd − 2csc x cot x
ssec xd − sec x tan x
scot xd − 2csc 2x
EXAMPLE 2 Differentiate f sxd −
of f have a horizontal tangent?
sec x
. For what values of x does the graph
1 1 tan x
SOLUTION The Quotient Rule gives
s1 1 tan xd
f 9sxd −
FIGURE 3
The horizontal tangents in Example 2
FIGURE 4
ssec xd 2 sec x
s1 1 tan xd
s1 1 tan xd2
s1 1 tan xd sec x tan x 2 sec x sec2x
s1 1 tan xd2
sec x stan x 1 tan2x 2 sec2xd
s1 1 tan xd2
sec x stan x 2 1d
ssec2 x − tan2 x 1 1d
s1 1 tan xd2
The graph of f has horizontal tangents when f 9sxd − 0. Because sec x is never 0, we
see that f 9sxd − 0 when tan x − 1, and this occurs when x − y4 1 n , where n is an
integer (see Figure 3).
Trigonometric functions are often used in modeling real-world phenomena. In particular, vibrations, waves, elastic motions, and other quantities that vary in a periodic
manner can be described using trigonometric functions. In the following example we
discuss an instance of simple harmonic motion.
EXAMPLE 3 An object fastened to the end of a vertical spring is stretched 4 cm
beyond its rest position and released at time t − 0. (See Figure 4 and note that the
downward direction is positive.) Its position at time t is
s − f std − 4 cos t
Find the velocity and acceleration at time t and use them to analyze the motion of the object.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.4Derivatives of Trigonometric Functions
SOLUTION The velocity and acceleration are
s4 cos td − 4
scos td − 24 sin t
s24 sin td − 24
ssin td − 24 cos t
2π t
FIGURE 5
The object oscillates from the lowest point ss − 4 cmd to the highest point
ss − 24 cmd. The period of the oscillation is 2, the period of cos t.
The speed is v − 4 sin t , which is greatest when sin t − 1, that is, when
cos t − 0. So the object moves fastest as it passes through its equilibrium position
ss − 0d. Its speed is 0 when sin t − 0, that is, at the highest and lowest points.
The acceleration a − 24 cos t − 0 when s − 0. It has greatest magnitude at the
highest and lowest points. See the graphs in Figure 5.
| |
EXAMPLE 4 Find the 27th derivative of cos x.
SOLUTION The first few derivatives of f sxd − cos x are as follows:
f 9sxd − 2sin x
f 0sxd − 2cos x
f -sxd − sin x
f s4dsxd − cos x
f s5dsxd − 2sin x
Look for a pattern.
We see that the successive derivatives occur in a cycle of length 4 and, in particular,
f sndsxd − cos x whenever n is a multiple of 4. Therefore
f s24dsxd − cos x
and, differentiating three more times, we have
f s27dsxd − sin x
■ Two Special Trigonometric Limits
In proving the formula for the derivative of sine we used two special limits, which we
now prove.
sin −1
PROOF Assume first that lies between 0 and y2. Figure 6 shows a sector of a circle
with center O, central angle , and radius 1. BC is drawn perpendicular to OA. By the
definition of radian measure, we have arc AB − . Also BC − OB sin − sin .
From the diagram we see that
| | |
| BC | , | AB | , arc AB
FIGURE 6
sin , sosin ,1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
Let the tangent lines at A and B intersect at E. You can see from Figure 7 that the
cir­cumference of a circle is smaller than the length of a circumscribed polygon, and
so arc AB , AE 1 EB . Thus
| | | |
| | | |
| | | |
− | AD | − | OA | tan − arc AB , AE 1 EB
, AE 1 ED
FIGURE 7
− tan (In Appendix F the inequality < tan is proved directly from the definition of the
length of an arc without resorting to geometric intuition as we did here.) Therefore
we have
sin ,
cos so
cos ,
sin ,1
We know that lim l 0 1 − 1 and lim l 0 cos − 1, so by the Squeeze Theorem, we
sin lim1
− 1s0 , , y2d
l 0
But the function ssin dy is an even function, so its right and left limits must be equal.
Hence, we have
sin lim
so we have proved Equation 5.
The first special limit we considered concerned the sine function. The following special limit involves cosine.
cos 2 1
PROOF We multiply numerator and denominator by cos 1 1 in order to put the
function in a form in which we can use limits that we know.
cos 2 1
− lim
− lim
2sin 2
− 2lim
scos 1 1d
− 2lim
− 21 D
cos 2 1 cos 1 1
cos 1 1
− lim
cos2 2 1
scos 1 1d
sin sin cos 1 1
sin sin lim
l 0 cos 1 1
S D
− 0(by Equation 5)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.4Derivatives of Trigonometric Functions
EXAMPLE 5 Find lim
sin 7x
SOLUTION In order to apply Equation 5, we first rewrite the function by multiplying
and dividing by 7:
sin 7x
Note that sin 7x ± 7 sin x.
S D
sin 7x
If we let − 7x, then l 0 as x l 0, so by Equation 5 we have
S D
sin 7x
sin 7x
− lim
sin 7
− 1− 4 l0 4
EXAMPLE 6 Calculate lim x cot x.
SOLUTION Here we divide numerator and denominator by x:
lim x cot x − lim
x cos x
sin x
lim cos x
cos x
− lim
sin x
x l 0 sin x
cos 0
(by the continuity of cosine and Equation 5)
EXAMPLE 7 Find lim
cos 2 1
sin SOLUTION In order to use Equations 5 and 6 we divide numerator and denominator by :
cos 2 1
− lim
sin cos 2 1
sin cos 2 1
− − 0
sin 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
1–22 Differentiate.
1. f sxd − 3 sin x 2 2 cos x
2. f sxd − tan x 2 4 sin x
3. y − x 1 cot x
4. y − 2 sec x 2 csc x
5. hsd − 2 sin 6. tsxd − 3x 1 x 2 cos x
7. y − sec tan 8. y − sin cos 9. f sd − s 2 cos d sin 10. f sxd − x cos x 1 2 tan x
11. Hstd − cos 2 t
12. y − usa cos u 1 b cot ud
sin 13. f sd −
1 1 cos 14. y −
15. y −
2 2 tan x
17. f swd −
19. y −
1 1 sec w
1 2 sec w
t sin t
cos x
1 2 sin x
16. f std −
18. y −
sin t
1 1 tan t
20. tszd −
21. f sd − cos sin cot t
sec z 1 tan z
22. y − x 2 sin x tan x
33. (a)If f sxd − sec x 2 x, find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
by graphing both f and f 9 for x , y2.
| |
34. (a)
If f sxd − sx sin x, find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
by graphing both f and f 9 for 0 < x < 2.
35. If tsd −
sin , find t9sd and t99sd.
36. If f std − sec t, find f 0sy4d.
37. (a)Use the Quotient Rule to differentiate the function
f sxd −
tan x 2 1
sec x
(b)Simplify the expression for f sxd by writing it in terms
of sin x and cos x, and then find f 9sxd.
(c)Show that your answers to parts (a) and (b) are
38. Suppose f sy3d − 4 and f 9sy3d − 22, and let
tsxd − f sxd sin x and hsxd − scos xdyf sxd. Find
(a) t9sy3d
(b) h9sy3d
23. Show that
scsc xd − 2csc x cot x.
24. Show that
ssec xd − sec x tan x.
40. Find the points on the curve y − scos xdys2 1 sin xd at
which the tangent is horizontal.
25. Show that
scot xd − 2csc 2x.
41. A
mass on a spring vibrates horizontally on a smooth
level surface (see the figure). Its equation of motion is
xstd − 8 sin t, where t is in seconds and x in centimeters.
(a) Find the velocity and acceleration at time t.
(b)Find the position, velocity, and acceleration of the mass
at time t − 2y3. In what direction is it moving at that
26. P
rove, using the definition of a derivative, that if
f sxd − cos x, then f 9sxd − 2sin x.
27–30 Find an equation of the tangent line to the curve at the
given point.
39. For what values of x does the graph of f sxd − x 1 2 sin x
have a horizontal tangent?
27. y − sin x 1 cos x, s0, 1d
28. y − x 1 sin x, s, d
29. y − x 1 tan x, s, d
30. y −
1 1 sin x
, s, 21d
cos x
31. (a)Find an equation of the tangent line to the curve
y − 2x sin x at the point sy2, d.
(b)Illustrate part (a) by graphing the curve and the tangent
line on the same screen.
32. (a)Find an equation of the tangent line to the curve
y − 3x 1 6 cos x at the point sy3, 1 3d.
(b)Illustrate part (a) by graphing the curve and the tangent
line on the same screen.
n elastic band is hung on a hook and a mass is hung on the
; 42. A
lower end of the band. When the mass is pulled downward
and then released, it vibrates vertically. The equation of
motion is s − 2 cos t 1 3 sin t, t > 0, where s is measured
in centi­meters and t in seconds. (Take the positive direction
to be downward.)
(a) Find the velocity and acceleration at time t.
(b) Graph the velocity and acceleration functions.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.4Derivatives of Trigonometric Functions
(c)When does the mass pass through the equilibrium
position for the first time?
(d)How far from its equilibrium position does the mass
(e)When is the speed the greatest?
43. A
ladder 6 m long rests against a vertical wall. Let be the
angle between the top of the ladder and the wall and let x be
the distance from the bottom of the ladder to the wall. If the
bottom of the ladder slides away from the wall, how fast
does x change with respect to when − y3?
44. A
n object with mass m is dragged along a horizontal plane
by a force acting along a rope attached to the object. If the
rope makes an angle with the plane, then the magnitude of
the force is
sin 1 cos where is a constant called the coefficient of friction.
(a) Find the rate of change of F with respect to .
(b) When is this rate of change equal to 0 ?
(c)If m − 20 kg and − 0.6, draw the graph of F as a
function of and use it to locate the value of for
which dFyd − 0. Is the value consistent with your
answer to part (b)?
63. Find constants A and B such that the function
y − A sin x 1 B cos x satisfies the differential equation
y99 1 y9 2 2y − sin x.
x sin
; 64. Evaluate xlim
xl 0
sin 5x
sin 3t
47. lim
t l 0 sin t
65. D
ifferentiate each trigonometric identity to obtain a new
(or familiar) identity.
(a) tan x −
46. lim
xl 0
1 2 sec x
51. lim
tan 2x
52. lim
sin tan 7
53. lim
sin 3x
5x 3 2 4x
54. lim
sin 3x sin 5x
55. lim
sin 1 tan 56. lim csc x sinssin xd
57. lim
cos 2 1
58. lim
sinsx 2 d
60. lim
sinsx 2 1d
x2 1 x 2 2
l 0
59. lim
x l y4
1 2 tan x
sin x 2 cos x
66. A
semicircle with diameter PQ sits on an isosceles triangle
PQR to form a region shaped like a two-dimensional icecream cone, as shown in the figure. If Asd is the area of the
semicircle and Bsd is the area of the triangle, find
61–62 Find the given derivative by finding the first few
derivatives and observing the pattern that occurs.
d 99
ssin xd
dx 99
d 35
sx sin xd
dx 35
l 01
A(¨ )
B(¨ )
10 cm
10 cm
67. T
he figure shows a circular arc of length s and a chord of
length d, both subtended by a central angle . Find
l 01
xl 0
cos x
1 1 cot x
csc x
sin2 3x
48. lim
50. lim
(b) sec x −
(c) sin x 1 cos x −
sin x
sin x
sin x 2 sin x cos x
sin x
cos x
49. lim
x l0
and illustrate by graphing
y − x sins1yxd
45–60 Find the limit.
45. lim
; 68. Let f sxd −
s1 2 cos 2x
(a)Graph f . What type of discontinuity does it appear to
have at 0?
(b)Calculate the left and right limits of f at 0. Do these
values confirm your answer to part (a)?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
2.5 The Chain Rule
Suppose you are asked to differentiate the function
Fsxd − sx 2 1 1
See Section 1.3 for a review of
composite functions.
The differentiation formulas you learned in the previous sections of this chapter do not
enable you to calculate F9sxd.
Observe that F is a composite function. In fact, if we let y − f sud − su and let
u − tsxd − x 2 1 1, then we can write y − Fsxd − f s tsxdd, that is, F − f 8 t. We know
how to differentiate both f and t, so it would be useful to have a rule that tells us how to
find the derivative of F − f 8 t in terms of the derivatives of f and t.
■ The Chain Rule
It turns out that the derivative of the composite function f 8 t is the product of the derivatives of f and t. This fact is one of the most important of the differentiation rules and is
called the Chain Rule. It seems plausible if we interpret derivatives as rates of change.
Regard duydx as the rate of change of u with respect to x, dyydu as the rate of change of
y with respect to u, and dyydx as the rate of change of y with respect to x. If u changes
twice as fast as x and y changes three times as fast as u, then it seems reasonable that
y changes six times as fast as x, and so we expect that dyydx is the product of dyydu
and duydx .
The Chain Rule If t is differentiable at x and f is differentiable at tsxd, then the
composite function F − f 8 t defined by Fsxd − f stsxdd is differentiable at x and
F9 is given by the product
F9sxd − f 9stsxdd ? t9sxd
In Leibniz notation, if y − f sud and u − tsxd are both differentiable functions, then
dy du
du dx
James Gregory
The first person to formulate the
Chain Rule was the Scottish mathematician James Gregory (1638 –1675),
who also designed the first practical
reflecting telescope. Gregory discovered the basic ideas of calculus at
about the same time as Newton. He
became the first Professor of Mathe­
matics at the University of St. Andrews
and later held the same position at
the University of Edinburgh. But one
year after accepting that position, he
died at the age of 36.
Formula 2 is easy to remember because if we think of dyydu and duydx as quotients,
then we could cancel du ; however, du has not been defined and duydx should not be
considered as an actual quotient.
COMMENTS ON THE PROOF OF THE CHAIN RULE Let Du be the change in u
corresponding to a change of Dx in x, that is,
Du − tsx 1 Dxd 2 tsxd
Then the corresponding change in y is
Dy − f su 1 Dud 2 f sud
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.5The Chain Rule
It is tempting to write
− lim
− lim
Dy Du
Du Dx
− lim
− lim
(Note that Du l 0 as Dx l 0
because t is continuous.)
Du Dx l 0 Dx
Dx l 0
Dx l 0
Du l 0
dy du
du dx
The only flaw in this reasoning is that in (3) it might happen that Du − 0 (even when
Dx ± 0) and, of course, we can’t divide by 0. Nonetheless, this reasoning does at least
suggest that the Chain Rule is true. A full proof of the Chain Rule is given at the end of
this section.
EXAMPLE 1 Find F9sxd if Fsxd − sx 2 1 1.
SOLUTION 1 (using Formula 1): At the beginning of this section we expressed F as
Fsxd − s f 8 tdsxd − f s tsxdd where f sud − su and tsxd − x 2 1 1. Since
f 9sud − 12 u21y2 −
2 su
andt9sxd − 2x
F9sxd − f 9stsxdd t9sxd
we have
2 sx 1 1
2x −
sx 1 1
SOLUTION 2 (using Formula 2): If we let u − x 2 1 1 and y − su , then
F9sxd −
dy du
s2xd −
s2xd −
du dx
2 su
2 sx 1 1
sx 1 1
When using Formula 2 we should bear in mind that dyydx refers to the derivative of
y when y is considered as a function of x (the derivative of y with respect to x), whereas
dyydu refers to the derivative of y when considered as a function of u (the derivative of
y with respect to u). For instance, in Example 1, y can be considered as a function of
x ( y − s x 2 1 1 ) and also as a function of u ( y − su ). Note that
− F9sxd −
whereas− f 9sud −
2 su
sx 2 1 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
NOTE In using the Chain Rule we work from the outside to the inside. Formula 1 says
that we differentiate the outer function f [at the inner function tsxd] and then we multiply
by the derivative of the inner function.
at inner
of outer
at inner
of inner
EXAMPLE 2 Differentiate (a) y − sinsx 2 d and (b) y − sin2x.
(a) If y − sinsx 2 d, then the outer function is the sine function and the inner function is
the squaring function, so the Chain Rule gives
sx 2 d
at inner
sx 2 d
of outer
at inner
of inner
− 2x cossx 2 d
(b) Note that sin2x − ssin xd2. Here the outer function is the squaring function and the
inner function is the sine function. So
ssin xd2
See Reference Page 2 or Appendix D.
ssin xd
cos x
derivative derivative
of outerofofof
at inner atatinner
inner of inner ofofof
function function
functionfunction function
The answer can be left as 2 sin x cos x or written as sin 2x (by a trigonometric identity
known as the double-angle formula).
In Example 2(a) we combined the Chain Rule with the rule for differentiating the sine
function. In general, if y − sin u, where u is a differentiable function of x, then, by the
Chain Rule,
dy du
− cos u
du dx
ssin ud − cos u
In a similar fashion, all of the formulas for differentiating trigonometric functions can
be combined with the Chain Rule.
Let’s make explicit the special case of the Chain Rule where the outer function f is a
power function. If y − ftsxdg n, then we can write y − f sud − u n where u − tsxd. By
using the Chain Rule and then the Power Rule, we get
dy du
− nu n21
− nf tsxdg n21 t9sxd
du dx
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.5The Chain Rule
4 The Power Rule Combined with the Chain Rule If n is any real number
and u − tsxd is differentiable, then
su n d − nu n21
ftsxdg n − n ftsxdg n21 t9sxd
Notice that the derivative we found in Example 1 could be calculated by taking
n − 12 in Rule 4.
EXAMPLE 3 Differentiate y − sx 3 2 1d100.
SOLUTION Taking u − tsxd − x 3 2 1 and n − 100 in (4), we have
sx 3 2 1d100 − 100sx 3 2 1d99
sx 3 2 1d
− 100sx 3 2 1d99 3x 2 − 300x 2sx 3 2 1d99
EXAMPLE 4 Find f 9sxd if f sxd −
sx 1 x 1 1
f sxd − sx 2 1 x 1 1d21y3
SOLUTION First rewrite f :
f 9sxd − 213 sx 2 1 x 1 1d24y3
sx 2 1 x 1 1d
− 213 sx 2 1 x 1 1d24y3s2x 1 1d
EXAMPLE 5 Find the derivative of the function
tstd −
S D
2t 1 1
SOLUTION Combining the Power Rule, Chain Rule, and Quotient Rule, we get
S D S D
S D
t9std − 9
2t 1 1
2t 1 1
s2t 1 1d 1 2 2st 2 2d
45st 2 2d8
s2t 1 1d2
s2t 1 1d10
2t 1 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
EXAMPLE 6 Differentiate y − s2x 1 1d5sx 3 2 x 1 1d4.
SOLUTION In this example we must use the Product Rule before using the Chain Rule:
The graphs of the functions y and y9
in Example 6 are shown in Figure 1.
Notice that y9 is large when y
increases rapidly and y9 − 0 when y
has a horizontal tangent. So our
answer appears to be reasonable.
− s2x 1 1d5
sx 3 2 x 1 1d4 1 sx 3 2 x 1 1d4
s2x 1 1d5
− s2x 1 1d5 4sx 3 2 x 1 1d3
sx 3 2 x 1 1d
1 sx 3 2 x 1 1d4 5s2x 1 1d4
s2x 1 1d
− 4s2x 1 1d5sx 3 2 x 1 1d3s3x 2 2 1d 1 5sx 3 2 x 1 1d4s2x 1 1d4 2
FIGURE 1
Noticing that each term has the common factor 2s2x 1 1d4sx 3 2 x 1 1d3, we could
factor it out and write the answer as
− 2s2x 1 1d4sx 3 2 x 1 1d3s17x 3 1 6x 2 2 9x 1 3d
The reason for the name “Chain Rule” becomes clear when we make a longer chain
by adding another link. Suppose that y − f sud, u − tsxd, and x − hstd where f , t, and h
are differentiable functions. Then, to compute the derivative of y with respect to t, we use
the Chain Rule twice:
dy dx
dy du dx
dx dt
du dx dt
EXAMPLE 7 If f sxd − sinscosstan xdd, then
f 9sxd − cosscosstan xdd
cosstan xd
stan xd
− 2cosscosstan xdd sinstan xd sec2x
− cosscosstan xdd f2sinstan xdg
Notice that we used the Chain Rule twice.
EXAMPLE 8 Differentiate y − ssec x 3 .
SOLUTION Here the outer function is the square root function, the middle function is
the secant function, and the inner function is the cubing function. So we have
ssec x 3 d
2 ssec x dx
sec x 3 tan x 3
sx 3 d
2 ssec x 3
3x 2 sec x 3 tan x 3
2 ssec x 3
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.5The Chain Rule
■ How to Prove the Chain Rule
Recall that if y − f sxd and x changes from a to a 1 Dx, we define the increment of y as
Dy − f sa 1 Dxd 2 f sad
According to the definition of a derivative, we have
Dx l 0
− f 9sad
So if we denote by « the difference between DyyDx and f 9sad, we obtain
lim « − lim
Dx l 0
Dx l 0
2 f 9sad − f 9sad 2 f 9sad − 0
2 f 9sad
Dy − f 9sad Dx 1 « Dx
If we define « to be 0 when Dx − 0, then « becomes a continuous function of Dx. Thus,
for a differentiable function f, we can write
Dy − f 9sad Dx 1 « Dx
« l 0 as Dx l 0
and « is a continuous function of Dx. This property of differentiable functions is what
enables us to prove the Chain Rule.
PROOF OF THE CHAIN RULE Suppose u − tsxd is differentiable at a and y − f sud is
differentiable at b − tsad. If Dx is an increment in x and Du and Dy are the corresponding increments in u and y, then we can use Equation 5 to write
Du − t9sad Dx 1 «1 Dx − ft9sad 1 «1 g Dx
where «1 l 0 as Dx l 0. Similarly
Dy − f 9sbd Du 1 «2 Du − f f 9sbd 1 «2 g Du
where «2 l 0 as Du l 0. If we now substitute the expression for Du from Equation 6
into Equation 7, we get
Dy − f f 9sbd 1 «2 gft9sad 1 «1 g Dx
− f f 9sbd 1 «2 gft9sad 1 «1 g
As Dx l 0, Equation 6 shows that Du l 0. Taking the limit as Dx l 0 , we get
− lim
− lim f f 9sbd 1 «2 gft9sad 1 «1 g
Dx l 0 Dx
Dx l 0
− f 9sbd t9sad − f 9stsadd t9sad
This proves the Chain Rule.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
1–6 Write the composite function in the form f s tsxdd.
[Identify the inner function u − tsxd and the outer function
y − f sud.] Then find the derivative dyydx.
1. y − s5 2 x 4d3
2. y − sx 3 1 2
3. y − sinscos xd
4. y − tansx 2 d
5. y − ssin x
6. y − sin sx
10. f sxd −
s2t 1 1d2
12. Fstd −
S D
2t 1 1
14. tsxd − s2 2 sin xd
15. f sd − coss 2 d
16. tsd − cos2 3
17. hsvd − v s
1 1 v2
18. f std − t sin t
20. Gszd − s1 2 4zd2sz 2 1 1
21. hstd − st 1 1d s2t 2 1d
25. tsud −
u3 2 1
u3 1 1
S D
24. y − x 1
26. sstd −
1 1 sin t
1 1 cos t
sr 2 2 1d 3
27. Hsrd −
s2r 1 1d 5
28. Fstd −
29. y − cosssec 4xd
30. Jsd − tan 2 snd
31. y −
33. y −
cos x
s1 1 sin x
1 2 cos 2x
1 1 cos 2x
st 3 1 1
32. hsd − tans 2 sin d
50. y − (1 1 sx )
51. y − scos x
52. y −
sx 1 1
53. y − s3x 2 1d26, s0, 1d
54. y − s1 1 x 3 , s2, 3d
56. y − sin 2 x cos x, sy2, 0d
57. (a)Find an equation of the tangent line to the curve
y − tansx 2y4d at the point s1, 1d.
part (a) by graphing the curve and the tangent
line on the same screen.
| |
22. Fstd − s3t 2 1d4 s2t 1 1d23
23. y −
49. y − cosssin 3d
55. y − sinssin xd, s, 0d
19. Fsxd − s4x 1 5d 3sx 2 2 2x 1 5d4
48. y − fx 1 sx 1 sin2 xd3 g 4
53–56 Find an equation of the tangent line to the curve at the
given point.
sx 2 1
13. Astd −
scos t 1 tan td 2
47. y − cos ssinstan xd
8. f sxd − sx 5 1 3x 2 2 xd50
9. f sxd − s5x 1 1
11. tstd −
46. y − sins 1 tans 1 cos dd
49–52 Find y9 and y 99.
7–48 Find the derivative of the function.
7. f sxd − s2x 3 2 5x 2 1 4d5
45. tsxd − s2r sin rx 1 nd p
34. y − x sin
35. f sxd − sin x coss1 2 x 2 d
36. y − sin ( t 1 cos st )
37. Fstd − tan s1 1 t 2
38. Gszd − s1 1 cos2zd3
39. y − sin2sx 2 1 1d
40. tsud − fsu 2 2 1d 6 2 3ug 4
41. y − cos 4ssin3 xd
42. y − sin3scossx 2 dd
43. f std − tanssecscos tdd
44. y − sx 1 sx 1 s x
58. (a)The curve y − x ys2 2 x 2 is called a bullet-nose
curve. Find an equation of the tangent line to this curve
at the point s1, 1d.
(b)Illustrate part (a) by graphing the curve and the tangent
line on the same screen.
59. (a)If f sxd − x s2 2 x 2 , find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
by comparing the graphs of f and f 9.
; 60. The function f sxd − sinsx 1 sin 2xd, 0 < x < , arises in
applications to frequency modulation (FM) synthesis.
(a)Use a graph of f produced by a calculator or computer
to make a rough sketch of the graph of f 9.
(b)Calculate f 9sxd and use this expression, with a
calculator or computer, to graph f 9. Compare with your
sketch in part (a).
61. F
ind all points on the graph of the function
f sxd − 2 sin x 1 sin2x at which the tangent line is
62. A
t what point on the curve y − s1 1 2x is the tangent line
perpendicular to the line 6x 1 2y − 1?
63. If Fsxd − f stsxdd, where f s22d − 8, f 9s22d − 4,
f 9s5d − 3, ts5d − 22, and t9s5d − 6, find F9s5d.
64. If hsxd − s4 1 3f sxd , where f s1d − 7 and f 9s1d − 4,
find h9s1d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.5The Chain Rule
65. A table of values for f , t, f 9, and t9 is given.
f sxd
f 9sxd
(a) If hsxd − f stsxdd, find h9s1d.
(b) If Hsxd − ts f sxdd, find H9s1d.
70. Suppose f is differentiable on R and is a real number.
Let Fsxd − f sx d and Gsxd − f f sxdg . Find expressions
for (a) F9sxd and (b) G9sxd.
71. Let rsxd − f s tshsxddd, where hs1d − 2, ts2d − 3, h9s1d − 4,
t9s2d − 5, and f 9s3d − 6. Find r9s1d.
72. If t is a twice differentiable function and f sxd − x tsx 2 d, find
f 99 in terms of t, t9, and t99.
73. I f Fsxd − f s3f s4 f sxddd, where f s0d − 0 and f 9s0d − 2,
find F9s0d.
66. Let f and t be the functions in Exercise 65.
(a) If Fsxd − f s f sxdd, find F9s2d.
(b) If Gsxd − tstsxdd, find G9s3d.
67. If f and t are the functions whose graphs are shown, let
usxd − f s tsxdd, vsxd − ts f sxdd, and w sxd − ts tsxdd. Find
each derivative, if it exists. If it does not exist, explain why.
(a) u9s1d(b) v9s1d(c) w9s1d
74. If Fsxd − f sx f sx f sxddd, where f s1d − 2, f s2d − 3,
f 9s1d − 4, f 9s2d − 5, and f 9s3d − 6, find F9s1d.
75–76 Find the given derivative by finding the first few deriva­
tives and observing the pattern that occurs.
75. D103 cos 2x
76. D 35 x sin x
77. T
he displacement of a particle on a vibrating string is given
by the equation
sstd − 10 1 14 sins10td
where s is measured in centimeters and t in seconds. Find the
velocity of the particle after t seconds.
68. If f is the function whose graph is shown, let hsxd − f s f sxdd
and tsxd − f sx 2 d. Use the graph of f to estimate the value
of each derivative.
(a) h9s2d
(b) t9s2d
78. I f the equation of motion of a particle is given by
s − A cosst 1 d, the particle is said to undergo simple
harmonic motion.
(a) Find the velocity of the particle at time t.
(b) When is the velocity 0 ?
79. A
Cepheid variable star is a star whose brightness alternately
increases and decreases. The most easily visible such star is
Delta Cephei, for which the interval between times of maximum brightness is 5.4 days. The average brightness of this
star is 4.0 and its brightness changes by 60.35. In view of
these data, the brightness of Delta Cephei at time t, where t is
measured in days, has been modeled by the function
S D
Bstd − 4.0 1 0.35 sin
69. If tsxd − sf sxd , where the graph of f is shown,
evaluate t9s3d.
(a) Find the rate of change of the brightness after t days.
(b)Find, correct to two decimal places, the rate of increase
after one day.
80. I n Example 1.3.4 we arrived at a model for the length of daylight (in hours) in Philadelphia on the t th day of the year:
Lstd − 12 1 2.8 sin
st 2 80d
Use this model to compare how the number of hours of
day­light is increasing in Philadelphia on March 21 st − 80d
and May 21 st − 141d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
(This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: the differentiation formulas would not be
as simple if we used degree measure.)
81. A
particle moves along a straight line with displacement
sstd, velocity vstd, and acceleration astd. Show that
astd − vstd
Explain the difference between the meanings of the derivatives dvydt and dvyds.
| |
86. (a) Write x − sx 2 and use the Chain Rule to show that
x −
| |
82. A
ir is being pumped into a spherical weather balloon. At any
time t, the volume of the balloon is Vstd and its radius is rstd.
(a)What do the derivatives dVydr and dVydt represent?
(b)Express dVydt in terms of drydt.
83. Use the Chain Rule to prove the following.
(a)The derivative of an even function is an odd function.
(b)The derivative of an odd function is an even function.
84. U
se the Chain Rule and the Product Rule to give an alter­
native proof of the Quotient Rule.
[Hint: Write f sxdytsxd − f sxdf tsxdg 21.]
85. U
se the Chain Rule to show that if is measured in degrees,
ssin d −
cos d
| |
(b)If f sxd − sin x , find f 9sxd and sketch the graphs of f
and f 9. Where is f not differentiable?
(c)If tsxd − sin x , find t9sxd and sketch the graphs of t
and t9. Where is t not differentiable?
| |
87. If F − f + t + h , where f , t, and h are differentiable functions, use the Chain Rule to show that
F9sxd − f 9s tshsxdd ? t9shsxdd ? h9sxd
88. If F − f + t, where f and t are twice differentiable functions, use the Chain Rule and the Product Rule to show that
the second derivative of F is given by
F 0sxd − f 0s tsxdd ? [ t9sxd] 2 1 f 9s tsxdd ? t 0sxd
An approach path for an aircraft landing is shown in the figure and satisfies the following
(i) The cruising altitude is h when descent starts at a horizontal distance , from touchdown at the origin.
(ii) The pilot must maintain a constant horizontal speed v throughout descent.
(iii) The absolute value of the vertical acceleration should not exceed a constant k (which
is much less than the acceleration due to gravity).
1. Find
a cubic polynomial Psxd − ax 3 1 bx 2 1 cx 1 d that satisfies condition (i) by
imposing suitable conditions on Psxd and P9sxd at the start of descent and at touchdown.
2. Use conditions (ii) and (iii) to show that
6h v 2
3. Suppose
that an airline decides not to allow vertical acceleration of a plane to exceed
k − 1385 kmyh2. If the cruising altitude of a plane is 11,000 m and the speed is 480 kmyh,
how far away from the airport should the pilot start descent?
; 4. Graph the approach path if the conditions stated in Problem 3 are satisfied.
2.6 Implicit Differentiation
■ Implicitly Defined Functions
The functions that we have met so far can be described by expressing one variable explicitly in terms of another variable —for example,
y − sx 3 1 1ory − x sin x
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.6Implicit Differentiation
or, in general, y − f sxd. Some functions, however, are defined implicitly by a relation
between x and y such as
x 2 1 y 2 − 25
x 3 1 y 3 − 6xy
In some cases it is possible to solve such an equation for y as an explicit function (or several functions) of x. For instance, if we solve Equation 1 for y, we get y − 6s25 2 x 2 ,
so two of the functions determined by the implicit Equation l are f sxd − s25 2 x 2 and
tsxd − 2s25 2 x 2 . The graphs of f and t are the upper and lower semicircles of the
cir­cle x 2 1 y 2 − 25. (See Figure 1.)
FIGURE 1
(a) ≈+¥=25
(b) ƒ=œ„„„„„„
(c) ©=_ œ„„„„„„
It’s not easy to solve Equation 2 for y explicitly as a function of x by hand. (Even if
we use technology, the resulting expressions are very complicated.) Nonetheless, (2) is
the equation of a curve called the folium of Descartes shown in Figure 2 and it implicitly
defines y as several functions of x. The graphs of three such functions are shown in Figure 3. When we say that f is a function defined implicitly by Equa­tion 2, we mean that
the equation
x 3 1 f f sxdg 3 − 6x f sxd
is true for all values of x in the domain of f .
FIGURE 2 The folium of Descartes
FIGURE 3 Graphs of three functions defined implicitly by the folium of Descartes
■ Implicit Differentiation
Fortunately, we don’t need to solve an equation for y in terms of x in order to find the
derivative of y. Instead we can use the method of implicit differentiation. This consists
of differentiating both sides of the equation with respect to x and then solving the resulting equation for dyydx. In the examples and exercises of this section it is always assumed
that the given equation determines y implicitly as a differentiable function of x so that the
method of implicit differentiation can be applied.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
EXAMPLE 1
. Then find an equation of the tangent to the circle
x 1 y − 25 at the point s3, 4d.
If x 2 1 y 2 − 25, find
Differentiate both sides of the equation x 2 1 y 2 − 25:
sx 2 1 y 2 d −
sx 2 d 1
sy 2 d − 0
Remembering that y is a function of x and using the Chain Rule, we have
sy 2 d −
sy 2 d
− 2y
2x 1 2y
Now we solve this equation for dyydx:
At the point s3, 4d we have x − 3 and y − 4, so
An equation of the tangent to the circle at s3, 4d is therefore
y 2 4 − 234 sx 2 3dor3x 1 4y − 25
Solving the equation x 2 1 y 2 − 25 for y, we get y − 6s25 2 x 2 . The point
s3, 4d lies on the upper semicircle y − s25 2 x 2 and so we consider the function
f sxd − s25 2 x 2 . Differentiating f using the Chain Rule, we have
f 9sxd − 12 s25 2 x 2 d21y2
s25 2 x 2 d
− 12 s25 2 x 2 d21y2s22xd − 2
s25 2 x 2
At the point s3, 4d we have
Example 1 illustrates that even when
it is possible to solve an equation
explicitly for y in terms of x, it may be
easier to use implicit differentiation.
f 9s3d − 2
s25 2 3
and, as in Solution 1, an equation of the tangent is 3x 1 4y − 25.
NOTE 1 The expression dyydx − 2xyy in Solution 1 gives the derivative in terms of
both x and y. It is correct no matter which function y is determined by the given equation.
For instance, for y − f sxd − s25 2 x 2 we have
−2 −2
s25 2 x 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.6Implicit Differentiation
whereas for y − tsxd − 2s25 2 x 2 we have
−2 −2
2s25 2 x
s25 2 x 2
EXAMPLE 2
(a) Find y9 if x 3 1 y 3 − 6xy.
(b) Find the tangent to the folium of Descartes x 3 1 y 3 − 6xy at the point s3, 3d.
(c) At what point in the first quadrant is the tangent line horizontal?
We can use either of the notations
dyydx or y9 for the derivative of y
with respect to x.
(a) Differentiating both sides of x 3 1 y 3 − 6xy with respect to x, regarding y as a
function of x, and using the Chain Rule on the term y 3 and the Product Rule on the
term 6xy, we get
3x 2 1 3y 2 y9 − 6xy9 1 6y
x 2 1 y 2 y9 − 2xy9 1 2y
We now solve for y9:
y 2 y9 2 2xy9 − 2y 2 x 2
sy 2 2 2xdy9 − 2y 2 x 2
(3, 3)
y9 −
2y 2 x 2
y 2 2 2x
(b) When x − y − 3,
y9 −
FIGURE 4
− 21
and a glance at Figure 4 confirms that this is a reasonable value for the slope at s3, 3d.
So an equation of the tangent to the folium at s3, 3d is
y 2 3 − 21sx 2 3dorx 1 y − 6
(c) The tangent line is horizontal if y9 − 0. Using the expression for y9 from part (a),
we see that y9 − 0 when 2y 2 x 2 − 0 (provided that y 2 2 2x ± 0d. Substituting
y − 12 x 2 in the equation of the curve, we get
x 3 1 ( 12 x 2) − 6x ( 12 x 2)
FIGURE 5
which simplifies to x 6 − 16x 3. Since x ± 0 in the first quadrant, we have x 3 − 16. If
x − 16 1y3 − 2 4y3, then y − 12 s2 8y3 d − 2 5y3. Thus the tangent is horizontal at s2 4y3, 2 5y3 d,
which is approximately (2.5198, 3.1748). Looking at Figure 5, we see that our answer
is reasonable.
NOTE 2 There is a formula for the three solutions of a cubic equation that is like the
quad­ratic formula but much more complicated. If we use this formula (or a computer) to
solve the equation x 3 1 y 3 − 6xy for y in terms of x, we get three functions determined
by the equations:
y − f sxd − s221 x 3 1 s14 x 6 2 8x 3 1 s221 x 3 2 s14 x 6 2 8x 3
y − 12 2f sxd 6 s23 (s221 x 3 1 s14 x 6 2 8x 3 2 s221 x 3 2 s14 x 6 2 8x 3
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
(These are the three functions whose graphs are shown in Figure 3.) You can see that the
method of implicit differentiation saves an enormous amount of work in cases such as
this. Moreover, implicit differentiation works just as easily for equations such as
Abel and Galois
The Norwegian mathematician Niels
Abel proved in 1824 that no general
formula in terms of radicals can be
given for the solutions of the equation
psxd − 0, where p is a polynomial of
degree 5 with integer coefficients.
Later the French mathematician
Evariste Galois proved that it is impos­
sible to find a general formula in
terms of radicals for the solutions of
an equation psxd − 0, where p is a
polynomial of degree n > 5 .
y 5 1 3x 2 y 2 1 5x 4 − 12
for which it is impossible to find an expression for y in terms of x.
EXAMPLE 3 Find y9 if sinsx 1 yd − y 2 cos x.
SOLUTION Differentiating implicitly with respect to x and remembering that y is a
function of x, we get
cossx 1 yd s1 1 y9d − y 2s2sin xd 1 scos xds2yy9d
(Note that we have used the Chain Rule on the left side and the Product Rule and Chain
Rule on the right side.) If we collect the terms that involve y9, we get
cossx 1 yd 1 y 2 sin x − s2y cos xdy9 2 cossx 1 yd y9
y9 −
y 2 sin x 1 cossx 1 yd
2y cos x 2 cossx 1 yd
Figure 6, drawn by a computer, shows part of the curve sinsx 1 yd − y 2 cos x. As a
check on our calculation, notice that y9 − 21 when x − y − 0 and it appears from the
graph that the slope is approximately 21 at the origin.
Figures 7, 8, and 9 show three more curves produced by a computer. In Exercises 43–44
you will have an opportunity to create and examine unusual curves of this nature.
FIGURE 6
FIGURE 7
sx 2 2 1dsx 2 2 4dsx 2 2 9d
− y 2 s y 2 2 4ds y 2 2 9d
FIGURE 8
cossx 2 sin yd − sins y 2 sin xd
FIGURE 9
sinsxyd − sin x 1 sin y
■ Second Derivatives of Implicit Functions
The following example shows how to find the second derivative of a function that is
defined implicitly.
EXAMPLE 4 Find y99 if x 4 1 y 4 − 16.
SOLUTION Differentiating the equation implicitly with respect to x, we get
4x 3 1 4y 3 y9 − 0
Solving for y9 gives
y9 − 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.6Implicit Differentiation
To find y99 we differentiate this expression for y9 using the Quotient Rule and remembering that y is a function of x:
S D
y 3 sdydxdsx 3 d 2 x 3 sdydxdsy 3 d
2 3 −2
sy 3 d2
y 3x 2 x s3y y9d
y99 −
If we now substitute Equation 3 into this expression, we get
S D
3x 2 y 3 2 3x 3 y 2 2
y99 − 2
3sx y 1 x 6 d
3x 2sy 4 1 x 4 d
But the values of x and y must satisfy the original equation x 4 1 y 4 − 16. So the
answer simplifies to
3x 2s16d
y99 − 2
Figure 10 shows the graph of the
curve x 4 1 y 4 − 16 of Example 4.
Notice that it’s a stretched and
flat­tened version of the circle
x 2 1 y 2 − 4. For this reason it’s
sometimes called a fat circle. It starts
out very steep on the left but quickly
becomes very flat. This can be seen
from the expression
y9 − 2
x $+y$ =16
2 x
FIGURE 10
(a) Find y9 by implicit differentiation.
(b) Solve the equation explicitly for y and differentiate to get y9
in terms of x.
(c) Check that your solutions to parts (a) and (b) are consistent
by substituting the expression for y into your solution for
part (a).
1. 5x 2 2 y 3 − 7
2. 6x 4 1 y 5 − 2x
3. sx 1 sy − 1
2 −4
5–20 Find dyydx by implicit differentiation.
5. x 2 2 4xy 1 y 2 − 4
6. 2x 2 1 xy 2 y 2 − 2
7. x 4 1 x 2 y 2 1 y 3 − 5
8. x 3 2 xy 2 1 y 3 − 1
− y2 1 1
10. y 5 1 x 2 y 3 − 1 1 x 4 y
11. sin x 1 cos y − 2x 2 3y
12. y sinsx 2 d − x sins y 2 d
13. sinsx 1 yd − cos x 1 cos y
14. tansx 2 yd − 2x y 3 1 1
15. tansxyyd − x 1 y
16. sinsxyd − cossx 1 yd
17. sx 1 y − x 4 1 y 4
18. sin x cos y − x 2 2 5y
19. sxy − 1 1 x 2 y
20. xy − sx 2 1 y 2
21. If f sxd 1 x 2 f f sxdg 3 − 10 and f s1d − 2, find f 9s1d.
22. If tsxd 1 x sin tsxd − x 2, find t9s0d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
23–24 Regard y as the independent variable and x as the
dependent variable and use implicit differentiation to find dxydy.
23. x 4y 2 2 x 3y 1 2xy 3 − 0
24. y sec x − x tan y
25–34 Use implicit differentiation to find an equation of the
tangent line to the curve at the given point.
25. y sin 2x − x cos 2y,sy2, y4d
sy8, y8d
26. tansx 1 yd 1 secsx 2 yd − 2,
27. x
− 4,
28. y s6 2 xd − x ,
(23 s3, 1) (astroid)
(2, s2 ) (cissoid of Diocles)
29. x 2 2 xy 2 y 2 − 1,
s2, 1d
30. x 1 2xy 1 4y − 12,
s2, 1d
31. x 2 1 y 2 − s2x 2 1 2y 2 2 xd 2,
(0, 12 )
35. (a)The curve with equation y 2 − 5x 4 2 x 2 is called a
kampyle of Eudoxus. Find an equation of the tangent
line to this curve at the point s1, 2d.
(b)Illustrate part (a) by graphing the curve and the tangent
line on a common screen. (Graph the implicitly defined
curve if possible, or you can graph the upper and lower
halves separately.)
36. (a)The curve with equation y 2 − x 3 1 3x 2 is called the
Tschirnhausen cubic. Find an equation of the tangent
line to this curve at the point s1, 22d.
(b)At what points does this curve have horizontal tangents?
(c)Illustrate parts (a) and (b) by graphing the curve and the
tangent lines on a common screen.
37–40 Find y99 by implicit differentiation. Simplify where
37. x 2 1 4y 2 − 4
38. x 2 1 xy 1 y 2 − 3
39. sin y 1 cos x − 1
40. x 3 2 y 3 − 7
41. If xy 1 y 3 − 1, find the value of y99 at the point where x − 0.
32. x y − s y 1 1d s4 2 y d,
(conchoid of Nicomedes)
42. If x 2 1 xy 1 y 3 − 1, find the value of y999 at the point
where x − 1.
(2 s3, 1)
ys y 2 2 1ds y 2 2d − xsx 2 1dsx 2 2d
At how many points does this curve have horizontal
tangents? Estimate the x-coordinates of these points.
(b)Find equations of the tangent lines at the points (0, 1)
and (0, 2).
(c)Find the exact x-coordinates of the points in part (a).
(d)Create even more fanciful curves by modifying the
equation in part (a).
33. 2sx 2 1 y 2 d2 − 25sx 2 2 y 2 d,
s3, 1d
; 44. (a) The curve with equation
2y 3 1 y 2 2 y 5 − x 4 2 2x 3 1 x 2
34. y 2s y 2 2 4d − x 2sx 2 2 5d,
anciful shapes can be created by using software that can
; 43. F
graph implicitly defined curves.
(a) Graph the curve with equation
s0, 22d
(devil’s curve)
has been likened to a bouncing wagon. Graph this curve
and discover why.
(b)At how many points does this curve have horizontal
tangent lines? Find the x-coordinates of these points.
45. F
ind the points on the lemniscate in Exercise 33 where the
tangent is horizontal.
46. S
how by implicit differentiation that the tangent line to
the ellipse
at the point sx 0 , y 0 d has equation
x0 x
y0 y
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.6Implicit Differentiation
47. Find an equation of the tangent line to the hyperbola
2 2 −1
at the point sx 0 , y 0 d.
48. Show that the sum of the x- and y-intercepts of any tangent
line to the curve sx 1 sy − sc is equal to c.
49. S
how, using implicit differentiation, that any tangent line at
a point P to a circle with center O is perpendicular to the
radius OP.
50. T
he Power Rule can be proved using implicit differentiation
for the case where n is a rational number, n − pyq, and
y − f sxd − x n is assumed beforehand to be a differentiable
function. If y − x pyq, then y q − x p. Use implicit differentiation to show that
p s pyqd21
y9 −
51–54 Orthogonal Trajectories Two curves are orthogonal if
their tangent lines are perpendicular at each point of intersection.
Show that the given families of curves are orthogonal trajectories
of each other; that is, every curve in one family is orthogonal to
every curve in the other family. Sketch both families of curves on
the same axes.
58. (a)Use implicit differentiation to find y9 if
x 2 1 xy 1 y 2 1 1 − 0
(b)Plot the curve in part (a). What do you see? Prove that
what you see is correct.
(c)In view of part (b), what can you say about the
expression for y9 that you found in part (a)?
59. T
he equation x 2 2 xy 1 y 2 − 3 represents a “rotated
ellipse,” that is, an ellipse whose axes are not parallel to the
coordinate axes. Find the points at which this ellipse crosses
the x-axis and show that the tangent lines at these points are
60. (a)Where does the normal line to the ellipse
x 2 2 xy 1 y 2 − 3 at the point s21, 1d intersect the
ellipse a second time?
(b)Illustrate part (a) by graphing the ellipse and the
normal line.
61. F
ind all points on the curve x 2 y 2 1 xy − 2 where the slope
of the tangent line is 21.
62. F
ind equations of both the tangent lines to the ellipse
x 2 1 4y 2 − 36 that pass through the point s12, 3d.
63. Use implicit differentiation to find dyydx for the equation
− y 2 1 1y ± 0
51. x 2 1 y 2 − r 2, ax 1 by − 0
52. x 1 y − ax, x 1 y − by
53. y − cx 2, x 2 1 2y 2 − k
and for the equivalent equation
54. y − ax 3, x 2 1 3y 2 − b
55. S
how that the ellipse x 2ya 2 1 y 2yb 2 − 1 and the hyperbola
x 2yA2 2 y 2yB 2 − 1 are orthogonal trajectories if A2 , a 2 and
a 2 2 b 2 − A2 1 B 2 (so the ellipse and hyperbola have the
same foci).
56. F
ind the value of the number a such that the families of
curves y − sx 1 cd21 and y − asx 1 kd1y3 are orthogonal
57. The van der Waals equation for n moles of a gas is
n 2a
P 1 2 sV 2 nbd − nRT
where P is the pressure, V is the volume, and T is the temperature of the gas. The constant R is the universal gas constant
and a and b are positive constants that are characteristic of a
particular gas.
(a)If T remains constant, use implicit differentiation to
find dVydP.
(b)Find the rate of change of volume with respect to
pressure of 1 mole of carbon dioxide at a volume of
V − 10 L and a pressure of P − 2.5 atm. Use
a − 3.592 L2-atmymole 2 and b − 0.04267 Lymole.
x − y 3 1 yy ± 0
Show that although the expressions you get for dyydx
look different, they agree for all points that satisfy the given
64. T
he Bessel function of order 0, y − J sxd, satisfies the differential equation xy99 1 y9 1 xy − 0 for all values of x and
its value at 0 is J s0d − 1.
(a)Find J9s0d.
(b)Use implicit differentiation to find J99s0d.
65. T
he figure shows a lamp located three units to the right of
the y-axis and a shadow created by the elliptical region
x 2 1 4y 2 < 5. If the point s25, 0d is on the edge of the
shadow, how far above the x-axis is the lamp located?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
In this project you will explore the changing shapes of implicitly defined curves as you vary
the constants in a family, and determine which features are common to all members of the
1. Consider the family of curves
y 2 2 2x 2 sx 1 8d − c fs y 1 1d2 s y 1 9d 2 x 2 g
(a)By graphing the curves with c − 0 and c − 2, determine how many points of intersection there are. (You might have to zoom in to find all of them.)
(b)Now add the curves with c − 5 and c − 10 to your graphs in part (a). What do you
notice? What about other values of c ?
2. (a) Graph several members of the family of curves
x 2 1 y 2 1 cx 2 y 2 − 1
Describe how the graph changes as you change the value of c.
(b)What happens to the curve when c − 21? Describe what appears on the screen.
Can you prove it algebraically?
(c)Find y9 by implicit differentiation. For the case c − 21, is your expression for y9 consistent with what you discovered in part (b)?
2.7 Rates of Change in the Natural and Social Sciences
We know that if y − f sxd, then the derivative dyydx can be interpreted as the rate of
change of y with respect to x. In this section we examine some of the applications of this
idea to physics, chemistry, biology, economics, and other sciences.
Let’s recall from Section 2.1 the basic idea behind rates of change. If x changes from
x 1 to x 2, then the change in x is
Dx − x 2 2 x 1
and the corresponding change in y is
Dy − f sx 2 d 2 f sx 1 d
Q {x™, ‡}
The difference quotient
f sx 2 d 2 f sx 1 d
x2 2 x1
P { ⁄, fl}
mPQ average rate of change
m=fª(⁄ )=instantaneous rate
of change
FIGURE 1
is the average rate of change of y with respect to x over the interval fx 1, x 2 g and can be
interpreted as the slope of the secant line PQ in Figure 1. Its limit as Dx l 0 is the derivative f 9sx 1 d, which can therefore be interpreted as the instantaneous rate of change of y
with respect to x or the slope of the tangent line at Psx 1, f sx 1 dd. Using Leibniz notation,
we write the process in the form
− lim
Whenever the function y − f sxd has a specific interpretation in one of the sciences,
its derivative will have a specific interpretation as a rate of change. (As we discussed in
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.7Rates of Change in the Natural and Social Sciences
Sec­tion 2.1, the units for dyydx are the units for y divided by the units for x.) We now
look at some of these interpretations in the natural and social sciences.
■ Physics
If s − f std is the position function of a particle that is moving in a straight line, then
DsyDt represents the average velocity over a time period Dt, and v − dsydt represents
the instantaneous velocity (the rate of change of displacement with respect to time).
The instantaneous rate of change of velocity with respect to time is acceleration:
astd − v9std − s99std. This was discussed in Sections 2.1 and 2.2, but now that we know
the differentiation formulas, we are able to more easily solve problems involving the
motion of objects.
EXAMPLE 1 The position of a particle is given by the equation
s − f std − t 3 2 6t 2 1 9t
where t is measured in seconds and s in meters.
(a) Find the velocity at time t.
(b) What is the velocity after 2 s? After 4 s?
(c) When is the particle at rest?
(d) When is the particle moving forward (that is, in the positive direction)?
(e) Draw a diagram to represent the motion of the particle.
(f ) Find the total distance traveled by the particle during the first five seconds.
(g) Find the acceleration at time t and after 4 s.
(h) Graph the position, velocity, and acceleration functions for 0 < t < 5.
(i) When is the particle speeding up? When is it slowing down?
(a) The velocity function is the derivative of the position function:
s − f std − t 3 2 6t 2 1 9t
vstd −
− 3t 2 2 12t 1 9
(b) The velocity after 2 s means the instantaneous velocity when t − 2, that is,
v s2d −
− 3s2d2 2 12s2d 1 9 − 23 mys
The velocity after 4 s is
v s4d − 3s4d2 2 12s4d 1 9 − 9 mys
(c) The particle is at rest when v std − 0, that is,
3t 2 2 12t 1 9 − 3st 2 2 4t 1 3d − 3st 2 1dst 2 3d − 0
and this is true when t − 1 or t − 3. Thus the particle is at rest after 1 s and after 3 s.
(d) The particle moves in the positive direction when v std . 0, that is,
3t 2 2 12t 1 9 − 3st 2 1dst 2 3d . 0
This inequality is true when both factors are positive st . 3d or when both factors are
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
FIGURE 2
negative st , 1d. Thus the particle moves in the positive direction in the time intervals
t , 1 and t . 3. It moves backward (in the negative direction) when 1 , t , 3.
(e) Using the information from part (d) we make a schematic sketch in Figure 2 of the
motion of the particle back and forth along a line (the s-axis).
(f ) Because of what we learned in parts (d) and (e), we need to calculate the distances
traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately.
The distance traveled in the first second is
| f s1d 2 f s0d | − | 4 2 0 | − 4 m
From t − 1 to t − 3 the distance traveled is
| f s3d 2 f s1d | − | 0 2 4 | − 4 m
From t − 3 to t − 5 the distance traveled is
| f s5d 2 f s3d | − | 20 2 0 | − 20 m
The total distance is 4 1 4 1 20 − 28 m.
(g) The acceleration is the derivative of the velocity function:
astd −
d 2s
− 6t 2 12
2 −
as4d − 6s4d 2 12 − 12 mys 2
FIGURE 3
(h) Figure 3 shows the graphs of s, v, and a.
( i) The particle speeds up when the velocity is positive and increasing (v and a are
both positive) and also when the velocity is negative and decreasing (v and a are both
negative). In other words, the particle speeds up when the velocity and acceleration
have the same sign. (The particle is pushed in the same direction it is moving.) From
Figure 3 we see that this happens when 1 , t , 2 and when t . 3. The particle slows
down when v and a have opposite signs, that is, when 0 < t , 1 and when 2 , t , 3.
Figure 4 summarizes the motion of the particle.
FIGURE 4
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.7Rates of Change in the Natural and Social Sciences
EXAMPLE 2 If a rod or piece of wire is homogeneous, then its linear density is
uniform and is defined as the mass per unit length s − myld and measured in
kilograms per meter. Suppose, however, that the rod is not homogeneous but that its
mass measured from its left end to a point x is m − f sxd, as shown in Figure 5.
FIGURE 5
This part of the rod has mass ƒ.
The mass of the part of the rod that lies between x − x 1 and x − x 2 is given by
Dm − f sx 2 d 2 f sx 1 d, so the average density of that part of the rod is
average density −
f sx 2 d 2 f sx 1 d
x2 2 x1
If we now let Dx l 0 (that is, x 2 l x 1), we are computing the average density over
smaller and smaller intervals. The linear density at x 1 is the limit of these average
densities as Dx l 0 ; that is, the linear density is the rate of change of mass with
respect to length. Symbolically,
− lim
Dx l 0
Thus the linear density of the rod is the derivative of mass with respect to length.
For instance, if m − f sxd − sx , where x is measured in meters and m in kilograms,
then the average density of the part of the rod given by 1 < x < 1.2 is
f s1.2d 2 f s1d
s1.2 2 1
< 0.48 kgym
1.2 2 1
while the density right at x − 1 is
FIGURE 6
− 0.50 kgym
EXAMPLE 3 A current exists whenever electric charges move. Figure 6 shows part
of a wire and electrons moving through a plane surface, shaded red. If DQ is the net
charge that passes through this surface during a time period Dt, then the average
current during this time interval is defined as
average current −
Q2 2 Q1
t2 2 t1
If we take the limit of this average current over smaller and smaller time intervals,
we get what is called the current I at a given time t1:
I − lim
Dt l 0
Thus the current is the rate at which charge flows through a surface. It is measured in
units of charge per unit time (often coulombs per second, called amperes).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
Velocity, density, and current are not the only rates of change that are important in
physics. Others include power (the rate at which work is done), the rate of heat flow,
temperature gradient (the rate of change of temperature with respect to position), and the
rate of decay of a radioactive substance in nuclear physics.
■ Chemistry
EXAMPLE 4 A chemical reaction results in the formation of one or more substances
(called products) from one or more starting materials (called reactants). For instance,
the “equation”
2H2 1 O2 l 2H2 O
indicates that two molecules of hydrogen and one molecule of oxygen form two
molecules of water. Let’s consider the reaction
where A and B are the reactants and C is the product. The concentration of a reactant A
is the number of moles (1 mole − 6.022 3 10 23 molecules) per liter and is denoted by
fAg. The concentration varies during a reaction, so fAg, fBg, and fCg are all functions of
time std. The average rate of reaction of the product C over a time interval t1 < t < t2 is
fCgst2 d 2 fCgst1 d
t2 2 t1
But chemists are interested in the instantaneous rate of reaction because it gives
information about the mechanism of the chemical reaction. The instantaneous rate of
reaction is obtained by taking the limit of the average rate of reaction as the time
interval Dt approaches 0 :
rate of reaction − lim
Dt l 0
Since the concentration of the product increases as the reaction proceeds, the derivative
dfCg ydt will be positive, and so the rate of reaction of C is positive. The concentrations
of the reactants, however, decrease during the reaction, so, to make the rates of reaction
of A and B positive numbers, we put minus signs in front of the derivatives dfAg ydt
and dfBg ydt. Since fAg and fBg each decrease at the same rate that fCg increases, we
rate of reaction −
More generally, it turns out that for a reaction of the form
aA 1 bB l cC 1 dD
we have
1 dfAg
1 dfBg
1 dfCg
1 dfDg
a dt
b dt
c dt
d dt
The rate of reaction can be determined from data and graphical methods. In some cases
there are explicit formulas for the concentrations as functions of time that enable us to
compute the rate of reaction (see Exercise 26).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.7Rates of Change in the Natural and Social Sciences
EXAMPLE 5 One of the quantities of interest in thermodynamics is compressibility. If
a given substance is kept at a constant temperature, then its volume V depends on its
pressure P. We can consider the rate of change of volume with respect to pressure—
namely, the derivative dVydP. As P increases, V decreases, so dVydP , 0. The compressibility is defined by introducing a minus sign and dividing this derivative by the
volume V:
isothermal compressibility − − 2
1 dV
V dP
Thus measures how fast, per unit volume, the volume of a substance decreases as the
pressure on it increases at constant temperature.
For instance, the volume V (in cubic meters) of a sample of air at 258C was found to
be related to the pressure P (in kilopascals) by the equation
The rate of change of V with respect to P when P − 50 kPa is
− 20.00212 m 3ykPa
The compressibility at that pressure is
1 dV
V dP
− 0.02 sm 3ykPadym 3
■ Biology
EXAMPLE 6 Let n − f std be the number of individuals in an animal or plant popu­la­
tion at time t. The change in the population size between the times t − t1 and t − t2
is Dn − f st2 d 2 f st1 d, and so the average rate of growth during the time period
t1 < t < t2 is
average rate of growth −
f st2 d 2 f st1 d
t2 2 t1
The instantaneous rate of growth is obtained from this average rate of growth by
letting the time period Dt approach 0:
growth rate − lim
Dt l 0
Strictly speaking, this is not quite accurate because the actual graph of a population function n − f std would be a step function that is discontinuous whenever a
birth or death occurs and therefore not differentiable. However, for a large animal
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
or plant population, we can replace the graph by a smooth approximating curve as
in Figure 7.
FIGURE 7
A smooth curve approximating
a growth function
Science Photo Library / Alamy Stock Photo
To be more specific, consider a population of bacteria in a homogeneous nutrient
medium. Suppose that by sampling the population at certain intervals it is determined
that the population doubles every hour. If the initial population is n0 and the time t is
measured in hours, then
f s1d − 2f s0d − 2n0
f s2d − 2f s1d − 2 2n0
f s3d − 2f s2d − 2 3n0
and, in general,
E. coli bacteria measure about
2 micrometers (mm) long and
0.75 mm wide. The image was
produced with a scanning electron
f std − 2 t n0
The population function is n − n0 2 t.
This an example of an exponential function. In Chapter 6 we will discuss expo­
nential functions in general; at that time we will be able to compute their derivatives
and thereby determine the rate of growth of the bacteria population.
EXAMPLE 7 When we consider the flow of blood through a blood vessel, such as a
vein or artery, we can model the shape of the blood vessel by a cylindrical tube with
radius R and length l as illustrated in Figure 8.
FIGURE 8
Blood flow in an artery
Because of friction at the walls of the tube, the velocity v of the blood is greatest
along the central axis of the tube and decreases as the distance r from the axis increases
until v becomes 0 at the wall. The relationship between v and r is given by the law of
laminar flow, which was experimentally derived by the French physicist Jean Léonard
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.7Rates of Change in the Natural and Social Sciences
Marie Poiseuille in 1838. This law states that
For more detailed information,
see W. Nichols, M. O’Rourke, and
C. Vlachopoulos (eds.), McDonald’s
Blood Flow in Arteries: Theoretical,
Experimental, and Clinical Principles,
6th ed. (Boca Raton, FL, 2011).
sR 2 2 r 2 d
where is the viscosity of the blood and P is the pressure difference between the ends
of the tube. If P and l are constant, then v is a function of r with domain f0, Rg.
The average rate of change of the velocity as we move from r − r1 outward to
r − r2 is given by
vsr2 d 2 vsr1 d
r2 2 r1
and if we let Dr l 0, we obtain the velocity gradient, that is, the instantaneous rate of
change of velocity with respect to r :
velocity gradient − lim
Dr l 0
Using Equation 1, we obtain
s0 2 2rd − 2
For one of the smaller human arteries we can take − 0.027, R − 0.008 cm,
l − 2 cm, and P − 4000 dynesycm2, which gives
s0.000064 2 r 2 d
< 1.85 3 10 4s6.4 3 10 25 2 r 2 d
At r − 0.002 cm the blood is flowing at a speed of
vs0.002d < 1.85 3 10 4s64 3 1026 2 4 3 10 26 d − 1.11 cmys
and the velocity gradient at that point is
< 274 scmysdycm
To get a feeling for what this statement means, let’s change our units from centimeters
to micrometers (1 cm − 10,000 mm). Then the radius of the artery is 80 mm. The
velocity at the central axis is 11,850 mmys, which decreases to 11,110 mmys at a
distance of r − 20 mm. The fact that dvydr − 274 (mmys)ymm means that, when
r − 20 mm, the velocity is decreasing at a rate of about 74 mmys for each micrometer
that we proceed away from the center.
■ Economics
EXAMPLE 8 Suppose Csxd is the total cost that a company incurs in producing
x units of a certain commodity. The function C is called a cost function. If the
number of items produced is increased from x 1 to x 2, then the additional cost is
DC − Csx 2 d 2 Csx 1 d, and the average rate of change of the cost is
Csx 2 d 2 Csx 1 d
Csx 1 1 Dxd 2 Csx 1 d
x2 2 x1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
The limit of this quantity as Dx l 0, that is, the instantaneous rate of change of
cost with respect to the number of items produced, is called the marginal cost by
marginal cost − lim
Dx l 0
[Since x often takes on only integer values, it may not make literal sense to let Dx
approach 0, but we can always replace Csxd by a smooth approximating function as in
Example 6.]
Taking Dx − 1 and n large (so that Dx is small compared to n), we have
C9snd < Csn 1 1d 2 Csnd
Thus the marginal cost of producing n units is approximately equal to the cost of
producing one more unit [the sn 1 1dst unit].
It is often appropriate to represent a total cost function by a polynomial
Csxd − a 1 bx 1 cx 2 1 dx 3
where a represents the overhead cost (rent, heat, maintenance) and the other terms
represent the cost of raw materials, labor, and so on. (The cost of raw materials may be
proportional to x, but labor costs might depend partly on higher powers of x because of
overtime costs and inefficiencies involved in large-scale operations.)
For instance, suppose a company has estimated that the cost (in dollars) of pro­
ducing x items is
Csxd − 10,000 1 5x 1 0.01x 2
Then the marginal cost function is
C9sxd − 5 1 0.02x
The marginal cost at the production level of 500 items is
C9s500d − 5 1 0.02s500d − $15yitem
This gives the rate at which costs are increasing with respect to the production level
when x − 500 and predicts the cost of the 501st item.
The actual cost of producing the 501st item is
Cs501d 2 Cs500d − f10,000 1 5s501d 1 0.01s501d2 g
2 f10,000 1 5s500d 1 0.01s500d2 g
− $15.01
Notice that C9s500d < Cs501d 2 Cs500d.
Economists also study marginal demand, marginal revenue, and marginal profit, the
derivatives of the demand, revenue, and profit functions. These will be considered in
Chapter 3 after we have developed techniques for finding the maximum and minimum
values of functions.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.7Rates of Change in the Natural and Social Sciences
■ Other Sciences
Rates of change occur in all the sciences. A geologist is interested in knowing the rate at
which an intruded body of molten rock cools by conduction of heat into surrounding
rocks. An engineer wants to know the rate at which water flows out of a reservoir. An
urban geographer is interested in the rate of change of the population density in a city as
the distance from the city center increases. A meteorologist is concerned with the rate of
change of atmospheric pressure with respect to height (see Exercise 6.5.19).
In psychology, those interested in learning theory study the learning curve, which
graphs the performance Pstd of someone learning a skill as a function of the training
time t. Of particular interest is the rate at which performance improves as time passes,
that is, dPydt. Psychologists have also studied the phenomenon of memory and have
developed models for the rate of memory retention (see Exercise 38). They also study the
difficulty involved in performing certain tasks and the rate at which difficulty increases
when a given parameter is changed (see Exercise 6.4.98).
In sociology, differential calculus is used in analyzing the spread of rumors (or innovations or fads or fashions). If pstd denotes the proportion of a population that knows a
rumor by time t, then the derivative dpydt represents the rate of spread of the rumor (see
Exercise 6.2.65).
■ A Single Idea, Many Interpretations
Velocity, density, current, power, and temperature gradient in physics; rate of reaction
and compressibility in chemistry; rate of growth and blood velocity gradient in biology;
marginal cost and marginal profit in economics; rate of heat flow in geology; rate of
improvement of performance in psychology; rate of spread of a rumor in sociology—
these are all special cases of a single mathematical concept, the derivative.
All of these different applications of the derivative illustrate the fact that part of the
power of mathematics lies in its abstractness. A single abstract mathematical concept
(such as the derivative) can have different interpretations in each of the sciences. When
we develop the properties of the mathematical concept once and for all, we can then turn
around and apply these results to all of the sciences. This is much more efficient than
developing properties of special concepts in each separate science. The French mathematician Joseph Fourier (1768–1830) put it succinctly: “Mathematics compares the most
diverse phenomena and discovers the secret analogies that unite them.”
1–4 A particle moves according to a law of motion s − f std,
t > 0, where t is measured in seconds and s in meters.
(a) Find the velocity at time t.
(b) What is the velocity after 1 second?
(c) When is the particle at rest?
(d) When is the particle moving in the positive direction?
(e) Find the total distance traveled during the first 6 seconds.
(f )Draw a diagram like Figure 2 to illustrate the motion of the
(g) Find the acceleration at time t and after 1 second.
; (h)Graph the position, velocity, and acceleration functions
for 0 < t < 6.
(i)When is the particle speeding up? When is it slowing down?
t2 1 9
4. f std − t 2e 2t
1. f std − t 3 2 8t 2 1 24t
2. f std −
3. f std − sinsty2d
5. G
raphs of the velocity functions of two particles are shown,
where t is measured in seconds. When is each particle
speeding up? When is it slowing down? Explain.
(a) √ √
(b) √ √
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
6. G
raphs of the position functions of two particles are
shown, where t is measured in seconds. When is the
velocity of each particle positive? When is it negative?
When is each particle speeding up? When is it slowing
down? Explain.
s s
(a)At what time does the particle have a velocity of 20 mys?
(b)At what time is the acceleration 0? What is the significance of this value of t ?
(b) s s
t t
t t
7. Suppose that the graph of the velocity function of a particle is
as shown in the figure, where t is measured in seconds. When
is the particle traveling forward (in the positive direction)?
When is it traveling backward? What is happening when
5 , t , 7?
8. For the particle described in Exercise 7, sketch a graph of the
acceleration function. When is the particle speeding up?
When is it slowing down? When is it traveling at a constant
9. T
he height (in meters) of a projectile shot vertically upward
from a point 2 m above ground level with an initial velocity
of 24.5 mys is h − 2 1 24.5t 2 4.9t 2 after t seconds.
(a) Find the velocity after 2 s and after 4 s.
(b) When does the projectile reach its maximum height?
(c) What is the maximum height?
(d) When does it hit the ground?
(e) With what velocity does it hit the ground?
10. I f a ball is thrown vertically upward with a velocity of
24.5 mys, then its height after t seconds is s − 24.5t 2 4.9t 2.
(a) What is the maximum height reached by the ball?
(b)What is the velocity of the ball when it is 29.4 above the
ground on its way up? On its way down?
11. I f a rock is thrown vertically upward from the surface
of Mars with velocity 15 mys, its height after t seconds
is h − 15t 2 1.86t 2.
(a) What is the velocity of the rock after 2 s?
(b)What is the velocity of the rock when its height is 25 m
on its way up? On its way down?
12. A particle moves with position function
s − t 4 2 4t 3 2 20t 2 1 20tt > 0
13. (a)A company makes computer chips from square wafers
of silicon. A process engineer wants to keep the side
length of a wafer very close to 15 mm and needs to know
how the area Asxd of a wafer changes when the side
length x changes. Find A9s15d and explain its meaning in
this situation.
(b)Show that the rate of change of the area of a square with
respect to its side length is half its perimeter. Try to
explain geometrically why this is true by drawing a
square whose side length x is increased by an amount Dx.
How can you approximate the resulting change in area
DA if Dx is small?
14. (a)Sodium chlorate crystals are easy to grow in the shape of
cubes by allowing a solution of water and sodium chlorate to evaporate slowly. If V is the volume of such a cube
with side length x, calculate dVydx when x − 3 mm and
explain its meaning.
(b)Show that the rate of change of the volume of a cube with
respect to its edge length is equal to half the surface area
of the cube. Explain geometrically why this result is true
by arguing by analogy with Exercise 13(b).
15. (a)Find the average rate of change of the area of a circle
with respect to its radius r as r changes from
(i) 2 to 3
(ii) 2 to 2.5
(iii) 2 to 2.1
(b) Find the instantaneous rate of change when r − 2.
(c)Show that the rate of change of the area of a circle with
respect to its radius (at any r) is equal to the circumference of the circle. Try to explain geometrically why this
is true by drawing a circle whose radius is increased by
an amount Dr. How can you approximate the resulting
change in area DA if Dr is small?
16. A
stone is dropped into a lake, creating a circular ripple that
travels outward at a speed of 60 cmys. Find the rate at which
the area within the circle is increasing after (a) 1 s, (b) 3 s,
and (c) 5 s. What can you conclude?
17. A
spherical balloon is being inflated. Find the rate of increase
of the surface area sS − 4r 2 d with respect to the radius r
when r is (a) 20 cm, (b) 40 cm, and (c) 60 cm. What conclusion can you make?
18. (a)The volume of a growing spherical cell is V − 43 r 3,
where the radius r is measured in micrometers
(1 μm − 1026 m). Find the average rate of change of V
with respect to r when r changes from
(i) 5 to 8 μm
(ii) 5 to 6 μm
(iii) 5 to 5.1 μm
(b)Find the instantaneous rate of change of V with respect
to r when r − 5 μm.
(c)Show that the rate of change of the volume of a sphere
with respect to its radius is equal to its surface area.
Explain geometrically why this result is true. Argue by
analogy with Exercise 15(c).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.7Rates of Change in the Natural and Social Sciences
19. T
he mass of the part of a metal rod that lies between its left
end and a point x meters to the right is 3x 2 kg. Find the linear density (see Example 2) when x is (a) 1 m, (b) 2 m, and
(c) 3 m. Where is the density the highest? The lowest?
How fast was the tide rising (or falling) at the following
(a) 3:00 am
(b) 6:00 am
(c) 9:00 am
(d) Noon
20. I f a cylindrical water tank holds 5000 liters, and the water
drains from the bottom of the tank in 40 minutes, then
Torricelli’s Law gives the volume V of water remaining in
the tank after t minutes as
25. B
oyle’s Law states that when a sample of gas is compressed
at a constant temperature, the product of the pressure and
the volume remains constant: PV − C.
(a)Find the rate of change of volume with respect to
(b)A sample of gas is in a container at low pressure
and is steadily compressed at constant temperature
for 10 min­utes. Is the volume decreasing more
rapidly at the beginning or the end of the 10 minutes?
(c)Prove that the isothermal compressibility (see
Example 5) is given by − 1yP.
V − 5000 (1 2 40
t) 0 < t < 40
Find the rate at which water is draining from the tank after
(a) 5 min, (b) 10 min, (c) 20 min, and (d) 40 min. At what
time is the water flowing out the fastest? The slowest? Summarize your findings.
21. T
he quantity of charge Q in coulombs (C) that has passed
through a point in a wire up to time t (measured in seconds)
is given by Qstd − t 3 2 2t 2 1 6t 1 2. Find the current
when (a) t − 0.5 s and (b) t − 1 s. (See Example 3. The
unit of current is an ampere [1 A − 1 Cys].) At what time is
the current lowest?
22. N
ewton’s Law of Gravitation says that the magnitude F
of the force exerted by a body of mass m on a body of
mass M is
where G is the gravitational constant and r is the distance
between the bodies.
(a)Find dFydr and explain its meaning. What does the
minus sign indicate?
(b)Suppose it is known that the earth attracts an object
with a force that decreases at the rate of 2 Nykm
when r − 20,000 km. How fast does this force change
when r − 10,000 km?
23. The force F acting on a body with mass m and velocity v is
the rate of change of momentum: F − sdydtd smvd. If m is
constant, this becomes F − ma, where a − dvydt is the
acceleration. But in the theory of relativity, the mass of a
particle varies with v as follows: m − m 0 ys1 2 v 2yc 2
where m 0 is the mass of the particle at rest and c is the speed
of light. Show that
s1 2 v 2yc 2 d3y2
24. S
ome of the highest tides in the world occur in the Bay of
Fundy on the Atlantic Coast of Canada. At Hopewell Cape
the water depth at low tide is about 2.0 m and at high tide it
is about 12.0 m. The natural period of oscillation is a little
more than 12 hours and on a day in June, high tide occurred
at 6:45 am. This helps explain the following model for the
water depth D (in meters) as a function of the time t (in
hours after midnight) on that day:
Dstd − 7 1 5 cosf0.503st 2 6.75dg
26. I f, in Example 4, one molecule of the product C is formed
from one molecule of the reactant A and one molecule
of the reactant B, and the initial concentrations of A
and B have a common value fAg − fBg − a molesyL,
fCg − a 2ktysakt 1 1d
where k is a constant.
(a) Find the rate of reaction at time t.
(b) Show that if x − fCg, then
− ksa 2 xd 2
27. T
he table gives the world population Pstd, in millions,
where t is measured in years and t − 0 corresponds to the
year 1900.
(a)Estimate the rate of population growth in 1920 and in
1980 by averaging the slopes of two secant lines.
(b)Use a graphing calculator or computer to find a cubic
function (a third-degree polynomial) that models the
(c)Use your model in part (b) to find a model for the rate
of population growth.
(d)Use part (c) to estimate the rates of growth in 1920
and 1980. Compare with your estimates in part (a).
(e) Estimate the rate of growth in 1985.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
28. T
he table shows how the average age of first marriage of
Japanese women has varied since 1950.
(a)Use a graphing calculator or computer to model these
data with a fourth-degree polynomial.
(b)Use part (a) to find a model for A9std.
(c)Estimate the rate of change of marriage age for women
in 1990.
(d)Graph the data points and the models for A and A9.
29. R
efer to the law of laminar flow given in Example 7. Consider a blood vessel with radius 0.01 cm, length 3 cm, pressure difference 3000 dynesycm2, and vis­cos­ity − 0.027.
(a)Find the velocity of the blood along the centerline
r − 0, at radius r − 0.005 cm, and at the wall
r − R − 0.01 cm.
(b)Find the velocity gradient at r − 0, r − 0.005,
and r − 0.01.
(c)Where is the velocity the greatest? Where is the velocity
changing most?
30. T
he frequency of vibrations of a vibrating violin string is
given by
where L is the length of the string, T is its tension, and is
its linear density. [See Chapter 11 in D. E. Hall, Musical
Acoustics, 3rd ed. (Pacific Grove, CA, 2002).]
(a) Find the rate of change of the frequency with respect to
(i) the length (when T and are constant),
(ii) the tension (when L and are constant), and
(iii) the linear density (when L and T are constant).
(b)The pitch of a note (how high or low the note sounds)
is determined by the frequency f . (The higher the
fre­quency, the higher the pitch.) Use the signs of the
derivatives in part (a) to determine what happens to the
pitch of a note
(i)when the effective length of a string is decreased
by placing a finger on the string so a shorter portion of the string vibrates,
(ii)when the tension is increased by turning a tuning
(iii)when the linear density is increased by switching
to another string.
31. S
uppose that the cost (in dollars) for a company to produce
x pairs of a new line of jeans is
Csxd − 2000 1 3x 1 0.01x 2 1 0.0002x 3
(a)Find the marginal cost function.
(b)Find C9s100d and explain its meaning. What does it
(c)Compare C9s100d with the cost of manufacturing the
101st pair of jeans.
32. The cost function for a certain commodity is
Csqd − 84 1 0.16q 2 0.0006q 2 1 0.000003q 3
(a) Find and interpret C9s100d.
(b)Compare C9s100d with the cost of producing the 101st
33. If psxd is the total value of the production when there are
x workers in a plant, then the average productivity of the
workforce at the plant is
Asxd −
(a)Find A9sxd. Why does the company want to hire more
workers if A9sxd . 0 ?
(b)Show that A9sxd . 0 if p9sxd is greater than the average
34. I f R denotes the reaction of the body to some stimulus of
strength x, the sensitivity S is defined to be the rate of change
of the reaction with respect to x. A particular example is that
when the brightness x of a light source is increased, the eye
reacts by decreasing the area R of the pupil. The experimental formula
40 1 24x 0.4
1 1 4x 0.4
has been used to model the dependence of R on x when R is
measured in square millimeters and x is measured in appropriate units of brightness.
(a) Find the sensitivity.
(b)Illustrate part (a) by graphing both R and S as functions
of x. Comment on the values of R and S at low levels of
brightness. Is this what you would expect?
35. T
he gas law for an ideal gas at absolute temperature T (in
kelvins), pressure ­P (in atmospheres), and volume V (in
liters) is PV − nRT, where n is the number of moles of the
gas and R − 0.0821 is the gas constant. Suppose that, at a
certain instant, P − 8.0 atm and is increasing at a rate of
0.10 atmymin and V − 10 L and is decreasing at a rate of
0.15 Lymin. Find the rate of change of T with respect to
time at that instant if n − 10 mol.
36. I nvasive species often display a wave of advance as
they colonize new areas. Mathematical models based on
random dispersal and reproduction have demonstrated that
the speed with which such waves move is given by the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.8Related Rates
function f srd − 2 sDr , where r is the reproductive rate
of individuals and D is a parameter quantifying dispersal.
Calculate the derivative of the wave speed with respect to the
reproductive rate r and explain its meaning.
37. I n the study of ecosystems, predator-prey models are often
used to study the interaction between species. Consider populations of tundra wolves, given by Wstd, and caribou, given by
Cstd, in northern Canada. The interaction has been modeled
by the equations
− aC 2 bCW− 2cW 1 dCW
(a)What values of dCydt and dWydt correspond to stable
(b)How would the statement “The caribou go extinct” be
represented mathematically?
(c)Suppose that a − 0.05, b − 0.001, c − 0.05, and
d − 0.0001. Find all population pairs sC, W d that lead to
stable populations. According to this model, is it possible
for the two species to live in balance or will one or both
species become extinct?
38. Hermann Ebbinghaus (1850 –1909) pioneered the study of
memory. A 2011 article in the Journal of Mathematical Psychology presents the mathematical model
Rstd − a 1 bs1 1 ctd2
for the Ebbinghaus forgetting curve, where Rstd is the fraction
of memory retained t days after learning a task; a, b, and c are
experimentally determined constants between 0 and 1; is a
positive constant; and Rs0d − 1. The constants depend on the
type of task being learned.
(a)What is the rate of change of retention t days after a task
is learned?
(b)Do you forget how to perform a task faster soon after
learning it or a long time after you have learned it?
(c)What fraction of memory is permanent?
2.8 Related Rates
If we are pumping air into a balloon, both the volume and the radius of the balloon are
increasing and their rates of increase are related to each other. But it is much easier to
measure directly the rate of increase of the volume than the rate of increase of the radius.
In a related rates problem the idea is to compute the rate of change of one quantity in
terms of the rate of change of another quantity (which may be more easily measured).
The procedure is to find an equation that relates the two quantities and then use the Chain
Rule to differentiate both sides with respect to time.
EXAMPLE 1 Air is being pumped into a spherical balloon so that its volume increases
at a rate of 100 cm3ys. How fast is the radius of the balloon increasing when the diameter is 50 cm?
PS According to the Principles of Problem
Solving discussed following Chapter 1, the
first step is to understand the problem. This
includes reading the problem carefully,
identifying the given and the unknown, and
introducing suitable notation.
SOLUTION We start by identifying two things:
the given information:
the rate of increase of the volume of air is 100 cm3ys
and the unknown:
the rate of increase of the radius when the diameter is 50 cm
In order to express these quantities mathematically, we introduce some suggestive
Let V be the volume of the balloon and let r be its radius.
The key thing to remember is that rates of change are derivatives. In this problem,
the volume and the radius are both functions of the time t. The rate of increase of the
volume with respect to time is the derivative dVydt, and the rate of increase of the
radius is drydt. We can therefore restate the given and the unknown as follows:
− 100 cm3ys
when r − 25 cm
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
PS The second stage of problem solving is to
think of a plan for connecting the given and
the unknown.
In order to connect dVydt and drydt, we first relate V and r by a formula—in this
case, the formula for the volume of a sphere:
V − 43 r 3
In order to use the given information, we differentiate each side of this equation with
respect to t. To differentiate the right side, we need to use the Chain Rule:
dV dr
− 4r 2
dr dt
Now we solve for the unknown quantity:
1 dV
4r 2 dt
Notice that, although dVydt is constant, drydt is not constant.
If we put r − 25 and dVydt − 100 in this equation, we obtain
2 100 −
The radius of the balloon is increasing at the rate of 1ys25d < 0.0127 cmys when the
diameter is 50 cm.
EXAMPLE 2 A ladder 5 m long rests against a vertical wall. If the bottom of the
ladder slides away from the wall at a rate of 1 mys, how fast is the top of the ladder
sliding down the wall when the bottom of the ladder is 3 m from the wall?
SOLUTION We first draw a diagram and label it as in Figure 1. Let x meters be the
distance from the bottom of the ladder to the wall and y meters the distance from the top
of the ladder to the ground. Note that x and y are both functions of t (time, measured
in seconds).
We are given that dxydt − 1 mys and we are asked to find dyydt when x − 3 m (see
Figure 2). In this problem, the relationship between x and y is given by the Pythagorean
x 2 1 y 2 − 25
FIGURE 1
Differentiating each side with respect to t using the Chain Rule, we have
1 2y
Solving this equation for the desired rate, we obtain
x dx
y dt
FIGURE 2
When x − 3, the Pythagorean Theorem gives y − 4 and so, substituting these values
and dxydt − 1, we have
− 2 s1d − 20.75 mys
The fact that dyydt is negative means that the distance from the top of the ladder to
the ground is decreasing at a rate of 0.75 mys. In other words, the top of the ladder is
sliding down the wall at a rate of 0.75 mys.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.8Related Rates
Problem Solving Strategy
It is useful to recall some of the problem-solving principles and adapt them to
related rates in light of our experience in Examples 1 and 2:
1. Read the problem carefully.
2. Draw a diagram if possible.
3. Introduce notation. Assign symbols to all quantities that are functions
of time.
4. Express the given information and the required rate in terms of derivatives.
5. Write an equation that relates the various quantities of the problem. If necessary,
use the geometry of the situation to eliminate one of the variables by substitution (see Example 3 below).
6. Use the Chain Rule to differentiate both sides of the equation with respect to t.
7. Substitute the given information into the resulting equation and solve for the
unknown rate.
See also Principles of Problem Solving following Chapter 1.
PS Look back: What have we learned from
Examples 1 and 2 that will help us solve
future problems?
WARNING A common error is
to substitute the given numerical
information (for quantities that vary
with time) too early. This should be
done only after the differentiation.
(Step 7 follows Step 6.) For instance,
in Example 1 we dealt with general
values of r until we finally substituted
r − 25 at the last step. (If we had put
r − 25 earlier, we would have gotten
dVydt − 0, which is clearly wrong.)
The following examples further illustrate this strategy.
EXAMPLE 3 A water tank has the shape of an inverted circular cone with base radius
2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m 3ymin, find
the rate at which the water level is rising when the water is 3 m deep.
SOLUTION We first sketch the cone and label it as in Figure 3. Let V, r, and h be the
volume of the water, the radius of the surface, and the height of the water at time t,
where t is measured in minutes.
We are given that dVydt − 2 m 3ymin and we are asked to find dhydt when h is 3 m.
The quantities V and h are related by the equation
V − 13 r 2h
FIGURE 3
but it is very useful to express V as a function of h alone. In order to eliminate r, we use
the similar triangles in Figure 3 to write
− r −
and the expression for V becomes
Now we can differentiate each side with respect to t :
2 dh
4 dV
h 2 dt
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
Substituting h − 3 m and dVydt − 2 m 3ymin, we have
2 2 −
The water level is rising at a rate of 8ys9d < 0.28 mymin.
EXAMPLE 4 Car A is traveling west at 80 kmyh and car B is traveling north at
100 kmyh. Both are headed for the intersection of the two roads. At what rate are the cars
approaching each other when car A is 0.3 km and car B is 0.4 km from the intersection?
SOLUTION We draw Figure 4, where C is the intersection of the roads. At a given
time t, let x be the distance from car A to C, let y be the distance from car B to C,
and let z be the distance between the cars, where x, y, and z are measured in miles.
We are given that dxydt − 280 kmyh and dyydt − 2700 kmyh. (The derivatives
are negative because x and y are decreasing.) We are asked to find dzydt. The equation
that relates x, y, and z is given by the Pythagorean Theorem:
z2 − x 2 1 y 2
FIGURE 4
Differentiating each side with respect to t, we have
− 2x
1 2y
When x − 0.3 km and y − 0.4 km, the Pythagorean Theorem gives z − 0.5 km, so
f0.3s280d 1 0.4s2100dg − 2128 kmyh
The cars are approaching each other at a rate of 128 kmyh.
EXAMPLE 5 A man walks along a straight path at a speed of 1 mys. A spotlight is located
on the ground 6 m from the path and is kept focused on the man. At what rate is the
spotlight rotating when the man is 4.5 m from the point on the path closest to the light?
SOLUTION We draw Figure 5 and let x be the distance from the man to the point on the
path closest to the spotlight. We let be the angle between the beam of the light and the
perpendicular to the path.
We are given that dxydt − 1 mys and are asked to find dydt when x − 4.5. The
equation that relates x and can be written from Figure 5:
− tan x − 6 tan 6
Differentiating each side with respect to t, we get
− 6 sec2
FIGURE 5
− cos2
cos2 s1d − cos2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.8Related Rates
When x − 4.5, the length of the beam is 7.5, so cos − 20
25 − 5 and
1 rotation
60 s
2 rad
1 min
< 1.02 rotations per min
− 0.107
The spotlight is rotating at a rate of 0.107 radys.
1. (a)If V is the volume of a cube with edge length x and the
cube expands as time passes, find dVydt in terms of
(b)If the length of the edge of a cube is increasing at a rate
of 4 cmys, how fast is the volume of the cube increasing
when the edge length is 15 cm?
2. (a)If A is the area of a circle with radius r and the circle
expands as time passes, find dAydt in terms of drydt.
(b)Suppose oil spills from a ruptured tanker and spreads in a
circular pattern. If the radius of the oil spill increases at a
constant rate of 2 mys, how fast is the area of the spill
increasing when the radius is 30 m?
3. E
ach side of a square is increasing at a rate of 6 cmys. At
what rate is the area of the square increasing when the area of
the square is 16 cm2 ?
4. T
he radius of a sphere is increasing at a rate of 4 mmys. How
fast is the volume increasing when the diameter is 80 mm?
5. T
he radius of a spherical ball is increasing at a rate of
2 cmymin. At what rate is the surface area of the ball increasing when the radius is 8 cm?
6. T
he length of a rectangle is increasing at a rate of 8 cmys and
its width is increasing at a rate of 3 cmys. When the length is
20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?
7. A
cylindrical tank with radius 5 m is being filled with water
at a rate of 3 m3ymin. How fast is the height of the water
8. T
he area of a triangle with sides of lengths a and b and
contained angle is A − 12 ab sin . (See Formula 6 in
Appendix D.)
(a)If a − 2 cm, b − 3 cm, and increases at a rate of
0.2 radymin, how fast is the area increasing when
− y3?
(b)If a − 2 cm, b increases at a rate of 1.5 cmymin, and increases at a rate of 0.2 radymin, how fast is the area
increasing when b − 3 cm and − y3?
(c)If a increases at a rate of 2.5 cmymin, b increases at a
rate of 1.5 cmymin, and increases at a rate of
0.2 radymin, how fast is the area increasing when
a − 2 cm, b − 3 cm, and − y3?
9. S
uppose 4x 2 1 9y 2 − 25, where x and y are functions of t.
(a)If dyydt − 13, find dxydt when x − 2 and y − 1.
(b)If dxydt − 3, find dy ydt when x − 22 and y − 1.
10. If x 2 1 y 2 1 z 2 − 9, dxydt − 5, and dyydt − 4, find dzydt
when sx, y, zd − s2, 2, 1d.
11. The weight w of an astronaut (in newtons) is related to her
height h above the surface of the earth (in kilometers) by
w − w0
6370 1 h
where w0 is the weight of the astronaut on the surface of the
earth. If the astronaut weighs 580 newtons on earth and is in
a rocket, being propelled upward at a speed of 19 kmys, find
the rate at which her weight is changing (in Nys) when she is
60 kilometers above the earth’s surface.
12. A
particle is moving along a hyperbola xy − 8. As it reaches
the point s4, 2d, the y-coordinate is decreasing at a rate of
3 cmys. How fast is the x-coordinate of the point changing at
that instant?
(a) What quantities are given in the problem?
(b) What is the unknown?
(c) Draw a picture of the situation for any time t.
(d) Write an equation that relates the quantities.
(e) Finish solving the problem.
13. A
plane flying horizontally at an altitude of 2 km and a speed
of 800 kmyh passes directly over a radar station. Find the rate
at which the distance from the plane to the station is increasing when the plane is 3 km away from the station.
14. If
a snowball melts so that its surface area decreases at a rate
of 1 cm2ymin, find the rate at which the diameter decreases
when the diameter is 10 cm.
15. A
street light is mounted at the top of a 6-meter-tall pole.
A man 2 m tall walks away from the pole with a speed of
1.5 mys along a straight path. How fast is the tip of his
shadow moving when he is 10 m from the pole?
16. At
noon, ship A is 150 km west of ship B. Ship A is sailing
east at 35 kmyh and ship B is sailing north at 25 kmyh. How
fast is the distance between the ships changing at 4:00 pm?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
17. T
wo cars start moving from the same point. One travels south
at 30 kmyh and the other travels west at 72 kmyh. At what rate
is the distance between the cars increasing two hours later?
18. A
spotlight on the ground shines on a wall 12 m away. If a
man 2 m tall walks from the spotlight toward the building at a
speed of 1.6 mys, how fast is the length of his shadow on the
building decreasing when he is 4 m from the building?
19. A
man starts walking north at 1.2 mys from a point P. Five
minutes later a woman starts walking south at 1.6 mys from a
point 200 m due east of P. At what rate are the people moving
apart 15 min after the woman starts walking?
20. A
baseball diamond is a square with side 18 m. A batter hits
the ball and runs toward first base with a speed of 7.5 mys.
(a)At what rate is his distance from second base decreasing
when he is halfway to first base?
(b)At what rate is his distance from third base increasing at
the same moment?
18 m
25. W
ater is leaking out of an inverted conical tank at a rate of
10,000 cm 3ymin at the same time that water is being pumped
into the tank at a constant rate. The tank has height 6 m and
the diameter at the top is 4 m. If the water level is rising at a
rate of 20 cmymin when the height of the water is 2 m, find
the rate at which water is being pumped into the tank.
26. A
particle moves along the curve y − 2 sinsxy2d. As the
particle passes through the point ( 13 , 1), its x-coordinate
increases at a rate of s10 cmys. How fast is the distance
from the particle to the origin changing at this instant?
27. A
water trough is 10 m long and a cross-section has the shape
of an isosceles trapezoid that is 30 cm wide at the bottom,
80 cm wide at the top, and has height 50 cm. If the trough is
being filled with water at the rate of 0.2 m 3ymin, how fast is
the water level rising when the water is 30 cm deep?
28. A
trough is 6 m long and its ends have the shape of isosceles triangles that are 1 m across at the top and have a height
of 50 cm. If the trough is being filled with water at a rate of
1.2 m 3ymin, how fast is the water level rising when the water
is 30 centimeters deep?
29. G
ravel is being dumped from a conveyor belt at a rate of
3 m 3ymin, and its coarseness is such that it forms a pile in the
shape of a cone whose base diameter and height are always
equal. How fast is the height of the pile increasing when the
pile is 3 m high?
21. T
he altitude of a triangle is increasing at a rate of 1 cmymin
while the area of the triangle is increasing at a rate of
2 cm 2ymin. At what rate is the base of the triangle changing
when the altitude is 10 cm and the area is 100 cm2 ?
22. A
boat is pulled into a dock by a rope attached to the bow of
the boat and passing through a pulley on the dock that is 1 m
higher than the bow of the boat. If the rope is pulled in at a
rate of 1 mys, how fast is the boat approaching the dock when
it is 8 m from the dock?
23–24 Use the fact that the distance (in meters) a dropped stone
falls after t seconds is d − 4.9t 2.
30. A
swimming pool is 5 m wide, 10 m long, 1 m deep at the
shallow end, and 3 m deep at its deepest point. A cross-section
is shown in the Large
If the
pool is being filled at a rate of
is preferable:
0.1 m 3ymin, how fast is the water level rising when the depth
at the deepest point is 1 m?
23. A woman stands near the edge of a cliff and drops a stone
over the edge. Exactly one second later she drops another
stone. One second after that, how fast is the distance between
the two stones changing?
24. Two men stand 10 m apart on level ground near the edge of a
cliff. One man drops a stone and one second later the other
man drops a stone. One second after that, how fast is the distance between the two stones changing?
31. T
he sides of an equilateral triangle are increasing at a rate of
10 cmymin. At what rate is the area of the triangle increasing
when the sides are 30 cm long?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.8Related Rates
32. A
kite 50 m above the ground moves horizontally at a speed
of 2 mys. At what rate is the angle between the string and the
horizontal decreasing when 100 m of string has been let out?
33. A car is traveling north on a straight road at 20 mys and a
drone is flying east at 6 mys at an elevation of 25 m. At one
instant the drone passes directly over the car. How fast is the
distance between the drone and the car changing 5 seconds
34. I f the minute hand of a clock has length r (in centimeters),
find the rate at which it sweeps out area as a function of r.
35. H
ow fast is the angle between the ladder and the ground
changing in Example 2 when the bottom of the ladder is
3 m from the wall?
36. A
ccording to the model we used to solve Example 2, what
happens as the top of the ladder approaches the ground? Is
the model appropriate for small values of y?
37. B
oyle’s Law states that when a sample of gas is compressed
at a constant temperature, the pressure P and volume V satisfy the equation PV − C , where C is a constant. Suppose
that at a certain instant the volume is 600 cm3, the pressure
is 150 kPa, and the pressure is increasing at a rate of
20 kPaymin. At what rate is the volume decreasing at this
faucet is filling a hemispherical basin of diameter 60 cm
; 38. A
with water at a rate of 2 Lymin. Find the rate at which the
water is rising in the basin when it is half full. [Use the following facts: 1 L is 1000 cm3. The volume of the portion
of a sphere with radius r from the bottom to a height h is
V − (rh 2 2 13 h 3), as we will show in Chapter 5.]
39. I f two resistors with resistances R1 and R2 are connected in
parallel, as shown in the figure, then the total resistance R,
measured in ohms (V), is given by
If R1 and R2 are increasing at rates of 0.3 Vys and 0.2 Vys,
respectively, how fast is R changing when R1 − 80 V
and R2 − 100 V?
40. W
hen air expands adiabatically (without gaining or losing
heat), its pressure P and volume V are related by the equation PV 1.4 − C , where C is a constant. Suppose that at a
certain instant the volume is 400 cm3 and the pressure is
80 kPa and is decreasing at a rate of 10 kPaymin. At what
rate is the volume increasing at this instant?
41. Two straight roads diverge from an intersection at an angle
of 60°. Two cars leave the intersection at the same time, the
first traveling down one road at 60 kmyh and the second
traveling down the other road at 100 kmyh. How fast is the
distance between the cars changing after half an hour? [Hint:
Use the Law of Cosines (Formula 21 in Appendix D).]
42. B
rain weight B as a function of body weight W in fish has
been modeled by the power function B − 0.007W 2y3, where
B and W are measured in grams. A model for body weight
as a function of body length L (measured in centimeters) is
W − 0.12L2.53. If, over 10 million years, the average length
of a certain species of fish evolved from 15 cm to 20 cm at a
constant rate, how fast was this species’ brain growing
when its average length was 18 cm?
43. T
wo sides of a triangle have lengths 12 m and 15 m. The
angle between them is increasing at a rate of 2 8ymin. How
fast is the length of the third side increasing when the angle
between the sides of fixed length is 60°? [Hint: Use the Law
of Cosines (Formula 21 in Appendix D).]
44. T
wo carts, A and B, are connected by a rope 12 m long that
passes over a pulley P. (See the figure.) The point Q is on the
floor 4 m directly beneath P and between the carts. Cart A is
being pulled away from Q at a speed of 0.5 mys. How fast is
cart B moving toward Q at the instant when cart A is 3 m
from Q?
45. A
television camera is positioned 1200 m from the base of a
rocket launching pad. The angle of elevation of the camera
has to change at the correct rate in order to keep the rocket
in sight. Also, the mechanism for focusing the camera has to
take into account the increasing distance from the camera to
the rising rocket. Let’s assume the rocket rises vertically and
its speed is 200 mys when it has risen 900 m.
(a)How fast is the distance from the television camera to
the rocket changing at that moment?
(b)If the television camera is always kept aimed at the
rocket, how fast is the camera’s angle of elevation
changing at that same moment?
46. A
lighthouse is located on a small island 3 km away from
the nearest point P on a straight shoreline and its light
makes four revolutions per minute. How fast is the beam of
light moving along the shoreline when it is 1 km from P ?
47. A
plane flies horizontally at an altitude of 5 km and passes
directly over a tracking telescope on the ground. When the
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
angle of elevation is y3, this angle is decreasing at a rate of
y6 radymin. How fast is the plane traveling at that time?
48. A
Ferris wheel with a radius of 10 m is rotating at a rate of
one revolution every 2 minutes. How fast is a rider rising
when his seat is 16 m above ground level?
49. A
plane flying with a constant speed of 300 kmyh passes over
a ground radar station at an altitude of 1 km and climbs at an
angle of 308. At what rate is the distance from the plane to the
radar station increasing a minute later?
50. T
wo people start from the same point. One walks east at
4 kmyh and the other walks northeast at 2 kmyh. How fast is
the distance between the people changing after 15 minutes?
51. A
runner sprints around a circular track of radius 100 m at
a constant speed of 7 mys. The runner’s friend is standing
at a distance 200 m from the center of the track. How fast is
the distance between the friends changing when the distance
between them is 200 m?
52. T
he minute hand on a watch is 8 mm long and the hour hand
is 4 mm long. How fast is the distance between the tips of the
hands changing at one o’clock?
53. Suppose that the volume V of a rolling snowball increases so
that dVydt is proportional to the surface area of the snowball
at time t. Show that the radius r increases at a constant rate,
that is, drydt is constant.
2.9 Linear Approximations and Differentials
We have seen that a curve lies very close to its tangent line near the point of tangency. In
fact, by zooming in toward a point on the graph of a differentiable function, we noticed
that the graph looks more and more like its tangent line. (See Figure 2.1.2.) This observation is the basis for a method of finding approximate values of functions.
■ Linearization and Approximation
It might be easy to calculate a value f sad of a function, but difficult (or even impossible)
to compute nearby values of f . So we settle for the easily computed values of the linear
function L whose graph is the tangent line of f at sa, f sadd. (See Figure 1.)
In other words, we use the tangent line at sa, f sadd as an approximation to the curve
y − f sxd when x is near a. An equation of this tangent line is
{a, f(a)}
y − f sad 1 f 9sadsx 2 ad
The linear function whose graph is this tangent line, that is,
FIGURE 1
Lsxd − f sad 1 f 9sadsx 2 ad
is called the linearization of f at a. The approximation f sxd < Lsxd or
f sxd < f sad 1 f 9sadsx 2 ad
is called the linear approximation or tangent line approximation of f at a.
EXAMPLE 1 Find the linearization of the function f sxd − sx 1 3 at a − 1 and use
it to approximate the numbers s3.98 and s4.05 . Are these approximations overestimates or underestimates?
SOLUTION The derivative of f sxd − sx 1 3d1y2 is
f 9sxd − 12 sx 1 3d21y2 −
and so we have f s1d − 2 and f 9s1d −
2sx 1 3
Putting these values into Equation 1, we see
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.9Linear Approximations and Differentials
that the linearization is
Lsxd − f s1d 1 f 9s1dsx 2 1d − 2 1 14 sx 2 1d −
The corresponding linear approximation (2) is
sx 1 3 <
1 (when x is near 1)
In particular, we have
s3.98 < 74 1 0.98
4 − 1.995ands4.05 < 4 1 4 − 2.0125
y= 4 + 4
FIGURE 2
(1, 2)
y= œ„„„„
The linear approximation is illustrated in Figure 2. We see that, indeed, the tangent line
approximation is a good approximation to the given function when x is near l. We also
see that our approximations are overestimates because the tangent line lies above the
Of course, a calculator could give us approximations for s3.98 and s4.05 , but the
linear approximation gives an approximation over an entire interval.
In the following table we compare the estimates from the linear approximation in
Example 1 with the true values. Notice from this table, and also from Figure 2, that the
tangent line approximation gives good estimates when x is close to 1 but the accuracy of
the approximation deteriorates when x is farther away from 1.
From Lsxd
Actual value
1.97484176. . .
1.99499373. . .
2.00000000. . .
2.01246117. . .
2.02484567. . .
2.23606797. . .
2.44948974. . .
How good is the approximation that we obtained in Example 1? The next example
shows that by using a graphing calculator or computer we can determine an interval
throughout which a linear approximation provides a specified accuracy.
EXAMPLE 2 For what values of x is the linear approximation
sx 1 3 <
accurate to within 0.5? What about accuracy to within 0.1?
SOLUTION Accuracy to within 0.5 means that the functions should differ by less
than 0.5:
, 0.5
sx 1 3 2
S DZ
Equivalently, we could write
sx 1 3 2 0.5 ,
1 , sx 1 3 1 0.5
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
y= œ„„„„
L (x)
This says that the linear approximation should lie between the curves obtained by shifting the curve y − sx 1 3 upward and downward by an amount 0.5. Figure 3 shows
the tangent line y − s7 1 xdy4 intersecting the upper curve y − sx 1 3 1 0.5 at P
and Q. We estimate that the x-coordinate of P is about 22.66 and the x-coordinate of Q
is about 8.66. Thus we see from the graph that the approximation
y= œ„„„„
sx 1 3 <
FIGURE 3
is accurate to within 0.5 when 22.6 , x , 8.6. (We have rounded the smaller value up
and the larger value down.)
Similarly, from Figure 4 we see that the approximation is accurate to within 0.1
when 21.1 , x , 3.9.
y= œ„„„„
■ Applications to Physics
y= œ„„„„
FIGURE 4
Linear approximations are often used in physics. In analyzing the consequences of an
equation, a physicist sometimes needs to simplify a function by replacing it with its linear approximation. For instance, in deriving a formula for the period of a pendulum,
physics textbooks obtain an expression involving sin and then replace sin by with
the remark that sin is very close to if is not too large. You can verify that the linearization of the function f sxd − sin x at a − 0 is Lsxd − x and so the lin­ear approximation
at 0 is
sin x < x
(see Exercise 48). So, in effect, the derivation of the formula for the period of a pendulum
uses the tangent line approximation for the sine function.
Another example occurs in the theory of optics, where light rays that arrive at shallow
angles relative to the optical axis are called paraxial rays. In paraxial (or Gaussian)
optics, both sin and cos are replaced by their linearizations. In other words, the linear
sin < andcos < 1
are used because is close to 0. The results of calculations made with these approximations became the basic theoretical tool used to design lenses. (See Optics, 5th ed., by
Eugene Hecht [Boston, 2017], p. 164.)
In Section 11.11 we will present several other applications of the idea of linear approximations to physics and engineering.
■ Differentials
If dx ± 0, we can divide both sides of
Equation 3 by dx to obtain
− f 9sxd
We have seen similar equations
before, but now the left side can genuinely be interpreted as a ratio of
The ideas behind linear approximations are sometimes formulated in the terminology
and notation of differentials. If y − f sxd, where f is a differentiable function, then the
differential dx is an independent variable; that is, dx can be given the value of any real
number. The differential dy is then defined in terms of dx by the equation
dy − f 9sxd dx
So dy is a dependent variable; it depends on the values of x and dx. If dx is given a specific value and x is taken to be some specific number in the domain of f , then the numerical value of dy is determined.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.9Linear Approximations and Differentials
The geometric meaning of differentials is shown in Figure 5. Let Psx, f sxdd and
Qsx 1 Dx, f sx 1 Dxdd be points on the graph of f and let dx − Dx. The corresponding
change in y is
Dy − f sx 1 Dxd 2 f sxd
The slope of the tangent line PR is the derivative f 9sxd. Thus the directed distance from
S to R is f 9sxd dx − dy. Therefore dy represents the amount that the tangent line rises or
falls (the change in the linearization), whereas Dy represents the amount that the curve
y − f sxd rises or falls when x changes
by an amount dx.
Prefer larger art, set in text:
dx=Î x
x+Î x
FIGURE 5
EXAMPLE 3 Compare the values of Dy and dy if y − f sxd − x 3 1 x 2 2 2x 1 1 and
x changes (a) from 2 to 2.05 and (b) from 2 to 2.01.
(a) We have
f s2d − 2 3 1 2 2 2 2s2d 1 1 − 9
f s2.05d − s2.05d3 1 s2.05d2 2 2s2.05d 1 1 − 9.717625
Figure 6 shows the function in
Example 3 and a comparison of dy
and Dy when a − 2. The viewing
rectangle is f1.8, 2.5g by f6, 18g.
Dy − f s2.05d 2 f s2d − 0.717625
dy − f 9sxd dx − s3xQ2 1R2x 2 2d dx
In general,
When x − 2 and dx − Dx − 0.05, this becomes
(2, 9)
FIGURE 6
dy − f3s2d2 1P2s2d 2 2g0.05 − 0.7
f s2.01d − s2.01d 1 s2.01d 2 2s2.01d 1 1 − 9.140701
x+Î x
Dy − f s2.01d 2 f s2d − 0.140701
When dx − Dx − 0.01,
dy − f3s2d2 1 2s2d 2 2g0.01 − 0.14
Notice that the approximation Dy < dy becomes better as Dx becomes smaller in
Example 3. Notice also that dy was easier to compute than Dy.
In the notation of differentials, the linear approximation f sxd < f sad 1 f 9sadsx 2 ad
can be written as
f sa 1 dxd < f sad 1 dy
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
by taking dx − x 2 a, so x − a 1 dx. For instance, for the function f sxd − sx 1 3 in
Example 1, we have
dy − f 9sxd dx −
2 sx 1 3
If a − 1 and dx − Dx − 0.05, then
dy −
2 s1 1 3
− 0.0125
s4.05 − f s1.05d − f s1 1 0.05d < f s1d 1 dy − 2.0125
just as we found in Example 1.
Our final example illustrates the use of differentials in estimating the errors that occur
because of approximate measurements.
EXAMPLE 4 The radius of a sphere was measured and found to be 21 cm with a
pos­sible error in measurement of at most 0.05 cm. What is the maximum error in using
this value of the radius to compute the volume of the sphere?
SOLUTION If the radius of the sphere is r, then its volume is V − 43 r 3. If the error in
the measured value of r is denoted by dr − Dr, then the corresponding error in the
calculated value of V is DV, which can be approximated by the differential
dV − 4r 2 dr
When r − 21 and dr − 0.05, this becomes
dV − 4s21d2 0.05 < 277
The maximum error in the calculated volume is about 277 cm3.
NOTE Although the possible error in Example 4 may appear to be rather large, a better
picture of the error is given by the relative error, which is computed by dividing the
error by the total volume:
4r 2 dr
− 4 3 −3
3 r
Thus the relative error in the volume is about three times the relative error in the radius.
In Example 4 the relative error in the radius is approximately dryr − 0.05y21 < 0.0024
and it produces a relative error of about 0.007 in the volume. The errors could also be
expressed as percentage errors of 0.24% in the radius and 0.7% in the volume.
1–4 Find the linearization Lsxd of the function at a.
1. f sxd − x 2 x 1 3, a − 22
2. f sxd − cos 2x,
a − y6
3. f sxd − s
x, a − 8
4. f sxd − 2ysx 2 2 5 , a − 3
ind the linear approximation of the function
; 5. F
f sxd − s1 2 x at a − 0 and use it to approximate the
numbers s0.9 and s0.99 . Illustrate by graphing f and the
tangent line.
ind the linear approximation of the function
; 6. F
tsxd − s
1 1 x at a − 0 and use it to approximate the
numbers s
0.95 and s
1.1 . Illustrate by graphing t and the
tangent line.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 2.9Linear Approximations and Differentials
; 7–10 Verify the given linear approximation at a − 0. Then
determine the values of x for which the linear approximation is
accurate to within 0.1.
7. s
1 1 2x < 1 1 12 x
9. 1ys1 1 2xd4 < 1 2 8x
8. s1 1 xd23 < 1 2 3x
10. tan x < x
11–18 Find the differential of the function.
12. y − s1 2 t 4
11. y − sx 2 2 3d 22
13. y −
1 1 2u
1 1 3u
14. y − 2 sin 2
15. y −
x 2 2 3x
16. y − s1 1 cos 17. y − st 2 cos t
18. y −
sin x
19–22 (a) Find the differential dy and (b) evaluate dy for the
given values of x and dx.
19. y − tan x,x − y4,dx − 20.1
20. y − cos x,x − 13,dx − 20.02
21. y − s3 1 x ,x − 1,dx − 20.1
,x − 2,dx − 0.05
23–26 Compute Dy and dy for the given values of x and
dx − Dx. Then sketch a diagram like Figure 5 showing the line
segments with lengths dx, dy, and Dy.
23. y − x 2 2 4x,
x − 3, Dx − 0.5
24. y − x 2 x 3,
x − 0, Dx − 20.3
25. y − sx 2 2 ,
x − 3, Dx − 0.8
37–38 Explain, in terms of linear approximations or differentials, why the approximation is reasonable.
37. sec 0.08 < 1
38. s4.02 < 2.005
39. T
he edge of a cube was found to be 30 cm with a possible
error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing (a) the volume of the cube and
(b) the sur­face area of the cube.
40. T
he radius of a circular disk is given as 24 cm with a maxi­
mum error in measurement of 0.2 cm.
(a)Use differentials to estimate the maximum error in the
calculated area of the disk.
(b) What is the relative error and the percentage error?
41. T
he circumference of a sphere was measured to be 84 cm
with a possible error of 0.5 cm.
(a)Use differentials to estimate the maximum error in the
calculated surface area. What is the relative error?
(b)Use differentials to estimate the maximum error in the
calculated volume. What is the relative error?
42. U
se differentials to estimate the amount of paint needed to
apply a coat of paint 0.05 cm thick to a hemispherical dome
with diameter 50 m.
22. y −
26. y − x 3,x − 1,Dx − 0.5
27–30 Compare the values of Dy and dy if x changes from 1 to
1.05. What if x changes from 1 to 1.01? Does the approximation
Dy < dy become better as D x gets smaller?
27. f sxd − x 4 2 x 1 1
28. f sxd − sx 3 1 3d2
29. f sxd − s5 2 x
30. f sxd −
x 11
31–36 Use a linear approximation (or differentials) to estimate
the given number.
31. s1.999d4
32. 1y4.002
33. s
34. s100.5
35. tan 2°
36. cos 29°
43. (a)Use differentials to find a formula for the approximate
volume of a thin cylindrical shell with height h, inner
radius r, and thickness Dr.
(b)What is the error involved in using the formula from
part (a)?
44. O
ne side of a right triangle is known to be 20 cm long and
the opposite angle is measured as 30°, with a possible error
of 61°.
(a)Use differentials to estimate the error in computing the
length of the hypotenuse.
(b)What is the percentage error?
45. I f a current I passes through a resistor with resistance R,
Ohm’s Law states that the voltage drop is V − RI. If V is
constant and R is measured with a certain error, use differentials to show that the relative error in calculating I is
approximately the same (in magnitude) as the relative
error in R.
46. W
hen blood flows along a blood vessel, the flux F (the volume of blood per unit time that flows past a given point) is
proportional to the fourth power of the radius R of the blood
F − kR 4
(This is known as Poiseuille’s Law; we will show why it
is true in Section 8.4.) A partially clogged artery can be
expanded by an operation called angioplasty, in which a
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
balloon-tipped catheter is inflated inside the artery in order
to widen it and restore normal blood flow.
Show that the relative change in F is about four times the
relative change in R. How will a 5% increase in the radius
affect the flow of blood?
47. E
stablish the following rules for working with differentials
(where c denotes a constant and u and v are functions of x).
(a) dc − 0
(b) dscud − c du
(c) dsu 1 vd − du 1 dv
(d) dsuvd − u dv 1 v du
(e) d
v du 2 u dv
(b)If − y18 (equivalent to 10°) and we approximate
sin by , what is the percentage error?
(c)Use a graph to determine the values of for which sin and differ by less than 2%. What are the values in
49. S
uppose that the only information we have about a function f
is that f s1d − 5 and the graph of its derivative is as shown.
(a)Use a linear approximation to estimate f s0.9d and f s1.1d.
(b)Are your estimates in part (a) too large or too small?
(f ) dsx n d − nx n21 dx
48. I n physics textbooks, the period T of a pendulum of length L
is often given as T < 2 sLyt , provided that the pendulum
swings through a relatively small arc. In the course of deriving this formula, the equation a T − 2t sin for the tangential
acceleration of the bob of the pendulum is obtained, and then
sin is replaced by with the remark that for small angles,
(in radians) is very close to sin .
(a)Verify the linear approximation at 0 for the sine function:
sin < DISCOVERY PROJECT
50. S
uppose that we don’t have a formula for tsxd but we know
that ts2d − 24 and t9sxd − sx 2 1 5 for all x.
(a)Use a linear approximation to estimate ts1.95d and ts2.05d.
(b)Are your estimates in part (a) too large or too small?
The tangent line approximation Lsxd is the best first-degree (linear) approximation to f sxd
near x − a because f sxd and Lsxd have the same rate of change (derivative) at a. For a better
approximation than a linear one, let’s try a second-degree (quadratic) approximation Psxd. In
other words, we approximate a curve by a parabola instead of by a straight line. To make sure
that the approximation is a good one, we stipulate the following:
(i) Psad − f sad
(P and f should have the same value at a.)
(ii) P9sad − f 9sad
(P and f should have the same rate of change at a.)
(iii) P99sad − f 99sad
(The slopes of P and f should change at the same rate at a.)
1. Find
the quadratic approximation Psxd − A 1 Bx 1 Cx 2 to the function f sxd − cos x that
satisfies conditions (i), (ii), and (iii) with a − 0. Graph P, f , and the linear approximation
Lsxd − 1 on a common screen. Comment on how well the functions P and L approximate f .
2. Determine
the values of x for which the quadratic approximation f sxd < Psxd in Problem 1
is accurate to within 0.1. [Hint: Graph y − Psxd, y − cos x 2 0.1, and y − cos x 1 0.1
on a common screen.]
3. To
approximate a function f by a quadratic function P near a number a, it is best to write P
in the form
Psxd − A 1 Bsx 2 ad 1 Csx 2 ad2
Show that the quadratic function that satisfies conditions (i), (ii), and (iii) is
Psxd − f sad 1 f 9sadsx 2 ad 1 12 f 99sadsx 2 ad2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2Review
4. Find
the quadratic approximation to f sxd − sx 1 3 near a − 1. Graph f , the quadratic
approximation, and the linear approximation from Example 2.9.2 on a common screen.
What do you conclude?
5. Instead
of being satisfied with a linear or quadratic approximation to f sxd near x − a,
let’s try to find better approximations with higher-degree polynomials. We look for an
nth-degree polynomial
Tnsxd − c0 1 c1 sx 2 ad 1 c2 sx 2 ad2 1 c3 sx 2 ad3 1 ∙ ∙ ∙ 1 cn sx 2 adn
such that Tn and its first n derivatives have the same values at x − a as f and its first n
derivatives. By differentiating repeatedly and setting x − a, show that these conditions are
satisfied if c0 − f sad, c1 − f 9sad, c2 − 12 f 99 sad, and in general
ck −
f skdsad
where k! − 1 ? 2 ? 3 ? 4 ? ∙ ∙ ∙ ? k. The resulting polynomial
Tn sxd − f sad 1 f 9sadsx 2 ad 1
f 99sad
f sndsad
sx 2 ad2 1 ∙ ∙ ∙ 1
sx 2 adn
is called the nth-degree Taylor polynomial of f centered at a. (We will study Taylor
polynomials in more detail in Chapter 11.)
6. Find
the 8th-degree Taylor polynomial centered at a − 0 for the function f sxd − cos x.
Graph f together with the Taylor polynomials T2 , T4 , T6 , T8 in the viewing rectangle
f25, 5g by f21.4, 1.4g and comment on how well they approximate f .
Answers to the Concept Check are available at StewartCalculus.com.
1. W
rite an expression for the slope of the tangent line to the
curve y − f sxd at the point sa, f sadd.
6. D
escribe several ways in which a function can fail to be
differentiable. Illustrate with sketches.
2. S
uppose an object moves along a straight line with position f std at time t. Write an expression for the instantaneous velocity of the object at time t − a. How can you
interpret this velocity in terms of the graph of f ?
7. W
hat are the second and third derivatives of a function f ?
If f is the position function of an object, how can you interpret f 0 and f -?
3. If y − f sxd and x changes from x 1 to x 2, write expressions for
the following.
(a)The average rate of change of y with respect to x over the
interval fx 1, x 2 g
(b)The instantaneous rate of change of y with respect to x
at x − x 1
4. D
efine the derivative f 9sad. Discuss two ways of interpreting
this number.
8. S
tate each differentiation rule both in symbols and in words.
(a) The Power Rule
(b) The Constant Multiple Rule
(c) The Sum Rule
(d) The Difference Rule
(e) The Product Rule
(f ) The Quotient Rule
(g) The Chain Rule
5. (a)What does it mean for f to be differentiable at a?
(b)What is the relation between the differentiability and
continuity of a function?
(c)Sketch the graph of a function that is continuous but not
differentiable at a − 2.
9. S
tate the derivative of each function.
(a) y − x n
(b) y − sin x
(d) y − tan x
(e) y − csc x
(g) y − cot x
(c) y − cos x
(f ) y − sec x
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
10. Explain how implicit differentiation works.
11. G
ive several examples of how the derivative can be interpreted as a rate of change in physics, chemistry, biology,
economics, or other sciences.
12. (a)Write an expression for the linearization of f at a.
(b)If y − f sxd, write an expression for the differential dy.
(c)If dx − Dx, draw a picture showing the geometric
meanings of Dy and dy.
Determine whether the statement is true or false. If it is true,
explain why. If it is false, explain why or give an example that
disproves the statement.
x 2 1 x − 2x 1 1
| |
8. If f 9srd exists, then lim x l r f sxd − f srd.
1. If f is continuous at a, then f is differentiable at a.
2. If f and t are differentiable, then
9. If tsxd − x 5, then lim
f f sxd 1 tsxdg − f 9sxd 1 t9sxd
d 2y
10. 2 −
3. If f and t are differentiable, then
f f sxd tsxdg − f 9sxd t9sxd
tsxd 2 ts2d
− 80.
S D
11. A
n equation of the tangent line to the parabola y − x 2
at s22, 4d is y 2 4 − 2xsx 1 2d.
4. If f and t are differentiable, then
f s tsxdd − f 9s tsxdd t9sxd
stan2xd −
ssec 2xd
13. The derivative of a polynomial is a polynomial.
f 9sxd
5. If f is differentiable, then
sf sxd −
2 sf sxd
14. The derivative of a rational function is a rational function.
| |
15. If f is differentiable at a, so is f .
f 9sxd
6. If f is differentiable, then
f (sx ) −
2 sx
16. If f sxd − sx 2 x d , then f
sxd − 0.
1. T
he displacement (in meters) of an object moving in a
straight line is given by s − 1 1 2t 1 14 t 2, where t is measured in seconds.
(a)Find the average velocity over each time period.
(i) f1, 3g
(ii) f1, 2g
(iii) f1, 1.5g
3–4 Trace or copy the graph of the function. Then sketch a graph
of its derivative directly beneath.
y y
y y
(iv) f1, 1.1g
(b) Find the instantaneous velocity when t − 1.
2. T
he graph of f is shown. State, with reasons, the numbers
at which f is not differentiable.
x x
x x
5. T
he figure shows the graphs of f , f 9, and f 0. Identify each
curve, and explain your choices.
_1 0
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2Review
6. Find a function f and a number a such that
s2 1 hd6 2 64
− f 9sad
h l0
7. T
he total cost of repaying a student loan at an interest rate
of r % per year is C − f srd.
(a)What is the meaning of the derivative f 9srd? What are its
(b) What does the statement f 9s10d − 1200 mean?
(c) Is f 9srd always positive or does it change sign?
8. The total fertility rate at time t, denoted by Fstd, is an estimate of the average number of children born to each woman
(assuming that current birth rates remain constant). The
graph of the total fertility rate in the United States shows
the fluctuations from 1940 to 2010.
(a)Estimate the values of F9s1950d, F9s1965d,
and F9s1987d.
(b) What are the meanings of these derivatives?
(c)Can you suggest reasons for the values of these
2010 t
9. Let Pstd be the percentage of Americans under the age of 18
at time t. The table gives values of this function in census
years from 1950 to 2010.
(a) What is the meaning of P9std? What are its units?
(b) Construct a table of estimated values for P9std.
(c) Graph P and P9.
(d)How would it be possible to get more accurate values
for P9std?
10–11Find f 9sxd from first principles, that is, directly from the
def­inition of a derivative.
10. f sxd −
13–40 Calculate y9.
13. y − sx 2 1 x 3 d4
15. y −
11. f sxd − x 3 1 5x 1 4
x2 2 x 1 2
17. y − x 2 sin x
19. y −
t4 2 1
t4 1 1
14. y −
16. y −
sx 3
tan x
1 1 cos x
S D
18. y − x 1
20. y − sinscos xd
sinsx 2 sin xd
21. y − tan s1 2 x
22. y −
23. xy 4 1 x 2 y − x 1 3y
24. y − secs1 1 x 2 d
25. y −
12. (a)If f sxd − s3 2 5x , use the definition of a derivative
to find f 9sxd.
(b) Find the domains of f and f 9.
(c)Graph f and f 9 on a common screen. Compare the
graphs to see whether your answer to part (a) is
sec 2
1 1 tan 2
26. y 1 x cos y − x 2 y
27. y − s1 2 x 21 d21
28. y − 1ys
x 1 sx
29. sinsxyd − x 2 2 y
30. y − ssin s x
31. y − cots3x 2 1 5d
32. y −
sx 1 d4
x 4 1 4
33. y − sx cos sx
34. y −
sin mx
35. y − tan2ssin d
36. x tan y − y 2 1
37. y − s
x tan x
38. y −
39. y − sin(tan s1 1 x 3 )
40. y − sin2 (cosssin x )
sx 2 1dsx 2 4d
sx 2 2dsx 2 3d
41. If f std − s4t 1 1, find f 99s2d.
42. If tsd − sin , find t99sy6d.
43. Find y99 if x 6 1 y 6 − 1.
44. Find f sndsxd if f sxd − 1ys2 2 xd.
45–46 Find the limit.
45. lim
sec x
1 2 sin x
46. lim
tan3 2t
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
47–48 Find an equation of the tangent line to the curve at the
given point.
47. y − 4 sin2 x,sy6, 1d
x2 2 1
,s0, 21d
x2 1 1
48. y −
49–50 Find equations of the tangent line and normal line to the
curve at the given point.
If Ssxd − f sxd 1 tsxd, find S9s1d.
If Psxd − f sxd tsxd, find P9s2d.
If Qsxd − f sxdytsxd, find Q9s1d.
If Csxd − f stsxdd, find C9s2d.
60. If f and t are the functions whose graphs are shown, let
Psxd − f sxd tsxd, Qsxd − f sxdytsxd, and Csxd − f s tsxdd.
Find (a) P9s2d, (b) Q9s2d, and (c) C9s2d.
49. y − s1 1 4 sin x ,s0, 1d
50. x 2 1 4xy 1 y 2 − 13,s2, 1d
51. (a) If f sxd − x s5 2 x , find f 9sxd.
(b)Find equations of the tangent lines to the curve
y − x s5 2 x at the points s1, 2d and s4, 4d.
(c)Illustrate part (b) by graphing the curve and tangent
lines on the same screen.
(d)Check to see that your answer to part (a) is reasonable
by comparing the graphs of f and f 9.
52. (a)If f sxd − 4x 2 tan x, 2y2 , x , y2, find f 9
and f 99.
(b)Check to see that your answers to part (a) are reason;
able by comparing the graphs of f , f 9, and f 99.
53. A
t what points on the curve y − sin x 1 cos x,
0 < x < 2, is the tangent line horizontal?
61–68 Find f 9 in terms of t9.
61. f sxd − x 2tsxd
62. f sxd − tsx 2 d
63. f sxd − f tsxdg 2
64. f sxd − x a tsx b d
65. f sxd − ts tsxdd
66. f sxd − sins tsxdd
67. f sxd − tssin xd
68. f sxd − tstan sx d
54. F
ind the points on the ellipse x 1 2y − 1 where the
tangent line has slope 1.
55. F
ind a parabola y − ax 2 1 bx 1 c that passes through the
point s1, 4d and whose tangent lines at x − 21 and x − 5
have slopes 6 and 22, respectively.
56. H
ow many tangent lines to the curve y − xysx 1 1) pass
through the point s1, 2d? At which points do these tangent
lines touch the curve?
57. If f sxd − sx 2 adsx 2 bdsx 2 cd, show that
f 9sxd
f sxd
58. (a) By differentiating the double-angle formula
cos 2x − cos2x 2 sin2x
obtain the double-angle formula for the sine function.
(b) By differentiating the addition formula
sinsx 1 ad − sin x cos a 1 cos x sin a
obtain the addition formula for the cosine function.
59. Suppose that
f s1d − 2f 9s1d − 3f s2d − 1f 9s2d − 2
ts1d − 3t9s1d − 1ts2d − 1t9s2d − 4
69–70 Find h9 in terms of f 9 and t9.
69. hsxd −
f sxd tsxd
f sxd 1 tsxd
70. hsxd − f s tssin 4xdd
; 71. (a)Graph the function f sxd − x 2 2 sin x in the viewing
rectangle f0, 8g by f22, 8g.
(b)On which interval is the average rate of change larger:
f1, 2g or f2, 3g?
(c)At which value of x is the instantaneous rate of change
larger: x − 2 or x − 5?
(d)Check your visual estimates in part (c) by computing
f 9sxd and comparing the numerical values of f 9s2d
and f 9s5d.
72. A
particle moves along a horizontal line so that its coordinate at time t is x − sb 2 1 c 2 t 2 , t > 0, where b and c
are positive constants.
(a) Find the velocity and acceleration functions.
(b)Show that the particle always moves in the positive
73. A
particle moves on a vertical line so that its coordinate at
time t is y − t 3 2 12t 1 3, t > 0.
(a) Find the velocity and acceleration functions.
(b)When is the particle moving upward and when is it
moving downward?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2Review
(c)Find the distance that the particle travels in the time
interval 0 < t < 3.
(d)Graph the position, velocity, and acceleration functions
for 0 < t < 3.
(e)When is the particle speeding up? When is it slowing
74. T
he volume of a right circular cone is V − 13 r 2h, where
r is the radius of the base and h is the height.
(a)Find the rate of change of the volume with respect to
the height if the radius is constant.
(b)Find the rate of change of the volume with respect to
the radius if the height is constant.
75. T
he mass of part of a wire is x (1 1 sx ) kilograms, where
x is measured in meters from one end of the wire. Find the
linear density of the wire when x − 4 m.
76. T
he cost, in dollars, of producing x units of a certain commodity is
Csxd − 920 1 2x 2 0.02x 2 1 0.00007x 3
(a) Find the marginal cost function.
(b) Find C9s100d and explain its meaning.
(c)Compare C9s100d with the cost of producing the 101st
77. T
he volume of a cube is increasing at a rate of 10 cm3ymin.
How fast is the surface area increasing when the length of
an edge is 30 cm?
78. A
paper cup has the shape of a cone with height 10 cm and
radius 3 cm (at the top). If water is poured into the cup at a
rate of 2 cm3ys, how fast is the water level rising when the
water is 5 cm deep?
79. A
balloon is rising at a constant speed of 2 mys. A boy is
cycling along a straight road at a speed of 5 mys. When he
passes under the balloon, it is 15 m above him. How fast is
the distance between the boy and the balloon increasing 3 s
80. A
waterskier skis over the ramp shown in the figure at a
speed of 10 mys. How fast is she rising as she leaves the
81. T
he angle of elevation of the sun is decreasing at a rate of
0.25 radyh. How fast is the shadow cast by a 400-meter-tall
building increasing when the angle of elevation of the sun
is y6?
; 82. (a)Find the linear approximation to f sxd − s25 2 x 2
near 3.
(b)Illustrate part (a) by graphing f and the linear
(c)For what values of x is the linear approximation
accurate to within 0.1?
83. (a)Find the linearization of f sxd − s
1 1 3x at a − 0.
State the corresponding linear approximation and
use it to give an approximate value for s
1.03 .
approximation given in part (a) is accurate to
within 0.1.
84. Evaluate dy if y − x 3 2 2x 2 1 1, x − 2, and dx − 0.2.
85. A
window has the shape of a square surmounted by a semi­circle. The base of the window is measured as having width
60 cm with a possible error in measurement of 0.1 cm. Use
differentials to estimate the maximum error possible in
computing the area of the window.
86–88 Express the limit as a derivative and evaluate.
86. lim
x l1
88. lim
x 17 2 1
87. lim
16 1 h 2 2
cos 2 0.5
2 y3
89. Evaluate lim
s1 1 tan x 2 s1 1 sin x
90. Suppose f is a differentiable function such that
f s tsxdd − x and f 9sxd − 1 1 f f sxdg 2. Show that
t9sxd − 1ys1 1 x 2 d.
91. Find f 9sxd if it is known that
f f s2xdg − x 2
92. S
how that the length of the portion of any tangent line to the
astroid x 2y3 1 y 2y3 − a 2y3 cut off by the coordinate axes is
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Problems Plus
Try to solve the following example yourself before reading the solution.
EXAMPLE How many lines are tangent to both of the parabolas y − 21 2 x 2 and
y − 1 1 x 2 ? Find the coordinates of the points at which these tangents touch the
FIGURE 1
SOLUTION To gain insight into this problem, it is essential to draw a diagram. So we
sketch the parabolas y − 1 1 x 2 (which is the standard parabola y − x 2 shifted 1 unit
upward) and y − 21 2 x 2 (which is obtained by reflecting the first parabola about
the x-axis). If we try to draw a line tangent to both parabolas, we soon discover that
there are only two possibilities, as illustrated in Figure 1.
Let P be a point at which one of these tangents touches the upper parabola and
let a be its x-coordinate. (The choice of notation for the unknown is important.
Of course we could have used b or c or x 0 or x1 instead of a. However, it’s not
advisable to use x in place of a because that x could be confused with the variable x
in the equation of the parabola.) Then, since P lies on the parabola y − 1 1 x 2, its
y-coordinate must be 1 1 a 2. Because of the symmetry shown in Figure 1, the
coordinates of the point Q where the tangent touches the lower parabola must
be s2a, 2s1 1 a 2 dd.
To use the given information that the line is a tangent, we equate the slope of the
line PQ to the slope of the tangent line at P. We have
mPQ −
1 1 a 2 2 s21 2 a 2 d
1 1 a2
a 2 s2ad
If f sxd − 1 1 x 2, then the slope of the tangent line at P is f 9sad − 2a. Thus the condition that we need to use is that
1 1 a2
− 2a
Solving this equation, we get 1 1 a 2 − 2a 2, so a 2 − 1 and a − 61. Therefore the
points are (1, 2) and s21, 22d. By symmetry, the two remaining points are s21, 2d
and s1, 22d.
1. Find points P and Q on the parabola y − 1 2 x 2 so that the triangle ABC formed by the
x-axis and the tangent lines at P and Q is an equilateral triangle (see the figure).
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
ind the point where the curves y − x 3 2 3x 1 4 and y − 3sx 2 2 xd are tangent to each
; 2. F
other, that is, have a common tangent line. Illustrate by sketching both curves and the
common tangent.
3. S
how that the tangent lines to the parabola y − ax 2 1 bx 1 c at any two points with
x-coordinates p and q must intersect at a point whose x-coordinate is halfway between p
and q.
4. Show that
5. If f sxd − lim
sin2 x
cos2 x
1 1 cot x
1 1 tan x
− 2cos 2x.
sec t 2 sec x
, find the value of f 9sy4d.
6. Find the values of the constants a and b such that
ax 1 b 2 2
7. Prove that
ssin4 x 1 cos4 xd − 4n21 coss4x 1 ny2d.
dx n
8. If f is differentiable at a, where a . 0, evaluate the following limit in terms of f 9sad:
f sxd 2 f sad
sx 2 sa
9. T
he figure shows a circle with radius 1 inscribed in the parabola y − x 2. Find the center of
the circle.
10. F
ind all values of c such that the parabolas y − 4x 2 and x − c 1 2y 2 intersect each other at
right angles.
11. H
ow many lines are tangent to both of the circles x 2 1 y 2 − 4 and x 2 1 s y 2 3d 2 − 1?
At what points do these tangent lines touch the circles?
12. If f sxd −
x 46 1 x 45 1 2
, calculate f s46ds3d. Express your answer using factorial notation:
n! − 1 2 3 ∙ ∙ ∙ sn 2 1d n
13. T
he figure shows a rotating wheel with radius 40 cm and a connecting rod AP with length
1.2 m. The pin P slides back and forth along the x-axis as the wheel rotates counter­
clockwise at a rate of 360 revolutions per minute.
(a)Find the angular velocity of the connecting rod, dydt, in radians per second,
when − y3.
(b) Express the distance x − OP in terms of .
(c) Find an expression for the velocity of the pin P in terms of .
P (x, 0)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14. Tangent lines T1 and T2 are drawn at two points P1 and P2 on the parabola y − x 2 and they
intersect at a point P. Another tangent line T is drawn at a point between P1 and P2; it
intersects T1 at Q1 and T2 at Q2. Show that
| PQ | 1 | PQ | − 1
| PP | | PP |
15. Let T and N be the tangent and normal lines to the ellipse x 2y9 1 y 2y4 − 1 at any point P
on the ellipse in the first quadrant. Let x T and yT be the x- and y-intercepts of T and x N and
yN be the intercepts of N. As P moves along the ellipse in the first quadrant (but not on the
axes), what values can x T , yT , x N, and yN take on? First try to guess the answers just by
looking at the figure. Then use calculus to solve the problem and see how good your
intuition is.
16. Evaluate lim
sins3 1 xd2 2 sin 9
17. (a)Use the identity for tansx 2 yd (see Equation 15b in Appendix D) to show that if two
lines L 1 and L 2 intersect at an angle , then
tan −
m 2 2 m1
1 1 m1 m 2
where m1 and m 2 are the slopes of L 1 and L 2, respectively.
(b)The angle between the curves C1 and C2 at a point of intersection P is defined to be
the angle between the tangent lines to C1 and C2 at P (if these tangent lines exist). Use
part (a) to find, correct to the nearest degree, the angle between each pair of curves at
each point of intersection.
(i) y − x 2andy − sx 2 2d2
(ii) x 2 2 y 2 − 3andx 2 2 4x 1 y 2 1 3 − 0
18. Let Psx 1, y1d be a point on the parabola y 2 − 4px with focus Fs p, 0d. Let be the angle
between the parabola and the line segment FP, and let be the angle between the horizontal line y − y1 and the parabola as in the figure. Prove that − . (Thus, by a prin­ciple
P(⁄, ›)
F( p, 0)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
of geometrical optics, light from a source placed at F will be reflected along a line parallel
to the x-axis. This explains why paraboloids, the surfaces obtained by rotating parabolas
about their axes, are used as the shape of some automobile headlights and mirrors for
19. S
uppose that we replace the parabolic mirror of Problem 18 by a spherical mirror. Although
the mirror has no focus, we can show the existence of an approximate focus. In the figure,
C is a semicircle with center O. A ray of light coming in toward the mirror parallel to the
axis along the line PQ will be reflected to the point R on the axis so that /PQO − /OQR
(the angle of incidence is equal to the angle of reflection). What happens to the point R as
P is taken closer and closer to the axis?
20. If f and t are differentiable functions with f s0d − ts0d − 0 and t9s0d ± 0, show that
f sxd
f 9s0d
21. Evaluate the following limit:
sinsa 1 2xd 2 2 sinsa 1 xd 1 sin a
22. G
iven an ellipse x 2ya 2 1 y 2yb 2 − 1, where a ± b, find the equation of the set of all points
from which there are two tangents to the curve whose slopes are (a) reciprocals and
(b) negative reciprocals.
23. Find the two points on the curve y − x 4 2 2x 2 2 x that have a common tangent line.
24. S
uppose that three points on the parabola y − x 2 have the property that their normal lines
intersect at a common point. Show that the sum of their x-coordinates is 0.
25. A lattice point in the plane is a point with integer coordinates. Suppose that circles with
radius r are drawn using all lattice points as centers. Find the smallest value of r such that
any line with slope 25 intersects some of these circles.
26. A
cone of radius r centimeters and height h centimeters is lowered point first at a rate of
1 cmys into a tall cylinder of radius R centimeters that is partially filled with water. How fast
is the water level rising at the instant the cone is completely submerged?
27. A
container in the shape of an inverted cone has height 16 cm and radius 5 cm at the top. It
is partially filled with a liquid that oozes through the sides at a rate proportional to the area
of the container that is in contact with the liquid. (The surface area of a cone is rl, where
r is the radius and l is the slant height.) If we pour the liquid into the container at a rate of
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2 cm3ymin, then the height of the liquid decreases at a rate of 0.3 cmymin when the height
is 10 cm. If our goal is to keep the liquid at a constant height of 10 cm, at what rate should
we pour the liquid into the container?
28. (a)The cubic function f sxd − xsx 2 2dsx 2 6d has three distinct zeros: 0, 2, and 6. Graph
f and its tangent lines at the average of each pair of zeros. What do you notice?
(b)Suppose the cubic function f sxd − sx 2 adsx 2 bdsx 2 cd has three distinct zeros:
a, b, and c. Prove, with the help of a computer algebra system, that a tangent line drawn
at the average of the zeros a and b intersects the graph of f at the third zero.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
The great mathematician Leonard Euler observed “… nothing at all takes place in the universe in which some rule of maximum or
minimum does not appear.” In Exercise 3.7.53 you will use calculus to show that bees construct the cells in their hive in a shape that
minimizes surface area.
Kostiantyn Kravchenko / Shutterstock.com
Applications of Differentiation
WE HAVE ALREADY INVESTIGATED SOME of the applications of derivatives, but now that we
know the differen­tiation rules we are in a better position to pursue the applications of differentiation in greater depth. Here we learn what derivatives tell us about the shape of a graph of a function and, in particular, how they help us locate maximum and minimum values of functions. Many
practical problems require us to minimize a cost or maximize an area or somehow find the best
possible outcome of a situation. In particular, we will be able to investigate the optimal shape of a
can and to explain the location of rainbows in the sky.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
3.1 Maximum and Minimum Values
Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. Here
are examples of such problems that we will solve in this chapter:
What is the shape of a can that minimizes manufacturing costs?
What is the maximum acceleration of a spacecraft? (This is an important question for
the astronauts who have to withstand the effects of acceleration.)
What is the radius of a contracted windpipe that expels air most rapidly during a cough?
At what angle should blood vessels branch so as to minimize the energy expended
by the heart in pumping blood?
These problems can be reduced to finding the maximum or minimum values of a function. Let’s first explain exactly what we mean by maximum and minimum values.
■ Absolute and Local Extreme Values
We see that the highest point on the graph of the function f shown in Figure 1 is the point
s3, 5d. In other words, the largest value of f is f s3d − 5. Likewise, the smallest value is
f s6d − 2. We say that f s3d − 5 is the absolute maximum of f and f s6d − 2 is the absolute minimum. In general, we use the following definition.
1 Definition Let c be a number in the domain D of a function f. Then f scd is the
● absolute maximum value of f on D if f scd > f sxd for all x in D.
● absolute minimum value of f on D if f scd < f sxd for all x in D.
FIGURE 1
FIGURE 2
Abs min f sad, abs max f sd d,
loc min f scd, f sed, loc max f sbd, f sd d
An absolute maximum or minimum is sometimes called a global maximum or minimum. The maximum and minimum values of f are called extreme values of f.
Figure 2 shows the graph of a function f with absolute maximum at d and absolute
minimum at a. Note that sd, f sddd is the highest point on the graph and sa, f sadd is the
lowest point. In Figure 2, if we consider only values of x near b [for instance, if we
restrict our attention to the interval sa, cd], then f sbd is the largest of those values of f sxd
and is called a local maximum value of f. Likewise, f scd is called a local minimum value
of f because f scd < f sxd for x near c [in the interval sb, dd, for instance]. The function f
also has a local minimum at e. In general, we have the following definition.
2 Definition The number f scd is a
● local maximum value of f if f scd > f sxd when x is near c.
● local minimum value of f if f scd < f sxd when x is near c.
FIGURE 3
In Definition 2 (and elsewhere), if we say that something is true near c, we mean that
it is true on some open interval containing c. (Thus a local maximum or minimum can’t
occur at an endpoint.) For instance, in Figure 3 we see that f s4d − 5 is a local minimum
because it’s the smallest value of f on the interval I. It’s not the absolute minimum
because f sxd takes on smaller values when x is near 12 (in the interval K, for instance).
In fact f s12d − 3 is both a local minimum and the absolute minimum. Similarly,
f s8d − 7 is a local maximum, but not the absolute maximum because f takes on larger
values near 1.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.1Maximum and Minimum Values
EXAMPLE 1 The graph of the function
(_1, 37)
f sxd − 3x 4 2 16x 3 1 18x 221 < x < 4
(1, 5)
(3, _27)
is shown in Figure 4. You can see that f s1d − 5 is a local maximum, whereas the
absolute maximum is f s21d − 37. (This absolute maximum is not a local maximum
because it occurs at an endpoint.) Also, f s0d − 0 is a local minimum and f s3d − 227
is both a local and an absolute minimum. Note that f has neither a local nor an absolute
maximum at x − 4.
EXAMPLE 2 The function f sxd − cos x takes on its (local and absolute) maximum value of 1 infinitely many times, because cos 2n − 1 for any integer n and
21 < cos x < 1 for all x. (See Figure 5.) Likewise, coss2n 1 1d − 21 is its
minimum value, where n is any integer.
FIGURE 4
Local and absolute maximum
FIGURE 5
f s0d − 0 is the absolute (and local) minimum value of f . This corresponds to the fact
that the origin is the lowest point on the parabola y − x 2. (See Figure 6.) However,
there is no highest point on the parabola and so this function has no maximum value. n
EXAMPLE 4 From the graph of the function f sxd − x 3, shown in Figure 7, we see that
FIGURE 6
Mimimum value 0, no maximum
this function has neither an absolute maximum value nor an absolute minimum value.
In fact, it has no local extreme values either.
We have seen that some functions have extreme values, whereas others do not. The
following theorem gives conditions under which a function is guaranteed to possess
extreme values.
EXAMPLE 3 If f sxd − x 2, then f sxd > f s0d because x 2 > 0 for all x. Therefore
Local and absolute minimum
y − cos x
3 The Extreme Value Theorem If f is continuous on a closed interval fa, bg,
then f attains an absolute maximum value f scd and an absolute minimum value
f sdd at some numbers c and d in fa, bg.
FIGURE 7
No mimimum, no maximum
The Extreme Value Theorem is illustrated in Figure 8. Note that an extreme value can
be taken on more than once. Although the Extreme Value Theorem is intuitively very
plausible, it is difficult to prove and so we omit the proof.
FIGURE 8
Functions continuous on a closed
interval always attain extreme values.
d b
a c¡
c™ b
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
Figures 9 and 10 show that a function need not possess extreme values if either
hypothe­sis (continuity or closed interval) is omitted from the Extreme Value Theorem.
FIGURE 9
This function has minimum value
f(2)=0, but no maximum value.
FIGURE 10
This continuous function g has
no maximum or minimum.
The function f whose graph is shown in Figure 9 is defined on the closed interval
[0, 2] but has no maximum value. (Notice that the range of f is [0, 3d. The function takes
on val­ues arbitrarily close to 3, but never actually attains the value 3.) This does not contradict the Extreme Value Theorem because f is not continuous. [Nonetheless, a discontinuous function could have maximum and minimum values. See Exercise 13(b).]
The function t shown in Figure 10 is continuous on the open interval s0, 2d but has
neither a maximum nor a minimum value. [The range of t is s1, `d. The function takes
on arbitrarily large values.] This does not contradict the Extreme Value Theorem because
the interval s0, 2d is not closed.
■ Critical Numbers and the Closed Interval Method
The Extreme Value Theorem says that a continuous function on a closed interval has a
maximum value and a minimum value, but it does not tell us how to find these extreme
values. Notice in Figure 8 that the absolute maximum and minimum values that are
between a and b occur at local maximum or minimum values, so we start by looking for
local extreme values.
Figure 11 shows the graph of a function f with a local maximum at c and a local
minimum at d. It appears that at the maximum and minimum points the tangent lines are
hor­izontal and therefore each has slope 0. We know that the derivative is the slope of the
tan­gent line, so it appears that f 9scd − 0 and f 9sdd − 0. The following theorem says that
this is always true for differentiable functions.
{c, f(c)}
{d, f (d )}
FIGURE 11
4 Fermat’s Theorem If f has a local maximum or minimum at c, and if f 9scd
exists, then f 9scd − 0.
PROOF Suppose, for the sake of definiteness, that f has a local maximum at c. Then,
according to Definition 2, f scd > f sxd if x is sufficiently close to c. This implies that if
h is sufficiently close to 0, with h being positive or negative, then
f scd > f sc 1 hd
and therefore
f sc 1 hd 2 f scd < 0
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.1Maximum and Minimum Values
Fermat’s Theorem is named after
Pierre Fermat (1601–1665), a French
lawyer who took up mathematics as
a hobby. Despite his amateur status,
Fermat was one of the two inventors
of analytic geometry (Descartes was
the other). His methods for finding
tangents to curves and maximum
and minimum values (before the
invention of limits and derivatives)
made him a forerunner of Newton in
the creation of differ­ential calculus.
We can divide both sides of an inequality by a positive number. Thus, if h . 0 and h is
sufficiently small, we have
f sc 1 hd 2 f scd
Taking the right-hand limit of both sides of this inequality (using Theorem 1.6.2), we get
hl 01
f sc 1 hd 2 f scd
< lim1 0 − 0
h l0
But since f 9scd exists, we have
f 9scd − lim
f sc 1 hd 2 f scd
f sc 1 hd 2 f scd
− lim1
and so we have shown that f 9scd < 0.
If h , 0, then the direction of the inequality (5) is reversed when we divide by h:
f sc 1 hd 2 f scd
So, taking the left-hand limit, we have
f 9scd − lim
f sc 1 hd 2 f scd
f sc 1 hd 2 f scd
− lim2
h l0
We have shown that f 9scd > 0 and also that f 9scd < 0. Since both of these inequalities
must be true, the only possibility is that f 9scd − 0.
We have proved Fermat’s Theorem for the case of a local maximum. The case of a
local minimum can be proved in a similar manner, or see Exercise 73 for an alternate
The following examples caution us against reading too much into Fermat’s Theorem:
we can’t expect to locate extreme values simply by setting f 9sxd − 0 and solving for x.
EXAMPLE 5 If f sxd − x 3, then f 9sxd − 3x 2, so f 9s0d − 0. But f has no maximum or
FIGURE 12
If f sxd − x 3, then f 9s0d − 0, but
f has no maximum or minimum.
y=| x|
FIGURE 13
| |
If f sxd − x , then f s0d − 0 is a minimum value, but f 9s0d does not exist.
minimum at 0, as you can see from its graph in Figure 12. (Or observe that x 3 . 0 for
x . 0 but x 3 , 0 for x , 0.) The fact that f 9s0d − 0 simply means that the curve
y − x 3 has a horizontal tangent at s0, 0d. Instead of having a maximum or minimum at
s0, 0d, the curve crosses its horizontal tangent there.
| |
EXAMPLE 6 The function f sxd − x has its (local and absolute) minimum value
at 0, but that value can’t be found by setting f 9sxd − 0 because, as was shown in
Example 2.2.5, f 9s0d does not exist. (See Figure 13.)
WARNING Examples 5 and 6 show that we must be careful when using Fermat’s
Theorem. Example 5 demonstrates that even when f 9scd − 0 there need not be a maximum or minimum at c. (In other words, the converse of Fermat’s Theorem is false in
general.) Fur­thermore, there may be an extreme value even when f 9scd does not exist
(as in Example 6).
Fermat’s Theorem does suggest that we should at least start looking for extreme values of f at the numbers c where f 9scd − 0 or where f 9scd does not exist. Such numbers
are given a special name.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
6 Definition A critical number of a function f is a number c in the domain of
f such that either f 9scd − 0 or f 9scd does not exist.
EXAMPLE 7 Find the critical numbers of (a) f sxd − x 3 2 3x 2 1 1 and
(b) f sxd − x 3y5s4 2 xd.
(a) The derivative of f is f 9sxd − 3x 2 2 6x − 3xsx 2 2d. Since f 9sxd exists for all x,
the only critical numbers of f occur when f 9sxd − 0, that is, when x − 0 or x − 2.
(b) First note that the domain of f is R. The Product Rule gives
Figure 14 shows a graph of the
function f in Example 7(b). It supports our answer because there is a
horizontal tangent when x − 1.5
fwhere f 9sxd − 0g and a vertical
tangent when x − 0 fwhere f 9sxd
is undefinedg.
f 9sxd − x 3y5s21d 1 s4 2 xd(53 x22y5) − 2x 3y5 1
3s4 2 xd
5x 2 y5
25x 1 3s4 2 xd
12 2 8x
5x 2y5
[The same result could be obtained by first writing f sxd − 4x 3y5 2 x 8y5.] Therefore
f 9sxd − 0 if 12 2 8x − 0, that is, x − 32, and f 9sxd does not exist when x − 0. Thus the
critical numbers are 32 and 0.
In terms of critical numbers, Fermat’s Theorem can be rephrased as follows (compare
Definition 6 with Theorem 4):
FIGURE 14
7 If f has a local maximum or minimum at c, then c is a critical number of f.
To find an absolute maximum or minimum of a continuous function on a closed interval, we note that either it is local [in which case it occurs at a critical number by (7)] or
it occurs at an endpoint of the interval, as we see from the examples in Figure 8. Thus the
following three-step procedure always works.
The Closed Interval Method To find the absolute maximum and minimum
values of a continuous function f on a closed interval fa, bg:
1. Find the values of f at the critical numbers of f in sa, bd.
2. Find the values of f at the endpoints of the interval.
3. The largest of the values from Steps 1 and 2 is the absolute maximum value;
the smallest of these values is the absolute minimum value.
EXAMPLE 8 Find the absolute maximum and minimum values of the function
f sxd − x 3 2 3x 2 1 1212 < x < 4
SOLUTION Since f is continuous on 212, 4 , we can use the Closed Interval Method.
In Example 7(a) we found the critical numbers x − 0 and x − 2. Notice that each of
these critical numbers lies in the interval (212 , 4). The values of f at these critical
numbers are
f s0d − 1f s2d − 23
The values of f at the endpoints of the interval are
f (212 ) − 18f s4d − 17
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.1Maximum and Minimum Values
(4, 17)
_1 0
EXAMPLE 9
(a) Use a calculator or computer to estimate the absolute minimum and maximum
values of the function f sxd − x 2 2 sin x, 0 < x < 2.
(b) Use calculus to find the exact minimum and maximum values.
FIGURE 15
Comparing these four numbers, we see that the absolute maximum value is f s4d − 17
and the absolute minimum value is f s2d − 23.
In this example the absolute maximum occurs at an endpoint, whereas the absolute
minimum occurs at a critical number. The graph of f is sketched in Figure 15.
With graphing software or a graphing calculator it is possible to estimate maximum
and minimum values very easily. But, as the next example shows, calculus is needed to
find the exact values.
(2, _3)
(a) Figure 16 shows a graph of f in the viewing rectangle f0, 2g by f21, 8g. The
absolute maximum value is about 6.97 and it occurs when x < 5.24. Similarly, the
absolute minimum value is about 20.68 and it occurs when x < 1.05. It is possible
to get more accurate numerical estimates, but for exact values we must use calculus.
(b) The function f sxd − x 2 2 sin x is continuous on f0, 2g. Since
f 9sxd − 1 2 2 cos x, we have f 9sxd − 0 when cos x − 12 and this occurs when
x − y3 or 5y3. The values of f at these critical numbers are
FIGURE 16
f sy3d −
f s5y3d −
2 2 sin
2 s3 < 20.684853
2 2 sin
1 s3 < 6.968039
The values of f at the endpoints are
f s0d − 0f s2d − 2 < 6.28
Comparing these four numbers and using the Closed Interval Method, we see that the
absolute minimum value is f sy3d − y3 2 s3 and the absolute maximum value is
f s5y3d − 5y3 1 s3 . The values from part (a) serve as a check on our work.
EXAMPLE 10 The Hubble Space Telescope was deployed on April 24, 1990, by the
space shuttle Discovery. A model for the velocity of the shuttle during this mission,
from liftoff at t − 0 until the solid rocket boosters were jettisoned at t − 126 seconds,
is given by
vstd − 0.000397t 3 2 0.02752t 2 1 7.196t 2 0.9397
(in meters per second). Using this model, estimate the absolute maximum and minimum
values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters.
SOLUTION We are asked for the extreme values not of the given velocity function,
but rather of the acceleration function. So we first need to differentiate to find the
astd − v9std −
s0.000397t 3 2 0.02752t 2 1 7.196t 2 0.9397d
− 0.001191t 2 2 0.05504t 1 7.196
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
We now apply the Closed Interval Method to the continuous function a on the interval
0 < t < 126. Its derivative is
a9std − 0.0023808t 2 0.05504
The only critical number occurs when a9std − 0:
t1 −
< 23.12
Evaluating astd at the critical number and at the endpoints, we have
as0d − 7.196ast1 d − as23.12d − 6.56as126d < 19.16
So the maximum acceleration is about 19.16 mys2 and the minimum acceleration is
about 6.56 mys2.
1. E
xplain the difference between an absolute minimum and
a local minimum.
2. S
uppose f is a continuous function defined on a closed
interval fa, bg.
(a)What theorem guarantees the existence of an absolute max­imum value and an absolute minimum value
for f ?
(b)What steps would you take to find those maximum
and minimum values?
3–4 For each of the numbers a, b, c, d, r, and s, state whether the
function whose graph is shown has an absolute maximum or minimum, a local maximum or minimum, or neither a maximum nor
a minimum.
5–6 Use the graph to state the absolute and local maximum and
minimum values of the function.
7–10 Sketch the graph of a function f that is continuous on [1, 5]
and has the given properties.
7. A
bsolute maximum at 5, absolute minimum at 2,
local maximum at 3, local minima at 2 and 4
8. A
bsolute maximum at 4, absolute minimum at 5,
local maximum at 2, local minimum at 3
0 a
10. A
bsolute maximum at 2, absolute minimum at 5,
4 is a critical number but there is no local maximum
or minimum there.
4. y
9. A
bsolute minimum at 3, absolute maximum at 4,
local maximum at 2
11. (a)Sketch the graph of a function that has a local maximum
at 2 and is differentiable at 2.
(b)Sketch the graph of a function that has a local maximum
at 2 and is continuous but not differentiable at 2.
(c)Sketch the graph of a function that has a local maximum
at 2 and is not continuous at 2.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.1Maximum and Minimum Values
12. (a)Sketch the graph of a function on [21, 2] that has an
absolute maximum but no local maximum.
(b)Sketch the graph of a function on [21, 2] that has a
local maximum but no absolute maximum.
13. (a)Sketch the graph of a function on [21, 2] that has an
absolute maximum but no absolute minimum.
(b)Sketch the graph of a function on [21, 2] that is
discontinuous but has both an absolute maximum and
an absolute minimum.
14. (a)Sketch the graph of a function that has two local maxima, one local minimum, and no absolute minimum.
(b)Sketch the graph of a function that has three local
minima, two local maxima, and seven critical numbers.
15–28 Sketch the graph of f by hand and use your sketch to
find the absolute and local maximum and minimum values of f.
(Use the graphs and transformations of Sections 1.2 and 1.3.)
15. f sxd − 3 2 2x,
37. psxd −
x2 1 2
2x 2 1
t2 1 9
t2 2 9
39. hstd − t 3y4 2 2 t 1y4
40. tsxd − s
4 2 x2
41. Fsxd − x 4y5sx 2 4d 2
42. hsxd − x21y3sx 2 2d
43. f sxd − x 1y3s4 2 xd2y3
44. f sd − 1 s2 cos 45. f sd − 2 cos 1 sin2 46. tsxd − s1 2 x 2
; 47–48 A formula for the derivative of a function f is given.
How many critical numbers does f have?
47. f 9sxd − 1 1
210 sin x
x 2 2 6x 1 10
100 cos 2 x
10 1 x 2
48. f 9sxd −
x > 21
16. f sxd − x 2, 21 < x , 2
38. qstd −
17. f sxd − 1yx,x > 1
49–60 Find the absolute maximum and absolute minimum
values of f on the given interval.
18. f sxd − 1yx,1 , x , 3
49. f sxd − 12 1 4x 2 x 2,f0, 5g
19. f sxd − sin x,0 < x , y2
50. f sxd − 5 1 54x 2 2x 3,f0, 4g
20. f sxd − sin x,0 , x < y2
51. f sxd − 2x 3 2 3x 2 2 12x 1 1,f22, 3g
21. f sxd − sin x,2y2 < x < y2
52. f sxd − x 3 2 6x 2 1 5,f23, 5g
22. f std − cos t,23y2 < t < 3y2
53. f sxd − 3x 4 2 4x 3 2 12x 2 1 1,f22, 3g
23. f sxd − 1 1 sx 1 1d 2,22 < x , 5
54. f std − st 2 2 4d 3,f22, 3g
| |
24. f sxd − x
55. f sxd − x 1
25. f sxd − 1 2 sx
26. f sxd − 1 2 x 3
27. f sxd −
28. f sxd −
56. f sxd −
if 21 < x < 0
2 2 3x if 0 , x < 1
2x 1 1 if 0 < x , 1
4 2 2x if 1 < x < 3
58. f std −
30. tsvd − v 3 2 12v 1 4
31. f sxd − 3x 4 1 8x 3 2 48x 2
32. f sxd − 2x 3 1 x 2 1 8x
33. tstd − t 1 5t 1 50t
34. Asxd − 3 2 2x
35. tsyd −
y2 2 y 1 1
36. hs pd −
,f0, 3g
x2 2 x 1 1
,f0, 2g
1 1 t2
59. f std − 2 cos t 1 sin 2t,f0, y2g
29. f sxd − 3x 2 1 x 2 2
,f0.2, 4g
57. f std − t 2 s
t ,f21, 4g
29–46 Find the critical numbers of the function.
p2 1 4
60. f sd − 1 1 cos2 , [y4, ]
61. I f a and b are positive numbers, find the maximum value
of f sxd − x as1 2 xd b, 0 < x < 1.
se a graph to estimate the critical numbers of
; 62. U
f sxd − 1 1 5x 2 x 3 correct to one decimal place.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
; 63–66
(a)Use a graph to estimate the absolute maximum and
minimum values of the function to two decimal places.
(b)Use calculus to find the exact maximum and minimum
63. f sxd − x 5 2 x 3 1 2, 21 < x < 1
64. f sxd − x 4 2 3x 3 1 3x 2 2 x, 0 < x < 2
65. f sxd − x sx 2 x 2
66. f sxd − x 2 2 cos x, 22 < x < 0
67. Between 0°C and 30°C, the volume V (in cubic centimeters)
of 1 kg of water at a temperature T is given approximately
by the formula
V − 999.87 2 0.06426T 1 0.0085043T 2 2 0.0000679T 3
Find the temperature at which water has its maximum density.
68. A
n object with weight W is dragged along a horizontal plane
by a force acting along a rope attached to the object. If the
rope makes an angle with the plane, then the magnitude of
the force is
sin 1 cos where is a positive constant called the coefficient of
friction and where 0 < < y2. Show that F is minimized
when tan − .
69. T
he water level, measured in meters above mean sea level,
of Lake Lanier in Georgia, USA, during 2012 can be modeled by the function
Lstd − 0.00439t 3 2 0.1273t 2 1 0.8239t 1 323.1
where t is measured in months since January 1, 2012. Estimate when the water level was highest during 2012.
70. I n 1992 the space shuttle Endeavour was launched on mission STS-49 in order to install a new perigee kick motor in
an Intelsat communications satellite. The table gives the
velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters.
Begin roll maneuver
End roll maneuver
Throttle to 89%
Throttle to 67%
Throttle to 104%
Maximum dynamic pressure
Solid rocket booster separation
Time (s) Velocity (mys)
(a)Use a graphing calculator or computer to find the
cubic polynomial that best models the velocity of the
shuttle for the time interval t [ f0, 125g. Then graph
this polynomial.
(b)Find a model for the acceleration of the shuttle and
use it to estimate the maximum and minimum values of
the acceleration during the first 125 seconds.
71. W
hen a foreign object lodged in the trachea forces a person
to cough, the diaphragm thrusts upward, causing an increase
in pressure in the lungs. This is accompanied by a contraction of the trachea, making a narrower channel for the
expelled air to flow through. For a given amount of air to
escape in a fixed time, it must move faster through the narrower channel than the wider one. The greater the velocity
of the airstream, the greater the force on the foreign object.
X-rays show that the radius of the circular tracheal tube
contracts to about two-thirds of its normal radius during a
cough. According to a mathematical model of coughing,
the velocity v of the airstream is related to the radius r of the
trachea by the equation
vsrd − ksr0 2 rdr 212 r0 < r < r0
where k is a constant and r0 is the normal radius of the
trachea. The restriction on r is due to the fact that the
tracheal wall stiffens under pressure and a contraction
greater than 12 r0 is prevented (otherwise the person would
(a)Determine the value of r in the interval 12 r 0 , r 0 at
which v has an absolute maximum. How does this
compare with experimental evidence?
(b)What is the absolute maximum value of v on the
(c) Sketch the graph of v on the interval f0, r0 g.
72. Prove that the function
f sxd − x 101 1 x 51 1 x 1 1
has neither a local maximum nor a local minimum.
73. (a) If f has a local minimum value at c, show that the function tsxd − 2f sxd has a local maximum value at c.
(b)Use part (a) to prove Fermat’s Theorem for the case in
which f has a local minimum at c.
74. A
cubic function is a polynomial of degree 3; that is, it has
the form f sxd − ax 3 1 bx 2 1 cx 1 d, where a ± 0.
(a)Show that a cubic function can have two, one, or no
critical number(s). Give examples and sketches to
illustrate the three possibilities.
(b)How many local extreme values can a cubic function
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPLIED PROJECTThe Calculus of Rainbows
Rainbows are created when raindrops scatter sunlight. They have fascinated mankind since
ancient times and have inspired attempts at scientific explanation since the time of Aristotle.
In this project we use the ideas of Descartes and Newton to explain the shape, location, and
colors of rainbows.
å A
D(å )
Formation of the primary rainbow
1. The
figure shows a ray of sunlight entering a spherical raindrop at A. Some of the light
is reflected, but the line AB shows the path of the part that enters the drop. Notice that
the light is refracted toward the normal line AO and in fact Snell’s Law says that
sin − k sin , where is the angle of incidence, is the angle of refraction, and k < 43
is the index of refraction for water. At B some of the light passes through the drop and is
refracted into the air, but the line BC shows the part that is reflected. (The angle of incidence equals the angle of reflection.) When the ray reaches C, part of it is reflected, but for
the time being we are more interested in the part that leaves the raindrop at C. (Notice that
it is refracted away from the normal line.) The angle of deviation Dsd is the amount of
clockwise rotation that the ray has undergone during this three-stage process. Thus
Dsd − s 2 d 1 s 2 2d 1 s 2 d − 1 2 2 4
how that the minimum value of the deviation is Dsd < 1388 and occurs when
< 59.48.
The significance of the minimum deviation is that when < 59.48 we have D9sd < 0,
so DDyD < 0. This means that many rays with < 59.48 become deviated by approximately the same amount. It is the concentration of rays coming from near the direction of
minimum deviation that creates the brightness of the primary rainbow. The figure at the
left shows that the angle of elevation from the observer up to the highest point on the rainbow is 180 8 2 1388 − 428. (This angle is called the rainbow angle.)
rays from sun
rays from sun
2. Problem
1 explains the location of the primary rainbow, but how do we explain the colors? Sunlight comprises a range of wavelengths, from the red range through orange, yellow, green, blue, indigo, and violet. As Newton discovered in his prism experiments of
1666, the index of refraction is different for each color. (The effect is called dispersion.)
For red light the refractive index is k < 1.3318, whereas for violet light it is k < 1.3435.
By repeating the calculation of Problem 1 for these values of k, show that the rainbow
angle is about 42.38 for the red bow and 40.68 for the violet bow. So the rainbow really
consists of seven individual bows corresponding to the seven colors.
3. Perhaps
you have seen a fainter secondary rainbow above the primary bow. That results
from the part of a ray that enters a raindrop and is refracted at A, reflected twice (at B and
C), and refracted as it leaves the drop at D (see the figure at the left). This time the deviation angle Dsd is the total amount of counterclockwise rotation that the ray undergoes in
this four-stage process. Show that
Formation of the secondary rainbow
Dsd − 2 2 6 1 2
and Dsad has a minimum value when
cos −
k2 2 1
(continued )
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
Taking k − 43, show that the minimum deviation is about 1298 and so the rainbow angle for
the secondary rainbow is about 518, as shown in the figure at the left.
Leonid Andronov / Shutterstock.com
4. Show
that the colors in the secondary rainbow appear in the opposite order from those in
the primary rainbow.
42° 51°
3.2 The Mean Value Theorem
We will see that many of the results of this chapter depend on one central fact, which is
called the Mean Value Theorem.
■ Rolle’s Theorem
To arrive at the Mean Value Theorem, we first need the following result.
Rolle’s Theorem Let f be a function that satisfies the following three hypotheses:
1.f is continuous on the closed interval fa, bg.
2. f is differentiable on the open interval sa, bd.
3. f sad − f sbd
Rolle’s Theorem was first published in
1691 by the French mathematician
Michel Rolle (1652–1719) in a book
entitled Méthode pour resoudre les
egalitez. He was a vocal critic of the
methods of his day and attacked calculus as being a “collection of ingenious
fallacies.” Later, however, he became
convinced of the essential correctness
of the methods of calculus.
Then there is a number c in sa, bd such that f 9scd − 0.
Before giving the proof let’s take a look at the graphs of some typical functions that
satisfy the three hypotheses. Figure 1 shows the graphs of four such functions. In each
case it appears that there is at least one point sc, f scdd on the graph where the tangent is
hori­zontal and therefore f 9scd − 0. Thus Rolle’s Theorem is plausible.
c™ b
FIGURE 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.2The Mean Value Theorem
PROOF There are three cases:
CASE I f sxd − k, a constant
Take cases
Then f 9sxd − 0, so the number c can be taken to be any number in sa, bd.
CASE II f sxd . f sad for some x in sa, bd [as in Figure 1(b) or (c)]
By the Extreme Value Theorem (which we can apply by hypothesis 1), f has a maximum value somewhere in fa, bg. Since f sad − f sbd, it must attain this maximum value
at a number c in the open interval sa, bd. Then f has a local maximum at c and, by
hypothesis 2, f is differentiable at c. Therefore f 9scd − 0 by Fermat’s Theorem.
CASE III f sxd , f sad for some x in sa, bd [as in Figure 1(c) or (d)]
By the Extreme Value Theorem, f has a minimum value in fa, bg and, since
f sad − f sbd, it attains this minimum value at a number c in sa, bd. Again f 9scd − 0 by
Fermat’s Theorem.
EXAMPLE 1 Let’s apply Rolle’s Theorem to the position function s − f std of a
moving object. If the object is in the same place at two different instants t − a and
t − b, then f sad − f sbd. Rolle’s Theorem says that there is some instant of time t − c
between a and b when f 9scd − 0; that is, the velocity is 0. (In particular, you can see
that this is true when a ball is thrown directly upward.)
Figure 2 shows a graph of the function f sxd − x 3 1 x 2 1 discussed in
Example 2. Rolle’s Theorem shows
that, no matter how much we enlarge
the viewing rectangle, we can never
find a second x-intercept.
EXAMPLE 2 Prove that the equation x 3 1 x 2 1 − 0 has exactly one real solution.
SOLUTION First we use the Intermediate Value Theorem (1.8.10) to show that a
solution exists. Let f sxd − x 3 1 x 2 1. Then f s0d − 21 , 0 and f s1d − 1 . 0.
Since f is a polynomial, it is continuous, so the Intermediate Value Theorem states
that there is a number c between 0 and 1 such that f scd − 0. Thus the given equation
has a solution.
To show that the equation has no other real solution, we use Rolle’s Theorem and
argue by contradiction. Suppose that it had two solutions a and b. Then f sad − 0 − f sbd
and, since f is a polynomial, it is differentiable on sa, bd and continuous on fa, bg. Thus,
by Rolle’s Theorem, there is a number c between a and b such that f 9scd − 0. But
f 9sxd − 3x 2 1 1 > 1for all x
FIGURE 2
(since x 2 > 0) so f 9sxd can never be 0. This gives a contradiction. Therefore the
equation can’t have two real solutions.
■ The Mean Value Theorem
Our main use of Rolle’s Theorem is in proving the following important theorem, which
was first stated by another French mathematician, Joseph-Louis Lagrange.
The Mean Value Theorem is an
example of what is called an existence
theorem. Like the Intermediate Value
Theorem, the Extreme Value Theorem, and Rolle’s Theorem, it guarantees that there exists a number with a
certain property, but it doesn’t tell us
how to find the number.
The Mean Value Theorem Let f be a function that satisfies the following
1. f is continuous on the closed interval fa, bg.
2. f is differentiable on the open interval sa, bd.
Then there is a number c in sa, bd such that
f 9scd −
f sbd 2 f sad
or, equivalently,
f sbd 2 f sad − f 9scdsb 2 ad
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
Before proving this theorem, we can see that it is reasonable by interpreting it geomet­
rically. Figures 3 and 4 show the points Asa, f sadd and Bsb, f sbdd on the graphs of two
dif­ferentiable functions. The slope of the secant line AB is
mAB −
f sbd 2 f sad
which is the same expression as on the right side of Equation 1. Since f 9scd is the slope of
the tangent line at the point sc, f scdd, the Mean Value Theorem, in the form given by Equa­
tion 1, says that there is at least one point Psc, f scdd on the graph where the slope of the
tangent line is the same as the slope of the secant line AB. In other words, there is a point
P where the tangent line is parallel to the secant line AB. (Imagine a line far away that
stays parallel to AB while moving toward AB until it touches the graph for the first time.)
P { c, f(c)}
A{ a, f(a)}
B { b, f(b)}
FIGURE 3
FIGURE 5
FIGURE 4
PROOF We apply Rolle’s Theorem to a new function h defined as the difference
between f and the function whose graph is the secant line AB. Using Equation 3
and the point-slope equation of a line, we see that the equation of the line AB can
be written as
or as
y 2 f sad −
f sbd 2 f sad
sx 2 ad
y − f sad 1
f sbd 2 f sad
sx 2 ad
So, as shown in Figure 5,
hsxd − f sxd 2 f sad 2
f sbd 2 f sad
sx 2 ad
First we must verify that h satisfies the three hypotheses of Rolle’s Theorem.
1. T
he function h is continuous on fa, bg because it is the sum of f and a firstdegree polynomial, both of which are continuous.
The function h is differentiable on sa, bd because both f and the first-degree
polynomial are differentiable. In fact, we can compute h9 directly from Equation 4:
h9sxd − f 9sxd 2
f sbd 2 f sad
(Note that f sad and f f sbd 2 f sadgysb 2 ad are constants.)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.2The Mean Value Theorem
Lagrange and the
Mean Value Theorem
The Mean Value Theorem was first formulated by Joseph-Louis Lagrange
(1736–1813), born in Italy of a French
father and an Italian mother. He was a
child prodigy and became a professor
in Turin at the age of 19. Lagrange
made great contributions to number
theory, theory of functions, theory of
equations, and analytical and celestial
mechanics. In particular, he applied calculus to the analysis of the stability of
the solar system. At the invitation of
Frederick the Great, he succeeded Euler
at the Berlin Academy and, when
Frederick died, Lagrange accepted King
Louis XVI’s invitation to Paris, where he
was given apartments in the Louvre
and became a professor at the Ecole
Polytechnique. Despite all the trappings of luxury and fame, he was a kind
and quiet man, living only for science.
f sbd 2 f sad
sa 2 ad − 0
hsbd − f sbd 2 f sad 2
f sbd 2 f sad
sb 2 ad
− f sbd 2 f sad 2 f f sbd 2 f sadg − 0
Therefore hsad − hsbd.
Since h satisfies all the hypotheses of Rolle’s Theorem, that theorem says there is a
number c in sa, bd such that h9scd − 0. Therefore
0 − h9scd − f 9scd 2
and so
f 9scd −
f sbd 2 f sad
f sbd 2 f sad
EXAMPLE 3 To illustrate the Mean Value Theorem with a specific function, let’s
consider f sxd − x 3 2 x, a − 0, b − 2. Since f is a polynomial, it is continuous and
differentiable for all x, so it is certainly continuous on f0, 2g and differentiable on s0, 2d.
Therefore, by the Mean Value Theorem, there is a number c in s0, 2d such that
f s2d 2 f s0d − f 9scds2 2 0d
Now f s2d − 6, f s0d − 0, and f 9sxd − 3x 2 2 1, so the above equation becomes
6 − s3c 2 2 1d2 − 6c 2 2 2
FIGURE 6 hsad − f sad 2 f sad 2
which gives c 2 − 43 , that is, c − 62ys3. But c must lie in s0, 2d, so c − 2ys3.
Figure 6 illustrates this calculation: the tangent line at this value of c is parallel to the
secant line OB. n
EXAMPLE 4 If an object moves in a straight line with position function s − f std, then
the average velocity between t − a and t − b is
f sbd 2 f sad
and the velocity at t − c is f 9scd. Thus the Mean Value Theorem (in the form of
Equation 1) tells us that at some time t − c between a and b the instantaneous velocity
f 9scd is equal to that average velocity. For instance, if a car traveled 180 km in 2 hours,
then the speedometer must have read 90 kmyh at least once.
In general, the Mean Value Theorem can be interpreted as saying that there is a
number at which the instantaneous rate of change is equal to the average rate of change
over an interval.
The main significance of the Mean Value Theorem is that it enables us to obtain information about a function from information about its derivative. The next example provides an instance of this principle.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
EXAMPLE 5 Suppose that f s0d − 23 and f 9sxd < 5 for all values of x. How large can
f s2d possibly be?
SOLUTION We are given that f is differentiable (and therefore continuous) everywhere.
In particular, we can apply the Mean Value Theorem on the interval f0, 2g. There exists
a number c such that
f s2d 2 f s0d − f 9scds2 2 0d
f s2d − f s0d 1 2f 9scd − 23 1 2f 9scd
We are given that f 9sxd < 5 for all x, so in particular we know that f 9scd < 5. Multiplying both sides of this inequality by 2, we have 2f 9scd < 10, so
f s2d − 23 1 2f 9scd < 23 1 10 − 7
The largest possible value for f s2d is 7.
The Mean Value Theorem can be used to establish some of the basic facts of differential calculus. One of these basic facts is the following theorem. Others will be discussed
in the following sections.
5 Theorem If f 9sxd − 0 for all x in an interval sa, bd, then f is constant on sa, bd.
PROOF Let x 1 and x 2 be any two numbers in sa, bd with x 1 , x 2 . Since f is differentiable on sa, bd, it must be differentiable on sx 1, x 2 d and continuous on fx 1, x 2 g. By
applying the Mean Value Theorem to f on the interval fx 1, x 2 g, we get a number c such
that x 1 , c , x 2 and
f sx 2 d 2 f sx 1d − f 9scdsx 2 2 x 1d
Since f 9sxd − 0 for all x, we have f 9scd − 0, and so Equation 6 becomes
f sx 2 d 2 f sx 1 d − 0orf sx 2 d − f sx 1 d
Therefore f has the same value at any two numbers x 1 and x 2 in sa, bd. This means that
f is constant on sa, bd.
Corollary 7 says that if two functions
have the same derivatives on an interval, then their graphs must be vertical
translations of each other there. In
other words, the graphs have the
same shape, but they could be shifted
up or down.
7 Corollary If f 9sxd − t9sxd for all x in an interval sa, bd, then f 2 t is constant
on sa, bd; that is, f sxd − tsxd 1 c where c is a constant.
PROOF Let Fsxd − f sxd 2 tsxd. Then
F9sxd − f 9sxd 2 t9sxd − 0
for all x in sa, bd. Thus, by Theorem 5, F is constant; that is, f 2 t is constant.
NOTE Care must be taken in applying Theorem 5. Let
f sxd −
if x . 0
21 if x , 0
| |
The domain of f is D − hx x ± 0j and f 9sxd − 0 for all x in D. But f is obviously not
a constant function. This does not contradict Theorem 5 because D is not an interval.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.2The Mean Value Theorem
1. T
he graph of a function f is shown. Verify that f satisfies the
hypotheses of Rolle’s Theorem on the interval f0, 8g. Then
estimate the value(s) of c that satisfy the conclusion of Rolle’s
Theorem on that interval.
10. f sxd − x 3 2 2x 2 2 4x 1 2,f22, 2g
11. f sxd − sins xy2d,fy2, 3y2g
12. f sxd − x 1 1y x,12 , 2
13. Let f sxd − 1 2 x 2y3. Show that f s21d − f s1d but there is
no number c in s21, 1d such that f 9scd − 0. Why does this
not contradict Rolle’s Theorem?
14. Let f sxd − tan x. Show that f s0d − f sd but there is no
number c in s0, d such that f 9scd − 0. Why does this not
contradict Rolle’s Theorem?
(a)Verify that t satisfies the hypotheses of the Mean Value
Theorem on the interval f0, 8g.
(b)Estimate the value(s) of c that satisfy the conclusion of
the Mean Value Theorem on the interval f0, 8g.
(c)Estimate the value(s) of c that satisfy the conclusion of
the Mean Value Theorem on the interval f2, 6g.
4. D
raw the graph of a function that is continuous on f0, 8g
where f s0d − 1 and f s8d − 4 and that does not satisfy the
conclusion of the Mean Value Theorem on f0, 8g.
5–8 The graph of a function f is shown. Does f satisfy the
hypotheses of the Mean Value Theorem on the interval f0, 5g ? If
so, find a value c that satisfies the conclusion of the Mean Value
Theorem on that interval.
15–18 Verify that the function satisfies the hypotheses of the
Mean Value Theorem on the given interval. Then find all
num­bers c that satisfy the conclusion of the Mean Value
15. f sxd − 2x 2 2 3x 1 1,f0, 2g
16. f sxd − x 3 2 3x 1 2,f22, 2g
17. f sxd − s
x ,f0, 1g
18. f sxd − 1yx,f1, 3g
; 19–20 Find the number c that satisfies the conclusion of the
Mean Value Theorem on the given interval. Graph the function, the secant line through the endpoints, and the tangent line
at sc, f scdd. Are the secant line and the tangent line parallel?
19. f sxd − sx ,f0, 4g
20. f sxd − x 3 2 2x,f22, 2g
21. Let f sxd − s x 2 3d22. Show that there is no value of c in
s1, 4d such that f s4d 2 f s1d − f 9scds4 2 1d. Why does this
not contradict the Mean Value Theorem?
9. f sxd − 2 x 2 2 4 x 1 5,f21, 3g
3. The graph of a function t is shown.
9–12 Verify that the function satisfies the three hypotheses of
Rolle’s Theorem on the given interval. Then find all numbers c
that satisfy the conclusion of Rolle’s Theorem.
2. D
raw the graph of a function defined on f0, 8g such that
f s0d − f s8d − 3 and the function does not satisfy the conclusion of Rolle’s Theorem on f0, 8g.
22. Let f sxd − 2 2 2 x 2 1 . Show that there is no value of c
such that f s3d 2 f s0d − f 9scds3 2 0d. Why does this not
contradict the Mean Value Theorem?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
23–24 Show that the equation has exactly one real solution.
33. Show that sin x , x if 0 , x , 2.
23. 2 x 1 cos x − 0
34. S
uppose f is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists
a number c in s2b, bd such that f 9scd − f sbdyb.
24. 2x 2 1 2 sin x − 0
25. S
how that the equation x 3 2 15x 1 c − 0 has at most one
solution in the interval f22, 2g.
35. Use the Mean Value Theorem to prove the inequality
| sin a 2 sin b | < | a 2 b |for all a and b
26. S
how that the equation x 1 4x 1 c − 0 has at most two
real solutions.
27. (a)Show that a polynomial of degree 3 has at most three real
(b)Show that a polynomial of degree n has at most n real
28. (a)Suppose that f is differentiable on R and has two zeros.
Show that f 9 has at least one zero.
(b)Suppose f is twice differentiable on R and has three
zeros. Show that f 0 has at least one real zero.
(c) Can you generalize parts (a) and (b)?
29. If f s1d − 10 and f 9sxd > 2 for 1 < x < 4, how small can
f s4d possibly be?
30. Suppose that 3 < f 9sxd < 5 for all values of x. Show that
18 < f s8d 2 f s2d < 30.
31. D
oes there exist a function f such that f s0d − 21, f s2d − 4,
and f 9sxd < 2 for all x ?
32. Suppose that f and t are continuous on fa, bg and differenti­able on sa, bd. Suppose also that f sad − tsad and
f 9sxd , t9sxd for a , x , b. Prove that f sbd , tsbd.
[Hint: Apply the Mean Value Theorem to the function
h − f 2 t.]
36. If f 9sxd − c (c a constant) for all x, use Corollary 7 to show
that f sxd − cx 1 d for some constant d.
37. Let f sxd − 1yx and
tsxd −
if x . 0
if x , 0
Show that f 9sxd − t9sxd for all x in their domains. Can we
conclude from Corollary 7 that f 2 t is constant?
38. At 2:00 pm a car’s speedometer reads 50 kmyh. At 2:10 pm it
reads 65 kmyh. Show that at some time between 2:00 and
2:10 the acceleration is exactly 90 kmyh2.
39. T
wo runners start a race at the same time and finish in a tie.
Prove that at some time during the race they have the same
speed. [Hint: Consider f std − tstd 2 hstd, where t and h are
the position functions of the two runners.]
40. F ixed Points A number a is called a fixed point of a function
f if f sad − a. Prove that if f 9sxd ± 1 for all real numbers x,
then f has at most one fixed point.
3.3 What Derivatives Tell Us about the Shape of a Graph
Many of the applications of calculus depend on our ability to deduce facts about a function f from information concerning its derivatives. Because f 9sxd represents the slope of
the curve y − f sxd at the point sx, f sxdd, it tells us the direction in which the curve proceeds at each point. So it is reasonable to expect that information about f 9sxd will provide
us with information about f sxd.
■ What Does f 9 Say about f ?
FIGURE 1
Let’s abbreviate the name of this test
to the I/D Test.
To see how the derivative of f can tell us where a function is increasing or decreasing,
look at Figure 1. (Increasing functions and decreasing functions were defined in Section 1.1.) Between A and ­B and between C and D, the tangent lines have positive slope and
so f 9sxd . 0. Between B and C, the tangent lines have negative slope and so f 9sxd , 0.
Thus it appears that f increases when f 9sxd is positive and decreases when f 9sxd is negative. To prove that this is always the case, we use the Mean Value Theorem.
Increasing/Decreasing Test
(a) If f 9sxd . 0 on an interval, then f is increasing on that interval.
(b) If f 9sxd , 0 on an interval, then f is decreasing on that interval.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.3What Derivatives Tell Us about the Shape of a Graph
(a) Let x 1 and x 2 be any two numbers in the interval with x1 , x2. According
to the definition of an increasing function (Section 1.1), we have to show that
f sx1 d , f sx2 d.
Because we are given that f 9sxd . 0, we know that f is differentiable on fx1, x2 g.
So, by the Mean Value Theorem, there is a number c between x1 and x2 such that
f sx 2 d 2 f sx 1 d − f 9scdsx 2 2 x 1 d
Now f 9scd . 0 by assumption and x 2 2 x 1 . 0 because x 1 , x 2. Thus the right side of
Equation 1 is positive, and so
f sx 2 d 2 f sx 1 d . 0orf sx 1 d , f sx 2 d
This shows that f is increasing.
Part (b) is proved similarly.
EXAMPLE 1 Find where the function f sxd − 3x 4 2 4x 3 2 12x 2 1 5 is increasing
and where it is decreasing.
SOLUTION We start by differentiating f :
f 9sxd − 12x 3 2 12x 2 2 24x − 12xsx 2 2dsx 1 1d
FIGURE 2
FIGURE 3
To use the IyD Test we have to know where f 9sxd . 0 and where f 9sxd , 0. To solve
these inequalities we first find where f 9sxd − 0, namely at x − 0, 2, and 21. These
are the critical numbers of f , and they divide the domain into four intervals (see the
number line in Figure 2). Within each interval, f 9sxd must be always positive or
always negative. (See Examples 3 and 4 in Appendix A.) We can determine which is
the case for each interval from the signs of the three factors of f 9sxd, namely 12x,
x 2 2, and x 1 1, as shown in the following chart. A plus sign indicates that the
given expression is positive, and a minus sign indicates that it is negative. The last
column of the chart gives the conclusion based on the IyD Test. For instance,
f 9sxd , 0 for 0 , x , 2, so f is decreasing on (0, 2). (It would also be true to say
that f is decreasing on the closed interval f0, 2g.)
f 9sxd
x , 21
21 , x , 0
decreasing on s2`, 21d
increasing on s21, 0d
decreasing on s0, 2d
increasing on s2, `d
The graph of f shown in Figure 3 confirms the information in the chart.
■ The First Derivative Test
Recall from Section 3.1 that if f has a local maximum or minimum at c, then c must be
a critical number of f (by Fermat’s Theorem), but not every critical number gives rise
to a maximum or a minimum. We therefore need a test that will tell us whether or not
f has a local maximum or minimum at a critical number.
You can see from Figure 3 that for the function f in Example 1, f s0d − 5 is a local
maximum value of f because f increases on s21, 0d and decreases on s0, 2d. Or, in terms
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
of derivatives, f 9sxd . 0 for 21 , x , 0 and f 9sxd , 0 for 0 , x , 2. In other words,
the sign of f 9sxd changes from positive to negative at 0. This observation is the basis of
the following test.
The First Derivative Test Suppose that c is a critical number of a continuous
function f.
(a)If f 9 changes from positive to negative at c, then f has a local maximum at c.
(b)If f 9 changes from negative to positive at c, then f has a local minimum at c.
(c)If f 9 is positive to the left and right of c, or negative to the left and right of c,
then f has no local maximum or minimum at c.
The First Derivative Test is a consequence of the IyD Test. In part (a), for instance,
because the sign of f 9sxd changes from positive to negative at c, f is increasing to the left
of c and decreasing to the right of c. It follows that f has a local maximum at c.
It is easy to remember the First Derivative Test by visualizing diagrams such as those
in Figure 4.
(a) Local maximum at c
(b) Local minimum at c
(c) No maximum or minimum at c
(d) No maximum or minimum at c
FIGURE 4
EXAMPLE 2 Find the local minimum and maximum values of the function f in
Example 1.
SOLUTION From the chart in the solution to Example 1 we see that f 9sxd changes
from negative to positive at 21, so f s21d − 0 is a local minimum value by the First
Derivative Test. Similarly, f 9 changes from negative to positive at 2, so f s2d − 227 is
also a local minimum value. As noted previously, f s0d − 5 is a local maximum value
because f 9sxd changes from positive to negative at 0.
EXAMPLE 3 Find the local maximum and minimum values of the function
tsxd − x 1 2 sin x0 < x < 2
SOLUTION As in Example 1, we start by finding the critical numbers. The derivative is
t9sxd − 1 1 2 cos x
so t9sxd − 0 when cos x − 212. The solutions of this equation are 2y3 and 4y3.
Because t is differentiable everywhere, the only critical numbers are 2y3 and 4y3.
We split the domain into intervals according to the critical numbers. Within each
interval, t9sxd is either always positive or always negative and so we analyze t in the
following chart.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.3What Derivatives Tell Us about the Shape of a Graph
The 1 signs in the chart come from
the fact that t9sxd . 0 when
cos x . 212 . From the graph of
y − cos x, this is true in the indicated
intervals. Alternatively, we can choose
a test value within each interval and
check the sign of t9sxd using that value.
t9sxd − 1 1 2 cos x
0 , x , 2y3
increasing on s0, 2y3d
2y3 , x , 4y3
decreasing on s2y3, 4y3d
4y3 , x , 2
increasing on s4y3, 2d
Because t9sxd changes from positive to negative at 2y3, the First Derivative Test
tells us that there is a local maximum at 2y3 and the local maximum value is
ts2y3d −
S D
1 2 sin
1 s3 < 3.83
Likewise, t9sxd changes from negative to positive at 4y3 and so
FIGURE 5 tsxd − x 1 2 sin x
ts4y3d −
S D
1 2 sin
2 s3 < 2.46
is a local minimum value. The graph of t in Figure 5 supports our conclusion.
■ What Does f 99 Say about f ?
Figure 6 shows the graphs of two increasing functions on sa, bd. Both graphs join point A
to point B but they look different because they bend in different directions. How can
we dis­tinguish between these two types of behavior?
FIGURE 6x
In Figure 7 tangents to these curves have been drawn at several points. In part (a)
the curve lies above the tangents and f is called concave upward on sa, bd. In part (b) the
curve lies below the tangents and t is called concave downward on sa, bd.
FIGURE 7
(a) Concave upward
(b) Concave downward
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
We use the abbreviation CU for concave upward and CD for concave
Definition If the graph of f lies above all of its tangents on an interval I, then f
is called concave upward on I. If the graph of f lies below all of its tangents on I,
then f is called concave downward on I.
Figure 8 shows the graph of a function that is concave upward (CU) on the intervals sb, cd, sd, ed, and se, pd and concave downward (CD) on the intervals sa, bd, sc, dd,
and sp, qd.
0 a
FIGURE 8
Let’s see how the second derivative helps determine the intervals of concavity. Look­
ing at Figure 7(a), you can see that, going from left to right, the slope of the tangent
increas­es. This means that the derivative f 9 is an increasing function and therefore its
derivative f 0 is positive. Likewise, in Figure 7(b) the slope of the tangent decreases from
left to right, so f 9 decreases and therefore f 0 is negative. This reasoning can be reversed
and suggests that the following theorem is true. A proof is given in Appendix F with the
help of the Mean Value Theorem.
Concavity Test
(a) If f 0sxd . 0 on an interval I, then the graph of f is concave upward on I.
(b) If f 0sxd , 0 on an interval I, then the graph of f is concave downward on I.
EXAMPLE 4 Figure 9 shows a population graph for honeybees raised in an apiary.
How does the rate of population growth change over time? When is this rate highest?
Over what intervals is P concave upward or concave downward?
Number of bees
(in thousands)
FIGURE 9
Time (in weeks)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.3What Derivatives Tell Us about the Shape of a Graph
SOLUTION By looking at the slope of the curve as t increases, we see that the rate of
growth of the population is initially very small, then gets larger until it reaches a
maximum at about t − 12 weeks, and decreases as the population begins to level off.
As the population approaches its maximum value of about 75,000 (called the carrying
capacity), the rate of growth, P9std, approaches 0. The curve appears to be concave
upward on (0, 12) and concave downward on (12, 18).
In Example 4, the population curve changed from concave upward to concave downward at approximately the point (12, 38,000). This point is called an inflection point of
the curve. The significance of this point is that the rate of population increase has its
maximum value there. In general, an inflection point is a point where a curve changes its
direction of concavity.
Definition A point P on a curve y − f sxd is called an inflection point if f is continuous there and the curve changes from concave upward to concave downward
or from concave downward to concave upward at P.
For instance, in Figure 8, B, C, D, and P are the points of inflection. Notice that if a
curve has a tangent at a point of inflection, then the curve crosses its tangent there.
In view of the Concavity Test, there is a point of inflection at any point where the
function is continuous and the second derivative changes sign.
(4, 6)
(2, 3)
FIGURE 10 EXAMPLE 5 Sketch a possible graph of a function f that satisfies the following
(i) f s0d − 0, f s2d − 3, f s4d − 6, f 9s0d − f 9s4d − 0
(ii) f 9sxd . 0 for 0 , x , 4, f 9sxd , 0 for x , 0 and for x . 4
(iii) f 0sxd . 0 for x , 2, f 0sxd , 0 for x . 2
SOLUTION Condition (i) tells us that the graph has horizontal tangents at the points
s0, 0d and s4, 6d. Condition (ii) says that f is increasing on the interval s0, 4d and
decreasing on the intervals s2`, 0d and s4, `d. It follows from the I/D Test that
f s0d − 0 is a local minimum and f s4d − 6 is a local maximum.
Condition (iii) says that the graph is concave upward on the interval s2`, 2d and
concave downward on s2, `d. Because the curve changes from concave upward to
concave downward when x − 2, the point s2, 3d is an inflection point.
We use this information to sketch the graph of f in Figure 10. Notice that we made
the curve bend upward when x , 2 and bend downward when x . 2.
■ The Second Derivative Test
Another application of the second derivative is the following test for identifying local
maximum and minimum values. It is a consequence of the Concavity Test, and it serves
as an alternative to the First Derivative Test.
The Second Derivative Test Suppose f 0 is continuous near c.
(a) If f 9scd − 0 and f 0scd . 0, then f has a local minimum at c.
(b) If f 9scd − 0 and f 0scd , 0, then f has a local maximum at c.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
For instance, part (a) is true because f 0sxd . 0 near c and so f is concave upward
near c. This means that the graph of f lies above its horizontal tangent at c and so f has
a local minimum at c. (See Figure 11.)
NOTE The Second Derivative Test is inconclusive when f 0scd − 0. In other words, at
such a point there might be a maximum, there might be a minimum, or there might be
neither. This test also fails when f 0scd does not exist. In such cases the First Derivative
Test must be used. In fact, even when both tests apply, the First Derivative Test is often
the easier one to use.
f ª(c)=0
FIGURE 11
f 99scd . 0, f is concave upward
EXAMPLE 6 Discuss the curve y − x 4 2 4x 3 with respect to concavity, points of
inflection, and local maxima and minima.
SOLUTION If f sxd − x 4 2 4x 3, then
f 9sxd − 4x 3 2 12x 2 − 4x 2sx 2 3d
f 0sxd − 12x 2 2 24x − 12xsx 2 2d
To find the critical numbers we set f 9sxd − 0 and obtain x − 0 and x − 3. (Note that f 9
is a polynomial and hence defined everywhere.) To use the Second Derivative Test we
evaluate f 0 at these critical numbers:
f 0s0d − 0f 0s3d − 36 . 0
Since f 9s3d − 0 and f 0s3d . 0, the Second Derivative Test tells us that f s3d − 227
is a local minimum. Because f 0s0d − 0, the Second Derivative Test gives no information about the critical number 0. But since f 9sxd , 0 for x , 0 and also for
0 , x , 3, the First Derivative Test tells us that f does not have a local maximum or
minimum at 0.
Since f 0sxd − 0 when x − 0 or 2, we divide the real line into intervals with these
numbers as endpoints and complete the following chart.
FIGURE 12
y − x 4 2 4x 3
f 99sxd − 12 x sx 2 2d
s2`, 0d
s0, 2d
s2, `d
The point s0, 0d is an inflection point because the curve changes from concave
upward to concave downward there. Also s2, 216d is an inflection point because the
curve changes from concave downward to concave upward there.
The graph of y − x 4 2 4x 3 in Figure 12 supports our conclusions.
■ Curve Sketching
We now use the information we obtain from the first and second derivatives to sketch the
graph of a function.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.3What Derivatives Tell Us about the Shape of a Graph
EXAMPLE 7 Sketch the graph of the function f sxd − x 2y3s6 2 xd1y3.
Use the differentiation rules to check
these calculations.
SOLUTION First note that the domain of f is R. Calculation of the first two derivatives
f 9sxd − 1y3
x s6 2 xd2y3
f 0sxd −
x s6 2 xd5y3
Since f 9sxd − 0 when x − 4 and f 9sxd does not exist when x − 0 or x − 6, the critical
numbers are 0, 4, and 6.
x 1y3
s6 2 xd2y3
f 9sxd
decreasing on s2`, 0d
increasing on s0, 4d
decreasing on s4, 6d
decreasing on s6, `d
Try reproducing the graph in Fig­ure 13
with a graphing calculator or computer. Some machines produce the
complete graph, some produce only
the portion to the right of the y-axis,
and some produce only the portion
between x − 0 and x − 6. For an
explanation, see Example 7 in Graphing Calculators and Computers at
FIGURE 13
To find the local extreme values we use the First Derivative Test. Since f 9 changes
from negative to positive at 0, f s0d − 0 is a local minimum. Since f 9 changes from posi­tive to negative at 4, f s4d − 2 5y3 is a local maximum. The sign of f 9 does not change
at 6, so there is no minimum or maximum there. (The Second Derivative Test could be
used at 4 but not at 0 or 6 because f 0 does not exist at either of these numbers.)
Looking at the expression for f 0sxd and noting that x 4y3 > 0 for all x, we have
f 0sxd , 0 for x , 0 and for 0 , x , 6 and f 0sxd . 0 for x . 6. So f is concave
downward on s2`, 0d and s0, 6d and concave upward on s6, `d, and the only inflection
point is s6, 0d. Using all of the information we gathered about f from its first and
second derivatives, we sketch the graph in Figure 13. Note that the curve has vertical
tangents at s0, 0d and s6, 0d because f 9sxd l ` as x l 0 and as x l 6.
(4, 2%?# )
y=x @ ?#(6-x)!?#
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
1–2 Use the given graph of f to find the following.
(a) The open intervals on which f is increasing.
(b) The open intervals on which f is decreasing.
(c) The open intervals on which f is concave upward.
(d) The open intervals on which f is concave downward.
(e) The coordinates of the points of inflection.
3. Suppose you are given a formula for a function f.
(a)How do you determine where f is increasing or
(b)How do you determine where the graph of f is concave
upward or concave downward?
(c) How do you locate inflection points?
4. (a) State the First Derivative Test.
(b)State the Second Derivative Test. Under what circum­
stances is it inconclusive? What do you do if it fails?
5–6 The graph of the derivative f 9 of a function f is shown.
(a) On what intervals is f increasing? Decreasing?
(b)At what values of x does f have a local maximum? Local
8. T
he graph of the first derivative f 9 of a function f is shown.
(a) On what intervals is f increasing? Explain.
(b)At what values of x does f have a local maximum or
minimum? Explain.
(c)On what intervals is f concave upward or concave
down­ward? Explain.
(d)What are the x-coordinates of the inflection points of f ?
9–14 Find the intervals on which f is increasing or decreasing,
and find the local maximum and minimum values of f.
9. f sxd − 2x 3 2 15x 2 1 24x 2 5
10. f sxd − x 3 2 6x 2 2 135x
11. f sxd − 6x 4 2 16x 3 1 1
13. f sxd −
x 2 2 24
12. f sxd − x 2y3sx 2 3d
14. f sxd − x 1
15–18 Find the intervals on which f is concave upward or
concave downward, and find the inflection points of f .
15. f sxd − x 3 2 3x 2 2 9x 1 4
16. f sxd − 2x 3 2 9x 2 1 12x 2 3
17. f sxd − sin2x 2 cos 2x, 0 < x < y=fª(x)
18. f sxd − x s
7. I n each part use the given graph to state the x-coordinates of
the inflection points of f. Give reasons for your answers.
(a) The curve is the graph of f .
(b) The curve is the graph of f 9.
(c) The curve is the graph of f 0.
(a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f .
(c) Find the intervals of concavity and the inflection points.
19. f sxd − x 4 2 2x 2 1 3
20. f sxd − 2
x 11
21. f sxd − sin x 1 cos x, 0 < x < 2
22. f sxd − cos2 x 2 2 sin x,0 < x < 2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.3What Derivatives Tell Us about the Shape of a Graph
23–24 Find the local maximum and minimum values of f using
both the First and Second Derivative Tests. Which method do
you prefer?
23. f sxd − 1 1 3x 2 2 2x 3
24. f sxd −
36. T
he graph of a function y − f sxd is shown. At which point(s)
are the following true?
d 2y
are both positive.
dx 2
d 2y
(b) and
are both negative.
dx 2
25. Suppose the derivative of a function f is
f 9sxd − sx 2 4d2sx 1 3d7sx 2 5d8
d 2y
is negative but
is positive.
dx 2
On what interval(s) is f increasing?
26. (a) Find the critical numbers of f sxd − x 4sx 2 1d3.
(b)What does the Second Derivative Test tell you about the
behavior of f at these critical numbers?
(c) What does the First Derivative Test tell you?
27. Suppose f 0 is continuous on s2`, `d.
(a)If f 9s2d − 0 and f 0s2d − 25, what can you say about f ?
(b)If f 9s6d − 0 and f 0s6d − 0, what can you say about f ?
28–35 Sketch the graph of a function that satisfies all of the given
28. (a) f 9sxd , 0 and f 0sxd , 0 for all x
(b) f 9sxd . 0 and f 0sxd . 0 for all x
29. (a) f 9sxd . 0 and f 0sxd , 0 for all x
(b) f 9sxd , 0 and f 0sxd . 0 for all x
30. Vertical asymptote x − 0,f 9sxd . 0 if x , 22,
f 9sxd , 0 if x . 22 sx ± 0d,
f 0sxd , 0 if x , 0,f 0sxd . 0 if x . 0
33. f 9s5d − 0,f 9sxd , 0 when x , 5,
f 9sxd . 0 when x . 5,f 99s2d − 0,f 99s8d − 0,
f 99sxd , 0 when x , 2 or x . 8,
f 99sxd . 0 for 2 , x , 8
34. f 9s0d − f 9s4d − 0,f 9sxd − 1 if x , 21,
f 9sxd . 0 if 0 , x , 2,
f 9sxd , 0 if 21 , x , 0 or 2 , x , 4 or x . 4,
lim2 f 9sxd − `,lim1 f 9sxd − 2`,
x l2
x l2
f 99sxd . 0 if 21 , x , 2 or 2 , x , 4,f 0sxd , 0 if x . 4
35. f s0d − f 9s0d − f 9s2d − f 9s4d − f 9s6d − 0,
f 9sxd . 0 if 0 , x , 2 or 4 , x , 6,
f 9sxd , 0 if 2 , x , 4 or x . 6,
f 0sxd . 0 if 0 , x , 1 or 3 , x , 5,
f 0sxd , 0 if 1 , x , 3 or x . 5,f s2xd − f sxd
37–38 The graph of the derivative f 9 of a continuous function f
is shown.
(a) On what intervals is f increasing? Decreasing?
(b) At what values of x does f have a local maximum? Local
(c) On what intervals is f concave upward? Concave downward?
(d) State the x-coordinate(s) of the point(s) of inflection.
(e) Assuming that f s0d − 0, sketch a graph of f .
31. f 9s0d − f 9s2d − f 9s4d − 0,f 9sxd . 0 if x , 0 or 2 , x , 4,f 9sxd , 0 if 0 , x , 2 or x . 4,f 0sxd . 0 if 1 , x , 3,f 0sxd , 0 if x , 1 or x . 3
32. f 9sxd . 0 for all x ± 1,vertical asymptote x − 1,
f 99sxd . 0 if x , 1 or x . 3,f 99sxd , 0 if 1 , x , 3
8 x
8 x
(a) Find the intervals of increase or decrease.
(b) Find the local maximum and minimum values.
(c) Find the intervals of concavity and the inflection points.
(d) Use the information from parts (a) – (c) to sketch the graph.
You may want to check your work with a graphing calculator
or computer.
39. f sxd − x 3 2 3x 2 1 4
40. f sxd − 36x 1 3x 2 2 2x 3
41. f sxd − 12 x 4 2 4x 2 1 3
42. tsxd − 200 1 8x 3 1 x 4
43. tstd − 3t 4 2 8t 3 1 12
44. hsxd − 5x 3 2 3x 5
45. f szd − z 7 2 112z 2
46. f sxd − sx 2 2 4d3
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
47. Fsxd − x s6 2 x
48. G
­ sxd − 5x 2y3 2 2x 5y3
49. Csxd − x 1y3sx 1 4d
50. f sxd − 2 sx 2 4x 2
51. f sd − 2 cos 1 cos 2,0 < < 2
52. Ssxd − x 2 sin x,0 < x < 4
53–54 Use the methods of this section to sketch several
members of the given family of curves. What do the members
have in common? How do they differ from each other?
53. f sxd − x 4 2 cx,
54. f sxd − x 2 3c x 1 2c 3,
; 55–56
(a)Use a graph of f to estimate the maximum and minimum
values. Then find the exact values.
(b)Estimate the value of x at which f increases most rapidly.
Then find the exact value.
55. f sxd −
sx 2 1 1
56. f sxd − x 1 2 cos x,0 < x < 2
; 57–58
(a)Use a graph of f to give a rough estimate of the intervals of
concavity and the coordinates of the points of inflection.
(b) Use a graph of f 0 to give better estimates.
62. I n an episode of The Simpsons television show, Homer
reads from a newspaper and announces “Here’s good news!
According to this eye-catching article, SAT scores are
declining at a slower rate.” Interpret Homer’s statement in
terms of a function and its first and second derivatives.
63. T
he president announces that the national deficit is increasing, but at a decreasing rate. Interpret this statement in
terms of a function and its first and second derivatives.
64. L
et f std be the temperature at time t where you live and
suppose that at time t − 3 you feel uncomfortably hot. How
do you feel about the given data in each case?
(a) f 9s3d − 2, f 0s3d − 4
(b) f 9s3d − 2, f 0s3d − 24
(c) f 9s3d − 22, f 0s3d − 4
(d) f 9s3d − 22, f 0s3d − 24
65. Let Kstd be a measure of the knowledge you gain by
studying for a test for t hours. Which do you think is larger,
Ks8d 2 Ks7d or Ks3d 2 Ks2d? Is the graph of K concave
upward or concave downward? Why?
66. C
offee is being poured into the mug shown in the figure
at a constant rate (measured in volume per unit time).
Sketch a rough graph of the depth of the coffee in the mug
as a function of time. Account for the shape of the graph
in terms of concavity. What is the significance of the
inflection point?
57. f sxd − sin 2x 1 sin 4x,0 < x < 58. f sxd − sx 2 1d2 sx 1 1d3
59–60 Estimate the intervals of concavity to one decimal place
by using a computer algebra system to compute and graph f 0.
59. f sxd −
x4 1 x3 1 1
60. f sxd −
sx 2 1 x 1 1
sx 1 1d3sx 2 1 5d
sx 3 1 1dsx 2 1 4d
61. A
graph of a population of yeast cells in a new laboratory
culture as a function of time is shown.
yeast cells
68. S
how that the curve y − s1 1 xdys1 1 x 2 d has three points
of inflection and they all lie on one straight line.
69. S
how that the inflection points of the curve y − x sin x lie
on the curve y 2sx 2 1 4d − 4x 2.
70–72 Assume that all of the functions are twice differentiable
and the second derivatives are never 0.
67. F
ind a cubic function f sxd − ax 3 1 bx 2 1 cx 1 d that
has a local maximum value of 3 at x − 22 and a local minimum value of 0 at x − 1.
Time (in hours)
(a) Describe how the rate of population increase varies.
(b) When is this rate highest?
(c)On what intervals is the population function concave
upward or downward?
(d) Estimate the coordinates of the inflection point.
70. (a)If f and t are concave upward on an interval I, show
that f 1 t is concave upward on I.
(b)If f is positive and concave upward on I, show that the
function tsxd − f f sxdg 2 is concave upward on I.
71. (a)If f and t are positive, increasing, concave upward
functions on an interval I, show that the product function f t is concave upward on I.
(b)Show that part (a) remains true if f and t are both
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.4Limits at Infinity; Horizontal Asymptotes
(c)Suppose f is increasing and t is decreasing. Show, by
giving three examples, that f t may be concave upward,
concave downward, or linear. Why doesn’t the argument in parts (a) and (b) work in this case?
72. S
uppose f and t are both concave upward on s2`, `d.
Under what condition on f will the composite function
hsxd − f s tsxdd be concave upward?
73. S
how that a cubic function (a third-degree polynomial)
always has exactly one point of inflection. If its graph has
three x-intercepts x 1 , x 2 , and x 3 , show that the x-coordinate
of the inflection point is sx 1 1 x 2 1 x 3 dy3.
or what values of c does the polynomial
; 74. F
Psxd − x 4 1 cx 3 1 x 2 have two inflection points? One
inflection point? None? Illustrate by graphing P for several
values of c. How does the graph change as c decreases?
75. P
rove that if sc, f scdd is a point of inflection of the graph
of f and f 0 exists in an open interval that contains c, then
f 0scd − 0. [Hint: Apply the First Derivative Test and
Fermat’s Theorem to the function t − f 9.]
76. S
how that if f sxd − x 4, then f 0s0d − 0, but s0, 0d is not an
inflection point of the graph of f .
| |
77. S
how that the function tsxd − x x has an inflection point
at s0, 0d but t0s0d does not exist.
78. Suppose that f 09 is continuous and f 9scd − f 0scd − 0, but
f -scd . 0. Does f have a local maximum or minimum
at c ? Does f have a point of inflection at c?
79. S
uppose f is differentiable on an interval I and f 9sxd . 0
for all numbers x in I except for a single number c. Prove
that f is increasing on the entire interval I.
80. For what values of c is the function
f sxd − cx 1
x2 1 3
increasing on s2`, `d?
81. T
he three cases in the First Derivative Test cover the situations commonly encountered but do not exhaust all pos­
sibilities. Consider the functions f, t, and h whose values
at 0 are all 0 and, for x ± 0,
f sxd − x 4 sin
tsxd − x 4 2 1 sin
hsxd − x 4 22 1 sin
(a)Show that 0 is a critical number of all three functions
but their derivatives change sign infinitely often on both
sides of 0.
(b)Show that f has neither a local maximum nor a local
min­imum at 0, t has a local minimum, and h has a local
3.4 Limits at Infinity; Horizontal Asymptotes
In Sections 1.5 and 1.7 we investigated infinite limits and vertical asymptotes of a curve
y − f sxd. There we let x approach a number and the result was that the values of y
became arbitrarily large (positive or negative). In this section we let x become arbitrarily
large ( positive or negative) and see what happens to y. We will find it very useful to
consider this so-called end behavior when sketching graphs.
■ Limits at Infinity and Horizontal Asymptotes
Let’s begin by investigating the behavior of the function f defined by
f sxd −
f sxd
x2 2 1
x2 1 1
as x becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of f has been drawn by a computer in Figure 1.
FIGURE 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
You can see that as x grows larger and larger, the values of f sxd get closer and closer
to 1. (The graph of f approaches the horizontal line y − 1 as we look to the right.) In
fact, it seems that we can make the values of f sxd as close as we like to 1 by taking x
sufficiently large. This situation is expressed symbolically by writing
x2 2 1
x2 1 1
In general, we use the notation
lim f sxd − L
to indicate that the values of f sxd approach L as x becomes larger and larger.
1 Intuitive Definition of a Limit at Infinity Let f be a function defined on
some interval sa, `d. Then
lim f sxd − L
means that the values of f sxd can be made arbitrarily close to L by requiring x to
be sufficiently large.
Another notation for lim x l ` f sxd − L is
f sxd l Lasx l `
The symbol ` does not represent a number. Nonetheless, the expression lim f sxd − L is
x l`
often read as
“the limit of f sxd, as x approaches infinity, is L”
“the limit of f sxd, as x becomes infinite, is L”
“the limit of f sxd, as x increases without bound, is L”
The meaning of such phrases is given by Definition 1. A more precise definition, similar
to the «, definition of Section 1.7, is given at the end of this section.
Geometric illustrations of Definition 1 are shown in Figure 2. Notice that there are
many ways for the graph of f to approach the line y − L (which is called a horizontal
asymptote) as we look to the far right of each graph.
Referring back to Figure 1, we see that for numerically large negative values of x, the
values of f sxd are close to 1. By letting x decrease through negative values without
bound, we can make f sxd as close to 1 as we like. This is expressed by writing
x l2`
x2 2 1
x2 1 1
The general definition is as follows.
2 Definition Let f be a function defined on some interval s2`, ad. Then
lim f sxd − L
FIGURE 2
x l 2`
Examples illustrating lim f sxd − L
means that the values of f sxd can be made arbitrarily close to L by requiring x to
be sufficiently large negative.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.4Limits at Infinity; Horizontal Asymptotes
Again, the symbol 2` does not represent a number, but the expression lim f sxd − L
x l 2`
is often read as
“the limit of f sxd, as x approaches negative infinity, is L”
Definition 2 is illustrated in Figure 3. Notice that the graph approaches the line y − L as
we look to the far left of each graph.
3 Definition The line y − L is called a horizontal asymptote of the curve
y − f sxd if either
lim f sxd − Lorlim f sxd − L
x l`
FIGURE 3
Examples illustrating lim f sxd − L
x l 2`
x l 2`
For instance, the curve illustrated in Figure 1 has the line y − 1 as a horizontal asymp­
tote because
x2 2 1
lim 2
xl` x 1 1
The curve y − f sxd sketched in Figure 4 has both y − 21 and y − 2 as horizontal asymptotes because
lim f sxd − 21
lim f sxd − 2
x l2`
FIGURE 4
EXAMPLE 1 Find the infinite limits, limits at infinity, and asymptotes for the function
f whose graph is shown in Figure 5.
SOLUTION We see that the values of f sxd become large as x l 21 from both sides, so
lim f sxd − `
x l21
Notice that f sxd becomes large negative as x approaches 2 from the left, but large
positive as x approaches 2 from the right. So
lim f sxd − 2`andlim1 f sxd − `
x l 22
FIGURE 5
x l2
Thus both of the lines x − 21 and x − 2 are vertical asymptotes.
As x becomes large, it appears that f sxd approaches 4. But as x decreases through
negative values, f sxd approaches 2. So
lim f sxd − 4andlim f sxd − 2
x l2`
This means that both y − 4 and y − 2 are horizontal asymptotes.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
EXAMPLE 2 Find lim
and lim .
x l2` x
SOLUTION Observe that when x is large, 1yx is small. For instance,
− 0.01− 0.0001− 0.000001
FIGURE 6
− 0, lim
x l2` x
In fact, by taking x large enough, we can make 1yx as close to 0 as we please. Therefore, according to Definition 1, we have
Similar reasoning shows that when x is large negative, 1yx is small negative, so we also
x l2` x
It follows that the line y − 0 (the x-axis) is a horizontal asymptote of the curve
y − 1yx. (This is a hyperbola; see Figure 6.)
■ Evaluating Limits at Infinity
Most of the Limit Laws that were given in Section 1.6 also hold for limits at infinity. It
can be proved that the Limit Laws listed in Section 1.6 (with the exception of Laws 10 and
11) are also valid if “x l a” is replaced by “x l `” or “x l 2`.” In particular, if we
combine Laws 6 and 7 with the results of Example 2, we obtain the following important
rule for calculating limits.
4 Theorem If r . 0 is a rational number, then
If r . 0 is a rational number such that x r is defined for all x, then
x l2`
EXAMPLE 3 Evaluate the following limit and indicate which properties of limits are
used at each stage.
3x 2 2 x 2 2
x l ` 5x 2 1 4x 1 1
SOLUTION As x becomes large, both numerator and denominator become large, so it
isn’t obvious what happens to their ratio. We need to do some preliminary algebra.
To evaluate the limit at infinity of any rational function, we first divide both the
numerator and denominator by the highest power of x that occurs in the denominator.
(We may assume that x ± 0, since we are interested only in large values of x.) In this
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.4Limits at Infinity; Horizontal Asymptotes
case the highest power of x in the denominator is x 2, so we have
3x 2 2 x 2 2
3x 2 x 2 2
− lim
− lim
x l ` 5x 2 1 4x 1 1
x l ` 5x 2 1 4x 1 1
x l`
lim 5 1 1 2
lim 3 2
lim 3 2 lim
x l`
FIGURE 7
3x 2 x 2 2
5x 2 1 4x 1 1
(by Limit Law 5)
2 2 lim
x l`
lim 5 1 4 lim 1 lim
x l`
x l` x
x l`
x l`
(by 1, 2, and 3)
(by 8 and Theorem 4)
A similar calculation shows that the limit as x l 2` is also 35. Figure 7 illustrates the
results of these calculations by showing how the graph of the given rational function
approaches the horizontal asymptote y − 35 − 0.6.
EXAMPLE 4 Find the horizontal asymptotes of the graph of the function
f sxd −
s2x 2 1 1
3x 2 5
SOLUTION Dividing both numerator and denominator by x (which is the highest power
of x in the denominator) and using the properties of limits, we have
s2x 2 1 1
s2x 2 1 1
− lim
− lim
x l ` 3x 2 5
3x 2 5
S D
2x 2 1 1
(since sx 2 − x for x . 0)
3x 2 5
lim 3 2 5 lim
xl` x
lim 2 1 lim
s2 1 0
Therefore the line y − s2 y3 is a horizontal asymptote of the graph of f.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
In computing the limit as x l 2`, we must remember that for x , 0, we have
sx 2 − x − 2x. So when we divide the numerator by x, for x , 0 we get
| |
s2x 2 1 1
s2x 2 1 1
2 sx 2
y= 3
y=_ 3
s2x 2 1 1
− lim
x l2`
x l2`
3x 2 5
s 2x 2 1 1
3x 2 5
3 2 5 lim
x l2` x
2 1 lim
x l2`
EXAMPLE 5 Compute lim (sx 2 1 1 2 x).
x l`
We can think of the given function as
having a denominator of 1.
SOLUTION Because both sx 2 1 1 and x are large when x is large, it’s difficult to see
what happens to their difference, so we use algebra to rewrite the function. We first
multiply numerator and denominator by the conjugate radical:
lim (sx 2 1 1 2 x) − lim (sx 2 1 1 2 x) x l`
FIGURE 9
sx 2 1 1 1 x
sx 2 1 1 1 x
sx 1 1d 2 x
− lim
− lim
x l ` sx 2 1 1 1 x
x l ` sx 2 1 1 1 x
x l`
Thus the line y − 2s2 y3 is also a horizontal asymptote. See Figure 8.
FIGURE 8
2x 2 1 1
Notice that the denominator of this last expression (sx 2 1 1 1 x) becomes large as
x l ` (it’s bigger than x). So
lim (sx 2 1 1 2 x) − lim
x l`
x l ` sx 2 1 1 1 x
Figure 9 illustrates this result.
EXAMPLE 6 Evaluate lim sin .
PS The problem-solving strategy for
SOLUTION If we let t − 1yx, then t l 01 as x l `. Therefore
Example 6 is introducing something extra
(see Principles of Problem Solving following
Chapter 1). Here, the something extra, the
auxiliary aid, is the new variable t.
lim sin
− lim1 sin t − 0
(See Exercise 75.)
EXAMPLE 7 Evaluate lim sin x.
SOLUTION As x increases, the values of sin x oscillate between 1 and 21 infinitely often
and so they don’t approach any definite number. Thus lim x l` sin x does not exist.
■ Infinite Limits at Infinity
The notation
lim f sxd − `
x l`
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.4Limits at Infinity; Horizontal Asymptotes
is used to indicate that the values of f sxd become large as x becomes large. Similar meanings are attached to the following symbols:
lim f sxd − `lim f sxd − 2`lim f sxd − 2`
x l 2`
x l`
x l 2`
EXAMPLE 8 Find lim x 3 and lim x 3.
x l2`
SOLUTION When x becomes large, x 3 also becomes large. For instance,
10 3 − 1000100 3 − 1,000,0001000 3 − 1,000,000,000
In fact, we can make x 3 as big as we like by requiring x to be large enough. Therefore
we can write
lim x 3 − `
Similarly, when x is large negative, so is x 3. Thus
lim x 3 − 2`
x l2`
FIGURE 10
lim x 3 − `, lim x 3 − 2`
x l2`
These limit statements can also be seen from the graph of y − x 3 in Figure 10.
EXAMPLE 9 Find lim sx 2 2 xd.
x l`
SOLUTION Limit Law 2 says that the limit of a difference is the difference of the
limits, provided that these limits exist. We cannot use Law 2 here because
lim x 2 − `andlim x − `
x l`
x l`
In general, the Limit Laws can’t be applied to infinite limits because ` is not a number
(` 2 ` can’t be defined). However, we can write
lim sx 2 2 xd − lim xsx 2 1d − `
x l`
x l`
because both x and x 2 1 become arbitrarily large and so their product does too.
EXAMPLE 10 Find lim
x2 1 x
SOLUTION As in Example 3, we divide the numerator and denominator by the highest
power of x in the denominator, which is simply x:
x l`
x2 1 x
− lim
− 2`
because x 1 1 l ` and 3yx 2 1 l 0 2 1 − 21 as x l `.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
The next example shows that by using infinite limits at infinity, together with intercepts, we can get a rough idea of the graph of a polynomial without computing
EXAMPLE 11 Sketch the graph of y − sx 2 2d4sx 1 1d3sx 2 1d by finding its
intercepts and its limits as x l ` and as x l 2`.
SOLUTION The y-intercept is f s0d − s22d4s1d3s21d − 216 and the x-intercepts are
found by setting y − 0: x − 2, 21, 1. Notice that since sx 2 2d4 is never negative, the
function doesn’t change sign at 2; thus the graph doesn’t cross the x-axis at 2. The
graph crosses the axis at 21 and 1.
When x is large positive, all three factors are large, so
lim sx 2 2d4sx 1 1d3sx 2 1d − `
When x is large negative, the first factor is large positive and the second and third
factors are both large negative, so
lim sx 2 2d4sx 1 1d3sx 2 1d − `
x l2`
FIGURE 11
y − sx 2 2d4 sx 1 1d3 sx 2 1d
Combining this information, we give a rough sketch of the graph in Figure 11.
■ Precise Definitions
Definition 1 can be stated precisely as follows.
5 Precise Definition of a Limit at Infinity Let f be a function defined on
some interval sa, `d. Then
lim f sxd − L
means that for every « . 0 there is a corresponding number N such that
ifx . Nthenf sxd 2 L , «
In words, this says that the values of f sxd can be made arbitrarily close to L (within a
distance «, where « is any positive number) by requiring x to be sufficiently large (larger
than N, where N depends on «). Graphically, it says that by keeping x large enough
(larger than some number N) we can make the graph of f lie between the given horizontal lines y − L 2 « and y − L 1 « as in Figure 12. This must be true no matter how
small we choose «.
L ∑
FIGURE 12
lim f sxd − L
ƒ is
in here
when x is in here
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.4Limits at Infinity; Horizontal Asymptotes
Figure 13 shows that if a smaller value of « is chosen, then a larger value of N may
be required.
FIGURE 13
lim f sxd − L
Similarly, a precise version of Definition 2 is given by Definition 6, which is illustrated in Figure 14.
6 Definition Let f be a function defined on some interval s2`, ad. Then
lim f sxd − L
x l 2`
means that for every « . 0 there is a corresponding number N such that
ifx , Nthenf sxd 2 L , «
FIGURE 14
lim f sxd − L
x l2`
In Example 3 we calculated that
3x 2 2 x 2 2
5x 2 1 4x 1 1
In the next example we use a calculator (or computer) to relate this statement to Definition 5 with L − 35 − 0.6 and « − 0.1.
EXAMPLE 12 Use a graph to find a number N such that
ifx . NthenZ
3x 2 2 x 2 2
2 0.6
5x 2 1 4x 1 1
, 0.1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
SOLUTION We rewrite the given inequality as
3x 2 2 x 2 2
, 0.7
5x 2 1 4x 1 1
0.5 ,
We need to determine the values of x for which the given curve lies between the horizontal lines y − 0.5 and y − 0.7. So we graph the curve and these lines in Figure 15. Then
we use the graph to estimate that the curve crosses the line y − 0.5 when x < 6.7. To the
right of this number it seems that the curve stays between the lines y − 0.5 and y − 0.7.
Rounding up to be safe, we can say that
FIGURE 15
ifx . 7thenZ
3x 2 2 x 2 2
2 0.6
5x 2 1 4x 1 1
, 0.1
In other words, for « − 0.1 we can choose N − 7 (or any larger number) in Defini­­tion 5.
− 0.
EXAMPLE 13 Use Definition 5 to prove that lim
SOLUTION Given « . 0, we want to find N such that
ifx . Nthenn
In computing the limit we may assume that x . 0. Then 1yx , « &? x . 1y«. Let’s
choose N − 1y«. So
ifx . N −
Therefore, by Definition 5,
Figure 16 illustrates the proof by showing some values of « and the corresponding
values of N.
FIGURE 16
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.4Limits at Infinity; Horizontal Asymptotes
Finally we note that an infinite limit at infinity can be defined as follows. The geometric illustration is given in Figure 17.
7 Precise Definition of an Infinite Limit at Infinity Let f be a function
defined on some interval sa, `d. Then
lim f sxd − `
FIGURE 17
means that for every positive number M there is a corresponding positive number
N such that
lim f sxd − `
ifx . Nthenf sxd . M
Similar definitions apply when the symbol ` is replaced by 2`. (See Exercise 76.)
1. E
xplain in your own words the meaning of each of the
(a) lim f sxd − 5
(b) lim f sxd − 3
x l 2`
2. (a)Can the graph of y − f sxd intersect a vertical asymptote? Can it intersect a horizontal asymptote? Illustrate
by sketching graphs.
(b)How many horizontal asymptotes can the graph
of y − f sxd have? Sketch graphs to illustrate the
3. For the function f whose graph is given, state the following.
(a) lim f sxd
(b) lim f sxd
x l`
; 5. Guess the value of the limit
x l 2`
(c) lim f sxd
x l`
(d) lim f sxd
x l1
x l3
; 6. (a) Use a graph of
S D
f sxd − 1 2
4. For the function t whose graph is given, state the following.
(a) lim tsxd
(b) lim tsxd
x l`
x l 2`
(c) lim tsxd
(d) lim2 tsxd
(e) lim1 tsxd
(f ) The equations of the asymptotes
x l2
by evaluating the function f sxd − x 2y2 x for x − 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 20, 50, and 100. Then use a graph of f to
support your guess.
(e) The equations of the asymptotes
to estimate the value of lim x l ` f sxd correct to two
decimal places.
(b)Use a table of values of f sxd to estimate the limit to
four decimal places.
7–8 Evaluate the limit and justify each step by indicating the
appropriate properties of limits.
7. lim
2x 2 2 7
5x 2 1 x 2 3
8. lim
9x 3 1 8x 2 4
3 2 5x 1 x 3
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
9–32 Find the limit or show that it does not exist.
9. lim
4x 1 3
5x 2 1
10. lim
3t 2 1 t
t 2 4t 1 1
11. lim
t l 2`
r 2 r3
13. lim
r l ` 2 2 r 2 1 3r 3
15. lim
4 2 sx
2 1 sx
17. lim
sx 1 3x 2
4x 2 1
19. lim
s1 1 4x
2 2 x3
5 1 4x
36. y −
2x 2 1 1
3x 1 2x 2 1
3x 3 2 8x 1 2
14. lim
x l ` 4x 3 2 5x 2 2 2
37. y −
2x 2 1 x 2 1
x2 1 x 2 2
38. y −
1 1 x4
x2 2 x4
su 2 1 1ds2u 2 2 1d
16. lim
u l 2`
su 2 1 2d2
39. y −
x3 2 x
x 2 6x 1 5
40. y −
t l 2`
x l 2`
6t 2 1 t 2 5
9 2 2t 2
s1 1 4x
2 2 x3
22. lim
q 1 6q 2 4
4q 2 2 3q 1 3
23. lim cos x
1 1 x6
24. lim 4
x l 2` x 1 1
25. lim (s25t 2 1 2 2 5t)
x l`
27. lim (sx 2 1 ax 2 sx 2 1 bx
x l`
28. lim ( x 2 sx
lim (s4x 2 1 3x 1 2 x )
x l2`
sin x
x2 1 1
32. lim sx sin
; 33. (a) Estimate the value of
f sxd − s3x 2 1 8x 1 6 2 s3x 2 1 3x 1 1
to estimate the value of lim x l ` f sxd to one decimal
(b)Use a table of values of f sxd to estimate the limit to
four decimal places.
(c)Find the exact value of the limit.
s2x 2 1 1
s2x 2 1 1
x l 2`
3x 2 5
3x 2 5
(b)By calculating values of f sxd, give numerical estimates
of the limits in part (a).
(c)Calculate the exact values of the limits in part (a). Did
you get the same value or different values for these two
limits? [In view of your answer to part (a), you might
have to check your calculation for the second limit.]
43. Let P and Q be polynomials. Find
lim (sx 2 1 x 1 1 1 x)
; 34. (a) Use a graph of
s2x 2 1 1
3x 2 5
How many horizontal and vertical asymptotes do you
observe? Use the graph to estimate the values of the
x l 2`
by graphing the function f sxd − sx 2 1 x 1 1 1 x.
(b)Use a table of values of f sxd to guess the value of the
(c)Prove that your guess is correct.
3x 3 1 500x 2
x 1 500x 2 1 100x 1 2000
f sxd −
x l`
30. lim
s4x 1 3x 1 2
; 42. (a) Graph the function
x l 2`
by graphing f for 210 < x < 10. Then calculate the equation of the asymptote by evaluating the limit. How do you
explain the discrepancy?
29. lim sx 2 1 2x 7 d
31. lim x sin
f sxd −
2x 2 x
x4 1 3
; 41. Estimate the horizontal asymptote of the function
s2t 2 2 1
x l 2`
21. lim
3x 1 7
35. y −
18. lim
35–40 Find the horizontal and vertical asymptotes of each
curve. You may want to use a graphing calculator (or computer)
to check your work by graphing the curve and estimating the
if the degree of P is (a) less than the degree of Q and
(b) greater than the degree of Q.
44. M
ake a rough sketch of the curve y − x n (n an integer)
for the following five cases:
(i) n − 0
(iii) n . 0, n even
(v) n , 0, n even
(ii) n . 0, n odd
(iv) n , 0, n odd
Then use these sketches to find the following limits.
(a) lim1 x n
(b) lim2 x n
(c) lim x n
x l0
x l`
x l0
lim x n
x l 2`
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.4Limits at Infinity; Horizontal Asymptotes
61. f s1d − f 9s1d − 0,lim x l21 f sxd − `,45. F
ind a formula for a function f that satisfies the following
lim x l22 f sxd − 2`,lim x l 0 f sxd − 2`,lim f sxd − 0,lim f sxd − 2`,f s2d − 0,
lim x l2` f sxd − `,lim x l ` f sxd − 0,
lim f sxd − `,lim1 f sxd − 2`
f 0sxd . 0 for x . 2,f 0sxd , 0 for x , 0 and for
x l 6`
x l0
x l 32
x l3
46. F
ind a formula for a function that has vertical asymptotes
x − 1 and x − 3 and horizontal asymptote y − 1.
62.ts0d − 0,t0sxd , 0 for x ± 0,lim x l2` tsxd − `,
47. A function f is a ratio of quadratic functions and has a
vertical asymptote x − 4 and just one x-intercept, x − 1.
It is known that f has a removable discontinuity at x − 21
and lim x l21 f sxd − 2. Evaluate
(a)f s0d
(b) lim f sxd
lim x l ` tsxd − 2`,lim x l 02 t9sxd − 2`,
48–51 Find the horizontal asymptotes of the curve and use
them, together with concavity and intervals of increase and
decrease, to sketch the curve.
63. (a)Use the Squeeze Theorem to evaluate lim
lim x l 01 t9sxd − `
48. y −
50. y −
1 1 2x 2
1 1 x2
sx 2 1 1
49. y −
51. y −
x2 1 1
52–56 Find the limits as x l ` and as x l 2`. Use this information, together with intercepts, to give a rough sketch of the
graph as in Example 11.
52. y − 2x 3 2 x 4
53. y − x 4 2 x 6
54. y − x 3sx 1 2d 2sx 2 1d
sin x
(b)Graph f sxd − ssin xdyx. How many times does the
graph cross the asymptote?
; 64. E nd Behavior of a Function By the end behavior of a
function we mean the behavior of its values as x l ` and
as x l 2`.
(a)Describe and compare the end behavior of the functions
Psxd − 3x 5 2 5x 3 1 2xQsxd − 3x 5
by graphing both functions in the viewing rectangles
f22, 2g by f22, 2g and f210, 10g by f210,000, 10,000g.
(b)Two functions are said to have the same end behavior if
their ratio approaches 1 as x l `. Show that P and Q
have the same end behavior.
65. Find lim x l ` f sxd if
55. y − s3 2 xds1 1 xd 2s1 2 xd 44x 2 1
4x 2 1 3x
, f sxd ,
56. y − x 2sx 2 2 1d 2sx 1 2d
57–62 Sketch the graph of a function that satisfies all of the
given conditions.
57. f s2d − 4,f s22d − 24,lim f sxd − 0,lim f sxd − 2
x l 2`
x l`
58. lim f sxd − 2`,lim 2 f sxd − `,lim 1 f sxd − 2`,
x l 2`
x l 22
x l 22
for all x . 5.
66. (a)A tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water is pumped into the
tank at a rate of 25 Lymin. Show that the concentration
of salt after t minutes (in grams per liter) is
lim f sxd − `,lim f sxd − `
x l2
Cstd −
x l`
59. f 9s2d − 0,f s2d − 21,f s0d − 0,
f 9sxd , 0 if 0 , x , 2,f 9sxd . 0 if x . 2,
f 0sxd , 0 if 0 < x , 1 or if x . 4,
f 0sxd . 0 if 1 , x , 4,lim x l ` f sxd − 1,
f s2xd − f sxd for all x
60.f 9s2d − 0,f 9s0d − 1,f 9sxd . 0 if 0 , x , 2,
f 9sxd , 0 if x . 2,f 0sxd , 0 if 0 , x , 4,
f 0sxd . 0 if x . 4,lim x l ` f sxd − 0,
f s2xd − 2f sxd for all x
200 1 t
(b) What happens to the concentration as t l ` ?
; 67. Use a graph to find a number N such that
ifx . NthenZ
3x2 1 1
2 1.5 , 0.05
2x 2 1 x 1 1
; 68. For the limit
1 2 3x
sx 2 1 1
− 23
illustrate Definition 5 by finding values of N that correspond
to « − 0.1 and « − 0.05.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
; 69. For the limit
73. Use Definition 6 to prove that lim
1 2 3x
sx 2 1 1
illustrate Definition 6 by finding values of N that correspond
to « − 0.1 and « − 0.05.
x l2`
− 0.
x l2`
; 70. For the limit
74. Prove, using Definition 7, that lim x 3 − `.
75. (a) Prove that
lim f sxd − lim1 f s1ytd
sx 2 3
illustrate Definition 7 by finding a value of N that corresponds to M − 100.
71. ( a)How large do we have to take x so that 1yx , 0.0001?
(b) Taking r − 2 in Theorem 4, we have the statement
assuming that these limits exist.
lim x sin
x l 01
Prove this directly using Definition 5.
76. Formulate a precise definition of
72. (a)How large do we have to take x so that 1ysx , 0.0001?
(b) Taking r − 12 in Theorem 4, we have the statement
f s1ytd
lim f sxd − t lim
l 02
(b) Use part (a) and Exercise 63 to find
lim f sxd − 2`
x l2`
Then use your definition to prove that
Prove this directly using Definition 5.
lim s1 1 x 3 d − 2`
x l2`
3.5 Summary of Curve Sketching
FIGURE 1
FIGURE 2
So far we have been concerned with some particular aspects of curve sketching: domain,
range, symmetry, limits, continuity, and vertical asymptotes in Chapter 1; deriva­tives and
tangents in Chapter 2; and extreme values, intervals of increase and decrease, concavity,
points of inflection, and horizontal asymptotes in this chapter. It is now time to put all of
this information together to sketch graphs that reveal the important features of functions.
You might ask: Why don’t we just use a graphing calculator or computer to graph a
curve? Why do we need to use calculus?
It’s true that technology is capable of producing very accurate graphs. But even the
best graphing devices have to be used intelligently. It is easy to arrive at a misleading
graph, or to miss important details of a curve, when relying solely on technology. (See
the topic Graphing Calculators and Computers at www.StewartCalculus.com, especially
Examples 1, 3, 4, and 5. See also Section 3.6.) The use of calculus enables us to discover
the most interesting aspects of graphs and in many cases to calculate maximum and
minimum points and inflection points exactly instead of approximately.
For instance, Figure 1 shows the graph of f sxd − 8x 3 2 21x 2 1 18x 1 2. At first
glance it seems reasonable: it has the same shape as cubic curves like y − x 3, and it
appears to have no maximum or minimum point. But if you compute the derivative, you
will see that there is a maximum when x − 0.75 and a minimum when x − 1. Indeed,
if we zoom in to this portion of the graph, we see that behavior exhibited in Figure 2.
Without calculus, we could easily have overlooked it.
In the next section we will graph functions by using the interaction between calculus
and technology. In this section we draw graphs (by hand) by first considering the following information. A graph produced by a calculator or computer can serve as a check on
your work.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.5Summary of Curve Sketching
■ Guidelines for Sketching a Curve
The following checklist is intended as a guide to sketching a curve y − f sxd by hand. Not
every item is relevant to every function. (For instance, a given curve might not have an
asymptote or possess symmetry.) But the guidelines provide all the information you need
to make a sketch that displays the most important aspects of the function.
A. Domain It’s often useful to start by determining the domain D of f , that is, the set
of values of x for which f sxd is defined.
(a) Even function: reflectional symmetry
(b) Odd function: rotational symmetry
FIGURE 3
B.Intercepts The y-intercept is f s0d and this tells us where the curve intersects the
y-axis. To find the x-intercepts, we set y − 0 and solve for x. (You can omit this
step if the equa­tion is difficult to solve.)
(i) If f s2xd − f sxd for all x in D, that is, the equation of the curve is unchanged
when x is replaced by 2x, then f is an even function and the curve is symmetric about
the y-axis. (See Section 1.1.) This means that our work is cut in half. If we know what
the curve looks like for x > 0, then we need only reflect about the y-axis to obtain the
complete curve [see Figure 3(a)]. Here are some examples: y − x 2, y − x 4, y − x ,
and y − cos x.
(ii) If f s2xd − 2f sxd for all x in D, then f is an odd function and the curve
is sym­metric about the origin. Again we can obtain the complete curve if we know
what it looks like for x > 0. [Rotate 180° about the origin; see Figure 3(b).] Some
simple examples of odd functions are y − x, y − x 3, y − 1yx, and y − sin x.
(iii) If f sx 1 pd − f sxd for all x in D, where p is a positive constant, then f is a
periodic function and the smallest such number p is called the period. For instance,
y − sin x has period 2 and y − tan x has period . If we know what the graph looks
like in an interval of length p, then we can use translation to visualize the entire graph
(see Figure 4).
| |
period p
FIGURE 4
Periodic function:
translational symmetry
D. Asymptotes
(i) Horizontal Asymptotes. Recall from Section 3.4 that if either lim x l ` f sxd − L
or lim x l2 ` f sxd − L, then the line y − L is a horizontal asymptote of the curve
y − f sxd. If it turns out that lim x l ` f sxd − ` (or 2`), then we do not have an
asymptote to the right, but this fact is still useful information for sketching the curve.
(ii) Vertical Asymptotes. Recall from Section 1.5 that the line x − a is a vertical
asymptote if at least one of the following statements is true:
lim f sxd − `
x l a1
lim f sxd − 2`
x l a1
lim f sxd − `
x l a2
lim f sxd − 2`
x l a2
(For rational functions you can locate the vertical asymptotes by equating the denominator to 0 after canceling any common factors. But for other functions this method
does not apply.) Furthermore, in sketching the curve it is useful to know exactly
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
which of the statements in (1) is true. If f sad is not defined but a is an endpoint of the
domain of f, then you should compute lim xl a f sxd or lim xl a f sxd, whether or not
this limit is infinite.
(iii) Slant Asymptotes. These are discussed at the end of this section.
E.Intervals of Increase or Decrease Use the I/D Test. Compute f 9sxd and find the
intervals on which f 9sxd is positive (f is increasing) and the intervals on which f 9sxd
is negative (f is decreasing).
F. L
ocal Maximum or Minimum Values Find the critical numbers of f [the numbers c where f 9scd − 0 or f 9scd does not exist]. Then use the First Derivative Test. If
f 9 changes from positive to negative at a critical number c, then f scd is a local maximum. If f 9 changes from negative to positive at c, then f scd is a local minimum.
Although it is usually prefer­able to use the First Derivative Test, you can use the
Second Derivative Test if f 9scd − 0 and f 0scd ± 0. Then f 0scd . 0 implies that
f scd is a local minimum, whereas f 0scd , 0 implies that f scd is a local maximum.
G.Concavity and Points of Inflection Compute f 0sxd and use the Concavity Test.
The curve is concave upward where f 0sxd . 0 and concave downward where
f 0sxd , 0. Inflection points occur where the direction of concavity changes.
H.Sketch the Curve Using the information in items A–G, draw the graph. Sketch the
asymptotes as dashed lines. Plot the intercepts, maximum and minimum points, and
inflection points. Then make the curve pass through these points, rising and falling
according to E, with concavity according to G, and approaching the asymptotes.
If additional accuracy is desired near any point, you can compute the value of the
derivative there. The tangent indicates the direction in which the curve proceeds.
EXAMPLE 1 Use the guidelines to sketch the curve y −
A. Domain The domain is
2 1 ± 0j − hx
2x 2
x2 2 1
| x ± 61j − s2`, 21d ø s21, 1d ø s1, `d
B. Intercepts The x- and y-intercepts are both 0.
C. Symmetry Since f s2xd − f sxd, the function f is even. The curve is symmetric
about the y-axis.
2x 2
D. Asymptotes
lim 2
− lim
x l6` x 2 1
x l6` 1 2 1yx 2
Therefore the line y − 2 is a horizontal asymptote (at both the left and right).
Since the denominator is 0 when x − 61, we compute the following limits:
x l1
FIGURE 5
Preliminary sketch
We have shown the curve approaching its horizontal asymptote from
above in Figure 5. This is confirmed by
the intervals of increase and decrease.
2x 2
x2 2 1
2x 2
− 2`
x 21
lim 2
2x 2
x2 2 1
x l1
lim 1
x l 21
− 2`
x 21
x l 21
Therefore the lines x − 1 and x − 21 are vertical asymptotes. This information
about limits and asymptotes enables us to draw the preliminary sketch in Figure 5,
showing the parts of the curve near the asymptotes.
E. Intervals of Increase or Decrease
f 9sxd −
sx 2 2 1ds4xd 2 2x 2 2x
− 2
sx 2 1d
sx 2 1d2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.5Summary of Curve Sketching
Since f 9sxd . 0 when x , 0 sx ± 21d and f 9sxd , 0 when x . 0 sx ± 1d, f is
increasing on s2`, 21d and s21, 0d and decreasing on s0, 1d and s1, `d.
F. Local Maximum or Minimum Values The only critical number is x − 0. Since f 9
changes from positive to negative at 0, f s0d − 0 is a local maximum by the First
Derivative Test.
G. Concavity and Points of Inflection
f 0sxd −
Since 12x 2 1 4 . 0 for all x, we have
FIGURE 6
Finished sketch of y −
2x 2
x 21
| |
f 0sxd . 0&?x 2 2 1 . 0&?x . 1
sx 2 2 1d2 s24d 1 4x 2sx 2 2 1d2x
12x 2 1 4
− 2
sx 2 1d
sx 2 1d3
| |
and f 0sxd , 0 &? x , 1. Thus the curve is concave upward on the intervals
s2`, 21d and s1, `d and concave downward on s21, 1d. It has no point of inflection because 1 and 21 are not in the domain of f.
H. Sketch the Curve Using the information in E–G, we finish the sketch in Figure 6. n
EXAMPLE 2 Sketch the graph of f sxd −
A. Domain The domain is hx
sx 1 1
| x 1 1 . 0j − hx | x . 21j − s21, `d.
B. Intercepts The x- and y-intercepts are both 0.
C. Symmetry None
D. Asymptotes Since
sx 1 1
there is no horizontal asymptote. Since sx 1 1 l 0 as x l 211 and f sxd is
always positive, we have
lim 1
x l 21 sx 1 1
and so the line x − 21 is a vertical asymptote.
E. Intervals of Increase or Decrease
f 9sxd −
3x 2 1 4x
xs3x 1 4d
sx 1 1 s2xd 2 x 2 1y( 2sx 1 1 )
3y2 −
2sx 1 1d
2sx 1 1d3y2
e see that f 9sxd − 0 when x − 0 (notice that 243 is not in the domain of f ), so the
only critical number is 0. Since f 9sxd , 0 when 21 , x , 0 and f 9sxd . 0 when
x . 0, f is decreasing on s21, 0d and increasing on s0, `d.
F.Local Maximum or Minimum Values Since f 9s0d − 0 and f 9 changes from negative to positive at 0, f s0d − 0 is a local (and absolute) minimum by the First Derivative Test.
G. Concavity and Points of Inflection
f 0sxd −
2sx 1 1d3y2s6x 1 4d 2 s3x 2 1 4xd3sx 1 1d1y2
3x 2 1 8x 1 8
4sx 1 1d
4sx 1 1d5y2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
ote that the denominator is always positive. The numerator is the quadratic
3x 2 1 8x 1 8, which is always positive because its discriminant is
b 2 2 4ac − 232, which is negative, and the coefficient of x 2 is positive. Thus
f 0sxd . 0 for all x in the domain of f , which means that f is concave upward
on s21, `d and there is no point of inflection.
H. Sketch the Curve The curve is sketched in Figure 7.
EXAMPLE 3 Sketch the graph of f sxd −
A. Domain The domain is R.
FIGURE 7
cos x
2 1 sin x
B. Intercepts The y-intercept is f s0d − 12. The x-intercepts occur when cos x − 0, that
is, x − sy2d 1 n, where n is an integer.
C. Symmetry f is neither even nor odd, but f sx 1 2d − f sxd for all x and so
f is periodic and has period 2. Thus, in what follows, we need to consider only
0 < x < 2 and then extend the curve by translation in part H.
D. Asymptotes None
E. Intervals of Increase or Decrease
f 9sxd −
s2 1 sin xds2sin xd 2 cos x scos xd
2 sin x 1 1
s2 1 sin xd 2
s2 1 sin xd 2
The denominator is always positive, so f 9sxd . 0 when 2 sin x 1 1 , 0 &?
sin x , 221 &? 7y6 , x , 11y6. So f is increasing on s7y6, 11y6d and
decreasing on s0, 7y6d and s11y6, 2d.
F.Local Maximum or Minimum Values From part E and the First Derivative Test,
we see that the local minimum value is f s7y6d − 21ys3 and the local maximum
value is f s11y6d − 1ys3.
G. Concavity and Points of Inflection If we use the Quotient Rule again and simplify, we get
2 cos x s1 2 sin xd
f 0sxd − 2
s2 1 sin xd 3
Because s2 1 sin xd 3 . 0 and 1 2 sin x > 0 for all x, we know that f 0sxd . 0
when cos x , 0, that is, y2 , x , 3y2. So f is concave upward on sy2, 3y2d
and concave downward on s0, y2d and s3y2, 2d. The inflection points are
sy2, 0d and s3y2, 0d.
H. Sketch the Curve The graph of the function restricted to 0 < x < 2 is shown in
Figure 8. Then we extend it, using periodicity, to arrive at the graph in Figure 9.
11π 1
6 , œ„
2π x
- ’
” 7π
6 , œ„3
FIGURE 8
FIGURE 9
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.5Summary of Curve Sketching
■ Slant Asymptotes
Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical. If
lim f f sxd 2 smx 1 bdg − 0
where m ± 0, then the line y − mx 1 b is called a slant asymptote because the ver­
tical distance between the curve y − f sxd and the line y − mx 1 b approaches 0, as
in Fig­ure 10. (A similar situation exists if we let x l 2`.) In the case of rational functions, slant asymp­totes occur when the degree of the numerator is one more than the
degree of the denominator. In such a case the equation of the slant asymptote can be
found by long division as in the following example.
FIGURE 10
EXAMPLE 4 Sketch the graph of f sxd −
x 11
A. Domain The domain is R.
B. Intercepts The x- and y-intercepts are both 0.
C. Symmetry Since f s2xd − 2f sxd, f is odd and its graph is symmetric about the
D. Asymptotes Since x 2 1 1 is never 0, there is no vertical asymptote. Since
f sxd l ` as x l ` and f sxd l 2` as x l 2`, there is no horizontal asymptote.
But long division gives
f sxd −
−x2 2
x 11
x 11
This equation suggests that y − x is a candidate for a slant asymptote. In fact,
f sxd 2 x − 2 2
x 11
l 0asx l 6`
So the line y − x is indeed a slant asymptote.
E. Intervals of Increase or Decrease
f 9sxd −
sx 2 1 1ds3x 2 d 2 x 3 2x
x 2sx 2 1 3d
sx 1 1d
sx 2 1 1d2
Since f 9sxd . 0 for all x (except 0), f is increasing on s2`, `d.
F. Local Maximum or Minimum Values Although f 9s0d − 0, f 9 does not change
sign at 0, so there is no local maximum or minimum.
G. Concavity and Points of Inflection
f 0sxd −
sx 2 1 1d2 s4x 3 1 6xd 2 sx 4 1 3x 2 d 2sx 2 1 1d2x
2xs3 2 x 2 d
sx 2 1 1d4
sx 2 1 1d3
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
Since f 0sxd − 0 when x − 0 or x − 6s3, we set up the following chart:
”_œ„3, _
3 2 x2
sx 2 1 1d3
f 0sxd
CU on (2`, 2s3 )
CD on (2s3, 0)
0 , x , s3
CU on ( 0, s3 )
x . s3
x , 2s3
2s3 , x , 0
The points of inflection are (2s3, 243 s3 ), s0, 0d, and (s3, 34 s3 ).
FIGURE 11
H. Sketch the Curve The graph of f is sketched in Figure 11.
1. y − x 3 1 3x 2
2. y − 2x 3 2 12x 2 1 18x
3. y − x 4 2 4x
4. y − x 4 2 8x 2 1 8
5. y − xsx 2 4d
7. y −
9. y −
2x 1 3
1 16x
x 2 x2
11. y −
2 2 3x 1 x 2
x2 2 4
15. y − 2
x 13
10. y −
x 2 1 5x
25 2 x 2
12. y − 1 1 1 2
14. y −
x2 2 4
sx 2 1d2
16. y − 2
x 11
19. y −
x 11
20. y −
22. y − sx 2 4dsx
23. y − sx 2 1 x 2 2
24. y − sx 2 1 x 2 x
s1 2 x 2
33. y − sin3 x
34. y − x 1 cos x
37. y − sin x 1 s3 cos x,22 < x < 2
38. y − csc x 2 2sin x,0 , x , 39. y −
sin x
1 1 cos x
sin x
2 1 cos x
41–44 The graph of a function f is shown. (The dashed lines
indicate horizontal asymptotes.) Find each of the following for the
given function t.
(a) The domains of t and t9
(b) The critical numbers of t
(c) The approximate value of t9s6d
(d) All vertical and horizontal asymptotes of t
26. y − x s2 2 x 2
28. y −
40. y −
21. y − sx 2 3dsx
27. y −
32. y − s
x3 1 1
36. y − 2x 2 tan x,2y2 , x , y2
x 21
sx 2 1 1
31. y − s
x2 2 1
8. y − s4 2 x d
18. y −
30. y − x 5y3 2 5x 2y3
35. y − x tan x,2y2 , x , y2
25. y −
29. y − x 2 3x 1y3
6. y − x 2 5x
17. y −
1–40 Use the guidelines of this section to sketch the curve.
13. y −
CD on (s3, `)
sx 2 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.5Summary of Curve Sketching
41. tsxd − s f sxd
43. tsxd − f sxd
42. tsxd − s
f sxd
where k is a positive constant. Sketch the graph of the net
force function. What does the graph say about the force?
44. tsxd − 1yf sxd
45. In the theory of relativity, the mass of a particle is
49–52 Find an equation of the slant asymptote. Do not sketch the
s1 2 v 2yc 2
where m 0 is the rest mass of the particle, m is the mass when
the particle moves with speed v relative to the observer, and c
is the speed of light. Sketch the graph of m as a function of v.
46. In the theory of relativity, the energy of a particle is
E − sm 02 c 4 1 h 2 c 2y2
where m 0 is the rest mass of the particle, is its wave length,
and h is Planck’s constant. Sketch the graph of E as a function of . What does the graph say about the energy?
47. T
he figure shows a beam of length L embedded in concrete
walls. If a constant load W is distributed evenly along its
length, the beam takes the shape of the deflection curve
W 4
WL 3
WL 2 2
x 1
x 2
where E and I are positive constants. (E is Young’s modulus of elasticity and I is the moment of inertia of a crosssection of the beam.) Sketch the graph of the deflection
49. y −
x2 1 1
50. y −
4x 3 2 10x 2 2 11x 1 1
x 2 2 3x
51. y −
2x 3 2 5x 2 1 3x
x2 2 x 2 2
52. y −
26x 4 1 2x 3 1 3
2x 3 2 x
53–58 Use the guidelines of this section to sketch the curve. In
guideline D, find an equation of the slant asymptote.
53. y −
54. y −
1 1 5x 2 2x 2
55. y −
x3 1 4
56. y −
sx 1 1d2
57. y −
2x 3 1 x 2 1 1
x2 1 1
58. y −
sx 1 1d3
sx 2 1d2
59. S
how that the curve y − s4x 2 1 9 has two slant asymptotes:
y − 2x and y − 22x. Use this fact to help sketch the curve.
60. S
how that the curve y − sx 2 1 4x has two slant asymptotes:
y − x 1 2 and y − 2x 2 2. Use this fact to help sketch the
61. S
how that the lines y − sbyadx and y − 2sbyadx are slant
asymptotes of the hyperbola sx 2ya 2 d 2 s y 2yb 2 d − 1.
62. Let f sxd − sx 3 1 1dyx. Show that
lim f f sxd 2 x 2 g − 0
x l 6`
48. C
oulomb’s Law states that the force of attraction between
two charged particles is directly proportional to the product
of the charges and inversely proportional to the square of
the distance between them. The figure shows particles with
charge 1 located at positions 0 and 2 on a coordinate line and
a particle with charge 21 at a position x between them. It
follows from Coulomb’s Law that the net force acting on the
middle particle is
Fsxd − 2
sx 2 2d2
This shows that the graph of f approaches the graph of y − x 2,
and we say that the curve y − f sxd is asymptotic to the parabola y − x 2. Use this fact to help sketch the graph of f .
63. D
iscuss the asymptotic behavior of f sxd − sx 4 1 1dyx in
the same manner as in Exercise 62. Then use your results to
help sketch the graph of f .
64. U
se the asymptotic behavior of f sxd − cos x 1 1yx 2 to
sketch its graph without going through the curve-sketching
procedure of this section.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
3.6 Graphing with Calculus and Technology
You may want to read Graphing
Calculators and Computers at
www.StewartCalculus.com if you
haven’t already. In particular, it
explains how to avoid some of the pitfalls of graphing devices by choosing
appropriate viewing rectangles.
The method we used to sketch curves in the preceding section was a culmination of much
of our study of differential calculus. The graph was the final object that we produced. In
this section our point of view is completely different. Here we start with a graph produced by a graphing calculator or computer and then we refine it. We use calculus to
make sure that we reveal all the important aspects of the curve. And with the use of
graphing devices we can tackle curves that would be far too complicated to consider
without technology. The theme is the interaction between calculus and technology.
EXAMPLE 1 Graph the polynomial f sxd − 2x 6 1 3x 5 1 3x 3 2 2x 2. Use the graphs
of f 9 and f 0 to estimate all maximum and minimum points and intervals of concavity.
FIGURE 1
SOLUTION If we specify a domain but not a range, graphing software will often deduce
a suitable range from the values computed. Figure 1 shows a plot that may result if we
specify that 25 < x < 5. Although this viewing rectangle is useful for showing that the
asymptotic behavior (or end behavior) is the same as for y − 2x 6, it is obviously hiding
some finer detail. So we change to the viewing rectangle f23, 2g by f250, 100g in
Figure 2.
Most graphing calculators and graphing software allow us to “trace” along a curve
and see approximate coordinates of points. (Some also have features to identify the
approximate locations of local maximum and minimum points.) Here it appears that
there is an absolute minimum value of about 215.33 when x < 21.62 and f is
decreasing on s2`, 21.62d and increasing on s21.62, `d. Also, there appears to be a
horizontal tangent at the origin and inflection points when x − 0 and when x is
somewhere between 22 and 21.
Now let’s try to confirm these impressions using calculus. We differentiate and get
f 9sxd − 12x 5 1 15x 4 1 9x 2 2 4x
f 0sxd − 60x 4 1 60x 3 1 18x 2 4
FIGURE 2
When we graph f 9 in Figure 3 we see that f 9sxd changes from negative to positive when
x < 21.62; this confirms (by the First Derivative Test) the minimum value that we
found earlier. But, perhaps to our surprise, we also notice that f 9sxd changes from
positive to negative when x − 0 and from negative to positive when x < 0.35. This
means that f has a local maximum at 0 and a local minimum when x < 0.35, but these
were hidden in Figure 2. Indeed, if we now zoom in toward the origin in Figure 4, we
see what we missed before: a local maximum value of 0 when x − 0 and a local
minimum value of about 20.1 when x < 0.35.
FIGURE 3
FIGURE 4
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.6Graphing with Calculus and Technology
What about concavity and inflection points? From Figures 2 and 4 there appear to
be inflection points when x is a little to the left of 21 and when x is a little to the
right of 0. But it’s difficult to determine inflection points from the graph of f , so we
graph the second derivative f 0 in Figure 5. We see that f 0 changes from positive
to negative when x < 21.23 and from negative to positive when x < 0.19. So, correct to two decimal places, f is concave upward on s2`, 21.23d and s0.19, `d and
concave downward on s21.23, 0.19d. The inflection points are s21.23, 210.18d
and s0.19, 20.05d.
We have discovered that no single graph reveals all the important features of this
polynomial. But Figures 2 and 4, when taken together, do provide an accurate picture.
FIGURE 5
EXAMPLE 2 Draw the graph of the function
f sxd −
x 2 1 7x 1 3
in a viewing rectangle that shows all the important features of the function. Estimate the
local maximum and minimum values and the intervals of concavity. Then use calculus to
find these quantities exactly.
SOLUTION Figure 6 — produced by graphing software with automatic scaling—is
a disaster. Some graphing calculators use f210, 10g by f210, 10g as the default
viewing rectangle, so let’s try it. We get the graph shown in Figure 7; it’s a major
3 10!*
FIGURE 6
FIGURE 7
The y-axis appears to be a vertical asymptote and indeed it is because
Figure 7 also allows us to estimate the x-intercepts: about 20.5 and 26.5. The
exact values are obtained by using the quadratic formula to solve the equation
x 2 1 7x 1 3 − 0; we get x − (27 6 s37 )y2.
To get a better look at horizontal asymptotes, we change to the viewing rectangle
f220, 20g by f25, 10g in Figure 8. It appears that y − 1 is the horizontal asymptote
and this is easily confirmed:
FIGURE 8
x 2 1 7x 1 3
x l 6`
x 2 1 7x 1 3
− lim 1 1 1 2
x l 6`
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
To estimate the minimum value we zoom in to the viewing rectangle f23, 0g by
f24, 2g in Figure 9. We find that the absolute minimum value is about 23.1 when
x < 20.9, and we see that the function decreases on s2`, 20.9d and s0, `d and
increases on s20.9, 0d. The exact values are obtained by differentiating:
f 9sxd − 2
7x 1 6
2 3 −2
This shows that f 9sxd . 0 when 267 , x , 0 and f 9sxd , 0 when x , 267 and when
< 23.08.
x . 0. The exact minimum value is f (2 67 ) − 2 12
Figure 9 also shows that an inflection point occurs somewhere between x − 21 and
x − 22. We could estimate it much more accurately using the graph of the second
deriv­ative, but in this case it’s just as easy to find exact values. Since
FIGURE 9
f 0sxd −
2s7x 1 9d
4 −
we see that f 0sxd . 0 when x . 297 sx ± 0d and f 0sxd , 0 when x , 297. So f is
concave upward on (297 , 0) and s0, `d and concave downward on (2`, 297 ). The
inflection point is (297 , 271
27 ).
The analysis using the first two derivatives shows that Figure 8 displays all the
major aspects of the curve.
EXAMPLE 3 Graph the function f sxd −
x 2sx 1 1d3
sx 2 2d2sx 2 4d4
SOLUTION Drawing on our experience with a rational function in Example 2, let’s
start by graphing f in the viewing rectangle f210, 10g by f210, 10g. From Figure 10
we have the feeling that we are going to have to zoom in to see some finer detail and
also zoom out to see the larger picture. But, as a guide to intelligent zooming, let’s
first take a close look at the expression for f sxd. Because of the factors sx 2 2d2 and
sx 2 4d4 in the denominator, we expect x − 2 and x − 4 to be the vertical asymptotes.
FIGURE 10
x l2
x 2sx 1 1d3
x 2sx 1 1d3
x l 4 sx 2 2d2sx 2 4d4
sx 2 2d2sx 2 4d4
To find the horizontal asymptotes, we divide numerator and denominator by x 6:
x 2 sx 1 1d3
x 2sx 1 1d3
sx 2 2d sx 2 4d
sx 2 2d
sx 2 4d4
FIGURE 11
S D
S DS D
This shows that f sxd l 0 as x l 6`, so the x-axis is a horizontal asymptote.
It is also very useful to consider the behavior of the graph near the x-intercepts using
an analysis like that in Example 3.4.11. Since x 2 is positive, f sxd does not change sign
at 0 and so its graph doesn’t cross the x-axis at 0. But, because of the factor sx 1 1d3,
the graph does cross the x-axis at 21 and has a horizontal tangent there. Putting all this
information together, but without using derivatives, we see that the curve has to look
something like the one in Figure 11.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.6Graphing with Calculus and Technology
Now that we know what to look for, we zoom in (several times) to produce the
graphs in Figures 12 and 13 and zoom out (several times) to get Figure 14.
FIGURE 12
FIGURE 13
FIGURE 14
We can read from these graphs that the absolute minimum is about 20.02 and
occurs when x < 220. There is also a local maximum <0.00002 when x < 20.3 and
a local minimum <211 when x < 2.5. These graphs also show three inflection points
near 235, 25, and 21 and two between 21 and 0. To estimate the inflection points
closely we would need to graph f 0, but to compute f 0 by hand is an unreasonable
chore. If you have a computer algebra system, then it’s easy to do (see Exercise 13).
We have seen that, for this particular function, three graphs (Figures 12, 13, and 14)
are necessary to convey all the useful information. The only way to display all these
features of the function on a single graph is to draw it by hand. Despite the exaggerations and distortions, Figure 11 does manage to summarize the essential nature of the
EXAMPLE 4 Graph the function f sxd − sinsx 1 sin 2xd. For 0 < x < , estimate
all maximum and minimum values, intervals of increase and decrease, and inflection
SOLUTION We first note that f is periodic with period 2. Also, f is odd and
f sxd < 1 for all x. So the choice of a viewing rectangle is not a problem for this
function: we start with f0, g by f21.1, 1.1g. (See Figure 15.) It appears that there are
three local maximum values and two local minimum values in that window. To confirm
these values and locate them more accurately, we calculate that
f 9sxd − cossx 1 sin 2xd s1 1 2 cos 2xd
and graph both f and f 9 in Figure 16.
After estimating the values of the x-intercepts of f 9, we use the First Derivative Test
to find the following approximate values:
FIGURE 15
y=f ª(x)
FIGURE 16
Intervals of increase:
s0, 0.6d, s1.0, 1.6d, s2.1, 2.5d
Intervals of decrease:
s0.6, 1.0d, s1.6, 2.1d, s2.5, d
Local maximum values: f s0.6d < 1, f s1.6d < 1, f s2.5d < 1
Local minimum values:
f s1.0d < 0.94, f s2.1d < 0.94
The second derivative is
f 0sxd − 2s1 1 2 cos 2xd2 sinsx 1 sin 2xd 2 4 sin 2x cossx 1 sin 2xd
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
Graphing both f and f 0 in Figure 17, we obtain the following approximate values:
s0.8, 1.3d, s1.8, 2.3d
Concave upward on:
Concave downward on: s0, 0.8d, s1.3, 1.8d, s2.3, d
FIGURE 17
s0, 0d, s0.8, 0.97d, s1.3, 0.97d, s1.8, 0.97d, s2.3, 0.97d
Inflection points:
Having checked that Figure 15 does indeed represent f accurately for 0 < x < ,
we can state that the extended graph in Figure 18 represents f accurately for
22 < x < 2.
The family of functions
f sxd − sinsx 1 sin cxd
where c is a constant, occurs in applications to frequency modulation (FM)
synthesis. A sine wave is modulated
by a wave with a different frequency
ssin cxd. The case where c − 2 is studied in Example 4. Exercise 19 explores
another special case.
FIGURE 18
Our final example is concerned with families of functions. This means that the functions in the family are related to each other by a formula that contains one or more arbitrary constants. Each value of the constant gives rise to a member of the family and the
idea is to see how the graph of the function changes as the constant changes.
EXAMPLE 5 How does the graph of f sxd − 1ysx 2 1 2x 1 cd vary as c varies?
SOLUTION The graphs in Figures 19 and 20 (the special cases c − 2 and c − 22)
show two very different-looking curves.
FIGURE 19
FIGURE 20
c − 22
Before drawing any more graphs, let’s see what members of this family have in
common. Since
lim 2
x l 6` x 1 2x 1 c
for any value of c, they all have the x-axis as a horizontal asymptote. A vertical asymptote will occur when x 2 1 2x 1 c − 0. Solving this quadratic equation, we get
x − 21 6 s1 2 c . When c . 1, there is no vertical asymptote (as in Figure 19).
When c − 1, the graph has a single vertical asymptote x − 21 because
x l21
− lim
x l21 sx 1 1d2
x 2 1 2x 1 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.6Graphing with Calculus and Technology
When c , 1, there are two vertical asymptotes: x − 21 6 s1 2 c (as in Figure 20).
Now we compute the derivative:
2x 1 2
f 9sxd − 2 2
sx 1 2x 1 cd2
This shows that f 9sxd − 0 when x − 21 (if c ± 1), f 9sxd . 0 when x , 21, and
f 9sxd , 0 when x . 21. For c > 1, this means that f increases on s2`, 21d
and decreases on s21, `d. For c . 1, there is an absolute maximum value
f s21d − 1ysc 2 1d. For c , 1, f s21d − 1ysc 2 1d is a local maximum value and the
intervals of increase and decrease are interrupted at the vertical asymptotes.
Figure 21 is a “slide show” displaying five members of the family, all graphed in the
viewing rectangle f25, 4g by f22, 2g. As predicted, a transition takes place from two
vertical asymptotes to one for c − 1, and then to none for c . 1. As c increases from 1,
we see that the maximum point becomes lower; this is explained by the fact that
1ysc 2 1d l 0 as c l `. As c decreases from 1, the vertical asymptotes become more
widely separated because the distance between them is 2 s 1 2 c , which becomes large
as c l 2`. Again, the maximum point approaches the x-axis because 1ysc 2 1d l 0
as c l 2`.
FIGURE 21
There is clearly no inflection point when c < 1. For c . 1 we calculate that
The family of functions
f sxd − 1ysx 2 1 2x 1 cd
f 0sxd −
2s3x 2 1 6x 1 4 2 cd
sx 2 1 2x 1 cd3
and deduce that inflection points occur when x − 21 6 s3sc 2 1dy3. So the inflection
points become more spread out as c increases and this seems plausible from the last
two graphs of Figure 21.
1–8 Produce graphs of f that reveal all the important aspects of
the curve. In particular, you should use graphs of f 9 and f 0 to
estimate the intervals of increase and decrease, extreme values,
intervals of concavity, and inflection points.
1. f sxd − x 5 2 5x 4 2 x 3 1 28x 2 2 2x
2. f sxd − 22x 1 5x 1 140x 2 110x
3. f sxd − x 6 2 5x 5 1 25x 3 2 6x 2 2 48x
x4 2 x 3 2 8
x2 2 x 2 6
5. f sxd − 3
x 1 x2 1 1
4. f sxd −
7. f sxd − 6 sin x 1 cot x,2 < x < 8. f sxd −
sin x
,22 < x < 2
9–10 Produce graphs of f that reveal all the important aspects
of the curve. Estimate the intervals of increase and decrease
and intervals of concavity, and use calculus to find these intervals exactly.
9. f sxd − 1 1
10. f sxd −
6. f sxd − 6 sin x 2 x 2,25 < x < 3
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
11–12 Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to
producing graphs using a calculator or computer that display
the major features of the curve. Use these graphs to estimate the
maximum and minimum values.
11. f sxd −
sx 1 4dsx 2 3d
x 4sx 2 1d
12. f sxd −
s2 x 1 3d 2 sx 2 2d 5
x 3 sx 2 5d 2
13. F
or the function f of Example 3, use a com­puter algebra
system to calculate f 9 and then graph it to confirm that all
the maximum and minimum values are as given in the
example. Calculate f 0 and use it to estimate the intervals of
concavity and inflection points.
14. F
or the function f of Exercise 12, use a computer algebra
system to find f 9 and f 0 and use their graphs to estimate the
intervals of increase and decrease and concavity of f .
15–18 Use a computer algebra system to graph f and to find f 9
and f 0. Use graphs of these derivatives to estimate the intervals
of increase and decrease, extreme values, intervals of concavity,
and inflection points of f .
x 3 1 5x 2 1 1
15. f sxd − 4
x 1 x3 2 x2 1 2
should also identify any transitional values of c at which the
basic shape of the curve changes.
20. f sxd − x 3 1 cx
21. f sxd − x 2 1 6 x 1 cyx (trident of Newton)
22. f sxd − x sc 2 2 x 2
24. f sxd −
23. f sxd −
sin x
c 1 cos x
1 1 c 2x 2
25. f sxd − cx 1 sin x
26. The figure shows graphs (in blue) of several members of the
family of polynomials f sxd − cx 4 2 4x 2 1 1.
(a)For which values of c does the curve have minimum
(b)Show that the minimum and maximum points of every
curve in the family lie on the parabola y − 22x 2 1 1
(shown in red).
x 2y3
16. f sxd −
1 1 x 1 x4
17. f sxd − sx 1 5 sin x ,x < 20
18. f sxd −
2x 2 1
x4 1 x 1 1
19. I n Example 4 we considered a member of the family of
functions f sxd − sinsx 1 sin cxd that occur in FM synthesis. Here we investigate the function with c − 3. Start by
graphing f in the viewing rectangle f0, g by f21.2, 1.2g.
How many local maximum points do you see? The graph
has more than are visible to the naked eye. To discover the
hidden maximum and minimum points you will need to
examine the graph of f 9 very carefully. In fact, it helps to
look at the graph of f 0 at the same time. Find all the maximum and minimum values and inflection points. Then
graph f in the viewing rectangle f22, 2g by f21.2, 1.2g
and comment on symmetry.
20–25 Describe how the graph of f varies as c varies. Graph
several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and
minimum points and inflection points move when c changes. You
27. I nvestigate the family of curves given by the equation
f sxd − x 4 1 cx 2 1 x. Start by determining the transitional value of c at which the number of inflection points
changes. Then graph several members of the family to see
what shapes are possible. There is another transitional
value of c at which the number of critical numbers
changes. Try to discover it graphically. Then prove what
you have discovered.
28. (a)Investigate the family of polynomials given by the
f sxd − 2x 3 1 cx 2 1 2 x
For what values of c does the curve have maximum and
minimum points?
(b)Show that the minimum and maximum points of
every curve in the family lie on the curve y − x 2 x 3.
Illustrate by graphing this curve and several members
of the family.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.7Optimization Problems
3.7 Optimization Problems
The methods we have learned in this chapter for finding extreme values have practical
applications in many areas of life: A businessperson wants to minimize costs and maximize
profits. A traveler wants to minimize transportation time. Fermat’s Principle in optics states
that light follows the path that takes the least time. In this section we solve such problems
as maximizing areas, volumes, and profits and minimizing distances, times, and costs.
In solving such practical problems the greatest challenge is often to convert the word
problem into a mathematical optimization problem by setting up the function that is to
be maximized or minimized. Let’s recall the problem-solving principles discussed in the
Principles of Problem Solving following Chapter 1 and adapt them to this situation:
Steps In Solving Optimization Problems
1. Understand the Problem The first step is to read the problem carefully until
it is clearly understood. Ask yourself: What is the unknown? What are the given
quantities? What are the given conditions?
2.Draw a Diagram In most problems it is useful to draw a diagram and identify
the given and required quantities on the diagram.
3.Introduce Notation Assign a symbol to the quantity that is to be maximized or
minimized (let’s call it Q for now). Also select symbols sa, b, c, . . . , x, yd for other
unknown quantities and label the diagram with these symbols. It may help to use
initials as suggestive symbols—for example, A for area, h for height, t for time.
4. Express Q in terms of some of the other symbols from Step 3.
5. If Q has been expressed as a function of more than one variable in Step 4, use
the given information to find relationships (in the form of equations) among
these variables. Then use these equations to eliminate all but one of the variables in the expression for Q. Thus Q will be expressed as a function of one
variable x, say, Q − f sxd. Write the domain of this function in the given context.
6. Use the methods of Sections 3.1 and 3.3 to find the absolute maximum or
minimum value of f. In particular, if the domain of f is a closed interval, then
the Closed Interval Method in Section 3.1 can be used.
EXAMPLE 1 A farmer has 1200 m of fencing and wants to fence off a rectangular
field that borders a straight river. He needs no fence along the river. What are the
dimensions of the field that has the largest area?
SOLUTION In order to get a feeling for what is happening in this problem, let’s experiment with some specific cases. Figure 1 (not to scale) shows three possible ways of
laying out the 1200 m of fencing.
PS Understand the problem
PS Analogy: Try special cases
PS Draw diagrams
200 m
400 m
100 m
1000 m
100 m
Area=100 · 1000=100,000 m@
400 m
400 m
Area=400 · 400=160,000 m@
500 m
500 m
Area=500 · 200=100,000 m@
FIGURE 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
We see that when we try shallow, wide fields or deep, narrow fields, we get relatively small areas. It seems plausible that there is some intermediate configuration that
produces the largest area.
Figure 2 illustrates the general case. We wish to maximize the area A of the rectangle. Let x and y be the depth and width of the rectangle (in meters). Then we express
A in terms of x and y:
PS Introduce notation
A − xy
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the
fencing is1200 m. Thus
2x 1 y − 1200
From this equation we have y − 1200 2 2x, which gives
FIGURE 2
A − xy − xs1200 2 2xd − 1200x 2 2x 2
Note that the largest x can be is 600 (this uses all the fence for the depth and none for
the width) and x can’t be negative, so the function that we wish to maximize is
Asxd − 1200x 2 2x 20 < x < 600
The derivative is A9sxd − 1200 2 4x, so to find the critical numbers we solve the
1200 2 4x − 0
which gives x − 300. The maximum value of A must occur either at this critical
number or at an endpoint of the interval. Since As0d − 0, As300d − 180,000,
and As600d − 0, the Closed Interval Method gives the maximum value as
As300d − 180,000.
[Alternatively, we could have observed that A0sxd − 24 , 0 for all x, so A is
always concave downward and the local maximum at x − 300 must be an absolute
The corresponding y-value is y − 1200 2 2s300d − 600, so the rectangular field
should be 300 m deep and 600 m wide. n
EXAMPLE 2 A cylindrical can is to be made to hold 1 L of oil. Find the dimensions
that will minimize the cost of the metal to manufacture the can.
SOLUTION Draw a diagram as in Figure 3, where r is the radius and h the height (both
in centimeters). In order to minimize the cost of the metal, we minimize the total surface
area of the cylinder (top, bottom, and sides). From Figure 4 we see that the sides are
made from a rectangular sheet with dimensions 2r and h. So the surface area is
FIGURE 3
A − 2r 2 1 2rh
We would like to express A in terms of one variable, r. To eliminate h we use the
fact that the volume is given as 1 L, which is equivalent to 1000 cm3. Thus
r 2h − 1000
which gives h − 1000ysr 2 d. Substitution of this into the expression for A gives
Area 2{πr@}
FIGURE 4
Area (2πr)h
A − 2r 2 1 2r
S D
r 2
− 2r 2 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.7Optimization Problems
We know that r must be positive, and there are no limitations on how large r can be.
Therefore the function that we want to minimize is
Asrd − 2r 2 1
r . 0
To find the critical numbers, we differentiate:
A9srd − 4r 2
FIGURE 5
In the Applied Project following this
section we investigate the most economical shape for a can by taking into
account other manufacturing costs.
4sr 3 2 500d
Then A9srd − 0 when r 3 − 500, so the only critical number is r − s
500y .
Since the domain of A is s0, `d, we can’t use the argument of Example 1 concerning
endpoints. But we can observe that A9srd , 0 for r , s
500y and A9srd . 0 for
r . s500y , so A is decreasing for all r to the left of the critical number and increas3
ing for all r to the right. Thus r − s
500y must give rise to an absolute minimum.
[Alternatively, we could argue that Asrd l ` as r l 0 1 and Asrd l ` as r l `, so
there must be a minimum value of Asrd, which must occur at the critical number. See
Figure 5.]
The value of h corresponding to r − s
500y is
2 −
− 2r
Thus, to minimize the cost of the can, the radius should be s
500y cm and the height
should be equal to twice the radius, namely, the diameter.
NOTE 1 The argument used in Example 2 to justify the absolute minimum is a variant
of the First Derivative Test (which applies only to local maximum or minimum values)
and is stated here for future reference.
First Derivative Test for Absolute Extreme Values Suppose that c is a critical
number of a continuous function f defined on an interval.
(a)If f 9sxd . 0 for all x , c and f 9sxd , 0 for all x . c, then f scd is the absolute maximum value of f .
(b)If f 9sxd , 0 for all x , c and f 9sxd . 0 for all x . c, then f scd is the absolute minimum value of f .
NOTE 2 An alternative method for solving optimization problems is to use implicit
differentiation. Let’s look at Example 2 again to illustrate the method. We work with the
same equations
A − 2r 2 1 2rhr 2h − 1000
but instead of eliminating h, we differentiate both equations implicitly with respect to r
(treating both A and h as functions of r):
A9 − 4r 1 2rh9 1 2hr 2h9 1 2rh − 0
The minimum occurs at a critical number, so we set A9 − 0, simplify, and arrive at the
2r 1 rh9 1 h − 0rh9 1 2h − 0
Subtraction of these two equations gives 2r 2 h − 0, or h − 2r.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
EXAMPLE 3 Find the point on the parabola y 2 − 2x that is closest to the point s1, 4d.
d − ssx 2 1d 2 1 sy 2 4d 2
(x, y)
SOLUTION The distance between the point s1, 4d and the point sx, yd is
(1, 4)
(See Figure 6.) But if sx, yd lies on the parabola, then x − 12 y 2, so the expression for d
d − s ( 12 y 2 2 1) 2 1 sy 2 4d 2
(Alternatively, we could have substituted y − s2x to get d in terms of x alone.)
Instead of minimizing d, we minimize its square:
FIGURE 6
d 2 − f syd − ( 21 y 2 2 1) 2 1 s y 2 4d 2
(You should convince yourself that the minimum of d occurs at the same point as the
minimum of d 2, but d 2 is easier to work with.) Note that there is no restriction on y, so
the domain is all real numbers. Differentiating, we obtain
f 9syd − 2( 12 y 2 2 1) y 1 2sy 2 4d − y 3 2 8
so f 9syd − 0 when y − 2. Observe that f 9syd , 0 when y , 2 and f 9s yd . 0 when
y . 2, so by the First Derivative Test for Absolute Extreme Values, the absolute
minimum occurs when y − 2. (Or we could simply say that because of the geometric
nature of the problem, it’s obvious that there is a closest point but not a farthest point.)
The corresponding value of x is x − 12 y 2 − 2. Thus the point on y 2 − 2x closest to
s1, 4d is s2, 2d. [The distance between the points is d − sf s2d − s5 .]
EXAMPLE 4 A woman launches her boat from point A on a bank of a straight river,
3 km wide, and wants to reach point B, 8 km downstream on the opposite bank, as
quickly as possible (see Figure 7). She could row her boat directly across the river to
point C and then run to B, or she could row directly to B, or she could row to some
point D between C and B and then run to B. If she can row 6 kmyh and run 8 kmyh,
where should she land to reach B as soon as possible? (We assume that the speed of the
water is negligible compared with the speed at which the woman rows.)
3 km
8 km
SOLUTION If we let x be the distance from C to D, then the running distance is
DB − 8 2 x and the Pythagorean Theorem gives the rowing distance as
AD − sx 2 1 9 . We use the equation
time −
FIGURE 7
Then the rowing time is sx 2 1 9 y6 and the running time is s8 2 xdy8, so the total time
T as a function of x is
sx 2 1 9
Tsxd −
The domain of this function T is f0, 8g. Notice that if x − 0, she rows to C and if
x − 8, she rows directly to B. The derivative of T is
T9sxd −
6 sx 2 1 9
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.7Optimization Problems
Thus, using the fact that x > 0, we have
T9sxd − 0 &?
6 sx 1 9
&? 16x 2 − 9sx 2 1 9d
Ts0d − 1.5T
&? 7x 2 − 81 &?
The only critical number is x − 9ys7 . To see whether the minimum occurs at this
critical number or at an endpoint of the domain f0, 8g, we follow the Closed Interval
Method by evaluating T at all three points:
&? 4x − 3sx 2 1 9
FIGURE 8
S D
< 1.33Ts8d −
< 1.42
Since the smallest of these values of T occurs when x − 9ys7 , the absolute minimum
value of T must occur there. Figure 8 illustrates this calculation by showing the graph
of T.
Thus the woman should land the boat at a point 9ys7 km (<3.4 km) downstream
from her starting point.
EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle
of radius r.
(x, y)
FIGURE 9
SOLUTION 1 Let’s take the semicircle to be the upper half of the circle x 2 1 y 2 − r 2
with center the origin. Then the word inscribed means that the rectangle has two
vertices on the semicircle and two vertices on the x-axis as shown in Figure 9.
Let sx, yd be the vertex that lies in the first quadrant. Then the rectangle has sides of
lengths 2x and y, so its area is
A − 2xy
To eliminate y we use the fact that sx, yd lies on the circle x 2 1 y 2 − r 2 and so
y − sr 2 2 x 2 . Thus
A − 2xsr 2 2 x 2
The domain of this function is 0 < x < r. Its derivative is
A9 − 2sr 2 2 x 2 2
2x 2
sr 2 2 x 2
2sr 2 2 2x 2d
sr 2 2 x 2
which is 0 when 2x 2 − r 2, that is, x − rys2 (since x > 0). This value of x gives a
maximum value of A since As0d − 0 and Asrd − 0. Therefore the area of the largest
inscribed rectangle is
S D
r2 2
− r2
SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable.
Let be the angle shown in Figure 10. Then the area of the rectangle is
r cos ¨
FIGURE 10
r sin ¨
Asd − s2r cos dsr sin d − r 2s2 sin cos d − r 2 sin 2
We know that sin 2 has a maximum value of 1 and it occurs when 2 − y2. So Asd
has a maximum value of r 2 and it occurs when − y4.
Notice that this trigonometric solution doesn’t involve differentiation. In fact, we
didn’t need to use calculus at all.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
■ Applications to Business and Economics
In Section 2.7 we introduced the idea of marginal cost. Recall that if Csxd, the cost function, is the cost of producing x units of a certain product, then the marginal cost is the
rate of change of C with respect to x. In other words, the marginal cost function is the
derivative, C9sxd, of the cost function.
Now let’s consider marketing. Let psxd be the price per unit that the company can
charge if it sells x units. Then p is called the demand function (or price function) and
we would expect it to be a decreasing function of x. (More units sold corresponds to a
lower price.) If x units are sold and the price per unit is psxd, then the total revenue is
Rsxd − quantity 3 price − xpsxd
and R is called the revenue function. The derivative R9 of the revenue function is called
the marginal revenue function and it is the rate of change of revenue with respect to the
num­ber of units sold.
If x units are sold, then the total profit is
Psxd − Rsxd 2 Csxd
and P is called the profit function. The marginal profit function is P9, the derivative of
the profit function. In Exercises 65 – 69 you are asked to use the marginal cost, revenue,
and profit functions to minimize costs and maximize revenues and profits.
EXAMPLE 6 A store has been selling 200 TV monitors a week at $350 each. A market
survey indicates that for each $10 rebate offered to buyers, the number of monitors sold
will increase by 20 a week. Find the demand function and the revenue function. How
large a rebate should the store offer to maximize revenue?
SOLUTION If x is the number of monitors sold per week, then the weekly increase in
sales is x 2 200. For each increase of 20 units sold, the price is decreased by $10. So
3 10 and the demand
for each additional unit sold, the decrease in price will be 20
function is
psxd − 350 2 10
20 sx 2 200d − 450 2 2 x
The revenue function is
Rsxd − xpsxd − 450x 2 12 x 2
Since R9sxd − 450 2 x, we see that R9sxd − 0 when x − 450. This value of x gives an
absolute maximum by the First Derivative Test (or simply by observing that the graph
of R is a parabola that opens downward). The corresponding price is
ps450d − 450 2 12 s450d − 225
and the rebate is 350 2 225 − 125. Therefore, to maximize revenue, the store should
offer a rebate of $125.
1. C
onsider the following problem: find two numbers whose
sum is 23 and whose product is a maximum.
(a)Make a table of values, like the one at the right, so that
the sum of the numbers in the first two columns is always
23. On the basis of the evidence in your table, estimate
the answer to the problem.
(b)Use calculus to solve the problem and compare with your
answer to part (a).
First number
Second number
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.7Optimization Problems
2. F
ind two numbers whose difference is 100 and whose product
is a minimum.
3. F
ind two positive numbers whose product is 100 and whose
sum is a minimum.
4. T
he sum of two positive numbers is 16. What is the smallest
possible value of the sum of their squares?
5. What
is the maximum vertical distance between the line
y − x 1 2 and the parabola y − x 2 for 21 < x < 2?
6. What
is the minimum vertical distance between the parabolas
y − x 2 1 1 and y − x 2 x 2 ?
7. F
ind the dimensions of a rectangle with perimeter 100 m
whose area is as large as possible.
8. F
ind the dimensions of a rectangle with area 1000 m2 whose
perimeter is as small as possible.
9. A
model used for the yield Y of an agricultural crop as a
function of the nitrogen level N in the soil (measured in
appropriate units) is
1 1 N2
where k is a positive constant. What nitrogen level gives the
best yield?
12. C
onsider the following problem: a box with an open top is to
be constructed from a square piece of cardboard, 3 m wide, by
cutting out a square from each of the four corners and bending
up the sides. Find the largest volume that such a box can have.
(a)Draw several diagrams to illustrate the situation, some
short boxes with large bases and some tall boxes with
small bases. Find the volumes of several such boxes.
Does it appear that there is a maximum volume? If so,
estimate it.
(b)Draw a diagram illustrating the general situation.
Introduce notation and label the diagram with your
(c) Write an expression for the volume.
(d)Use the given information to write an equation that
relates the variables.
(e)Use part (d) to write the volume as a function of one
(f)Finish solving the problem and compare the answer with
your estimate in part (a).
13. A
farmer wants to fence in an area of 15,000 m2 in a rectangular field and then divide it in half with a fence parallel to
one of the sides of the rectangle. How can he do this so as to
minimize the cost of the fence?
14. A farmer has 400 m of fencing for enclosing a trapezoidal
field along a river as shown. One of the parallel sides is three
times longer than the other. No fencing is needed along the
river. Find the largest area the farmer can enclose.
10. T
he rate sin mg carbonym 3yhd at which photosynthesis takes
place for a species of phytoplankton is modeled by the function
100 I
I2 1 I 1 4
where I is the light intensity (measured in thousands of footcandles). For what light intensity is P a maximum?
11. C
onsider the following problem: a farmer with 300 m of
fencing wants to enclose a rectangular area and then divide it
into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?
(a)Draw several diagrams illustrating the situation, some
with shallow, wide pens and some with deep, narrow
pens. Find the total areas of these configurations. Does it
appear that there is a maximum area? If so, estimate it.
(b)Draw a diagram illustrating the general situation.
Introduce notation and label the diagram with your
(c) Write an expression for the total area.
(d)Use the given information to write an equation that
relates the variables.
(e)Use part (d) to write the total area as a function of one
(f)Finish solving the problem and compare the answer with
your estimate in part (a).
15. A
farmer wants to fence in a rectangular plot of land adjacent
to the north wall of his barn. No fencing is needed along the
barn, and the fencing along the west side of the plot is shared
with a neighbor who will split the cost of that portion of the
fence. If the fencing costs $30 per linear meter to install and
the farmer is not willing to spend more than $1800, find the
dimensions for the plot that would enclose the most area.
16. I f the farmer in Exercise 15 wants to enclose 750 square
meters of land, what dimensions will minimize the cost of the
17. (a)Show that of all the rectangles with a given area, the one
with smallest perimeter is a square.
(b)Show that of all the rectangles with a given perimeter, the
one with greatest area is a square.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
18. A
box with a square base and open top must have a volume
of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used.
19. I f 1200 cm2 of material is available to make a box with a
square base and an open top, find the largest possible volume of the box.
20. A box with an open top is to be constructed from a 2 m by
1 m rectangular piece of cardboard by cutting out squares or
rectangles from each of the four corners, as shown in the
figure, and bending up the sides. One of the longer sides of
the box is to have a double layer of cardboard, which is
obtained by folding the side twice. Find the largest volume
that such a box can have.
30. F
ind the area of the largest rectangle that can be inscribed in
the ellipse x 2ya 2 1 y 2yb 2 − 1.
31. F
ind the dimensions of the rectangle of largest area that can
be inscribed in an equilateral triangle of side L if one side of
the rectangle lies on the base of the triangle.
32. F
ind the area of the largest trapezoid that can be inscribed in
a circle of radius 1 and whose base is a diameter of the circle.
35. If one side of a triangle has length a and another has length
2a, show that the largest possible area of the triangle is a 2.
22. Rework Exercise 21 assuming the container has a lid that is
made from the same material as the sides.
23. A package to be mailed using the US postal service may not
measure more than 274 cm in length plus girth. (Length is
the longest dimension and girth is the largest distance
around the package, perpendicular to the length.) Find the
dimensions of the rectangular box with square base of
greatest volume that may be mailed.
29. F
ind the dimensions of the rectangle of largest area that can
be inscribed in a circle of radius r.
34. I f the two equal sides of an isosceles triangle have length a,
find the length of the third side that maximizes the area of
the triangle.
21. A
rectangular storage container without a lid is to have a
volume of 10 m3. The length of its base is twice the width.
Material for the base costs $10 per square meter. Material
for the sides costs $6 per square meter. Find the cost of
materials for the least expensive such container.
ind, correct to two decimal places, the coordinates of the
; 28. F
point on the curve y − sin x that is closest to the point s4, 2d.
33. F
ind the dimensions of the isosceles triangle of largest area
that can be inscribed in a circle of radius r.
27. F
ind the points on the ellipse 4x 2 1 y 2 − 4 that are farthest
away from the point s1, 0d.
36. A rectangle has its base on the x-axis and its upper two vertices on the parabola y − 4 2 x 2. What is the largest possible area of the rectangle?
37. A
right circular cylinder is inscribed in a sphere of radius r.
Find the largest possible volume of such a cylinder.
38. A
right circular cylinder is inscribed in a cone with height h
and base radius r. Find the largest possible volume of such
a cylinder.
39. A
right circular cylinder is inscribed in a sphere of radius r.
Find the largest possible surface area of such a cylinder.
40. A
Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the
semicircle is equal to the width of the rectangle. See Exercise 1.1.72.) If the perimeter of the window is 10 m, find the
dimensions of the window so that the greatest possible
amount of light is admitted.
24. Refer to Exercise 23. Find the dimensions of the cylindrical
mailing tube of greatest volume that may be mailed using
the US postal service.
25. F
ind the point on the line y − 2x 1 3 that is closest to the
26. F
ind the point on the curve y − sx that is closest to the
point s3, 0d.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.7Optimization Problems
41. T
he top and bottom margins of a poster are each 6 cm and
the side margins are each 4 cm. If the area of printed material
on the poster is fixed at 384 cm2, find the dimensions of the
poster with the smallest area.
48. A
cone-shaped paper drinking cup is to be made to hold
27 cm3 of water. Find the height and radius of the cup that
will use the smallest amount of paper.
49. A
cone with height h is inscribed in a larger cone with
height H so that its vertex is at the center of the base of the
larger cone. Show that the inner cone has maximum volume
when h − 13 H.
50. A
n object with weight W is dragged along a horizontal plane
by a force acting along a rope attached to the object. If the
rope makes an angle with a plane, then the magnitude of the
force is
42. A
poster is to have an area of 900 cm2 with 2.5 cm margins at
the bottom and sides and a 5 cm margin at the top. What
dimensions will give the largest printed area?
43. A
piece of wire 10 m long is cut into two pieces. One piece is
bent into a square and the other is bent into an equilateral
triangle. How should the wire be cut so that the total area
enclosed is (a) a maximum? (b) A minimum?
44. A
nswer Exercise 43 if one piece is bent into a square and the
other into a circle.
45. I f you are offered one slice from a round pizza (in other
words, a sector of a circle) and the slice must have a perimeter of 60 cm, what diameter pizza will reward you with the
largest slice?
46. A
fence 2 m tall runs parallel to a tall building at a distance of
1 m from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall
of the building?
47. A
cone-shaped drinking cup is made from a circular piece of
paper of radius R by cutting out a sector and joining the edges
CA and CB. Find the maximum capacity of such a cup.
sin 1 cos where is a constant called the coefficient of friction. For
what value of is F smallest?
51. I f a resistor of R ohms is connected across a battery of
E volts with internal resistance r ohms, then the power
(in watts) in the external resistor is
E 2R
sR 1 rd 2
If E and r are fixed but R varies, what is the maximum value
of the power?
52. F
or a fish swimming at a speed v relative to the water, the
energy expenditure per unit time is proportional to v 3. It is
believed that migrating fish try to minimize the total energy
required to swim a fixed distance. If the fish are swimming
against a current u su , vd, then the time required to swim a
distance L is Lysv 2 ud and the total energy E required to
swim the distance is given by
Esvd − av 3 L
where a is the proportionality constant.
(a) Determine the value of v that minimizes E.
(b) Sketch the graph of E.
Note: This result has been verified experimentally; migrating
fish swim against a current at a speed 50% greater than the
current speed.
53. In a beehive, each cell is a regular hexagonal prism, open at
one end; the other end is capped by three congruent rhombi
forming a trihedral angle at the apex, as in the figure on
page 274. Let be the angle at which each rhombus meets
the altitude, s the side length of the hexagon, and h the length
of the longer base of the trapezoids on the sides of the cell. It
can be shown that if s and h are held fixed, then the volume of
the cell is constant (independent of ), and for a given value
of the surface area S of the cell is
S − 6sh 2 32 s 2 cot 1 32 s3 s 2 csc It is believed that bees form their cells in such a way as to
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
minimize surface area, thus using the least amount of wax
in cell construction.
(a) Calculate dSyd.
(b) What angle should the bees prefer?
(c)Determine the minimum surface area of the cell in
terms of s and h.
Note: Actual measurements of the angle in beehives have
been made, and the measures of these angles seldom differ
from the calculated value by more than 28.
59. T
he illumination of an object by a light source is directly
proportional to the strength of the source and inversely proportional to the square of the distance from the source. If
two light sources, one three times as strong as the other,
are placed 4 m apart, where should an object be placed
on the line between the sources so as to receive the least
60. F
ind an equation of the line through the point s3, 5d that
cuts off the least area from the first quadrant.
61. L
et a and b be positive numbers. Find the length of the
shortest line segment that is cut off by the first quadrant and
passes through the point sa, bd.
62. A
t which points on the curve y − 1 1 40x 3 2 3x 5 does the
tangent line have the largest slope?
63. W
hat is the shortest possible length of the line segment that
is cut off by the first quadrant and is tangent to the curve
y − 3yx at some point?
open end
54. A
boat leaves a dock at 2:00 pm and travels due south at a
speed of 20 kmyh. Another boat has been heading due east
at 15 kmyh and reaches the same dock at 3:00 pm. At what
time were the two boats closest together?
55. S
olve the problem in Example 4 if the river is 5 km wide
and point B is only 5 km downstream from A.
56. A
woman at a point A on the shore of a circular lake with
radius 3 km wants to arrive at the point C diametrically
opposite A on the other side of the lake in the shortest
possible time (see the figure). She can walk at the rate of
6 kmyh and row a boat at 3 kmyh. How should she proceed?
57. A
n oil refinery is located on the north bank of a straight
river that is 2 km wide. A pipeline is to be constructed from
the refinery to storage tanks located on the south bank of the
river 6 km east of the refinery. The cost of laying pipe is
$400,000ykm over land to a point P on the north bank and
$800,000ykm under the river to the tanks. To minimize the
cost of the pipeline, where should P be located?
58. S
uppose the refinery in Exercise 57 is located 1 km north
of the river. Where should P be located?
64. W
hat is the smallest possible area of the triangle that is cut
off by the first quadrant and whose hypotenuse is tangent to
the parabola y − 4 2 x 2 at some point?
65. (a)If Csxd is the cost of producing x units of a commodity,
then the average cost per unit is csxd − Csxdyx. Show
that if the average cost is a minimum, then the marginal
cost equals the average cost.
(b)If Csxd − 16,000 1 200x 1 4x 3y2, in dollars, find
(i) the cost, average cost, and marginal cost at a production level of 1000 units; (ii) the production level
that will minimize the average cost; and (iii) the minimum average cost.
66. (a)Show that if the profit Psxd is a maximum, then the
marginal revenue equals the marginal cost.
(b)If Csxd − 16,000 1 500x 2 1.6x 2 1 0.004x 3 is the
cost function and psxd − 1700 2 7x is the demand
function, find the production level that will maximize
67. A
baseball team plays in a stadium that seats 55,000 spectators. With ticket prices at $10, the average attendance had
been 27,000. When ticket prices were lowered to $8, the
average attendance rose to 33,000.
(a) Find the demand function, assuming that it is linear.
(b) How should ticket prices be set to maximize revenue?
68. D
uring the summer months Terry makes and sells necklaces
on the beach. Last summer she sold the necklaces for $10
each and her sales averaged 20 per day. When she increased
the price by $1, she found that the average sales decreased
by two per day.
(a) Find the demand function, assuming that it is linear.
(b)If the material for each necklace costs $6, what selling
price should Terry set to maximize her profit?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.7Optimization Problems
69. A
retailer has been selling 1200 tablet computers a week at
$350 each. The marketing department estimates that an
additional 80 tablets will sell each week for every $10 that
the price is lowered.
(a) Find the demand function.
(b)What should the price be set at in order to maximize
(c)If the retailer’s weekly cost function is
and C is minimized (see the figure). Express L as a function
of x − AP and use the graphs of L and dLydx to estimate
the minimum value of L.
| |
Csxd − 35,000 1 120x
what price should it choose in order to maximize its
70. A
company operates 16 oil wells in a designated area. Each
pump, on average, extracts 240 barrels of oil daily. The
company can add more wells but every added well reduces
the average daily ouput of each of the wells by 8 barrels.
How many wells should the company add in order to maximize daily production?
71. S
how that of all the isosceles triangles with a given perimeter, the one with the greatest area is equilateral.
72. C
onsider the situation in Exercise 57 if the cost of laying
pipe under the river is considerably higher than the cost of
laying pipe over land ($400,000ykm). You may suspect that
in some instances, the minimum distance possible under
the river should be used, and P should be located 6 km from
the refinery, directly across from the storage tanks. Show
that this is never the case, no matter what the “under river”
cost is.
73. C
onsider the tangent line to the ellipse 2 1 2 − 1 at
a point s p, qd in the first quadrant.
(a)Show that the tangent line has x-intercept a 2yp and
y-intercept b 2yq.
(b)Show that the portion of the tangent line cut off by the
coordinate axes has minimum length a 1 b.
(c)Show that the triangle formed by the tangent line and
the coordinate axes has minimum area ab.
74. T
he frame for a kite is to be made from six pieces of wood.
The four exterior pieces have been cut with the lengths indicated in the figure. To maximize the area of the kite, how
long should the diagonal pieces be?
76. T
he graph shows the fuel consumption c of a car (measured
in liters per hour) as a function of the speed v of the car.
At very low speeds the engine runs inefficiently, so initially
c decreases as the speed increases. But at high speeds the
fuel consumption increases. You can see that csvd is minimized for this car when v < 48 kmyh. However, for fuel
efficiency, what must be minimized is not the consumption
in gallons per hour but rather the fuel consumption in liters
per kilometer. Let’s call this consumption G. Using the
graph, estimate the speed at which G has its minimum
77. L
et v1 be the velocity of light in air and v2 the velocity of
light in water. According to Fermat’s Principle, a ray of light
will travel from a point A in the air to a point B in the water
by a path ACB that minimizes the time taken. Show that
sin 1
sin 2
where 1 (the angle of incidence) and 2 (the angle of refraction) are as shown. This equation is known as Snell’s Law.
; 75. A point P needs to be located somewhere on the line AD so
that the total length L of cables linking P to the points A, B,
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
78. T
wo vertical poles PQ and ST are secured by a rope PRS
going from the top of the first pole to a point R on the ground
between the poles and then to the top of the second pole as in
the figure. Show that the shortest length of such a rope occurs
when 1 − 2.
other. Find the maximum value of the observer’s angle of
sight between the runners. [Hint: Maximize tan .]
82. A
rain gutter is to be constructed from a metal sheet of width
30 cm by bending up one-third of the sheet on each side
through an angle . How should be chosen so that the gutter
will carry the maximum amount of water?
79. T
he upper right-hand corner of a piece of paper, 30 centimeters by 20 centimeters , as in the figure, is folded over to the
bottom edge. How would you fold the paper so as to minimize the length of the fold? In other words, how would you
choose x to minimize y?
10 cm
10 cm
10 cm
83. F
ind the maximum area of a rectangle that can be circum­
scribed about a given rectangle with length L and width W.
[Hint: Express the area as a function of an angle .]
80. A
steel pipe is being carried down a hallway that is 3 m
wide. At the end of the hall there is a right-angled turn into
a narrower hallway, 2 m wide. What is the length of the
longest pipe that can be carried horizontally around the
81. A
n observer stands at a point P, one unit away from a track.
Two runners start at the point S in the following figure and
run along the track. One runner runs three times as fast as the
84. T
he blood vascular system consists of blood vessels (arteries,
arterioles, capillaries, and veins) that convey blood from the
heart to the organs and back to the heart. This system should
work so as to minimize the energy expended by the heart in
pumping the blood. In particular, this energy is reduced when
the resistance of the blood is lowered. One of Poiseuille’s
Laws gives the resistance R of the blood as
where L is the length of the blood vessel, r is the radius, and
C is a positive constant determined by the viscosity of the
blood. (Poiseuille established this law experimentally, but it
also follows from Equation 8.4.2.) The following figure
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.7Optimization Problems
shows a main blood vessel with radius r1 branching at an
angle into a smaller vessel with radius r 2.
(a)Use Poiseuille’s Law to show that the total resistance of
the blood along the path ABC is
a 2 b cot b csc 1
where a and b are the distances shown in the figure.
(b) Prove that this resistance is minimized when
cos −
(c)What should the value of WyL be in order for the bird to
fly directly to its nesting area D? What should the value
of WyL be for the bird to fly to B and then along the
shore to D?
(d)If the ornithologists observe that birds of a certain
species reach the shore at a point 4 km from B, how
many times more energy does it take a bird to fly over
water than over land?
(c)Find the optimal branching angle (correct to the nearest
degree) when the radius of the smaller blood vessel is
two-thirds the radius of the larger vessel.
5 km
13 km
85. O
rnithologists have determined that some species of birds
tend to avoid flights over large bodies of water during daylight hours. It is believed that more energy is required to fly
over water than over land because air generally rises over
land and falls over water during the day. A bird with these
tendencies is released from an island that is 5 km from the
nearest point B on a straight shoreline, flies to a point C on
the shoreline, and then flies along the shoreline to its nesting
area D as in the figure. Assume that the bird instinctively
chooses a path that will minimize its energy expenditure.
Points B and D are 13 km apart.
(a)In general, if it takes 1.4 times as much energy to fly
over water as it does over land, to what point C should
the bird fly in order to minimize the total energy
expended in returning to its nesting area?
(b)Let W and L denote the energy (in joules) per kilometer
flown over water and land, respectively. What would a
large value of the ratio WyL mean in terms of the bird’s
flight? What would a small value mean? Determine the
ratio WyL corresponding to the minimum expenditure
of energy.
wo light sources of identical strength are placed 10 m
; 86. T
apart. An object is to be placed at a point P on a line ,,
paral­lel to the line joining the light sources and at a distance
d meters from it (see the figure). We want to locate P on ,
so that the intensity of illumination is minimized. We need
to use the fact that the intensity of illumination for a single
source is directly proportional to the strength of the source
and inversely proportional to the square of the distance from
the source.
(a)Find an expression for the intensity Isxd at the point P.
(b)If d − 5 m, use graphs of Isxd and I9sxd to show that the
intensity is minimized when x − 5 m, that is, when P is
at the midpoint of ,.
(c)If d − 10 m, show that the intensity (perhaps surpris­
ingly) is not minimized at the midpoint.
(d)Somewhere between d − 5 m and d − 10 m there is a
transitional value of d at which the point of minimal
illumination abruptly changes. Estimate this value
of d by graphical methods. Then find the exact value
of d.
10 m
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
THE SHAPE OF A CAN
In this project we investigate the most economical shape for a can. We first interpret this to
mean that the volume V of a cylindrical can is given and we need to find the height h and
radius r that minimize the cost of the metal to construct the can (see the figure). If we disregard any waste metal in the manufacturing process, then the problem is to minimize the surface area of the cylinder. We solved this problem in Example 3.7.2 and we found that h − 2r ;
that is, the height should be the same as the diameter. But if you go to your cupboard or your
supermarket with a ruler, you will discover that the height is usually greater than the diameter
and the ratio hyr varies from 2 up to about 3.8. Let’s see if we can explain this phenomenon.
1. The material for the cans is cut from sheets of metal. The cylindrical sides are formed by
bending rectangles; these rectangles are cut from the sheet with little or no waste. But if the
top and bottom discs are cut from squares of side 2r (as in the figure), this leaves considerable waste metal, which may be recycled but has little or no value to the can makers. If this
is the case, show that the amount of metal used is minimized when
< 2.55
Discs cut from squares
2. A
more efficient packing of the discs is obtained by dividing the metal sheet into hexagons
and cutting the circular lids and bases from the hexagons (see the figure). Show that if this
strategy is adopted, then
4 s3
< 2.21
Discs cut from hexagons
3. T
he values of hyr that we found in Problems 1 and 2 are a little closer to the ones that
actually occur on supermarket shelves, but they still don’t account for everything. If
we look more closely at some real cans, we see that the lid and the base are formed from
discs with radius larger than r that are bent over the ends of the can. If we allow for this
we would increase hyr. More significantly, in addition to the cost of the metal we need to
incorporate the manufacturing of the can into the cost. Let’s assume that most of the
expense is incurred in joining the sides to the rims of the cans. If we cut the discs from
hexagons as in Problem 2, then the total cost is proportional to
4 s3 r 2 1 2rh 1 ks4r 1 hd
where k is the reciprocal of the length that can be joined for the cost of one unit area of
metal. Show that this expression is minimized when
2 2 hyr
hyr 2 4 s3
lot s
V yk as a function of x − hyr and use your graph to argue that when a can is large or
; 4. P
joining is cheap, we should make hyr approximately 2.21 (as in Problem 2). But when the
can is small or joining is costly, hyr should be substantially larger.
5. O
ur analysis shows that large cans should be almost square but small cans should be tall
and thin. Take a look at the relative shapes of the cans in a supermarket. Is our conclusion
usually true in practice? Are there exceptions? Can you suggest reasons why small cans are
not always tall and thin?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPLIED PROJECTPlanes and Birds: Minimizing Energy
Targn Pleiades
Pleiades // Shutterstock.com
Small birds like finches alternate between flapping their wings and keeping them folded while
gliding (see Figure 1). In this project we analyze this phenomenon and try to determine how
frequently a bird should flap its wings. Some of the principles are the same as for fixed-wing
aircraft and so we begin by considering how required power and energy depend on the speed
of airplanes.1
FIGURE 1
1. The power needed to propel an airplane forward at velocity v is
P − Av 3 1
BL 2
where A and B are positive constants specific to the particular aircraft and L is the lift, the
upward force supporting the weight of the plane. Find the speed that minimizes the
required power.
2. The speed found in Problem 1 minimizes power but a faster speed might use less fuel. The
energy needed to propel the airplane a unit distance is E − Pyv. At what speed is energy
3. How much faster is the speed for minimum energy than the speed for minimum power?
4. I n applying the equation of Problem 1 to bird flight we split the term Av 3 into two parts:
Ab v 3 for the bird’s body and Aw v 3 for its wings. Let x be the fraction of flying time spent in
flapping mode. If m is the bird’s mass and all the lift occurs during flapping, then the lift is
mtyx and so the power needed during flapping is
Pflap − sA b 1 Awdv 3 1
The power while wings are folded is Pfold − A b v 3. Show that the average power over an
entire flight cycle is
P − x Pflap 1 s1 2 xdPfold − A b v 3 1 x Aw v 3 1
Bm2t 2
5. F
or what value of x is the average power a minimum? What can you conclude if the bird
flies slowly? What can you conclude if the bird flies faster and faster?
6. The average energy over a cycle is E − Pyv. What value of x minimizes E?
1. Adapted from R. McNeill Alexander, Optima for Animals (Princeton, NJ: Princeton University
Press, 1996.)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
3.8 Newton’s Method
Suppose that a car dealer offers to sell you a car for $18,000 or for payments of $375 per
month for five years. You would like to know what monthly interest rate the dealer is, in
effect, charging you. To find the answer, you have to solve the equation
FIGURE 1
Try to solve Equation 1 numerically
using a calculator or computer. Some
machines are not able to solve it.
Others are successful but require you
to specify a starting point for the
{x ¡, f(x¡)}
FIGURE 2
x™ x¡
48xs1 1 xd60 2 s1 1 xd60 1 1 − 0
(The details are explained in Exercise 39.) How would you solve such an equation?
For a quadratic equation ax 2 1 bx 1 c − 0 there is a well-known formula for the
solutions. For third- and fourth-degree equations there are also formulas for the solutions, but they are extremely complicated. If f is a polynomial of degree 5 or higher,
there is no such formula. Likewise, there is no formula that will enable us to find the
exact solutions of a transcendental equation such as cos x − x.
We can find an approximate solution to Equation 1 by plotting the left side of the
equation and finding the x-intercepts. Using a graphing calculator (or computer), and
after experimenting with viewing rectangles, we pro­duce the graph in Figure 1.
We see that in addition to the solution x − 0, which doesn’t interest us, there is a solution between 0.007 and 0.008. Zooming in shows that the x-intercept is approximately
0.0076. If we need more accuracy than graphing provides, we can use a calculator or
computer algebra system to solve the equation numerically. If we do so, we find that the
solution, correct to nine decimal places, is 0.007628603.
How do these devices solve equations? They use a variety of methods, but most of
them make some use of Newton’s method, also called the Newton-Raphson method.
We will explain how this method works, partly to show what happens inside a calculator
or computer, and partly as an application of the idea of linear approximation.
The geometry behind Newton’s method is shown in Figure 2. We wish to solve an equation of the form f sxd − 0, so the solutions of the equation correspond to the x-intercepts of
the graph of f. The solution that we are trying to find is labeled r in the figure. We start with
a first approximation x 1, which is obtained by guess­ing, or from a rough sketch of the graph
of f, or from a computer-generated graph of f. Consider the tangent line L to the curve
y − f sxd at the point sx 1, f sx 1dd and look at the x-intercept of L, labeled x 2. The idea behind
Newton’s method is that the tangent line is close to the curve and so its x-intercept, x2, is
close to the x-intercept of the curve (namely, the solution r that we are seeking). Because
the tangent is a line, we can easily find its x-intercept.
To find a formula for x2 in terms of x1 we use the fact that the slope of L is f 9sx1 d, so
its equation is
y 2 f sx 1 d − f 9sx 1 dsx 2 x 1 d
Since the x-intercept of L is x 2, we know that the point sx 2 , 0d is on the line, and so
0 2 f sx 1 d − f 9sx 1 dsx 2 2 x 1 d
If f 9sx 1d ± 0, we can solve this equation for x 2:
x2 − x1 2
f sx 1 d
f 9sx 1 d
We use x 2 as a second approximation to r.
Next we repeat this procedure with x 1 replaced by the second approximation x 2, using
the tangent line at sx 2 , f sx 2 dd. This gives a third approximation:
x3 − x2 2
f sx 2 d
f 9sx 2 d
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.8Newton’s Method
If we keep repeating this process, we obtain a sequence of approximations x 1, x 2, x 3, x 4, . . .
as shown in Figure 3. In general, if the nth approximation is x n and f 9sx n d ± 0, then
the next approximation is given by
{x¡, f(x¡)}
{x™, f(x™)}
x™ x¡
x n11 − x n 2
lim x n − r
Sequences are discussed in more
detail in Section 11.1.
lthough the sequence of successive approximations converges to the desired solution
for fun­ctions of the type illustrated in Figure 3, in certain circumstances the sequence
may not converge. For example, consider the situation shown in Figure 4. You can see
that x 2 is a worse approximation than x 1. This is likely to be the case when f 9sx 1d is close
to 0. It might even happen that an approximation (such as x 3 in Figure 4) falls outside the
domain of f. Then Newton’s method fails and a better initial approximation x 1 should be
chosen. See Exercises 29–32 for specific examples in which Newton’s method works
very slowly or does not work at all.
f sx n d
f 9sx n d
If the numbers x n become closer and closer to r as n becomes large, then we say that
the sequence converges to r and we write
FIGURE 3
EXAMPLE 1 Starting with x 1 − 2, find the third approximation x 3 to the solution of
the equation x 3 2 2x 2 5 − 0.
FIGURE 4
SOLUTION We apply Newton’s method with
f sxd − x 3 2 2x 2 5andf 9sxd − 3x 2 2 2
Figure 5 shows the geometry behind
the first step in Newton’s method in
Example 1. Since f 9s2d − 10, the tangent line to y − x 3 2 2x 2 5 at
s2, 21d has equation y − 10x 2 21
so its x-intercept is x 2 − 2.1.
Newton himself used this equation to illustrate his method and he chose x 1 − 2 after
some experimentation because f s1d − 26, f s2d − 21, and f s3d − 16. Equation 2
f sx nd
x n3 2 2x n 2 5
x n11 − x n 2
− xn 2
f 9sx nd
3x n2 2 2
With n − 1 we have
x2 − x1 2
FIGURE 5
f sx1 d
x13 2 2x 1 2 5
− x1 2
f 9sx1d
3x12 2 2
2 3 2 2s2d 2 5
− 2.1
3s2d2 2 2
Then with n − 2 we obtain
x3 − x2 2
x 23 2 2x 2 2 5
s2.1d3 2 2s2.1d 2 5
< 2.0946
3x 22 2 2
3s2.1d2 2 2
It turns out that this third approximation x 3 < 2.0946 is accurate to four decimal
Suppose that we want to achieve a given accuracy, say to eight decimal places,
using Newton’s method. How do we know when to stop? The rule of thumb that is
generally used is: stop when successive approximations x n and x n11 agree to eight decimal places. (A precise statement concerning accuracy in Newton’s method will be
given in Exercise 11.11.39.)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
Notice that the procedure in going from n to n 1 1 is the same for all values of n.
(It is called an iterative process.) This means that Newton’s method is particularly convenient for use with a programmable calculator or a computer.
2 correct to eight decimal places.
EXAMPLE 2 Use Newton’s method to find s
2 is equivalent to finding the positive
SOLUTION First we observe that finding s
solution of the equation
x6 2 2 − 0
so we take f sxd − x 6 2 2. Then f 9sxd − 6x 5 and Formula 2 (Newton’s method)
fsx nd
x n6 2 2
x n11 − x n 2
− xn 2
f 9sx nd
6x n5
If we choose x 1 − 1 as the initial approximation, then we obtain
x 2 < 1.16666667
x 3 < 1.12644368
x 4 < 1.12249707
x 5 < 1.12246205
x 6 < 1.12246205
Since x 5 and x 6 agree to eight decimal places, we conclude that
2 < 1.12246205
to eight decimal places.
EXAMPLE 3 Find, correct to six decimal places, the solution of the equation cos x − x.
SOLUTION We first rewrite the equation in standard form: cos x 2 x − 0. Therefore
we let f sxd − cos x 2 x. Then f 9sxd − 2sin x 2 1, so Formula 2 becomes
x n11 − x n 2
y=cos x
FIGURE 6
cos x n 2 x n
cos x n 2 x n
− xn 1
2sin x n 2 1
sin x n 1 1
In order to guess a suitable value for x 1 we sketch the graphs of y − cos x and y − x in
Figure 6. It appears that they intersect at a point whose x-coordinate is somewhat less
than 1, so let’s take x 1 − 1 as a convenient first approximation. Then, remembering to
put our calculator in radian mode, we get
x 2 < 0.75036387
x 3 < 0.73911289
x 4 < 0.73908513
x 5 < 0.73908513
Since x 4 and x 5 agree to six decimal places (eight, in fact), we conclude that the
solution of the equation, correct to six decimal places, is 0.739085.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.8Newton’s Method
Instead of using the rough sketch in Figure 6 to get a starting approximation for
Newton’s method in Example 3, we could have used the more accurate graph that a calculator or computer provides. Figure 7 suggests that we use x1 − 0.75 as the initial
approximation. Then Newton’s method gives
y=cos x
x 2 < 0.73911114x 3 < 0.739085131
x 4 < 0.73908513
FIGURE 7
and so we obtain the same answer as before, but with one fewer step.
1. T
he figure shows the graph of a function f . Suppose that
Newton’s method is used to approximate the solution s of
the equation f sxd − 0 with initial approximation x 1 − 6.
(a)Draw the tangent lines that are used to find x 2 and x 3, and
esti­mate the numerical values of x 2 and x 3.
(b)Would x 1 − 8 be a better first approximation? Explain.
5. F
or which of the initial approximations x1 − a, b, c, and d
do you think Newton’s method will work and lead to the
solution of the equation f sxd − 0?
6. 2 x 3 2 3x 2 1 2 − 0,
2. F
ollow the instructions for Exercise 1(a) but use x 1 − 1
as the starting approximation for finding the solution r.
3. S
uppose the tangent line to the curve y − f sxd at the point
s2, 5d has the equation y − 9 2 2x. If Newton’s method is
used to locate a solution of the equation f sxd − 0 and the initial approximation is x1 − 2, find the second approximation x 2.
4. F
or each initial approximation, determine graphically what
happens if Newton’s method is used for the function whose
graph is shown.
(a) x1 − 0
(b) x1 − 1
(c) x1 − 3
(d) x1 − 4
(e) x1 − 5
6–8 Use Newton’s method with the specified initial approximation x 1 to find x 3, the third approximation to the solution of the
given equation. (Give your answer to four decimal places.)
2 x 2 1 1 − 0,
8. x 5 − x 2 1 1,
x 1 − 21
x1 − 2
x1 − 1
se Newton’s method with initial approximation x1 − 21
; 9. U
to find x 2, the second approximation to the solution of
the equation x 3 1 x 1 3 − 0. Explain how the method
works by first graphing the function and its tangent line
at s21, 1d.
se Newton’s method with initial approximation x1 − 1
; 10. U
to find x 2, the second approximation to the solution of
the equation x 4 2 x 2 1 − 0. Explain how the method
works by first graphing the function and its tangent line
at s1, 21d.
11–12 Use Newton’s method to approximate the given number
correct to eight decimal places.
11. s
12. s
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
30. (a)Use Newton’s method with x 1 − 1 to find the solution
of the equation x 3 2 x − 1 correct to six decimal
(b)Solve the equation in part (a) using x 1 − 0.6 as the
13. 3x 2 8x 1 2 − 0,f2, 3g
initial approximation.
14. 22 x 5 1 9x 4 2 7x 3 2 11x − 0,f3, 4g
(c)Solve the equation in part (a) using x 1 − 0.57. (You
definitely need a programmable calculator for this part.)
f sxd − x 3 2 x 2 1 and its tangent lines at
15–16 Use Newton’s method to approximate the indicated solux
0.6, and 0.57 to explain why Newton’s method
tion of the equation correct to six decimal places.
is so sensitive to the value of the initial approximation.
15. The negative solution of cos x − x 2 2 4
31. Explain why Newton’s method fails when applied to the
16. The positive solution of 3 sin x − x
equation s
x − 0 with any initial approximation x 1 ± 0.
Illustrate your explanation with a sketch.
17–22 Use Newton’s method to find all solutions of the
32. If
if x > 0
equation correct to six decimal places.
f sxd −
2s2x if x , 0
17. sin x − x 2 1
18. cos 2x − x 3
then the solution of the equation f sxd − 0 is x − 0. Explain
x 21
20. sx 2 1d 2 − sx
Newton’s method fails to find the solution no matter
which initial approximation x 1 ± 0 is used. Illustrate your
21. x 3 − cos x
22. x 3 − 5x 2 3
explanation with a sketch.
13–14 (a) Explain how we know that the given equation must
have a solution in the given interval. (b) Use Newton’s method
to approximate the solution correct to six decimal places.
; 23–26 Use Newton’s method to find all the solutions of the
equation correct to eight decimal places. Start by looking at a
graph to find initial approximations.
23. 22x 7 2 5x 4 1 9x 3 1 5 − 0
34. U
se Newton’s method to find the absolute maximum value
of the function f sxd − x cos x, 0 < x < , correct to six
decimal places.
24. x 5 2 3x 4 1 x 3 2 x 2 2 x 1 6 − 0
33. (a)Use Newton’s method to find the critical numbers of the
function f sxd − x 6 2 x 4 1 3x 3 2 2x correct to six
decimal places.
(b)Find the absolute minimum value of f correct to four
decimal places.
− s1 2 x
x2 1 1
26. cossx 2 2 xd − x 4
35. U
se Newton’s method to find the coordinates of the inflection point of the curve y − x 2 sin x, 0 < x < p, correct to
six decimal places.
27. (a)Apply Newton’s method to the equation x 2 2 a − 0 to
derive the following square-root algorithm (used by the
ancient Babylonians to compute sa ):
36. O
f the infinitely many lines that are tangent to the curve
y − 2sin x and pass through the origin, there is one that has
the largest slope. Use Newton’s method to find the slope of
that line correct to six decimal places.
x n11 −
xn 1
(b)Use part (a) to compute s1000 correct to six decimal
28. (a)Apply Newton’s method to the equation 1yx 2 a − 0 to
derive the following reciprocal algorithm:
37. U
se Newton’s method to find the coordinates, correct to six
decimal places, of the point on the parabola y − sx 2 1d 2
that is closest to the origin.
38. I n the figure, the length of the chord AB is 4 cm and the
length of the arc AB is 5 cm. Find the central angle ,
in radians, correct to four decimal places. Then give the
answer to the nearest degree.
x n11 − 2x n 2 ax n2
(This algorithm enables a computer to find reciprocals
without actually dividing.)
(b)Use part (a) to compute 1y1.6984 correct to six decimal
5 cm
4 cm
29. E
xplain why Newton’s method doesn’t work for finding the
solution of the equation x 3 2 3x 1 6 − 0 if the initial
approximation is chosen to be x 1 − 1.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.9Antiderivatives
39. A
car dealer sells a new car for $18,000. He also offers to sell
the same car for payments of $375 per month for five years.
What monthly interest rate is this dealer charging?
To solve this problem you will need to use the formula
for the present value A of an annuity consisting of n equal
payments of size R with interest rate i per time period:
f1 2 s1 1 i d2n g
points. (A solar research satellite has been placed at one of
these libration points.) If m1 is the mass of the sun, m 2 is the
mass of the earth, and r − m 2ysm1 1 m 2 d, it turns out that the
x-coordinate of L 1 is the unique solution of the fifth-degree
psxd − x 5 2 s2 1 rdx 4 1 s1 1 2rdx 3 2 s1 2 rdx 2
1 2s1 2 rdx 1 r 2 1 − 0
and the x-coordinate of L 2 is the solution of the equation
Replacing i by x, show that
psxd 2 2rx 2 − 0
48xs1 1 xd 2 s1 1 xd 1 1 − 0
Use Newton’s method to solve this equation.
Using the value r < 3.04042 3 10 26, find the locations of the
libration points (a) L 1 and (b) L 2.
40. T
he figure shows the sun located at the origin and the earth
at the point s1, 0d. (The unit here is the distance between
the centers of the earth and the sun, called an astronomical
unit: 1 AU < 1.496 3 10 8 km.) There are five locations
L 1, L 2, L 3, L 4, and L 5 in this plane of rotation of the earth
about the sun where a satellite remains motionless with
respect to the earth because the forces acting on the satellite
(including the gravitational attractions of the earth and the
sun) balance each other. These locations are called libration
3.9 Antiderivatives
A physicist who knows the velocity of a particle might wish to know its position at a
given time. An engineer who can measure the variable rate at which water is leaking
from a tank wants to know the amount leaked over a certain time period. A biologist who
knows the rate at which a bacteria population is increasing might want to deduce what
the size of the population will be at some future time. In each case, the problem is to find
a function whose derivative is a known function.
■ The Antiderivative of a Function
If we have a function F whose derivative is the function f, then F is called an antiderivative of f.
Definition A function F is called an antiderivative of f on an interval I if
F9sxd − f sxd for all x in I.
For instance, let f sxd − x 2. It isn’t difficult to discover an antiderivative of f if we
keep the Power Rule in mind. In fact, if Fsxd − 13 x 3, then F9sxd − x 2 − f sxd. But the
function Gsxd − 13 x 3 1 100 also satisfies G9sxd − x 2. Therefore both F and G are antiderivatives of f. Indeed, any function of the form Hsxd − 13 x 3 1 C, where C is a constant,
is an antiderivative of f. The question arises: are there any others?
To answer this question, recall that in Section 3.2 we used the Mean Value Theorem to
prove that if two functions have identical derivatives on an interval, then they must differ
by a constant (Corollary 3.2.7). Thus if F and G are any two antiderivatives of f , then
F9sxd − f sxd − G9sxd
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
y= 3 ˛+3
so Gsxd 2 Fsxd − C, where C is a constant. We can write this as Gsxd − Fsxd 1 C, so
we have the following result.
y= 3 ˛+2
y= 3 ˛+1
y= 3 ˛
y= 3 ˛-1
y= 3 ˛-2
FIGURE 1
Members of the family of
antiderivatives of f sxd − x 2
1 Theorem If F is an antiderivative of f on an interval I, then the most general
antiderivative of f on I is
Fsxd 1 C
where C is an arbitrary constant.
Going back to the function f sxd − x 2, we see that the general antiderivative of f is
1 C. By assigning specific values to the constant C, we obtain a family of functions
whose graphs are vertical translates of one another (see Figure 1). This makes sense
because each curve must have the same slope at any given value of x.
EXAMPLE 1 Find the most general antiderivative of each of the following functions.
(a) f sxd − sin x(b) f sxd − x n,n > 0(c) f sxd − x 23
(a) If Fsxd − 2cos x, then F9sxd − sin x, so an antiderivative of sin x is 2cos x. By
Theorem 1, the most general antiderivative is Gsxd − 2cos x 1 C.
(b) We use the Power Rule to discover an antiderivative of x n :
S D
x n11
sn 1 1dx n
− xn
Therefore the general antiderivative of f sxd − x n is
Fsxd −
x n11
This is valid for n > 0 because then f sxd − x n is defined on an interval.
(c) If we put n − 23 in the antiderivative formula from part (b), we get the particular
antiderivative Fsxd − x 22ys22d. But notice that f sxd − x 23 is not defined at x − 0.
Thus Theorem 1 tells us only that the general antiderivative of f is x 22ys22d 1 C on
any interval that does not contain 0. So the general antiderivative of f sxd − 1yx 3 is
1 C1 if x . 0
2x 2
F sxd −
2 2 1 C2 if x , 0
■ Antidifferentiation Formulas
As in Example 1, every differentiation formula, when read from right to left, gives rise to
an antidifferentiation formula. In Table 2 we list some particular antiderivatives. Each for­
mula in the table is true because the derivative of the function in the right column appears
in the left column. In particular, the first formula says that the antiderivative of a constant
times a function is the constant times the antiderivative of the function. The second formula says that the antiderivative of a sum is the sum of the antiderivatives. (We use the
notation F9− f , G9 − t.)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.9Antiderivatives
2 Table of Antidifferentiation Formulas
To obtain the most general anti­
derivative from the particular ones in
Table 2, we have to add a constant (or
constants), as in Example 1.
Particular antiderivative
Particular antiderivative
c f sxd
cos x
sin x
f sxd 1 tsxd
Fsxd 1 Gsxd
sin x
2cos x
x n sn ± 21d
x n11
sec x
tan x
sec x tan x
sec x
EXAMPLE 2 Find all functions t such that
t9sxd − 4 sin x 1
2x 5 2 sx
SOLUTION We first rewrite the given function as follows:
t9sxd − 4 sin x 1
2x 5
− 4 sin x 1 2x 4 2
Thus we want to find an antiderivative of
t9sxd − 4 sin x 1 2x 4 2 x21y2
Using the formulas in Table 2 together with Theorem 1, we obtain
We often use a capital letter F to represent an antiderivative of a function f.
If we begin with derivative notation, f 9,
an antiderivative is f, of course.
tsxd − 4s2cos xd 1 2
2 1 1C
− 24 cos x 1 25 x 5 2 2sx 1 C n
In applications of calculus it is very common to have a situation as in Example 2,
where it is required to find a function, given knowledge about its derivatives. An equation
that involves the derivatives of a function is called a differential equation. These will be
studied in some detail in Chapter 9, but for the present we can solve some elementary differential equations. The general solution of a differential equation involves an arbitrary con­stant (or constants) as in Example 2. However, there may be some extra conditions
given that will determine the constants and therefore uniquely specify the solution.
EXAMPLE 3 Find f if f 9sxd − xsx and f s1d − 2.
SOLUTION The general antiderivative of
f 9sxd − xsx − x 3y2
f sxd −
x 5y2
1 C − 25 x 5y2 1 C
To determine C we use the fact that f s1d − 2:
f s1d − 25 1 C − 2
Solving for C, we get C − 2 2 25 − 85, so the particular solution is
f sxd −
2x 5y2 1 8
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
EXAMPLE 4 Find f if f 0sxd − 12x 2 1 6x 2 4, f s0d − 4, and f s1d − 1.
SOLUTION The general antiderivative of f 0sxd − 12x 2 1 6x 2 4 is
f 9sxd − 12
2 4x 1 C − 4x 3 1 3x 2 2 4x 1 C
Using the antidifferentiation rules once more, we find that
f sxd − 4
1 Cx 1 D − x 4 1 x 3 2 2x 2 1 Cx 1 D
To determine C and D we use the given conditions that f s0d − 4 and f s1d − 1. Since
f s0d − 0 1 D − 4, we have D − 4. Since
f s1d − 1 1 1 2 2 1 C 1 4 − 1
we have C − 23. Therefore the required function is
f sxd − x 4 1 x 3 2 2x 2 2 3x 1 4
■ Graphing Antiderivatives
If we are given the graph of a function f, it seems reasonable that we should be able to
sketch the graph of an antiderivative F. Suppose, for instance, that we are given that
Fs0d − 1. Then we have a place to start, the point s0,1d, and the direction in which we
move our pencil is given at each stage by the derivative F9sxd − f sxd. In the next example
we use the principles of this chapter to show how to graph F even when we don’t have a
formula for f. This would be the case, for instance, when f sxd is determined by experimental data.
SOLUTION We are guided by the fact that the slope of y − Fsxd is f sxd. We start at
the point s0, 2d and draw F as an initially decreasing function since f sxd is negative
when 0 , x , 1. Notice that f s1d − f s3d − 0, so F has horizontal tangents when
x − 1 and x − 3. For 1 , x , 3, f sxd is positive and so F is increasing. We see that
F has a local minimum when x − 1 and a local maximum when x − 3. For x . 3,
f sxd is negative and so F is decreasing on s3, `d. Since f sxd l 0 as x l `, the graph
of F becomes flatter as x l `. Also notice that F0sxd − f 9sxd changes from positive
to negative at x − 2 and from negative to positive at x − 4, so F has inflection points
when x − 2 and x − 4. We use this information to sketch the graph of the antiderivative in Figure 3.
FIGURE 2
FIGURE 3
EXAMPLE 5 The graph of a function f is given in Figure 2. Make a rough sketch of
an antiderivative F, given that Fs0d − 2.
■ Linear Motion
Antidifferentiation is particularly useful in analyzing the motion of an object moving in
a straight line. Recall that if the object has position function s − f std, then the velocity
function is vstd − s9std. This means that the position function is an antiderivative of the
velocity function. Likewise, the acceleration function is astd − v9std, so the velocity
function is an antiderivative of the acceleration. If the acceleration and the initial values
ss0d and vs0d are known, then the position function can be found by antidifferentiating
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.9Antiderivatives
EXAMPLE 6 A particle moves in a straight line and has acceleration given by
astd − 6t 1 4. Its initial velocity is vs0d − 26 cmys and its initial displacement is
ss0d − 9 cm. Find its position function sstd.
SOLUTION Since v9std − astd − 6t 1 4, antidifferentiation gives
vstd − 6
1 4t 1 C − 3t 2 1 4t 1 C
Note that vs0d − C. But we are given that vs0d − 26, so C − 26 and
vstd − 3t 2 1 4t 2 6
Since vstd − s9std, s is the antiderivative of v :
sstd − 3
2 6t 1 D − t 3 1 2t 2 2 6t 1 D
This gives ss0d − D. We are given that ss0d − 9, so D − 9 and the required position
function is
sstd − t 3 1 2t 2 2 6t 1 9
An object near the surface of the earth is subject to a gravitational force that produces
a downward acceleration denoted by t. For motion close to the ground we may assume
that t is constant, its value being about 9.8 mys2 (or 32 ftys2 ). It is remarkable that from
the single fact that the acceleration due to gravity is constant, we can use calculus to
deduce the position and velocity of any object moving under the force of gravity, as
illustrated in the next example.
EXAMPLE 7 A ball is thrown upward with a speed of 15 mys from the edge of a cliff,
130 m above the ground. Find its height above the ground t seconds later. When does it
reach its maximum height? When does it hit the ground?
SOLUTION The motion is vertical and we choose the positive direction to be upward.
At time t the distance above the ground is sstd and the velocity vstd is decreasing.
Therefore the acceleration must be negative and we have
astd −
− 29.8
Taking antiderivatives, we have
vstd − 29.8t 1 C
To determine C we use the given information that vs0d − 15. This gives 15 − 0 1 C, so
vstd − 29.8t 1 15
The maximum height is reached when vstd − 0, that is, after 1.5 seconds. Since
s9std − vstd, we antidifferentiate again and obtain
sstd − 24.9t 2 1 15t 1 D
Using the fact that ss0d − 130, we have 130 − 0 1 D and so
sstd − 24.9t 2 1 15t 1 130
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
Figure 4 shows the position function
of the ball in Example 7. The graph
corroborates the con­clusions we
reached: the ball reaches its maximum height after 1.5 seconds and hits
the ground after about 6.9 seconds.
The expression for sstd is valid until the ball hits the ground. This happens when
sstd − 0, that is, when
24.9t 2 1 15t 1 130 − 0
or, equivalently,
4.9t 2 2 15t 2 130 − 0
Using the quadratic formula to solve this equation, we get
15 6 s2773
We reject the solution with the minus sign because it gives a negative value for t. Therefore the ball hits the ground after 15 1 s2773y9.8 < 6.9 seconds.
FIGURE 4
; 25–26 Find the antiderivative F of f that satisfies the given
condition. Check your answer by comparing the graphs of f
and F.
1–4 Find an antiderivative of the function.
1. (a) f sxd − 6
(b) tstd − 3t
2. (a) f sxd − 2x
(b) tsxd − 21yx 2
25. f sxd − 5x 4 2 2x 5, Fs0d − 4
3. (a) hsqd − cos q
(b) f sxd − sec x tan x
26. f sxd − x 1 2 sin x, Fs0d − 26
4. (a) tstd − sin t
(b) r sd − sec 2
27–48 Find f .
5–24 Find the most general antiderivative of the function.
(Check your answer by differentiation.)
5. f sxd − 4x 1 7
7. f sxd − 2x 2
6. f sxd − x 2 2 3x 1 2
1 5x
9. f sxd − xs12 x 1 8d
11. tsxd − 4x
2 2x
13. f sxd − 3sx 2 2 sx
17. f sxd −
10. f sxd − sx 2 5d 2
15. f std −
8. f sxd − 6x 2 8x 2 9x
2t 2 4 1 3st
12. hszd − 3z
30. f 0sxd − 6x 2 x 4 1 3x 5
31. f 0sxd − 4 2 s
32. f 0sxd − x 2y3 1 x 22y3
34. f -std − st 2 2 cos t
33. f -std − 12 1 sin t
36. f 9sxd − sx 2 2, f s9d − 4
14. tsxd − sx s2 2 x 1 6x d
37. f 9sxd − 5x 2y3,f s8d − 21
38. f 9std − t 1 1yt 3,t . 0,f s1d − 6
16. f sxd − s
5 2 4x 1 2x
29. f 0sxd − 4x 3 1 24x 2 1
35. f 9sxd − 5x 2 3x 1 4,f s21d − 2
18. tsxd −
28. f 0std − t 2 2 4
27. f 0sxd − 24x
39. f 9std − sec t ssec t 1 tan td,2y2 , t , y2,f sy4d − 21
40. f 9sxd − sx 1 1dysx ,f s1d − 5
19. f sd − 2 sin 2 3 sec tan 41. f 0sxd − 22 1 12x 2 12x 2, f s0d − 4,
20. f std − 3 cos t 2 4 sin t
21. hsd − 2 sin 2 sec 2 42. f 0sxd − 8x 3 1 5,
22. hsxd − sec 2x 1 cos x
23. tsvd − s
v 2 2 2 sec 2v
24. f sxd − 1 1 2 sin x 1 3ysx
f s1d − 0,
f 9s0d − 12
f 9s1d − 8
43. f 0sd − sin 1 cos ,f s0d − 3,f 9s0d − 4
44. f 0std − 4 2 6yt 4,f s1d − 6,f 9s2d − 9,
45. f 0sxd − 4 1 6x 1 24x ,f s0d − 3,f s1d − 10
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 3.9Antiderivatives
46. f 0sxd − 20x 3 1 12x 2 1 4,f s0d − 8,f s1d − 5
; 56. (a) Graph f sxd − 2x 2 3 sx .
(b)Starting with the graph in part (a), sketch a rough graph
of the antiderivative F that satisfies Fs0d − 1.
(c)Use the rules of this section to find an expression for Fsxd.
(d)Graph F using the expression in part (c). Compare with
your sketch in part (b).
47. f 0std − st 2 cos t,f s0d − 2,f s1d − 2
48. f -sxd − cos x,f s0d − 1,f 9s0d − 2,f 0s0d − 3
49. G
iven that the graph of f passes through the point (2, 5)
and that the slope of its tangent line at sx, f sxdd
is 3 2 4x, find f s1d.
; 57–58 Draw a graph of f and use it to make a rough sketch
of the antiderivative that passes through the origin.
50. F
ind a function f such that f 9sxd − x 3 and the line
x 1 y − 0 is tangent to the graph of f .
57. f sxd −
51–52 The graph of a function f is shown. Which graph is
an antiderivative of f and why?
58. f sxd − sx 4 2 2 x 2 1 2 2 2,23 < x < 3
51. y
52. y
59. v std − 2 cos t 1 4 sin t, ss0d − 3
62. astd − 3 cos t 2 2 sin t,ss0d − 0,v s0d − 4
54. T
he graph of the velocity function of a particle is shown
in the figure. Sketch the graph of a position function.
55. T
he graph of f 9 is shown in the figure. Sketch the graph
of f if f is continuous on f0, 3g and f s0d − 21.
63. astd − sin t 2 cos t, ss0d − 0, ssd − 6
64. astd − t 2 2 4t 1 6,ss0d − 0,ss1d − 20
65. A
stone is dropped from the upper observation deck (the
Space Deck) of the CN Tower, 450 m above the ground.
(a)Find the distance of the stone above ground level at time t.
(b) How long does it take the stone to reach the ground?
(c) With what velocity does it strike the ground?
(d)If the stone is thrown downward with a speed of 5 mys,
how long does it take to reach the ground?
66. S
how that for motion in a straight line with constant acceleration a, initial velocity v 0, and initial displacement s 0 , the
dis­placement after time t is s − 12 at 2 1 v 0 t 1 s 0.
67. A
n object is projected upward with initial velocity v 0 meters
per second from a point s0 meters above the ground. Show
fvstdg 2 − v02 2 19.6fsstd 2 s0 g
68. T
wo balls are thrown upward from the edge of the cliff in
Example 7. The first is thrown with a speed of 15 mys and
the other is thrown a second later with a speed of 8 mys. Do
the balls ever pass each other?
69. A stone was dropped off a cliff and hit the ground with a
speed of 40 mys. What is the height of the cliff ?
60. vstd − t 2 2 3 st , ss4d − 8
61. astd − 2t 1 1,ss0d − 3,v s0d − 22
53. T
he graph of a function is shown in the figure. Make a rough
sketch of an antiderivative F, given that Fs0d − 1.
sin x
,22 < x < 2
1 1 x2
59–64 A particle is moving with the given data. Find the
position of the particle.
70. I f a diver of mass m stands at the end of a diving board with
length L and linear density , then the board takes on the
shape of a curve y − f sxd, where
EI y 0 − mtsL 2 xd 1 12 tsL 2 xd2
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
E and I are positive constants that depend on the material of
the board and t s, 0d is the acceleration due to gravity.
(a)Find an expression for the shape of the curve.
(b)Use f sLd to estimate the distance below the horizontal at
the end of the board.
71. A
company estimates that the marginal cost (in dollars per
item) of producing x items is 1.92 2 0.002x. If the cost of producing one item is $562, find the cost of producing 100 items.
72. T
he linear density of a rod of length 1 m is given by
sxd − 1ysx , in grams per centimeter, where x is measured in centimeters from one end of the rod. Find the
mass of the rod.
73. S
ince raindrops grow as they fall, their surface area increases
and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 mys and its
downward acceleration is
9 2 0.9t if 0 < t < 10
if t . 10
If the raindrop forms 500 m above the ground, how long does
it take to fall?
74. A
car is traveling at 80 kmyh when the brakes are fully
applied, producing a constant deceleration of 7 mys2. What is
the distance traveled before the car comes to a stop?
75. W
hat constant acceleration is required to increase the speed
of a car from 50 kmyh to 80 kmyh in 5 seconds?
76. A
car braked with a constant deceleration of 5 mys2, pro­
ducing skid marks measuring 60 m before coming to a stop.
How fast was the car traveling when the brakes were first
77. A
car is traveling at 100 kmyh when the driver sees an
accident 80 m ahead and slams on the brakes. What constant deceleration is required to stop the car in time to
avoid a multicar pileup?
78. A
model rocket is fired vertically upward from rest. Its
acceler­ation for the first three seconds is astd − 18t, at
which time the fuel is exhausted and it becomes a freely
“falling” body. Fourteen seconds later, the rocket’s parachute opens, and the (downward) velocity slows linearly
to 25.5 mys in 5 seconds. The rocket then “floats” to the
ground at that rate.
(a)Determine the position function s and the velocity function v (for all times t). Sketch the graphs of s and v.
(b)At what time does the rocket reach its maximum height,
and what is that height?
(c) At what time does the rocket land?
79. A
particular bullet train accelerates and decelerates at the rate
of 1.2 mys2. Its maximum cruising speed is 145 kmyh.
(a)What is the maximum distance the train can travel if it
accelerates from rest until it reaches its cruising speed
and then runs at that speed for 20 minutes?
(b)Suppose that the train starts from rest and must come to
a complete stop in 20 minutes. What is the maximum
distance it can travel under these conditions?
(c)Find the minimum time that the train takes to travel
between two consecutive stations that are 72 km apart.
(d)The trip from one station to the next takes 37.5 minutes.
How far apart are the stations?
1. E
xplain the difference between an absolute maximum and
a local maximum. Illustrate with a sketch.
2. (a) What does the Extreme Value Theorem say?
(b) Explain how the Closed Interval Method works.
3. (a) State Fermat’s Theorem.
(b) Define a critical number of f .
4. (a) State Rolle’s Theorem.
(b)State the Mean Value Theorem and give a geometric
Answers to the Concept Check are available at StewartCalculus.com.
5. (a) State the Increasing/Decreasing Test.
(b)What does it mean to say that f is concave upward on an
interval I ?
(c) State the Concavity Test.
(d) What are inflection points? How do you find them?
6. (a) State the First Derivative Test.
(b) State the Second Derivative Test.
(c)What are the relative advantages and disadvantages of
these tests?
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3Review
7. Explain the meaning of each of the following statements.
(a) lim f sxd − L
(b) lim f sxd − L
(c) lim f sxd − `
x l 2`
(d)The curve y − f sxd has the horizontal asymptote y − L.
8. I f you have a graphing calculator or computer, why do you
need calculus to graph a function?
9. (a)Given an initial approximation x1 to a solution of the equation f sxd − 0, explain geometrically, with a diagram, how
the second approximation x 2 in Newton’s method is
(b)Write an expression for x 2 in terms of x1, f sx 1 d, and
f 9sx 1d.
(c)Write an expression for x n11 in terms of x n , f sx n d, and
f 9sx n d.
(d)Under what circumstances is Newton’s method likely to
fail or to work very slowly?
10. (a) What is an antiderivative of a function f ?
(b)Suppose F1 and F2 are both antiderivatives of f on an
interval I. How are F1 and F2 related?
Determine whether the statement is true or false. If it is true,
explain why. If it is false, explain why or give an example that
disproves the statement.
1. If f 9scd − 0, then f has a local maximum or minimum at c.
2. If f has an absolute minimum value at c, then f 9scd − 0.
3. If f is continuous on sa, bd, then f attains an absolute maximum value f scd and an absolute minimum value f sd d at some
numbers c and d in sa, bd.
10. T
here exists a function f such that f sxd , 0, f 9sxd , 0,
and f 0 sxd . 0 for all x.
11. If f and t are increasing on an interval I, then f 1 t is
increasing on I.
12. I f f and t are increasing on an interval I, then f 2 t is
increasing on I.
13. I f f and t are increasing on an interval I, then f t is
increasing on I.
4. If f is differentiable and f s21d − f s1d, then there is a number
c such that c , 1 and f 9scd − 0.
14. If f and t are positive increasing functions on an interval I,
then f t is increasing on I.
5. If f 9sxd , 0 for 1 , x , 6, then f is decreasing on (1, 6).
15. If f is increasing and f sxd . 0 on I, then tsxd − 1yf sxd is
decreasing on I.
6. If f 0s2d − 0, then s2, f s2dd is an inflection point of the
curve y − f sxd.
16. If f is even, then f 9 is even.
| |
7. If f 9sxd − t9sxd for 0 , x , 1, then f sxd − tsxd for
0 , x , 1.
17. If f is periodic, then f 9 is periodic.
18. T
he most general antiderivative of f sxd − x 22 is
Fsxd − 2s1yxd 1 C.
8. T
here exists a function f such that f s1d − 22, f s3d − 0,
and f 9sxd . 1 for all x.
19. If f 9sxd exists and is nonzero for all x, then f s1d ± f s0d.
9. T
here exists a function f such that f sxd . 0, f 9sxd , 0,
and f 0 sxd . 0 for all x.
20. If f has domain f0, `d and has no horizontal asymptote, then
lim x l ` f sxd − ` or lim x l ` f sxd − 2`.
1–6 Find the local and absolute extreme values of the function on
the given interval.
1. f sxd − x 2 9x 1 24 x 2 2,f0, 5g
6. f sxd − sin x 1 cos 2 x,f0, g
2. f sxd − x s1 2 x ,f21, 1g
3. f sxd −
5. f sxd − x 1 2 cos x,f2, g
3x 2 4
,f22, 2g
x2 1 1
4. f sxd − sx 2 1 x 1 1 ,f22, 1g
7–12 Find the limit.
7. lim
3x 4 1 x 2 5
6x 4 2 2x 2 1 1
8. lim
t3 2 t 1 2
s2 t 2 1dst 2 1 t 1 1d
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
9. lim
x l 2`
Applications of Differentiation
s4x 2 1 1
3x 2 1
11. lim ss4x 2 1 3x 2 2xd
lim sx 2 1 x 3 d
x l 2`
12. lim
13. f s0d − 0, f 9s22d − f 9s1d − f 9s9d − 0,
lim f sxd − 0,
lim f sxd − 2`,
f 9sxd , 0 on s2`, 22d, s1, 6d, and s9, `d,
f 9sxd . 0 on s22, 1d and s6, 9d,
f 0sxd , 0 on s0, 6d and s6, 12d
f 9sxd − 1 if x . 3
15. f is odd,f 9sxd , 0 for 0 , x , 2,f 9sxd . 0 for x . 2,f 0sxd . 0 for 0 , x , 3,
30. f sxd −
x3 1 1
x6 1 1
16. T
he figure shows the graph of the derivative f 9of a
function f.
(a) On what intervals is f increasing or decreasing?
(b)For what values of x does f have a local maximum or
(c) Sketch the graph of f 0.
(d) Sketch a possible graph of f.
34. S
uppose that f is continuous on f0, 4g, f s0d − 1, and
2 < f 9sxd < 5 for all x in s0, 4d. Show that 9 < f s4d < 21.
33 , 2.0125
36. F
or what values of the constants a and b is s1, 3d a point of
inflection of the curve y − ax 3 1 bx 2?
37. Let tsxd − f sx 2 d, where f is twice differentiable for all x,
f 9sxd . 0 for all x ± 0, and f is concave downward on
s2`, 0d and concave upward on s0, `d.
(a) At what numbers does t have an extreme value?
(b) Discuss the concavity of t.
38. F
ind two positive integers such that the sum of the first
number and four times the second number is 1000 and
the product of the numbers is as large as possible.
33. S
how that the equation 3x 1 2 cos x 1 5 − 0 has exactly
one real solution.
35. B
y applying the Mean Value Theorem to the function
f sxd − x 1y5 on the interval f32, 33g , show that
f 0sxd , 0 for x . 3,lim f sxd − 22
17–28 Use the guidelines of Section 3.5 to sketch the curve.
17. y − 2 2 2x 2 x 3
39. S
how that the shortest distance from the point sx 1, y1 d to the
straight line Ax 1 By 1 C − 0 is
| Ax
1 By1 1 C
sA 1 B
40. F
ind the point on the hyperbola x y − 8 that is closest to the
point s3, 0d.
18. y − 22 x 3 2 3x 2 1 12 x 1 5
xsx 2 3d2
x2 2 1
32. f sxd − x 2 1 6.5 sin x, 25 < x < 5
f 9sxd − 2x if 0 , x , 1, f 9sxd − 21 if 1 , x , 3,
21. y −
; 29–32 Produce graphs of f that reveal all the important aspects
of the curve. Use graphs of f 9 and f 0 to estimate the intervals of
increase and decrease, extreme values, intervals of concavity,
and inflection points. In Exercise 29 use calculus to find these
quantities exactly.
31. f sxd − 3x 6 2 5x 5 1 x 4 2 5x 3 2 2x 2 1 2
14. f s0d − 0,f is continuous and even,19. y − 3x 4 2 4 x 3 1 2
26. y − x 2y3sx 2 3d 2
28. y − 4x 2 tan x, 2y2 , x , y2
29. f sxd −
f 0sxd . 0 on s2`, 0d and s12, `d,
24. y − s1 2 x 1 s1 1 x
27. y − sin 2 x 2 2 cos x
13–15 Sketch the graph of a function that satisfies the given
sx 2 1d 3
25. y − x s2 1 x
sin 4 x
23. y −
20. y −
1 2 x2
41. F
ind the smallest possible area of an isosceles triangle that
is circumscribed about a circle of radius r.
22. y −
sx 2 2d 2
42. F
ind the volume of the largest circular cone that can be
inscribed in a sphere of radius r.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3Review
| |
43. I n D ABC, D lies on AB, CD AB, AD − BD − 4 cm,
and CD − 5 cm. Where should a point P be chosen on CD
so that the sum PA 1 PB 1 PC is a minimum?
57–60 Find f.
58. f 9sud −
| | | | | |
Solve Exercise 43 when | CD | − 2 cm.
45. The velocity of a wave of length L in deep water is
57. f 9std − 2t 2 3 sin t,f s0d − 5
u 2 1 su
,f s1d − 3
59. f 0sxd − 1 2 6x 1 48x 2,f s0d − 1,f 9s0d − 2
60. f 0sxd − 5x 3 1 6x 2 1 2,f s0d − 3,f s1d − 22
where K and C are known positive constants. What is the
length of the wave that gives the minimum velocity?
46. A metal storage tank with volume V is to be constructed in
the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of
47. A hockey team plays in an arena that seats 15,000 spectators. With the ticket price set at $12, average attendance at
a game has been 11,000. A market survey indicates that
for each dollar the ticket price is lowered, average attendance will increase by 1000. How should the owners of the
team set the ticket price to maximize their revenue from
ticket sales?
; 48. A manufacturer determines that the cost of making
x units of a commodity is
Csxd − 1800 1 25x 2 0.2x 2 1 0.001x 3
and the demand function is psxd − 48.2 2 0.03x.
(a)Graph the cost and revenue functions and use the
graphs to estimate the production level for maximum
(b)Use calculus to find the production level for maximum
(c)Estimate the production level that minimizes the
average cost.
49. Use Newton’s method to find the solution of the equation
x 5 2 x 4 1 3x 2 2 3x 2 2 − 0 in the interval f1, 2g correct
to six decimal places.
50. Use Newton’s method to find all solutions of the equation
sin x − x 2 2 3x 1 1 correct to six decimal places.
51. U
se Newton’s method to find the absolute maximum
value of the function f std − cos t 1 t 2 t 2 correct to eight
decimal places.
61–62 A particle is moving along a straight line with the given
data. Find the position of the particle.
61. vstd − 2t 2 sin t,ss0d − 3
62. astd − sin t 1 3 cos t,ss0d − 0,v s0d − 2
; 63. Graph the function f sxd − x sin sx d, 0 < x < , and use
that graph to sketch the antiderivative F of f that satisfies
the initial condition Fs0d − 0.
; 64. Investigate the family of curves given by
f sxd − x 4 1 x 3 1 cx 2
In particular you should determine the transitional value
of c at which the number of critical numbers changes and
the transitional value at which the number of inflection
points changes. Illustrate the various possible shapes with
65. A
canister is dropped from a helicopter hovering 500 m
above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of
100 mys. Will it burst?
66. I n an automobile race along a straight road, car A passed
car B twice. Prove that at some time during the race their
accelera­tions were equal. State the assumptions that you
67. A
rectangular beam will be cut from a cylindrical log of
radius 30 centimeters.
(a)Show that the beam of maximal cross-sectional area is a
52. Use the guidelines in Section 3.5 to sketch the curve
y − x sin x, 0 < x < 2. Use Newton’s method when
53–56 Find the most general antiderivative of the function.
53. f sxd − 4 sx 2 6x 1 3
54. tsxd − cos x 1 2 sec x
55. hstd − t 23 1 5 sin t
56. f sxd −
3x 5 2 4x 2 1 1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Applications of Differentiation
(b)Four rectangular planks will be cut from the four
sections of the log that remain after cutting the square
beam. Determine the dimensions of the planks that will
have maximal cross-sectional area.
(c)Suppose that the strength of a rectangular beam is
proportional to the product of its width and the square of
its depth. Find the dimensions of the strongest beam that
can be cut from the cylindrical log.
68. I f a projectile is fired with an initial velocity v at an angle of
inclination from the horizontal, then its trajectory, neglecting air resistance, is the parabola
y − stan dx 2
x 20 , ,
2v 2 cos 2
(a)Suppose the projectile is fired from the base of a plane
that is inclined at an angle , . 0, from the horizontal,
as shown in the figure. Show that the range of the projectile, measured up the slope, is given by
Rsd −
2v 2 cos sins 2 d
t cos2
(b) Determine so that R is a maximum.
(c)Suppose the plane is at an angle below the horizontal.
Determine the range R in this case, and determine the
angle at which the projectile should be fired to maximize R.
should the measure of angle be in order to maximize the
total area?
70. W
ater is flowing at a constant rate into a spherical tank. Let
Vstd be the volume of water in the tank and Hstd be the height
of the water in the tank at time t.
(a)What are the meanings of V9std and H9std? Are these
derivatives positive, negative, or zero?
(b)Is V 0std positive, negative, or zero? Explain.
(c)Let t1, t 2, and t 3 be the times when the tank is one-quarter
full, half full, and three-quarters full, respectively. Are
each of the values H 0st1d, H 0st 2 d, and H 0st 3 d positive,
negative, or zero? Why?
71. A
light is to be mounted atop a pole of height h feet to
illuminate a busy traffic circle, which has a radius of 20 m.
The intensity of illumination I at any point P on the circle
is directly proportional to the cosine of the angle (see
the figure) and inversely proportional to the square of the
distance d from the source.
(a) How tall should the light pole be to maximize I?
(b)Suppose that the light pole is h meters tall and that a
woman is walking away from the base of the pole at the
rate of 1 mys. At what rate is the intensity of the light at
the point on her back 1 m above the ground decreasing
when she reaches the outer edge of the traffic circle?
72. If f sxd −
69. The following figure shows an isosceles triangle with
equal sides of length a surmounted by a semicircle. What
cos 2 x
, 2 < x < , use the graphs
sx 2 1 x 1 1
of f, f 9, and f 0 to estimate the x-coordinates of the maximum
and minimum points and inflection points of f.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Problems Plus
One of the most important principles of problem solving is analogy (see Principles of
Problem Solving following Chapter 1). If you are having trouble getting started on a
problem, it is sometimes helpful to start by solving a similar, but simpler, problem. The
following example illustrates the principle. Cover up the solution and try solving it yourself first.
EXAMPLE 1 If x, y, and z are positive numbers, prove that
sx 2 1 1dsy 2 1 1dsz 2 1 1d
SOLUTION It may be difficult to get started on this problem. (Some students have
tackled it by multiplying out the numerator, but that just creates a mess.) Let’s try to
think of a similar, simpler problem. When several variables are involved, it’s often
helpful to think of an analogous problem with fewer variables. In the present case we
can reduce the number of variables from three to one and prove the analogous inequality
x2 1 1
for x . 0
In fact, if we are able to prove (1), then the desired inequality follows because
sx 2 1 1dsy 2 1 1dsz 2 1 1d
S DS DS D
x2 1 1
y2 1 1
z2 1 1
The key to proving (1) is to recognize that it is a disguised version of a minimum problem. If we let
f sxd −
x2 1 1
− x 1 x . 0
then f 9sxd − 1 2 s1yx 2 d, so f 9sxd − 0 when x − 1. Also, f 9sxd , 0 for 0 , x , 1 and
f 9sxd . 0 for x . 1. Therefore the absolute minimum value of f is f s1d − 2. This
means that
x2 1 1
PS Look Back
What have we learned from the solution to
this example?
● To solve a problem involving several variables, it might help to solve a similar
problem with just one variable.
● When trying to prove an inequality, it
might help to think of it as a maximum
or minimum problem.
for all positive values of x
and, as previously mentioned, the given inequality follows by multiplication.
The inequality in (1) could also be proved without calculus. In fact, if x . 0, we
x2 1 1
> 2 &?
x 2 1 1 > 2x &? x 2 2 2x 1 1 > 0
&? sx 2 1d2 > 0
Because the last inequality is obviously true, the first one is true too.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1. Show that sin x 2 cos x < s2 for all x.
| |
2. Show that x y s4 2 x ds4 2 y 2 d < 16 for all numbers x and y such that x < 2
and y < 2.
| |
3. Show that the inflection points of the curve y − ssin xdyx lie on the curve y 2 sx 4 1 4d − 4.
4. F
ind the point on the parabola y − 1 2 x 2 at which the tangent line cuts from the first
quadrant the triangle with the smallest area.
5. Find the highest and lowest points on the curve x 2 1 x y 1 y 2 − 12.
6. Find the absolute maximum value of the function
f sxd −
11 x
11 x22
| |
7. Show that if f is a differentiable function that satisfies
f sx 1 nd 2 f sxd
− f 9sxd
for all real numbers x and all positive integers n, then f is a linear function.
8. Find a function f such that f 9s21d − 12 , f 9s0d − 0, and f 0sxd . 0 for all x, or prove that
such a function cannot exist.
9. I f Psa, a 2 d is any first-quadrant point on the parabola y − x 2, let Q be the point where the
normal line at P intersects the parabola again (see the figure).
(a) Show that the y-coordinate of Q is smallest when a − 1ys2 .
(b) Show that the line segment PQ has the shortest possible length when a − 1ys2 .
10. A
n isosceles triangle is circumscribed about the unit circle so that the equal sides meet
at the point s0, ad on the y-axis (see the figure). Find the value of a that minimizes the
lengths of the equal sides. (You may be surprised that the result does not give an equilateral triangle.).
FIGURE FOR PROBLEM 11
11. The line y − mx 1 b intersects the parabola y − x 2 in points A and B. (See the figure.) Find the point P on the arc AOB of the parabola that maximizes the area of the
triangle PAB.
| |
12. Sketch
the graph of a function f such that f 9sxd , 0 for all x, f 0sxd . 0 for x . 1,
f 0sxd , 0 for x , 1, and lim x l6` f f sxd 1 xg − 0.
| |
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13. Determine the values of the number a for which the function f has no critical number:
f sxd − sa 2 1 a 2 6d cos 2x 1 sa 2 2dx 1 cos 1
14. Sketch the region in the plane consisting of all points sx, yd such that
2xy < x 2 y < x 2 1 y 2
| | |
15. Let ABC be a triangle with /BAC − 120° and AB AC − 1.
(a) Express the length of the angle bisector AD in terms of x − AB .
(b) Find the largest possible value of AD .
| |
16. (a)Let ABC be a triangle with right angle A and hypotenuse a − BC . (See the figure.)
If the inscribed circle touches the hypotenuse at D, show that
FIGURE FOR PROBLEM 16
| CD | − 12 ( | BC | 1 | AC | 2 | AB | )
(b)If − 12 /C, express the radius r of the inscribed circle in terms of a and .
(c) If a is fixed and varies, find the maximum value of r.
17. A
triangle with sides a, b, and c varies with time t, but its area never changes. Let be the
angle opposite the side of length a and suppose always remains acute.
(a) Express dydt in terms of b, c, , dbydt, and dcydt.
(b) Express daydt in terms of the quantities in part (a).
18. ABCD is a square piece of paper with sides of length 1 m. A quarter-circle is drawn from
B to D with center A. The piece of paper is folded along EF, with E on AB and F on AD,
so that A falls on the quarter-circle. Determine the maximum and minimum areas that the
triangle AEF can have.
19. T
he speeds of sound c1 in an upper layer and c2 in a lower layer of rock and the thick­ness h
of the upper layer can be determined by seismic exploration if the speed of sound in the
lower layer is greater than the speed in the upper layer. A dynamite charge is detonated at a
point P and the transmitted signals are recorded at a point Q, which is a distance D from P.
The first signal to arrive at Q travels along the surface and takes T1 seconds. The next signal
travels from P to a point R, from R to S in the lower layer, and then to Q, taking T2 seconds.
The third signal is reflected off the lower layer at the midpoint O of RS and takes T3 seconds
to reach Q. (See the figure.)
(a) Express T1, T2, and T3 in terms of D, h, c1, c2, and .
(b)Show that T2 is a minimum when sin − c1yc2 .
(c)Suppose that D − 1 km, T1 − 0.26 s, T2 − 0.32 s, and T3 − 0.34 s. Find c1, c2, and h.
Note: Geophysicists use this technique when studying the structure of the earth’s crust—
searching for oil or examining fault lines, for example.
speed of sound=c¡
speed of sound=c™
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
20. For what values of c is there a straight line that intersects the curve
y − x 4 1 cx 3 1 12x 2 2 5x 1 2
in four distinct points?
FIGURE FOR PROBLEM 21
21. O
ne of the problems posed by the Marquis de l’Hospital in his calculus textbook Analyse
des infiniment petits concerns a pulley that is attached to the ceiling of a room at a point C
by a rope of length r. At another point B on the ceiling, at a distance d from C (where
d . r), a rope of length , is attached and passed through the pulley at F and connected to a
weight W. The weight is released and comes to rest at its equilibrium position D. (See the
figure.) As l’Hospital argued, this happens when the distance ED is maximized. Show that
when the system reaches equilibrium, the value of x is
(r 1 sr 2 1 8d 2 )
Notice that this expression is independent of both W and ,.
22. G
iven a sphere with radius r, find the height of a pyramid of minimum volume whose base is
a square and whose base and triangular faces are all tangent to the sphere. What if the base
of the pyramid is a regular n-gon? (A regular n-gon is a polygon with n equal sides and
angles.) (Use the fact that the volume of a pyramid is 13 Ah, where A is the area of the base.)
23. Assume
that a snowball melts so that its volume decreases at a rate proportional to its
surface area. If it takes three hours for the snowball to decrease to half its original volume,
how much longer will it take for the snowball to melt completely?
24. A
hemispherical bubble is placed on a spherical bubble of radius 1. A smaller
hemispherical bubble is then placed on the first one. This process is continued until
n chambers, including the sphere, are formed. (The figure shows the case n − 4.) Use
mathematical induction to prove that the maximum height of any bubble tower with
n chambers is 1 1 sn .
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
In Exercise 4.4.73 we see how to use electric power consumption data and an integral to compute the amount of electric energy used in a
typical day in the New England states.
ixpert / Shutterstock.com
IN CHAPTER 2 WE USED the tangent and velocity problems to introduce the derivative. In this
chapter we use the area and distance problems to introduce the other central idea in calculus—the
integral. The all-important relationship between the derivative and the integral is expressed in the
Fundamental Theorem of Calculus, which says that differentiation and integration are in a sense
inverse processes. We learn in this chapter, and in Chapters 5 and 8, how integration can be used to
solve problems involving volumes, length of curves, population predictions, cardiac output, forces
on a dam, work, consumer surplus, and baseball, among many others.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 4
4.1 The Area and Distance Problems
Now is a good time to read (or reread)
A Preview of Calculus, which discusses
the unifying ideas of calculus and
helps put in perspec­tive where we
have been and where we are going.
FIGURE 1
In this section we discover that in trying to find the area under a curve or the distance
traveled by a car, we end up with the same special type of limit.
■ The Area Problem
We begin by attempting to solve the area problem: find the area of the region S that lies
under the curve y − f sxd from a to b. This means that S, illustrated in Figure 1, is bounded
by the graph of a continuous function f [where f sxd > 0], the vertical lines x − a and
x − b, and the x-axis.
In trying to solve the area problem we have to ask ourselves: what is the meaning of
the word area? This question is easy to answer for regions with straight sides. For a rectangle, the area is defined as the product of the length and the width. The area of a triangle
is half the base times the height. The area of a polygon is found by dividing it into triangles (as in Figure 2) and adding the areas of the triangles.
S − hsx, yd a < x < b, 0 < y < f sxdj
FIGURE 2
A= 21 bh
It isn’t so easy, however, to find the area of a region with curved sides. We all have an
intuitive idea of what the area of a region is. But part of the area problem is to make this
intuitive idea precise by giving an exact definition of area.
Recall that in defining a tangent we first approximated the slope of the tangent line by
slopes of secant lines and then we took the limit of these approximations. We pursue a
sim­ilar idea for areas. We first approximate the region S by rectangles and then we take
the limit of the sum of the areas of the approximating rectangles as we increase the number of rectangles. The follow­ing example illustrates the procedure.
EXAMPLE 1 Use rectangles to estimate the area under the parabola y − x 2 for
0 < x < 1 (the parabolic region S illustrated in Figure 3).
(1, 1)
SOLUTION We first notice that the area of S must be somewhere between 0 and 1
because S is contained in a square with side length 1, but we can certainly do better
than that. Suppose we divide S into four strips S1, S2, S3, and S4 by drawing the vertical
lines x − 14, x − 12, and x − 34 as in Figure 4(a).
(1, 1)
(1, 1)
FIGURE 3
FIGURE 4
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 4.1The Area and Distance Problems
We can approximate each strip by a rectangle that has the same base as the strip and
whose height is the same as the right edge of the strip [see Figure 4(b)]. In other words,
the heights of these rectangles are the values of the function f sxd − x 2 at the right end­-
f g f 41 , 12 g, f 12 , 34 g, and f 34 , 1g.
points of the subintervals 0, 14 ,
Each rectangle has width 14 and the heights are ( 14 ) , ( 12 ) , ( 34 ) , and 12. If we let R 4 be
the sum of the areas of these approximating rectangles, we get
R4 − 14 ( 14 ) 1 14 ( 12 ) 1 14 ( 43 ) 1 14 12 − 15
32 − 0.46875
From Figure 4(b) we see that the area A of S is less than R 4, so
A , 0.46875
Instead of using the rectangles in Figure 4(b) we could use the smaller rectangles in
Figure 5 whose heights are the values of f at the left endpoints of the subintervals. (The
leftmost rectangle has collapsed because its height is 0.) The sum of the areas of these
approximating rectangles is
(1, 1)
L 4 − 14 0 2 1 14 ( 14 ) 1 14 ( 12 ) 1 14 ( 34 ) − 32
− 0.21875
We see that the area of S is larger than L 4, so we have lower and upper estimates for A:
0.21875 , A , 0.46875
We can repeat this procedure with a larger number of strips. Figure 6 shows what
happens when we divide the region S into eight strips of equal width.
FIGURE 5
(1, 1)
FIGURE 6
Approximating S with eight rectangles
(a) Using left endpoints
(1, 1)
(b) Using right endpoints
By computing the sum of the areas of the smaller rectangles sL 8 d and the sum of the
areas of the larger rectangles sR 8 d, we obtain better lower and upper estimates for A:
0.2734375 , A , 0.3984375
So one possible answer to the question is to say that the true area of S lies somewhere
between 0.2734375 and 0.3984375.
We could obtain better estimates by increasing the number of strips. The table at
the left shows the results of similar calculations (with a computer) using n rectangles
whose heights are found with left endpoints sL n d or right endpoints sR n d. In particular,
we see by using 50 strips that the area lies between 0.3234 and 0.3434. With 1000 strips
we narrow it down even more: A lies between 0.3328335 and 0.3338335. A good estimate is obtained by averaging these numbers: A < 0.3333335.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 4
From the values listed in the table in Example 1, it looks as if R n is approaching 13 as
n increases. We confirm this in the next example.
EXAMPLE 2 For the region S in Example 1, show that the approximating sums R n
approach 13, that is,
lim R n − 13
SOLUTION R n is the sum of the areas of the n rectangles in Figure 7. Each rectangle
has width 1yn and the heights are the values of the function f sxd − x 2 at the points
1yn, 2yn, 3yn, . . . , nyn; that is, the heights are s1ynd2, s2ynd2, s3ynd2, . . . , snynd2. Thus
(1, 1)
Rn −
FIGURE 7
SD SD SD
SD SD SD
s1 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 d
n n2
s1 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 d
1 ∙∙∙ 1
1 ∙∙∙ 1
Here we need the formula for the sum of the squares of the first n positive integers:
nsn 1 1ds2n 1 1d
12 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 −
Perhaps you have seen this formula before. It is proved in Example 5 in Appendix E.
Putting Formula 1 into our expression for R n, we get
Rn −
Here we are computing the limit of
the sequence hR n j. Sequences and
their limits will be studied in detail in
Section 11.1. The idea is very similar to
a limit at infinity (Section 3.4) except
that in writing lim n l ` we restrict n to
be a positive integer. In particular, we
know that
lim − 0
nl ` n
When we write lim n l ` Rn − 13 we
mean that we can make Rn as close
to 13 as we like by taking n sufficiently
1 nsn 1 1ds2n 1 1d
sn 1 1ds2n 1 1d
6n 2
Thus we have
lim R n − lim
nl `
nl `
sn 1 1ds2n 1 1d
6n 2
− lim
− lim
S DS D
S DS D
2n 1 1
It can be shown that the approximating sums L n in Example 2 also approach 13, that is,
lim L n − 13
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 4.1The Area and Distance Problems
From Figures 8 and 9 it appears that as n increases, both L n and R n become better and
bet­ter approximations to the area of S. Therefore we define the area A to be the limit of
the sums of the areas of the approximating rectangles, that is,
A − lim R n − lim L n − 13
n=10 R¡¸=0.385
n=50 R∞¸=0.3434
n=30 R£¸Å0.3502
FIGURE 8 Right endpoints produce upper estimates because f sxd − x 2 is increasing.
n=10 L¡¸=0.285
n=50 L∞¸=0.3234
n=30 L£¸Å0.3169
FIGURE 9 Left endpoints produce lower estimates because f sxd − x 2 is increasing.
Let’s apply the idea of Examples 1 and 2 to the more general region S of Figure 1.
We start by subdividing S into n strips S1, S2 , . . . , Sn of equal width as in Figure 10.
FIGURE 10
. . . xi-1
. . . xn-1
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 4
The width of the interval fa, bg is b 2 a, so the width of each of the n strips is
Dx −
These strips divide the interval fa, bg into n subintervals
fx 0 , x 1 g, fx 1, x 2 g, fx 2 , x 3 g, . . . , fx n21, x n g
where x 0 − a and x n − b. The right endpoints of the subintervals are
x 1 − a 1 Dx,
x 2 − a 1 2 Dx,
x 3 − a 1 3 Dx,
and, in general, x i − a 1 i Dx. Now let’s approximate the ith strip Si by a rectangle with
width Dx and height f sx i d, which is the value of f at the right endpoint (see Figure 11).
Then the area of the ith rectangle is f sx i d Dx. What we think of intuitively as the area of
S is approximated by the sum of the areas of these rectangles, which is
R n − f sx 1 d Dx 1 f sx 2 d Dx 1 ∙ ∙ ∙ 1 f sx n d Dx
FIGURE 11
Figure 12 shows this approximation for n − 2, 4, 8, and 12. Notice that this approximation appears to become better and better as the number of strips increases, that is, as
n l `. Therefore we define the area A of the region S in the following way.
(a) n=2
b x
(b) n=4
(c) n=8
(d) n=12
FIGURE 12
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 4.1The Area and Distance Problems
2 Definition The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:
A − lim R n − lim f f sx 1 d Dx 1 f sx 2 d Dx 1 ∙ ∙ ∙ 1 f sx n d Dxg
It can be proved that the limit in Definition 2 always exists, since we are assuming
that f is continuous. It can also be shown that we get the same value if we use left
A − lim L n − lim f f sx 0 d Dx 1 f sx 1 d Dx 1 ∙ ∙ ∙ 1 f sx n21 d Dxg
In fact, instead of using left endpoints or right endpoints, we could take the height of
the ith rectangle to be the value of f at any number x*i in the ith subinterval fx i21, x i g.
We call the numbers x1*, x2*, . . . , x n* the sample points. Figure 13 shows approximating
rectangles when the sample points are not chosen to be endpoints. So a more general
expression for the area of S is
A − lim f f sx 1* d Dx 1 f sx 2* d Dx 1 ∙ ∙ ∙ 1 f sx *n d Dxg
f(x i*)
FIGURE 13
x *i
x n*
NOTE To approximate the area under the graph of f we can form lower sums (or upper
sums) by choosing the sample points x*i so that f sx*i d is the minimum (or maximum)
value of f on the ith subinterval (see Figure 14). [Since f is continuous, we know that the
minimum and maximum values of f exist on each subinterval by the Extreme Value
Theorem.] It can be shown that an equivalent definition of area is the following: A is the
unique number that is smaller than all the upper sums and bigger than all the lower sums.
(a) Lower sums
(b) Upper sums
(c) Upper and lower sums
FIGURE 14
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 4
We saw in Examples 1 and 2, for instance, that the area ( A − 13 ) is trapped between
all the left approximating sums L n and all the right approximating sums Rn. The function
in those examples, f sxd − x 2, happens to be increasing on f0, 1g and so the lower sums
arise from left endpoints and the upper sums from right endpoints. (See Figures 8 and 9.)
This tells us to
end with i=n.
This tells us
to add.
This tells us to
start with i=m.
We often use sigma notation to write sums with many terms more compactly. For
µ f(xi) Îx
o f sx i d Dx − f sx 1 d Dx 1 f sx 2 d Dx 1 ∙ ∙ ∙ 1 f sx n d Dx
So the expressions for area in Equations 2, 3, and 4 can be written as follows:
o f sx i d Dx
n l ` i−1
A − lim
If you need practice with sigma notation, look at the examples and try
some of the exercises in Appendix E.
o f sx i21 d Dx
n l ` i−1
A − lim
o f sx i*d Dx
n l ` i−1
A − lim
We can also rewrite Formula 1 in the following way:
o i2 −
nsn 1 1ds2n 1 1d
EXAMPLE 3 Let A be the area of the region that lies under the graph of f sxd − cos x
between x − 0 and x − b, where 0 < b < y2.
(a) Using right endpoints, find an expression for A as a limit. Do not evaluate the limit.
(b) Estimate the area for the case b − y2 by taking the sample points to be midpoints
and using four subintervals.
(a) Since a − 0, the width of a subinterval is
Dx −
So x 1 − byn, x 2 − 2byn, x 3 − 3byn, x i − ibyn, and x n − nbyn. The sum of the areas of
the approximating rectangles is
R n − f sx 1 d Dx 1 f sx 2 d Dx 1 ∙ ∙ ∙ 1 f sx n d Dx
− scos x 1 d Dx 1 scos x 2 d Dx 1 ∙ ∙ ∙ 1 scos x n d Dx
S D S D
− cos
S D
b b
2b b
nb b
1 cos
1 ∙ ∙ ∙ 1 cos
n n
n n
n n
According to Definition 2, the area is
A − lim R n − lim
1 cos
1 cos
1 ∙ ∙ ∙ 1 cos
Using sigma notation we could write
nl` n
A − lim
o cos
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 4.1The Area and Distance Problems
It is very difficult to evaluate this limit directly by hand, but with the aid of a computer
algebra system it isn’t hard (see Exercise 33). In Section 4.3 we will be able to find A
more easily using a different method.
(b) With n − 4 and b − y2 we have Dx − sy2dy4 − y8, so the subintervals are
f0, y8g, fy8, y4g, fy4, 3y8g, and f3y8, y2g. The midpoints of these subinter­
vals are
x1* −
x2* −
x3* −
x4* −
y=cos x
and the sum M 4 of the areas of the four approximating rectangles (see Figure 15) is
FIGURE 15
M4 −
o f sxi*d Dx
− f sy16d Dx 1 f s3y16d Dx 1 f s5y16d Dx 1 f s7y16d Dx
S D S D S D S D
− cos
1 cos
1 cos
1 cos
1 cos
1 cos
1 cos
< 1.006
So an estimate for the area is
A < 1.006
■ The Distance Problem
In Section 1.4 we considered the velocity problem: find the velocity of a moving object
at a given instant if the distance of the object (from a starting point) is known at all times.
Now let’s consider the distance problem: find the distance traveled by an object during a
certain time period if the velocity of the object is known at all times. (In a sense this is
the inverse problem of the velocity problem.) If the velocity remains constant, then the
distance problem is easy to solve by means of the formula
distance − velocity 3 time
But if the velocity varies, it’s not so easy to find the distance traveled. We investigate the
problem in the following example.
EXAMPLE 4 Suppose the odometer on our car is broken and we want to estimate the
distance driven over a 30-second time interval. We take speedometer readings every
five seconds and record them in the following table:
Time (s)
Velocity (kmyh)
In order to have the time and the velocity in consistent units, let’s convert the velocity
readings to meters per second (1 kmyh − 1000y3600 mys):
Time (s)
Velocity (mys)
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 4
During the first five seconds the velocity doesn’t change very much, so we can estimate
the distance traveled during that time by assuming that the velocity is constant. If we
take the velocity during that time interval to be the initial velocity (8 mys), then we
obtain the approximate distance traveled during the first five seconds:
8 mys 3 5 s − 40 m
9 mys 3 5 s − 45 m
If we add similar estimates for the other time intervals, we obtain an estimate for the
total distance traveled:
s8 3 5d 1 s9 3 5d 1 s10 3 5d 1 s12 3 5d 1 s13 3 5d 1 s12 3 5d − 320 m
We could just as well have used the velocity at the end of each time period instead
of the velocity at the beginning as our assumed constant velocity. Then our estimate
Similarly, during the second time interval the velocity is approximately constant and
we take it to be the velocity when t − 5 s. So our estimate for the distance traveled
from t − 5 s to t − 10 s is
FIGURE 16
s9 3 5d 1 s10 3 5d 1 s12 3 5d 1 s13 3 5d 1 s12 3 5d 1 s11 3 5d − 335 m
Now let’s sketch an approximate graph of the velocity function of the car along
with rectangles whose heights are the initial velocities for each time interval [see
Figure 16(a)]. The area of the first rectangle is 8 3 5 − 40, which is also our estimate
for the distance traveled in the first five seconds. In fact, the area of each rectangle can
be interpreted as a distance because the height represents velocity and the width
represents time. The sum of the areas of the rectangles in Figure 16(a) is L 6 − 320,
which is our initial estimate for the total distance traveled.
If we want a more accurate estimate, we could take velocity readings more often,
as illustrated in Figure 16(b). You can see that the more velocity readings we take, the
closer the sum of the areas of the rectangles gets to the exact area under the velocity
curve [see Figure 16(c)]. This suggests that the total distance traveled is equal to the
area under the velocity graph.
In general, suppose an object moves with velocity v − f std, where a < t < b and
f std > 0 (so the object always moves in the positive direction). We take velocity readings
at times t0 s− ad, t1, t2 , . . . , tn s− bd so that the velocity is approximately constant on
each subinterval. If these times are equally spaced, then the time between consecutive
readings is Dt − sb 2 adyn. During the first time interval the velocity is approximately
f st0 d and so the distance traveled is approximately f st0 d Dt. Similarly, the distance traveled during the second time interval is about f st1 d Dt and the total distance traveled during the time inter­val fa, bg is approximately
f st0 d Dt 1 f st1 d Dt 1 ∙ ∙ ∙ 1 f stn21 d Dt −
o f sti21 d Dt
If we use the velocity at right endpoints instead of left endpoints, our estimate for the
total distance becomes
f st1 d Dt 1 f st2 d Dt 1 ∙ ∙ ∙ 1 f stn d Dt −
o f sti d Dt
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 4.1The Area and Distance Problems
The more frequently we measure the velocity, the more accurate our estimates become,
so it seems plausible that the exact distance d traveled is the limit of such expressions:
lim o f sti d Dt
o f sti21 d Dt − nl`
nl` i−1
d − lim
We will see in Section 4.4 that this is indeed true.
Because Equation 5 has the same form as our expressions for area in Equations 2 and 3,
it follows that the distance traveled is equal to the area under the graph of the velocity
func­tion. In Chapters 5 and 8 we will see that other quantities of interest in the natural and
social sciences—such as the work done by a variable force or the cardiac output of the
heart—can also be interpreted as the area under a curve. So when we compute areas in this
chapter, bear in mind that they can be interpreted in a variety of practical ways.
1. (a) By reading values from the given graph of f , use five
rect­angles to find a lower estimate and an upper estimate
for the area under the given graph of f from x − 0 to
x − 10. In each case sketch the rectangles that you use.
(b) Find new estimates using ten rectangles in each case.
4. (a)Estimate the area under the graph of f sxd − sin x
from x − 0 to x − y2 using four approximating rect­
angles and right endpoints. Sketch the graph and the
rectangles. Is your estimate an underestimate or an
(b)Repeat part (a) using left endpoints.
2. (a)Use six rectangles to find estimates of each type for the
area under the given graph of f from x − 0 to x − 12.
(i) L 6 (sample points are left endpoints)
(ii) R 6 (sample points are right endpoints)
(iii) M6 (sample points are midpoints)
(b)Is L 6 an underestimate or overestimate of the true area?
(c) Is R 6 an underestimate or overestimate of the true area?
(d)Which of the numbers L 6, R 6, or M6 gives the best
estimate? Explain.
5. (a)Estimate the area under the graph of f sxd − 1 1 x 2 from
x − 21 to x − 2 using three rectangles and right end­
points. Then improve your estimate by using six rect­
angles. Sketch the curve and the approximating
(b) Repeat part (a) using left endpoints.
(c) Repeat part (a) using midpoints.
(d)From your sketches in parts (a) – (c), which estimate
appears to be the most accurate?
; 6. (a)Graph the function
f sxd − 1ys1 1 x 2 d
22 < x < 2
(b)Estimate the area under the graph of f using four
approximating rectangles and taking the sample points to
be (i) right endpoints and (ii) midpoints. In each case
sketch the curve and the rectangles.
(c)Improve your estimates in part (b) by using eight
3. (a)Estimate the area under the graph of f sxd − 1yx from
x − 1 to x − 2 using four approximating rectangles
and right endpoints. Sketch the graph and the rectangles.
Is your estimate an underestimate or an overestimate?
(b)Repeat part (a) using left endpoints.
12 x
7. E
valuate the upper and lower sums for f sxd − 6 2 x 2,
22 < x < 2, with n − 2, 4, and 8. Illustrate with diagrams
like Figure 14.
Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 4
the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Use these data to estimate the height above the earth’s surface of the Endeavour,
62 seconds after liftoff.
8. Evaluate the upper and lower sums for
f sxd − 1 1 cossxy2d2 < x < with n − 3, 4, and 6. Illustrate with diagrams like
Figure 14.
9. T
he speed of a runner increased steadily during the first
three seconds of a race. Her speed at half-second intervals is
given in the table. Find lower and upper estimates for the
distance that she traveled during these three seconds.
t (s)
v (mys)
10. T
he table shows speedometer readings at 10-second intervals during a 1-minute period for a car racing at the Daytona International Speedway in Florida.
(a)Estimate the distance the race car traveled during this
time period using the velocities at the beginning of the
time intervals.
(b)Give another estimate using the velocities at the end of
the time periods.
(c)Are your estimates in parts (a) and (b) upper and lower
estimates? Explain.
Velocity (miyh)
Begin roll maneuver
End roll maneuver
Throttle to 89%
Throttle to 67%
Throttle to 104%
Maximum dynamic pressure
Solid rocket booster separation
Time ssd
Velocity smysd
13. T
he velocity graph of a braking car is shown. Use it to esti­mate
the distance traveled by the car while the brakes are applied.
√ (m/s)
t (seconds)
14. T
he velocity graph of a car accelerating from rest to a speed
of 120 kmyh over a period of 30 seconds is shown. Estimate
the distance traveled during this period.
√ (km/h)
11. O
il leaked from a tank at a rate of rstd liters per hour. The
rate decreased as time passed and values of the rate at twohour time intervals are shown in the table. Find lower
and upper estimates for the total amount of oil that
leaked out.
t (h)
rstd (Lyh)
12. W
hen we estimate distances from velocity data, it is sometimes necessary to use times t0 , t1, t2 , t3 , . . . that are not
equally spaced. We can still estimate distances using the
time periods Dt i − t i 2 t i21. For example, in 1992 the space
shuttle Endeavour was launched on mission STS-49 in
order to install a new perigee kick motor in an Intelsat communications satellite. The table, provided by NASA, gives
t (seconds)
15. I n a person infected with measles, the virus level N (measured in number of infected cells per mL of blood plasma)
reaches a peak density at about t − 12 days (when a rash
appears) and then decreases fairly rapidly as a result of
immune response. The area under t | {"url":"https://studylib.net/doc/27142257/1.calculus--metric-edition-by-james-stewart--saleem-watso...","timestamp":"2024-11-14T20:42:27Z","content_type":"text/html","content_length":"1049521","record_id":"<urn:uuid:d66652cb-5a6c-47f4-87e3-6a78004297a1>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00606.warc.gz"} |
gravity is not a force
this paper better explains the effects of general relativity and has insights into dark matter and dark energy
this white paper has no math and should be easy to read and understand (it is only 4 pages with pictures)
i am interested in feedback as i have yet to find something to disprove this theory
(re-posted as original disappeared for unknown reasons)
Do you have math for it? Because without math there is no model, and certainly no theory -only hypothesis.
Einstein modeled gravity and time dilation as curvature in space not because he believed space existentially curved, but because it is a beautifully elegant and effective way to describe the behavior
we observe.
In fact, under Einstein's theory, gravity is indeed not a force. General relativity describes gravity as objects traveling along geodesics in curved spacetime, meaning gravity is a virtual force, or
a phenomenon that looks like a force in one context but is not in the system as a whole.
And no, we do not expect space to "flatten out" across large distances. In fact, modeling gravity as spacial curvature leads us to the exact opposite conclusion. We expect the curvature to become
less influential asymptotically, never reaching 0. Of course, gravity also has a speed, so what we actually expect is for the wave to travel outward without bound, but I'll leave it at that.
Spacetime is also not a field, it is a pseudo-Riemannian manifold called a Minkowski space. Fields propagate through space.
The most immediate issue I see with that paper's proposition is that it imposes a static reference. Space, under general relativity, has no sense of absolute location -you cannot move through space
itself. You can, however, move through space relative to another object. What the paper describes is an aether, where space itself is a physical entity reacting to movement and interacting with
For instance, under this model, time dilation occurs to an object as a result of moving through space. That does not describe what we observe at all. What we observe is a difference between the time
experienced by multiple observers moving relative to each other. There is no time dilation without an observer. You can, in fact, travel anywhere in the universe in an arbitrarily short amount of
time -but the planet you took off from will only see you moving arbitrarily close to the speed of light. There is no correct frame of reference, there are only different references.
Even if this model had no apparent flaws, I do not see what merit it would have over general relativity. A model is not measured by its existential correctness, as that is not something we can verify
-it is measured by its predictive power. Einstein's model has extraordinary predictive power, not only enabling far more accurate modeling of the interaction between spacial bodies and their effect
on light, but also paving the way to entirely new concepts whose existence we continue to verify, like gravitational waves. I do not see what this model accomplishes.
thank you for taking the time to read and reply to my paper. its clear that some things were not explained clearly as most of your arguments are based on a misunderstanding, it has been brought to my
attention that a similar theory already exists. Thad Roberts already had a theory of quanitized space and is about %80 similar to mine if you want a better idea of what i was trying to convey.
i will continue on the math and post again if i make any progress | {"url":"https://www.accursedfarms.com/forums/topic/5634-gravity-is-not-a-force/","timestamp":"2024-11-14T09:01:05Z","content_type":"text/html","content_length":"69231","record_id":"<urn:uuid:b0c65f62-a078-4577-bcb0-753dfb6aabc0>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00238.warc.gz"} |
TR06-105 | 23rd August 2006 00:00
Derandomizing the AW matrix-valued Chernoff bound using pessimistic estimators and applications
Ahlswede and Winter introduced a Chernoff bound for matrix-valued random variables, which is a non-trivial generalization of the usual Chernoff bound for real-valued random variables. We present an
efficient derandomization of their bound using the method of pessimistic estimators (see Raghavan). As a consequence, we derandomize a construction of Alon and Roichman (see also the alternate proof
given by Landau-Russell and Loh-Schulman) to efficiently construct an expanding Cayley graph of logarithmic degree on any (possibly non-abelian) group. This also gives an optimal solution to the
homomorphism testing problem of Shpilka and Wigderson. We also apply these pessimistic estimators to the problem of solving semi-definite covering problems, thus giving a deterministic algorithm for
the quantum hypergraph cover problem of Ahslwede and Winter.
The results above appear as theorems in the paper "A Randomness-Efficient Sampler for Matrix-Valued Functions and Applications" by the authors, as consequences to the main theorem of that paper: a
randomness efficient sampler for matrix valued functions via expander walks. However, we discovered an error in the proof of that main theorem (which we briefly describe in the appendix). That main
theorem stating that the expander walk sampler is good for matrix-valued functions thus remains open. One purpose of the current paper is to show that the applications in that paper hold true despite
our inability to prove the expander walk sampler theorem for matrix-valued functions. | {"url":"https://eccc.weizmann.ac.il/eccc-reports/2006/TR06-105/index.html","timestamp":"2024-11-14T21:40:22Z","content_type":"application/xhtml+xml","content_length":"21884","record_id":"<urn:uuid:5c0ce224-0adf-427d-af65-f093cd31aaf2>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00142.warc.gz"} |
Understanding interpolation analysis
Available with Spatial Analyst license.
Available with 3D Analyst license.
Interpolation predicts values for cells in a raster from a limited number of sample data points. It can be used to predict unknown values for any geographic point data, such as elevation, rainfall,
chemical concentrations, and noise levels.
Why interpolate to raster?
The assumption that makes interpolation a viable option is that spatially distributed objects are spatially correlated; in other words, things that are close together tend to have similar
characteristics. For instance, if it is raining on one side of the street, you can predict with a high level of confidence that it is raining on the other side of the street. You would be less
certain if it was raining across town and less confident still about the state of the weather in the next county.
Using the above analogy, it is easy to see that the values of points close to sampled points are more likely to be similar than those that are farther apart. This is the basis of interpolation. A
typical use for point interpolation is to create an elevation surface from a set of sample measurements. Geostatistical Analyst also provides and extensive collection of interpolation methods.
Examples of interpolation applications
Some typical examples of applications for the interpolation tools follow. The accompanying illustrations will show the distribution and values of sample points and the raster generated from them.
Interpolating a rainfall surface
The input here is a point dataset of known rainfall-level values, shown by the illustration on the left. The illustration on the right shows a raster interpolated from these points. The unknown
values are predicted with a mathematical formula that uses the values of nearby known points.
Input rainfall point data
Interpolated rainfall surface
Interpolating an elevation surface
A typical use for point interpolation is to create an elevation surface from a set of sample measurements.
In the following graphic, each symbol in the point layer represents a location where the elevation has been measured. By interpolating, the values for each cell between these input points will be
Input elevation point data
Interpolated elevation surface
Interpolating a concentration surface
In the example below, the interpolation tools were used to study the correlation of the ozone concentration on lung disease in California. The image on the left shows the locations of the ozone
monitoring stations. The image on the right displays the interpolated surface, providing predictions for each location in California. The surface was derived using kriging.
Point locations of ozone monitoring stations
Interpolated prediction surface | {"url":"https://desktop.arcgis.com/en/arcmap/latest/tools/3d-analyst-toolbox/understanding-interpolation-analysis.htm","timestamp":"2024-11-04T20:31:49Z","content_type":"text/html","content_length":"25268","record_id":"<urn:uuid:61f74788-be06-4145-9995-76f143eac970>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00326.warc.gz"} |
New paper: An estimated 13 million people worldwide killed by the COVID vaccines
That's twice as many as were killed by the virus. In the US, the estimates are 670K Americans killed. Perhaps it's time to declare that the vaccines are a worldwide emergency?
Executive summary
The paper suggests you can take the number of vaccine doses delivered, divide by 1,000 to get an estimate of the number of people killed by the COVID vaccine.
The tweet directly from the author sums it up:
The details
The paper: Age-stratified COVID-19 vaccine-dose fatality rate for Israel and Australia (ResearchGate removed it after 100K views; this is the original paper)
The full story about how universities undermine people like Denis Rancourt who tell the truth
The main result
The paper finds that the vDFR (vaccine-dose fatality rate) is exponential with respect to age.
The paper points out that “it is not unreasonable to assume an all-population global value of vDFR = 0.1 %”
For the US, 670M doses have been given, so the estimate is 670,000 people have been killed by the COVID vaccines in the US.
I have said for a long time that the URF in VAERS is 41, and there are 16,300 excess US deaths in VAERS (subtracting 250 background deaths per year which gets reported into VAERS) which comes to
668K is very close to 670K, isn’t it?
What an amazing “coincidence”!
A third method gives a similar result
The paper gives a simple way to estimate the number of people a country has killed by deploying the COVID vaccines: 0.001*# of doses.
In short, you can just take the number of vaccine doses in millions and just change “millions” to “thousands” to estimate the number of people killed by the vaccine.
Using data from Israel and Australia, the paper estimates 13 million deaths worldwide from the COVID vaccines:
The COVID-19 vaccines did not only not save lives but they are highly toxic.
On the global scale, given the 3.7 million fatalities in India alone, having vDFR = 1 % (Rancourt, 2022), and given the age-stratified vDFR results presented in this work, it is not unreasonable
to assume an all-population global value of vDFR = 0.1 %. Based on the global number of COVID-19 vaccine doses administered to date (13.25 billion 24 doses, up to 24 January 2023, Our World in
Data),3 this would correspond to 13 million deaths from the COVID-19 vaccines worldwide. By comparison, the official World Health Organization (WHO) number of COVID-19 deaths to date is 6.8
million (6,817,478 deaths, reported to WHO, as 3 February 2023),4 which are not detected as COVID-19 assignable deaths in ACM studies.
Given the available German numbers of number of vaccinations and the Danish study about severe adverse events of 0.23 percent and 1/1000 death rate (Steve Kirsch's estimiate). I then calculated the
death by mRNA vaccine as 200,000 citizens in Germany and another atl least 250,000 severely injured citizens. I used Bernoulli formula to account for mutiple shots per person and not reach excess
numbers. Still, nearly half a million people's lives were destroyed alone in Germany - including mine (heart attack from booster, father dead, brother cancer). And none of those criminals got
prosecuted. They instead push for new cancer "vaccines" in speed trials. Time to stand up worldwide for justice ad a Nuremberg 2!
Expand full comment
So my husband is reduced to a sad statistic. WHO do I sue?
Expand full comment
823 more comments... | {"url":"https://kirschsubstack.com/p/new-paper-an-estimated-13-million?utm_source=twitter&utm_campaign=auto_share&r=o7iqo","timestamp":"2024-11-14T13:38:39Z","content_type":"text/html","content_length":"231816","record_id":"<urn:uuid:145d3c1f-2619-4352-bc64-cee427a62e18>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00863.warc.gz"} |
Westlake Probability Day 2023
Time：April 1st 2023
Venue：E10-405, Yungu Campus, Westlake University
Speaker: Hao Wu, Tsinghua University
Title: Connection probabilities for random-cluster model and uniform spanning tree
Abstract: Conformal invariance of critical lattice models in two-dimensional has been vigorously studied for decades. In this talk, we focus on connection probabilities of critical lattice models in
polygons. This talk has two parts.
• In the first part, we consider critical random-cluster model with cluster weight $q\in (0,4)$ and give conjectural formulas for connection probabilities of multiple interfaces. The conjectural
formulas are proved for q=2, i.e. the FK-Ising model.
• In the second part, we consider uniform spanning tree (UST) and give formulas for connection probabilities of multiple Peano curves. UST can be viewed as the limit of random-cluster model as $q$
goes to 0. Its connection probabilities turn out to be related to logarithmic CFT.
This talk is based on joint works with Yu Feng, Mingchang Liu, and Eveliina Peltola.
Speaker: Alejandro Ramírez, NYU Shanghai
Title: Balanced excited random walk in two dimensions
Abstract: We give non-trivial upper and lower bounds on the range of the so-called Balanced Excited Random Walk in two dimensions, and verify a conjecture of Benjamini, Kozma and Schapira. These are
the first non-trivial results for this 2-dimensional model. This is a joint work with Omer Angel (University of British Columbia) and Mark Holmes (University of Melbourne).
Speaker: Elie Aidekon, Fudan University
Title: 1-d Brownian loop soup, Fleming--Viot processes and Bass--Burdzy flow
Abstract: We describe the connection between these three objects which appear in the problem of conditioning the so-called perturbed reflecting Brownian motion on its occupation field.
Joint work with Yueyun Hu and Zhan Shi.
Speaker: Dong Yao, Jiangsu Normal University
Title: Mean Field Behavior during the Big Bang regime for Coalescing Random Walks
Abstract: The talk is concerned with the coalescing random walk model on general graphs G. Initially every vertex of G has a particle. Each particle performs independent random walks. Whenever two
particles meet, they merge into one particle which continues to perform a random walk. We set up a unified framework to study the leading order of the decay rate of P(t), which is the expectation of
the fraction of occupied sites at time t, particularly for the ‘Big Bang’ regime where the time t<< T(coal):=E[inf{s: There is only one particle left at time s}].
Our results show that P(t) satisfies certain `mean field behavior', if the graphs satisfy certain ‘transience-like’ conditions. We apply this framework to two families of graphs: (1) graphs generated
by configuration model with degree at least 3, and (2) finite and infinite vertex-transitive graphs. In the first case, (tP(t))^{-1} is approximately the probability that two particles starting from
the root of the corresponding unimodular Galton-Watson tree never collide after one of them leaves the root. In the second case we establish similar results for finite ‘uniformly transient’ graphs
and infinite transient transitive unimodular graphs. Based on joint work with Jonathan Hermon, Shuangping Li and Lingfu Zhang. | {"url":"https://its.westlake.edu.cn/info/1131/1714.htm","timestamp":"2024-11-06T07:11:20Z","content_type":"text/html","content_length":"28891","record_id":"<urn:uuid:b121bbf7-3dce-430a-a9d6-c87e5f24d06a>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00010.warc.gz"} |
On This Day in Math - December 24
The 358th day of the year; 358 is twice a prime, and the sum of six consecutive primes, 47 + 53 + 59 + 61 + 67 + 71
The sum of the first 358 prime numbers is itself a prime number.
and in case you were curious, the 358th digit of pi (after the decimal point) is 3.
The product of the first 358 integers is not divisible by the sum of the first 358 integers. That means that for any numbers x, y both greater than 1, there is no solution to the equation x + xy + y
=158. If this is not a leap year, then this is the last year day this year with this quality. (younger students might try to show these interrelated ideas with a smaller number like 10.)
35+ 8 + 3 + 58 + 3 + 5 + 8 = 3 x 5 x 8 = 120
Derek Orr pointed out that 358 = 2 x 179, and 2+ 179 = 181 is prime, and 2 + 1 + 7 + 9 = 19 is also prime.
358 = 18^2 + 5^2 + 3^2 = 14^2 + 9^2 + 9^2
1734 A letter and drawing by Martha Gerrish, a woman of New England, described a Parhelion, (aka Sundog, an atmospheric optical phenomenon that consists of a bright spot to one or both sides of the
Sun) a rare astronomical phenomenon. Gerrish comments ‘if this came from a masculine hand, I believe it would be an acceptable present to the Royal Society’. This is the first letter to the Royal
Society known to be sent by a woman in her own name and demonstrates that women contributed to science even when their work was not made public.
The photograph shows very bright sun dogs in Fargo, North Dakota.*Wij
1754 Euler writes to Muller in St Petersburg and describes d'Alembert as "the most argumentative man in the world," and calls him, "hated by everyone in Paris." *Thomas L. Hankins, Jean d'Alembert:
science and the Enlightenment; pg 58
Bernard Bolzano was dismissed from his theological chair at the University of Prague and put under police supervision for his unorthodox religious views. In mathematics he helped remove the scandal
of infinitesimals from the calculus. *VFR
Gauss writes to the Astronomer Johann Franz Encke in response to Encke's remarks about the Frequency of Primes.
"Most Honored Friend!
. . .The kind communication of your remarks on the frequency of prime numbers
was interesting to me in more than one respect. You have reminded me of
my own pursuit of the same subject, whose first beginnings occurred a very long
time ago, in 1792 or 1793, when I had procured for myself Lambert’s supplement
to the table of logarithms. Before I had occupied myself with the finer
investigations of higher arithmetic, one of my first projects was to direct my
attention to the decreasing frequency of prime numbers, to which end I counted
them up in several chiliads (sets of a thousand) and recorded the results on one of the affixed white sheets. I soon recognized, that under all variations of this frequency, on average, it is
nearly inversely proportional to the logarithm..."
Gauss never published his results. The first published version of the Prime Number Theorem was by Legendre in 1798. *John Derbyshire, Prime Obsession
the Physico-Mathematic Society of Kazan (Russia) celebrated a Jubilee in honor of the twenty-fifth year of professional and scientific service of its President, Professor A. Vasiliev. It is also the
fifteenth year of his presidency. Professor Vasiliev has been an extraordinarily important figure in Russian science. Outside of Russia he has chiefly been known for his remarkable discourse on
Lobachevski. *The American Mathematical Monthly, Vol. 7, No. 1 (Jan., 1900), p. 30
First Long Range transmisson of voice: Reginald Aubrey Fessenden was a Canadian inventor and engineer with 300 patents. He broadcast the first program of voice and music. In 1893, Fessenden moved to
Pittsburgh as the head of electrical engineering at the university, Fessenden read of Marconi's work and began experimenting himself. Marconi could only transmit Morse code. But Fessenden's goal was
to transmit the human voice and music. He invented the "continuous wave": sound superimposed onto a radio wave for transmission. A radio receiver extracts the signal so the listener with the original
sound. Fessenden made the first long-range transmissions of voice on Christmas Eve 1906 from a station at Brant Rock, Massachusetts, heard hundreds of miles out in the Atlantic.*TIS He used a 42 kHz
radio frequency Alexanderson alternator which produced about 1kW of power. Although Fessenden's work made voice radio possible, it would take 10 years and the First World War before it became
commonplace. Throughout this period, radio was still seen primarily as point-to-point communication between transmitting stations--a sort of wireless telephone. *CHM
Irving Fisher (1867-1947), a Yale professor, patented an archiving system with index cards. On 1 Jul 1925, Fisher's own firm, the Index Visible Company, merged with its principal competitor to form
Kardex Rand Co., later Remington Rand, still later Sperry Rand. Fisher earned about
for the invention, which grew to the princely sum of
before being lost in the stock market crash of 1929. Fisher is widely regarded as the greatest economist America has produced, who made much use of mathematics in his work.*TIS
On Christmas Eve, the Apollo 8 astronauts saw the entirety of Earth for the first time. Turing their cameras to the Earth they took the first three pictures of the whole earth from space. The one
above, by William Anders, has been called "the most influential environmental photograph ever taken." *Wik, NASA
In 2004
, the Huygens probe began a 22-day descent towards Saturn's largest moon, Titan. It had been launched as part of the Cassini spacecraft in 1997, and together they entered Saturn's orbit in June 2004.
As the paths of the spacecraft and Titan converged, Cassini ejected the Huygens probe, sending it on a 22-day coast toward the cloud-covered moon. It landed 14 Jan 2005, and sent back photographs of
the moon's surface. Cassini will remain in orbit around Saturn until at least July 2008. The Cassini-Huygens mission to study Saturn and its 33 known moons resulted from an unprecedented cooperative
effort between the NASA of the United States, the European Space Agency and Italy's space program, at a cost of $3.3 billion. *TIS
1740 Anders Johan Lexell
(December 24, 1740 – December 11, 1784 (Julian calendar: November 30)) was a Swedish-born Russian astronomer, mathematician, and physicist who spent most of his life in Russia where he is known as
Andrei Ivanovich Leksel.
Lexell made important discoveries in polygonometry and celestial mechanics; the latter led to a comet named in his honor. La Grande Encyclopédie states that he was the prominent mathematician of his
time who contributed to the spherical trigonometry with new and interesting solutions, which he took as a basis for his research of comet and planet motion. His name was given to one of the theorems
about spherical triangles.
Lexell was one of the most prolific members of the Russian Academy of Sciences at that time, having published 66 papers in 16 years of his work there. A statement attributed to Leonhard Euler
expresses high approval of Lexell's works: "Besides Lexell, such a paper could only be written by D'Alambert or me". Daniel Bernoulli also praised his work, writing in a letter to Johann Euler "I
like Lexell's works, they are profound and interesting, and the value of them is increased even more because of his modesty, which adorns great men".
Lexell did not have a family and kept up a close friendship with Leonhard Euler and his family. He witnessed Euler's death at his house and succeeded him to the chair of the mathematics department at
the Russian Academy of Sciences, but died the following year. The asteroid 2004 Lexell is named in his honour, as is the lunar crater Lexell.*Wik
1818 James Prescott Joule
(24 Dec 1818; 11 Oct 1889) English physicist who established that the various forms of energy - mechanical, electrical, and heat - are basically the same and can be changed, one into another. Thus he
formed the basis of the law of conservation of energy, the first law of thermodynamics. He discovered (1840) the relationship between electric current, resistance, and the amount of heat produced. In
1849 he devised the kinetic theory of gases, and a year later announced the mechanical equivalent of heat. Later, with William Thomson (Lord Kelvin), he discovered the Joule-Thomson effect. The SI
unit of energy or work , the joule (symbol J), is named after him. It is defined as the work done when a force of 1 newton moves a distance of 1 metre in the direction of the force.*TIS
1822 Charles Hermite
(24 Dec 1822; 14 Jan 1901) French mathematician whose work in the theory of functions includes the application of elliptic functions to provide the first solution to the general equation of the fifth
degree, the quintic equation. In 1873 he published the first proof that e is a transcendental number. Hermite is known also for a number of mathematical entities that bear his name, Hermite
polynomials, Hermite's differential equation, Hermite's formula of interpolation and Hermitian matrices. Poincaré is the best known of Hermite's students.*TIS
1838 Thorvald Nicolai Thiele
(24 December 1838 – 26 September 1910) was a Danish astronomer, actuary and mathematician, most notable for his work in statistics, interpolation and the three-body problem. He was the first to
propose a mathematical theory of Brownian motion. Thiele introduced the cumulants and (in Danish) the likelihood function; these contributions were not credited to Thiele by Ronald A. Fisher, who
nevertheless named Thiele to his (short) list of the greatest statisticians of all time on the strength of Thiele's other contributions.
Thiele also was a founder and Mathematical Director of the Hafnia Insurance Company and led the founding of the Danish Society of Actuaries. It was through his insurance work that he came into
contact with fellow mathematician Jørgen Pedersen Gram. Asteroid 843 Nicolaia is named in his honor. *Wik
1868 Emanuel Lasker
(24 Dec 1868 in Berlinchen, Prussia (now Barlinek, Poland) - 11 Jan 1941 in New York, USA) Lasker became World Chess Champion in 1894 and held the championship until 1921. In mathematics he
introduced the notion of a primary ideal. *SAU
1904 Sir William Hunter McCrea FRS
(13 December 1904, Dublin – 25 April 1999) was an English astronomer and mathematician. He went to Trinity College, Cambridge in 1923 where he studied Mathematics, later gaining a PhD in 1929 under
Ralph H. Fowler. He was later appointed a lecturer of Mathematics at the University of Edinburgh in 1929. He also served as reader and assistant professor at Imperial College London. In 1936 he
became head of the mathematics department at the Queen's University of Belfast. After serving in the war, he joined the mathematics department at Royal Holloway College in 1944 (the McCrea Building
on Royal Holloway's campus is named after him). In 1965, McCrea created the astronomy centre of the physics department at the University of Sussex.
In 1928, he studied Albrecht Unsöld's hypothesis, and discovered that three quarters of the Sun is made of Hydrogen, and about one quarter is Helium, with 1% being other elements. Previous to this
many people thought the Sun consisted mostly of Iron. After this, people realized most stars consist of Hydrogen.
McCrea was president of the Royal Astronomical Society from 1961-3 and president of Section A of the British Association for the Advancement of Science from 1965-6. He was knighted in 1985. He won
the Gold Medal of the Royal Astronomical Society in 1976. McCrea died on April 25, 1999 in Lewes. *Wik
1910 William Hayward Pickering
(24 Dec 1910; 15 Mar 2004) Engineer and physicist, head of the team that developed Explorer 1, the first U.S. satellite. He collaborated with Neher and Robert Millikan on cosmic ray experiments in
the 1930s, taught electronics in the 1930s, and was at Caltech during the war. He spent the rest of his career with the Jet Propusion Laboratory, becoming its Director (1954) with responsibility for
the U.S. unmanned exploration of the planets and the solar system. Among these were the Mariner spacecraft to Venus and Mercury, and the Viking mission to Mars. The Voyager spacecraft yielded
stunning photographs of the planets Jupiter and Saturn.*TIS
How sad to be one of these people who died on Christmas Eve
1872 William John Macquorn Rankine
(5 Jul 1820, 24 Dec 1872) Scottish engineer and physicist and one of the founders of the science of thermodynamics, particularly in reference to steam-engine theory. As the chair (1855) of civil
engineering and mechanics at Glasgow, he developed methods to solve the force distribution in frame structures. Rankine also wrote on fatigue in the metal of railway axles, on Earth pressures in soil
mechanics and the stability of walls. He was elected a Fellow of the Royal Society in 1853. Among his most important works are Manual of Applied Mechanics (1858), Manual of the Steam Engine and Other
Prime Movers (1859) and On the Thermodynamic Theory of Waves of Finite Longitudinal Disturbance. *TIS (
Many students are not aware there is a absolute temperature scale called the Rankine, named for him
1882 Johann Benedict Listing
(25 July 1808 in Frankfurt am Main, Germany - 24 Dec 1882 in Göttingen, Germany)wrote one of the earliest texts on Topology. he studied the figure of the earth in minute detail; he made observations
in meteorology, terrestrial magnetism, and spectroscopy; he wrote on the quantitative determination of sugar in the urine of diabetics; he promoted the nascent optical industry in Germany and better
street lighting in Göttingen; he travelled to the world exhibitions in London 1851, Vienna 1873 and London 1876 as an observer for his government; he assisted in geodetic surveys; ... he invented a
good many terms [other than topology], some of which have became current: "entropic phenomenona", "nodal points", "homocentric light", "telescopic system", " geoid" ...he coined "one micron" for the
millionth of a metre ...*SAU
1927 William Henry Dines
(5 Aug 1855, 24 Dec 1927) was an English meteorologist (like his father) and inventor of related measurement instruments such as the Dines pressure tube anemometer (the first instrument to measure
both the velocity and direction of wind, 1901), a very lightweight meteorograph, and a radiometer (1920). He joined the Royal Meteorological Society study of the cause of the disastrous Tay Bridge
collapse of 1879. His measurements of upper air conditions, first with kites and later by balloon ascents (1907), brought an understanding of cyclones from dynamic processes in the lower stratosphere
rather than thermal effects nearer to the ground.*TIS
1962 Wilhelm Friedrich Ackermann
(29 March 1896 – 24 December 1962) was a German mathematician best known for the Ackermann function, an important example in the theory of computation.*Wik
1994 Alfred Leon Foster
(13 July 1904 in New York City, New York, USA - 24 Dec 1994 in Berkeley, California, USA) Foster went on to define the concept of a primal algebra generalising a Boolean algebra within the theory of
varieties of universal algebras. In 1953 showed that the variety generated by a primal algebra has the same essential structure as the variety of Boolean algebras. He continued devoting his efforts
to the structure theory of algebras that are generalizations of Boolean algebras and, more than ten years down the line in 1966, he published Families of algebras with unique (sub-)direct
factorization. Equational characterization of factorization in Mathematische Annalen.*SAU
2000 Laurence Chisholm Young
(14 July 1905 – 24 December 2000) was a mathematician known for his contributions to measure theory, the calculus of variations, optimal control theory, and potential theory. He is the son of William
Henry Young and Grace Chisholm Young, both prominent mathematicians.
The concept of Young measure is named after him. *Wik
Credits :
*CHM=Computer History Museum
*FFF=Kane, Famous First Facts
*NSEC= NASA Solar Eclipse Calendar
*RMAT= The Renaissance Mathematicus, Thony Christie
*SAU=St Andrews Univ. Math History
*TIA = Today in Astronomy
*TIS= Today in Science History
*VFR = V Frederick Rickey, USMA
*Wik = Wikipedia
*WM = Women of Mathematics, Grinstein & Campbell | {"url":"https://pballew.blogspot.com/2022/12/on-this-day-in-math-december-24.html","timestamp":"2024-11-04T23:26:31Z","content_type":"application/xhtml+xml","content_length":"162352","record_id":"<urn:uuid:0bbb1118-9cfe-4cc6-938d-5753ea3520df>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00622.warc.gz"} |
Find the 7th term in the following sequence:
1, 4, 7, 10, cdots ? | HIX Tutor
Find the 7th term in the following sequence: #1, 4, 7, 10, cdots #?
Answer 1
Sign up to view the whole answer
By signing up, you agree to our Terms of Service and Privacy Policy
Answer 2
$\textcolor{g r e e n}{19}$
The sequence appears to be an arithmetic sequence with a difference between terms of #color(blue)(3)#.
In general given a first term: #color(red)(a_1)# and a difference between successive terms: #a_n-a_(n-1)=color(blue)(d)#
the #n#th term can be calculated as #a_n=color(red)(a_0)+color(blue)(d) * (n-1)#
Therefore the #7#th term of the given sequence should be #color(white)("XXX")a_7=color(red)(1)+color(blue)(3) * (7-1) = 19#
Sign up to view the whole answer
By signing up, you agree to our Terms of Service and Privacy Policy
Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
Not the question you need?
HIX Tutor
Solve ANY homework problem with a smart AI
• 98% accuracy study help
• Covers math, physics, chemistry, biology, and more
• Step-by-step, in-depth guides
• Readily available 24/7 | {"url":"https://tutor.hix.ai/question/find-the-7th-term-in-the-following-sequence-1-4-7-10-cdots-8f9af8d739","timestamp":"2024-11-04T17:56:01Z","content_type":"text/html","content_length":"569707","record_id":"<urn:uuid:5f6db71f-48e5-4b60-9703-a80e7cd56db2>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00784.warc.gz"} |
140 Followers Problem
$\Large\lfloor \sqrt{x} \rfloor + \lfloor \sqrt{y} \rfloor + \lfloor \sqrt{z} \rfloor = 140$
Find the number of non-negative integral solutions to the equation above.
Note: $\lfloor k \rfloor$ represents the greatest integer less than or equal to $k$ .
You can use calculators to calculate the final sum.
Moderator note:
Great approach with converting it into a generating function and manipulating it from there.
Log in to reply
Log in to reply | {"url":"https://solve.club/problems/140-followers-problem/140-followers-problem.html","timestamp":"2024-11-10T11:51:20Z","content_type":"text/html","content_length":"311473","record_id":"<urn:uuid:c0d17187-f9c6-45ff-bad3-b2d1c586a768>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00571.warc.gz"} |
The Parable of the Golfers
Why a high match% to Rybka is usually not evidence of cheating, unless...
In the Gospels, a parable means a hypothetical example for express purpose of comparison to real-life situations.
According to the official study by mathematician Francis Scheid summarized here on Golf.About.com, the odds of a "low handicapper" golfer hitting a hole-in-one on a standard par-3 hole are about
1-in-5,000. Many of us hackers will never see one, let alone make one.
But if you assemble 10,000 such golfers to take a swing, you will expect to see not just one but two of them. And there will be nothing sinister about those who get them---it is just the run of
(Photo source; Payne Stewart.)
However, suppose you slip a black spot into the jacket pockets of 10 of the golfers, and one of them hits a hole-in-one. Now you have a coincidence! The odds are over 500-1 against, well within the
common civil court standard for statistical unlikelihood.
The application in chess starts with my having taken over 28,000 human performances, about 10,000 in recent years and a different 10,000 from grandmaster-level players of all years. Would you guess I
have seen some holes-in-one? You already know enough to justify saying that by pure chance, I probably must have...
So what can constitute evidence of cheating?
The answer is, what is a black spot?
The policy which I set on this site almost five years ago is that the spot can only be physical or observational evidence of cheating, something independent of the consideration of chess analysis and
statistical matching to a computer.
Thus the statistical analysis can only be supporting evidence of cheating, in cases that have some other concrete distinguishing mark. When such a mark is present, the stats can be effective, and can
meet civil court standards of evidence, which begin with a two-sigma deviation representing odds of roughly 40-1 (one-sided) or 20-1 (two-sided) against the "null hypothesis" of a chance occurrence.
One other thing the stats can do is quantify how much benefit in rating-point terms was obtained by the alleged means.
(I am told that in certain industrial contexts, a five-sigma or six-sigma deviation---talking odds over a million to one against---can be prima-facie evidence of malfeasance. Moreover as I noted here
, five-sigma is the particle-physics standard for discovery, while 3.5-sigma (roughly 2000-1 odds) enables one to claim evidence. There just aren't enough moves and games and players to get even a
sniff of five-sigma, but 3.5 sigma, that happens...) Update 1/13/13: Oh, 5-sigma happens too, but there still aren't enough games and players to compare against actual frequencies of such deviations.
One thing that does not constitute a black spot is an unfounded accusation of cheating. They are all too cheap, and also importantly based on cases since "Toiletgate" where move-matching has been the
alleged evidence, are not independent either.
The general statistical principle about holes-in-one described here is called Littlewood's Law. Another example is that if you play 40 games, chances are you'll have one where you would have the 40-1
statistical outlier situation. And every 250 players will have one of those in the last-round of a six-round Swiss. Note that you already need the full model described in my papers even just to judge
this kind of outlier---if you merely get a lot of matching, you may simply have played an unusually forcing game.
Still another example notes that my Intrinsic Ratings Compendium states "95% confidence intervals" for every IPR, both the "e"xpected one (theoretical assuming no error in the rest of my modeling)
and "a"ctual ones from preliminary field tests that are 1.4 times wider. And note that they are pretty wide, often over 100 Elo points both ways from the middle. The stated IPR's are "best
estimates", but individual ones can be off the mark from "true" values by the amounts allowed. In particular for every 40 data points---and there are already well ver 40 there---you can expect that
one of them will be gonzo-high above that range, and another will be gonzo-low, since 38/40 = 95%. This may be so in particular for the runs on some world championship matches. But when you take the
aggregate you see that the scheme is on the whole pretty close---the average is within 4 Elo of the players' ratings with each match at face value, and within 9 Elo if weighting by (games or) moves.
The Cat 20+ tournaments on the other hand, are systematically low. My current hypothesis is that this is due to a higher mean difference in ratings among players in tournaments, whereas the training
sets are all games with both players within 20 points of each other, and there as in matches the perception of skill parity may increase the circumspection of the play. That would cause higher IPR's
for both the training sets which set the scale, and the matches. But then again, the run on the 2011 Canadian Open with much greater game disparity came out systematically high, so other factors may
be present. Plus as I noted here, I will be upgrading the fitting procedure so numbers may change---though the last change mattered less than +- 5 Elo so I've ignored it. | {"url":"https://cse.buffalo.edu/faculty/regan/chess/fidelity/Golfers.html","timestamp":"2024-11-07T17:13:59Z","content_type":"text/html","content_length":"6909","record_id":"<urn:uuid:07901bbf-e8d3-4b85-a1a6-df66a7f7219c>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00791.warc.gz"} |
6326 (number)
Interesting facts about the number 6326
• (6326) Idamiyoshi is asteroid number 6326. It was discovered by A. Sugie from Dynic Astronomical Observatory on 3/18/1991.
• There is a 3,931 miles (6,326 km) direct distance between Brisbane (Australia) and Padang (Indonesia).
• There is a 6,326 miles (10,180 km) direct distance between Copenhagen (Denmark) and Duque de Caxias (Brazil).
• There is a 6,326 miles (10,180 km) direct distance between Donetsk (Ukraine) and Houston (USA).
• There is a 3,931 miles (6,326 km) direct distance between Karaj (Iran) and Suzhou (China).
• More distances ...
• There is a 6,326 miles (10,180 km) direct distance between Kazan (Russia) and Natal (Brazil).
• There is a 3,931 miles (6,326 km) direct distance between Kiev (Ukraine) and Lijiang (China).
• There is a 6,326 miles (10,180 km) direct distance between La Paz (Bolivia) and Lusaka (Zambia).
• There is a 3,931 miles (6,326 km) direct distance between Marrakesh (Morocco) and Santo Domingo (Dominican republic).
• There is a 3,931 miles (6,326 km) direct distance between Mosul (Iraq) and Tianjin (China).
• There is a 3,931 miles (6,326 km) direct distance between Nowrangapur (India) and Riga (Latvia).
• There is a 3,931 miles (6,326 km) direct distance between Rostov-na-Donu (Russia) and Xinyang (China).
• There is a 3,931 miles (6,326 km) direct distance between Sakai (Japan) and Thāne (India).
What is 6,326 in other units
The decimal (Arabic) number
converted to a
Roman number
Roman and decimal number conversions
The number 6326 converted to a Mayan number is
Decimal and Mayan number conversions.
Length conversion
6326 kilometers (km) equals to
miles (mi).
6326 miles (mi) equals to
kilometers (km).
6326 meters (m) equals to
feet (ft).
6326 feet (ft) equals
meters (m).
Time conversion
(hours, minutes, seconds, days, weeks)
6326 seconds equals to 1 hour, 45 minutes, 26 seconds
6326 minutes equals to 4 days, 9 hours, 26 minutes
Zip codes 6326
• Zip code 6326 ANGUIL, LA PAMPA, Argentina a map
• Zip code 6326 COLONIA ANGUIL, LA PAMPA, Argentina a map
• Zip code 6326 COLONIA SAN JUAN, LA PAMPA, Argentina a map
Zip code areas 6326
Number 6326 morse code:
-.... ...-- ..--- -....
Sign language for number 6326:
Number 6326 in braille:
Share in social networks
Advanced math operations
Is Prime?
The number 6326 is not a
prime number
. The closest prime numbers are
The 6326th prime number in order is
Factorization and factors (dividers)
The prime factors of 6326 are
2 * 3163 The factors of
6326 are
, 6326.
Total factors 4.
Sum of factors 9492 (3166).
Prime factor tree
The second power of 6326
is 40.018.276.
The third power of 6326
is 253.155.613.976.
The square root √
is 79,536155.
The cube root of
is 18,49452.
The natural logarithm of No. ln 6326 = log
6326 = 8,752423.
The logarithm to base 10 of No. log
6326 = 3,801129.
The Napierian logarithm of No. log
6326 = -8,752423.
Trigonometric functions
The cosine of 6326 is 0,392356.
The sine of 6326 is -0,919813.
The tangent of 6326 is -2,344331.
Properties of the number 6326
Number 6326 in Computer Science
Code type Code value
PIN 6326 It's recommended that you use 6326 as your password or PIN.
6326 Number of bytes 6.2KB
Unix time Unix time 6326 is equal to Thursday Jan. 1, 1970, 1:45:26 a.m. GMT
IPv4, IPv6 Number 6326 internet address in dotted format v4 0.0.24.182, v6 ::18b6
6326 Decimal = 1100010110110 Binary
6326 Decimal = 22200022 Ternary
6326 Decimal = 14266 Octal
6326 Decimal = 18B6 Hexadecimal (0x18b6 hex)
6326 BASE64 NjMyNg==
6326 MD5 17b65afe58c49edc1bdd812c554ee3bb
6326 SHA1 290b0c49057303b315fe94ca53f1f1aec1154b4f
6326 SHA224 5a86187cf4d39df69150056fc5057d161a0a482879f3187e2b97e200
6326 SHA256 cef142549ec7b7a05cc43a9f567e37abc3dbb19b496e656f66c3bffd4eb8cfea
6326 SHA384 f6d3d9fe66f0d10cc16cf34a6941e55af999be2c3e8ed5d438236d2fa3cc76fe27dbcd198c440b379f3522fae41722d2
More SHA codes related to the number 6326 ...
If you know something interesting about the 6326 number that you did not find on this page, do not hesitate to write us here.
Numerology 6326
The meaning of the number 6 (six), numerology 6
Character frequency 6: 2
The number 6 (six) denotes emotional responsibility, love, understanding and harmonic balance. The person with the personal number 6 must incorporate vision and acceptance in the world. Beauty,
tenderness, stable, responsible and understanding exchange, the sense of protection and availability also define the meaning of the number 6 (six).
More about the the number 6 (six), numerology 6 ...
The meaning of the number 3 (three), numerology 3
Character frequency 3: 1
The number three (3) came to share genuine expression and sensitivity with the world. People associated with this number need to connect with their deepest emotions. The number 3 is characterized by
its pragmatism, it is utilitarian, sagacious, dynamic, creative, it has objectives and it fulfills them. He/she is also self-expressive in many ways and with good communication skills.
More about the the number 3 (three), numerology 3 ...
The meaning of the number 2 (two), numerology 2
Character frequency 2: 1
The number two (2) needs above all to feel and to be. It represents the couple, duality, family, private and social life. He/she really enjoys home life and family gatherings. The number 2 denotes a
sociable, hospitable, friendly, caring and affectionate person. It is the sign of empathy, cooperation, adaptability, consideration for others, super-sensitivity towards the needs of others.
The number 2 (two) is also the symbol of balance, togetherness and receptivity. He/she is a good partner, colleague or companion; he/she also plays a wonderful role as a referee or mediator. Number 2
person is modest, sincere, spiritually influenced and a good diplomat. It represents intuition and vulnerability.
More about the the number 2 (two), numerology 2 ...
№ 6,326 in other languages
How to say or write the number six thousand, three hundred and twenty-six in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 6.326) seis mil trescientos veintiséis
German: 🔊 (Nummer 6.326) sechstausenddreihundertsechsundzwanzig
French: 🔊 (nombre 6 326) six mille trois cent vingt-six
Portuguese: 🔊 (número 6 326) seis mil, trezentos e vinte e seis
Hindi: 🔊 (संख्या 6 326) छः हज़ार, तीन सौ, छब्बीस
Chinese: 🔊 (数 6 326) 六千三百二十六
Arabian: 🔊 (عدد 6,326) ستة آلاف و ثلاثمائة و ستة و عشرون
Czech: 🔊 (číslo 6 326) šest tisíc třista dvacet šest
Korean: 🔊 (번호 6,326) 육천삼백이십육
Danish: 🔊 (nummer 6 326) sekstusinde og trehundrede og seksogtyve
Hebrew: (מספר 6,326) ששת אלפים שלש מאות עשרים ושש
Dutch: 🔊 (nummer 6 326) zesduizenddriehonderdzesentwintig
Japanese: 🔊 (数 6,326) 六千三百二十六
Indonesian: 🔊 (jumlah 6.326) enam ribu tiga ratus dua puluh enam
Italian: 🔊 (numero 6 326) seimilatrecentoventisei
Norwegian: 🔊 (nummer 6 326) seks tusen, tre hundre og tjue-seks
Polish: 🔊 (liczba 6 326) sześć tysięcy trzysta dwadzieścia sześć
Russian: 🔊 (номер 6 326) шесть тысяч триста двадцать шесть
Turkish: 🔊 (numara 6,326) altıbinüçyüzyirmialtı
Thai: 🔊 (จำนวน 6 326) หกพันสามร้อยยี่สิบหก
Ukrainian: 🔊 (номер 6 326) шість тисяч триста двадцять шість
Vietnamese: 🔊 (con số 6.326) sáu nghìn ba trăm hai mươi sáu
Other languages ...
News to email
If you know something interesting about the number 6326 or any other natural number (positive integer), please write to us here or on Facebook.
Legal Notices & Terms of Use
The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties.
Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic.
Privacy Policy
Frequently asked questions about the number 6326
• How do you write the number 6326 in words?
6326 can be written as "six thousand, three hundred and twenty-six".
What is your opinion? | {"url":"https://number.academy/6326","timestamp":"2024-11-09T03:44:16Z","content_type":"text/html","content_length":"41591","record_id":"<urn:uuid:34c21d77-3f5a-4e24-9c3d-06bdb4641786>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00737.warc.gz"} |
Factorization of MSI data - part 1
Factorization of MSI data - part 1
In this first post of the series, we will discuss three linear approaches to factorization of MSI data, namely Principal Component Analysis (PCA), PCA + Varimax and Independent Component Analysis
Access publication
This post is part of our series titled "Unsupervised learning on MSI data", which contains the following entries:
1. Factorization of MSI data - part 1 (current post)
2. Non-negative matrix factorization in MSI
Table of contents
Unsupervised data analysis is primarily targeted at exploring the content of the data, and extracting the trends present in a mostly unbiased way. In contrast to supervised methods, these methods do
not require any labelling or prior information on the data. When such information is available, (semi-)supervised methods are generally the recommended way to go. If you are not yet familiar with MSI
technology and its associated data, please refer to our introductory post on MSI data analysis.
We recently published a review paper on unsupervised analysis of MSI data together with profs. Raf Van de Plas (TU Delft) and Richard Caprioli (Vanderbilt University). For in-depth information about
the topics we touch upon in this blog series, please consult our review paper and its references.
A wide array of techniques has been used in the unsupervised analysis of MSI data, which can broadly be broken down into 3 main categories, namely factorization methods, clustering methods and
manifold learning or non-linear dimensionality reduction techniques, as shown in Figure 1.
Figure1: Illustration of some broad classes of unsupervised data analysis approaches within MSI data analysis. This figure was taken from our review paper.
In this post, we will focus on factorization methods without non-negativity constraints on the results. Furthermore, as Aspect uses Python as one of its main foundations, we will limit ourselves to
methods that are readily available in the Python ecosystem. We will focus on non-negative factorization methods in one of our following posts, so stay tuned for that.
We will demonstrate our selected methods on a MSI dataset acquired in human lymphoma tissue using a Bruker rapifleX MALDI Tissuetyper instrument. The experiments focused on lipids with 2,5-DHB matrix
deposited by sublimation. A sampling resolution of 10 μm was used, collecting around 500.000 pixels. We focus on the m/z range of ~ 600-1200 totalling 8000 ion images per tissue. Below, we will apply
PCA, PCA + Varimax, and ICA to this dataset.
Matrix factorization in a nutshell
Matrix factorization techniques are an important class of methods used in unsupervised MSI data analysis. Matrix factorization methods take a large and often high-dimensional dataset acquired by a
MSI experiment (or several experiments), and decompose it into a (typically reduced) number of trends that underlie the observed data.
These approaches are applied to the matrix representation commonly used for MSI data sets, as discussed in our introductory post on MSI data analysis. When representing MSI data in matrix form, D is
a data matrix where rows denote pixels and columns denote m/z bins. Hence, for a single MSI data set, m is the number of pixels and n is the number of spectral bins or m/z bins, i.e. each row of D
represents the mass spectrum of a pixel in the sample along a common m/z binning
Specifically, all matrix factorization approaches formulate a data matrix D∈R^m^×^n^ as the product of two factor matrices M∈R^m^×^p^ and N∈R^n^×^p^, such that:
where T denotes the matrix transpose operator. The differences between matrix factorization approaches stem from their underlying assumptions, potential constraints on M and N and how goodness-of-fit
is implicitly defined within the objective function underpinning each approach. Very often, MNT is chosen to be a low-rank representation of D, i.e., p≪m and p≪n, thus leading to a latent
dimensionality reduction and approximation of D.
The resulting reduced representation enables the analyst to gain visual insight into the underlying structure of the MSI data, and it often exposes the spatial and molecular signals that tend to
co-localize and correlate (usually under the assumption of linear mixing). Furthermore, as these techniques can provide a lower-dimensional and lower-complexity representation of the original data,
they regularly serve as a starting point for follow-up computational analysis as well.
Principal Component Analysis (PCA)
The first method we’ll show is Principal Component Analysis (PCA). PCA has been one of the bread and butter tools for the analysis of MSI from the start, and is still probably the most widely applied
factorization method in MSI (see review paper for an extensive list of references). PCA is a well-known factorization method that is widely used in other fields and is ubiquitous in machine learning
in general. Seeing the importance of this method, we’ll use it to lay down some of the groundwork concepts for the other methods.
The goal of PCA is to reduce the dimensionality of a dataset, i.e. describe the dataset with a lower number of variables, while retaining as much of the original variation as possible given the
chosen amount of variables. These new variables, called the principal components (PCs), are linear combinations of the original variables (intensities per m/z bin) and each PC is uncorrelated to all
The first PC is defined such that it captures the largest possible amount of variance in the data. Each subsequent PC is uncorrelated to the previous PCs and describes the largest possible variance
that remains in the data after removal of the preceding PCs. Roughly speaking, this means that the largest trend in the MSI data will be captured by the first PC, the second largest trend by the
second PC, etc.
Figure 2: Graphical representation of PCA as a matrix decomposition for an MSI data matrix
PCA analysis can be written in the form of a matrix decomposition. Let us take our MSI data matrix D. PCA then decomposes D as
S is an m×p matrix with orthogonal columns, often called the score matrix, and L is an n×p matrix with orthonormal columns, traditionally called the loading matrix. Note that in PCA the exact
equality holds as long as we retain all principal components, in which case there is no dimensionality reduction. We’ll explain how PCA works, and how it applies to MSI data below.
PCA applied to toy data
We’ll first illustrate PCA using a simple toy example. In Figure 3 we see a distribution of points in two dimensions. In this very simple example, let’s say that each point is a pixel in our MSI
image, and that we only measured two m/z bins, where axis 1 and axis 2 represent the intensities measured for m/z bins 1 and 2 respectively. From this image we can see that there is a relatively
large variance in the intensities along the axis of bin 2, and a smaller variance in the intensities along bin 1.
If we apply PCA to this simple example, we would get a new orthogonal set of axes, where the first axis (the first principal component or PC 1) follows the direction of the largest variance. In our
case, the largest variation occurs along m/z 2, but still contains some influence from m/z 1 as the two are correlated – it is a linear combination of the two, where m/z 2 has the largest
coefficient. PC 2 follows the direction of second largest variance, and must be orthogonal to PC1. In this simple 2D case, there is only one way the axis can be oriented due to the orthogonality
constrains (except for a sign change).
If we are interested in the large trends in the data, we can discard PC2, and still maintain a lot of that variability in our original data. This said, we will lose the “information” along PC2 if we
do so. In this simple case we can see that we probably would not lose a lot of information if PC2 is discarded, as that direction contains mostly noise.
Note that, while we started with all-positive data on axes 1 and 2 initially, the origin of the new orthogonal axes defined by PCA is placed at the center of where the variance occurs (mean
subtraction). This will result in negative values when we project our data onto these new axes, which will be an important point below.
Figure 3: PCA applied on a simple toy data set.
PCA on an image
Let’s now take an example a bit less abstract, namely a regular color image of another well known Principal. The image has 4 channels, namely RGB and a channel for transparency (A). This example can
be seen as an MSI experiment with 4 m/z bins.
If we apply PCA to this image we get the result in Figure 4, where each column represents a different principal component. On the top, we can see the scores or here the spatial trends for this image
(in false color images), and on the bottom we can see the loadings, which represent the relative importance of each channel.
From the first principal component, we can see that a lot of the relevant “information” in this image is immediately captured in the first principal component. Furthermore, we can see that the RGB
channels are nearly equally important in the loadings, but that the transparency channel is pretty much ignored.
Then, in PC2 we see essentially the varation in the image along the transparency channel, with a clear localization of where the transparency channel is high. This is something that is also observed
in MSI experiments where the area outside of the tissue is measured as well.
Finally, we see that PC3 and PC4 provide additional information, but focus increasingly more on smaller parts of the image. This also shows that the number of components that are relevant depend on
the the goal of the analysis. If the goal was to recognize the character, we would have enough with PC1 alone, and could discard the other components, but we’ll come back to that later.
Figure 4: Principal Skinner Component Analysis. On the left the original image, on the right the principal components with their score images (top) and loading vectors (bottom).
PCA applied to MSI data
When applying PCA to MSI data, the resulting score matrix S extracts trends in the pixel space, i.e. spatial trends (the singular vectors spanning the pixel space), whereas the loading matrix L
extracts spectral trends (singular vectors spanning the spectral space) from the data. Figure 5 shows the 10 first principal components, with the spatial expression on the left, and the matching
spectral expression on the right.
Figure 5: Graphical representation of the PCA decomposition, along with the resulting components for the lymphoma dataset. We calculate the first 200 principal components using the PCA implementation
in scikit-learn, one of the key machine learning libraries in Python.
Looking at these principal components, we can immediately see various large spatial trends in the data, with different regions in the tissue lighting up, such as the connective tissues, the lymph
nodes and necrotic regions, each with their respective spectral patterns. The spectral patterns are also sometimes referred to as pseudo-spectra, and represent the linear combinations of m/z bins
that constitute the principal component. This pseudo-spectrum shows the biomolecular ions that are involved in the expression of the spatial pattern on the left, where each region has its own
characteristic spectral fingerprint, i.e. its own set of biomolecular ions, which attests to the chemical specificity and richness of the biochemical information that is captured by MSI.
As a general rule, peaks that have a high absolute intensity in the pseudo-spectrum will have an important role in the spatial expression of the region highlighted by the principal component.
Some caveats must be made: ideally, this pseudo-spectrum would give a straightforward list of which bio-molecular ions are involved in each region from a biological sense. However, as we said before,
the goal of PCA is to capture and summarize as much of the variation in the data as possible per principal component. This is not necessarily the same as correctly modeling the underlying biology or
sample content. It’s a bit like you had only one sheet of paper to summarize an entire book and went a bit overboard. While it may capture the essence of the book, the end result might not be too
reader-friendly or clear on what everything means.
One example of interpretability issues is the negative peaks that appear in the pseudo-spectra. These can be difficult to interpret from a mass spectrometry perspective as the original MSI data
contains only positive values, namely the ion counts. Similarly, negative values appear in the spatial expressions of the components, making the information difficult to interpret. The issue of
negative values in the resulting components is handled by techniques such as non-negative matrix factorization, which we will treat in a future post.
Furthermore, due to its eagerness to summarize the data, PCA has a tendency to pack as many m/z bins in the pseudo-spectrum as possible, making the pseudo-spectra difficult to interpret. As we’ll see
below, techniques such as varimax can help in improving this issue.
Selecting the number of principal components
An important issue when using PCA is the number of components to retain. This is not just a problem for PCA, but for nearly any dimensionality reduction method, be it factorization or clustering.
When using factorization in the context of MSI, we hope to get a much more concise representation of the original data by grouping together m/z bins that operate together, such that we end up with
tens to hundreds of components compared to the original thousands to millions of m/z variables. In PCA, each subsequent principal component will capture less and less of the variation in the data
until we end up with components that just capture noise. Identifying that cut-off point is difficult and there is no clear cut answer to this problem, however, an often-used method is the cumulative
variance plot, which shows the percentage of variance that each additional component explains. This can be used to set a cut off point for how many components to retain, based on the amount of
variance that is captured.
Figure 6: Plot showing the explained variance by the PCA model in terms of the amount of principal components.
In figure 6, we see that for our lymphoma dataset, we can capture 95% of the variance by the first 50 principal components, which is a lot less than our original 8000 m/z bins. While this will
definitely capture the most important trends in the data, biological signals can be relatively localized or low intensity, thun contributing relatively little to the variance. As such, it is
generally recommended to visually check if there is still spatial structure in the spatial expression images after, in this case, the 50th component.
In summary, despite some of its shortcomings, PCA is a great tool to get a sense of the content of the MSI data, and is widely available in many different programming languages. Furthermore, it can
often serve as an important dimensionality reduction step, prior to applying other algorithms, such as e.g. independent component analysis (ICA) or clustering techniques, due to its ability to
capture as much of the original variability in the data as possible.
PCA, as well as many other factorization methods, suffer from what is known as ‘rotational ambiguity’. Above we showed that the result of a PCA analysis is a new set of orthogonal axes onto which the
data is projected. We can rotate these axes and still represent the same amount of information, i.e. there are an infinite number of ways in which we can orient the axes without losing any
“information”, as is illustrated in Figure 7.
Figure 8: Varimax applied to the first 50 principal components of our lymphoma dataset. We clearly see that many of the pseudo-spectra have fewer and larger peaks.
Applied to our MSI data, roughly speaking, maximizing the variance of the squared loadings, means that the algorithm will try to find a rotation, such that there are a lot of loadings with high
peaks, and that these high peaks are spread across the different loadings as much as possible. Compared to PCA, this means fewer m/z bins that are non-zero, which greatly improves the
interpretability of the loading components.Below, we have applied the Varimax rotation to the first 50 principal components, using the Varimax rotator function in the FactorAnalyzer module. Similar
to the PCA results, we plotted the 10 components that we get out of the Varimax rotation. We again get a spatial expression and a pseudo-spectrum, which we plotted for each. As anticipated, the
pseudo-spectra for many of the components clearly contain fewer and larger peaks, making the resulting components more readily interpretable. Some of these are now perhaps too sparse to give a proper
summarization of the data, however the pseudo-spectra are now much more likely to have the spatial distribution shown on the left.
Using PCA+Varimax involves the following steps:
1. Compute the PCA composition with a certain (relatively large) number of components, typically 100-500 though this depends on the dataset and use-case.
2. Determine the number of PCs that must be retained (N) to capture sufficient variation within the data, for instance using a Pareto chart as in Figure 5.
3. Compute the Varimax rotation on the first N PCs to obtain the new basis (which spans the same subspace as the original N PCs).
Note that steps 1 and 2 are identical to PCA, only the rotation in step 3 is new.
There are two important caveats with using Varimax. First off, unlike PCA, the components are no longer ranked by variance i.e. “importance”, so we would have to look at all 50 components to get a
good image of what is going on. Nonetheless, when applying a Varimax rotation to N principal components, the exact same amount of variance is explained as in PCA with N components. Secondly, unlike
PCA, Varimax does not have an analytical solution; it is an iterative algorithm, and a global optimum is not guaranteed. This means that running the algorithm multiple times will generally result in
different solutions of comparable quality.
Independent Component Analysis (ICA)
For the last algorithm in this post we’ll have a look at Independent Component Analysis, a matrix decomposition technique that originated from the area of blind source separation, and that aims to
retrieve the statistically independent components that underlie the observed data. ICA has been used extensively in a wide range of applications, such as facial recognition, fMRI, and remote sensing.
While PCA and ICA are similar in premise, ICA aims to find components that are statistically independent of each other, where PCA requires its components to be “merely” uncorrelated. The requirement
by ICA is stronger than that imposed by PCA: if two variables X and Y are statistically independent, they are also uncorrelated, however, uncorrelated variables are not necessarily independent. A
similar statistical peculiarity has been featured in xkcd.
As said, ICA originates in blind source separation theory, where a well known example to illustrate the goal at hand is the cocktail party problem. This example, featured in Figure 9, has the premise
that a party is held in a room, and microphones are placed around different locations in that room. The microphones all pick up the same party, but microphone 1 is located closer to group of guests 1
than microphone 2. The aim of the algorithm is to find the different “sources” (here e.g. the speakers) from which sound in the party is originating, and reconstruct their original signals (here e.g.
the conversations).
Figure 9: A casual cocktail party with microphones listening in on everybody.
This premise differs from that of PCA, which simply aims to find the find the largest variance in the signal picked up by the microphones, which may or may not be the same as separating the different
sources and their signals. If a song with a heavy bass is playing in the room, PCA will very likely separate that bass as a separate component, but it might also throw the loud and low-humming aircon
in the back together with that bass line in one component. This is not to say that this would never happen in ICA, but in principle it should at least try not to ;).
An additional motivation for using ICA over PCA is that one of the underlying assumptions of PCA is that the underlying variables have a normal distribution, which is generally not true for mass
spectrometry imaging data. The intensities of a peak in a mass spectrum are the result of a counting process, and are thus Poisson distributed, rather than normally distributed. Furthermore,
biological processes often give rise to non-normally distributed signals (e.g. feedback loops). In fact, most real-life data is not normally distributed, and ICA instead uses such non-normal features
of the data to find the underlying signals. ICA algorithms will often use PCA as a preprocessing step to perform dimensionality reduction, and to “whiten” the data, i.e. remove any correlations from
Caveats: Similar to Varimax, fastICA does not rank the components and does not have an analytical solution. Likewise a global optimum is not guaranteed and running the algorithm multiple times will
generally result in different solutions.
Below we’ve applied the FastICA algorithm available in scikit-learn to our dataset, setting the number of components to 50 based on our variance plot, and the number of iteration per component to
1000. Compared to PCA, the spatial expressions seem to be more focused to specific regions in the tissue. In our results, we saw some spatial structures that did not immediately show up in the PCA
and PCA+Varimax decompositions. Something that we’ve observed in the past is that often the ICA pseudo-spectra contain less of a mixture of negative and positive peaks, although this is not very
obvious in our current example. While it is difficult to say which decomposition is the better one, without a ground truth, in my PhD thesis (which you can find here). I did some experiments with
artificially constructed MSI datasets, where ICA was better at retrieving the original artificial spectra than PCA.
Figure 10: ICA applied to the lymphoma dataset. Compared to PCA, the spatial expressions seem to be more focused to specific regions in the tissue.
In conclusion, I hope I’ve shown you in this first part how factorization methods can help in extracting the underlying trends from complex mass spectrometry imaging datasets, making this information
ready for human consumption. If you have questions, want to know more, or have a project where we can help, please feel free to contact us. | {"url":"https://www.aspect-analytics.com/media-blog/factorization-of-msi-data-part-1-","timestamp":"2024-11-01T21:05:20Z","content_type":"text/html","content_length":"47017","record_id":"<urn:uuid:f44b6afb-d572-4c16-b298-ea8df75faa78>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00695.warc.gz"} |
Decision procedures for arithmetic with powers
Suitable for
The project will look at implication and satisfaction problems involving formulas that involve additive arithmetic, inequalities, and some form of exponentiations with a constant base. For example,
systems of inequalities with constant multiples of variables and constant multiple of expressions like 2^x. We will also look at first order logic built up from these inequalities. In the absence of
exponentiatial terms like 2^x, the theory is known as Presburger Arithmetic, and has been heavily investigated in both theory and practice. In the presence of expenontial terms to a fixed base, it is
known to be decidable, from work of Semenov in the late 70's and 80's. Recent work has shown fragments where the complexity is not too terrible, and in the process some new algorithms have been
The goal of the project is to implement decision procedures for expressions of this form. A theoretically-oriented student can also work on complexity analysis. A pre-requisite for the project is
good background in logic - Logic and Proof and preferably one of CAFV or KRR. | {"url":"https://www.cs.ox.ac.uk/teaching/studentprojects/878.html","timestamp":"2024-11-01T19:33:53Z","content_type":"text/html","content_length":"25184","record_id":"<urn:uuid:dcd7808d-a5a6-45da-b97e-0dacaa4dbc44>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00193.warc.gz"} |
Multiplication Chart Blank 12×12 | Multiplication Chart Printable
Multiplication Chart Blank 12×12
How To Use Times Tables Grids 11 Free Printable Times Tables Grids
Multiplication Chart Blank 12×12
Multiplication Chart Blank 12×12 – A Multiplication Chart is a valuable tool for youngsters to discover exactly how to multiply, divide, and discover the smallest number. There are lots of uses for a
Multiplication Chart. These helpful tools aid children comprehend the procedure behind multiplication by utilizing tinted paths and filling in the missing out on items. These charts are complimentary
to download and publish.
What is Multiplication Chart Printable?
A multiplication chart can be utilized to aid children learn their multiplication facts. Multiplication charts been available in several kinds, from full web page times tables to solitary page ones.
While private tables serve for offering chunks of info, a complete web page chart makes it less complicated to examine facts that have actually already been grasped.
The multiplication chart will usually feature a left column as well as a leading row. The leading row will certainly have a listing of products. When you wish to locate the product of 2 numbers,
select the first number from the left column as well as the second number from the top row. As soon as you have these numbers, move them along the row or down the column till you get to the square
where the two numbers fulfill. You will after that have your product.
Multiplication charts are helpful discovering tools for both grownups and youngsters. Multiplication Chart Blank 12×12 are available on the Internet and can be printed out and also laminated for
Why Do We Use a Multiplication Chart?
A multiplication chart is a layout that reveals how to multiply two numbers. You pick the first number in the left column, move it down the column, and also after that select the second number from
the leading row.
Multiplication charts are valuable for many factors, including aiding youngsters find out exactly how to separate and simplify portions. They can also help youngsters discover exactly how to pick a
reliable common measure. Due to the fact that they serve as a constant pointer of the pupil’s progress, multiplication charts can likewise be practical as desk sources. These tools help us develop
independent learners who recognize the basic principles of multiplication.
Multiplication charts are additionally helpful for helping students remember their times tables. They help them discover the numbers by reducing the number of steps needed to complete each operation.
One technique for memorizing these tables is to concentrate on a single row or column at a time, and then relocate onto the next one. Eventually, the entire chart will be committed to memory. Just
like any type of skill, remembering multiplication tables takes time and method.
Multiplication Chart Blank 12×12
12×12 Multiplication Chart Template Download Printable PDF Templateroller
Blank Printable Multiplication Table Of 12×12
Printable Blank Multiplication Table 12 12 Printable Multiplication
Multiplication Chart Blank 12×12
If you’re looking for Multiplication Chart Blank 12×12, you’ve come to the appropriate location. Multiplication charts are readily available in different styles, including full dimension, half
dimension, as well as a variety of charming designs.
Multiplication charts as well as tables are important tools for kids’s education. You can download and also publish them to make use of as a mentor help in your kid’s homeschool or classroom. You can
additionally laminate them for longevity. These charts are fantastic for use in homeschool mathematics binders or as class posters. They’re specifically beneficial for kids in the 2nd, 3rd, and 4th
A Multiplication Chart Blank 12×12 is a valuable tool to reinforce math facts and can help a kid learn multiplication swiftly. It’s also an excellent tool for miss checking and finding out the
moments tables.
Related For Multiplication Chart Blank 12×12 | {"url":"https://multiplicationchart-printable.com/multiplication-chart-blank-12x12/","timestamp":"2024-11-13T02:04:30Z","content_type":"text/html","content_length":"43317","record_id":"<urn:uuid:ba5f3cec-cfd4-479f-82f0-82399fd9a4f9>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00856.warc.gz"} |
Inquiry Maths - Completing the square
Completing the square inquiry
Mathematical inquiry processes: Explore; generate examples; conjecture; reason. Conceptual field of inquiry: Completing the square; graphs of quadratic functions; turning point; algebraic
manipulation .
Shawki Dayekh, a teacher of mathematics responsible for A-level teaching in his school, devised the prompt for his year 12 (grade 11) class. He wanted the students to explore minimum and maximum
turning points by graphing quadratic equations. Why is there a minimum point when a > 1 and a maximum point when a < 1? What is the relationship between the values of a, b and c and the y-coordinate
at the turning point? Why are only a and b involved in finding the x-coordinate?
The design of the prompt uses y(max), which, perhaps, is not typically the first case students would meet. Shawki explained that he was hoping the students themselves would initiate a line of inquiry
related to y(min). He also saw the potential for students to move from right to left as well as left to right. If y(max) is known, what does that imply about f(x)? Another line of inquiry involved
starting with higher order functions, such as ax^4 + bx^2 + c.
May 2021
Shawki reports on how the fast-moving inquiry developed:
The students' questions about the prompt (right) showed their existing level of knowledge. We had just covered completing the square and you can see a couple of questions about how the topic fitted
in with the prompt. They were trying to apply their knowledge to the new context.
The class was new to inquiry so I directed them a lot at first. We started with students exploring the turning points. They made up different values for a, b and c. I encouraged them to vary the
values systematically to see how the turning point changed. One pair, for example, used (1, 2, 3) for (a, b, c) and then (2, 4, 6), (3, 6, 9) and so on (see illustration below). They were getting
immediate results on the Desmos graphing calculator and then checking the y-value using the algebraic expression in the prompt.
Reporting back and discussion
After about 20 minutes, one pair reported on using negative values for a, b and c. This led on to a discussion about y(min) and y(max) and their relationship to the variables.
I'm not sure how, but we started to concentrate on the y-intercept. One student suggested that each y(max) or y(min) had to have a corresponding unique value of the y-intercept. This was really
exciting. I wrote the conjecture in formal language: "If y(max) or y(min) is given, then c is unique".
Students started to disagree, but based more on instincts and a partial visualisation than reasoning. We stopped the discussion to consider and come up with convincing reasons. After a couple of
minutes, one student used the algebraic expression to show us two examples:
a = 0.5, b = 2, c = 4 and y(min) = 2 and
a = 0.25, b= 2, c = 6 and y(min) = 2
These two cases with the same y(min) and different y-intercepts is a counter-example that shows the conjecture is false (see illustration).
I reminded the class that we were assuming that the prompt was correct. It was the perfect time to link the two sides of the prompt and we worked through the algebra by completing the square to
finish the lesson.
The next lesson students were still talking about the inquiry. One tried to remember the conjecture, but reversed it: "If c is given, then y(max) or y(min) is unique". This is also false and students
were quickly able to suggest equations for a demonstration on Desmos.
A further line of inquiry
Since the inquiry, I have been thinking about the link between the y-intercept and the turning point, rather than just the y-coordinate. If we are given one, does it imply the uniqueness of the
other? What if we use the condition that a,b ≠ 0? That is the beauty of inquiry. It opens the door to the depth of mathematics. | {"url":"https://www.inquirymaths.org/home/algebra-prompts/completing-the-square","timestamp":"2024-11-07T06:51:22Z","content_type":"text/html","content_length":"177470","record_id":"<urn:uuid:d90b8fd8-fb02-41a5-af89-fb6c6f87111e>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00817.warc.gz"} |
[EM] Approval Equilibrium
Forest W Simmons fsimmons at pcc.edu
Tue Jun 12 14:56:19 PDT 2007
What is an approval equilibrium?
Is it possible to deduce an approval equilibrium from sincere rankings
or ratings?
These questions are amazingly slippery!
I won't attempt to survey the many answers that have been proposed, but
I would like to share a line of thought that came to me after pondering
Lomax' recent post on majority ratification.
He wrote about trying to simulate or predict (from sincere range
ballots) what the outcome of an interactive process would be. Small
groups have the luxury of the interactive process, but it becomes
expensive for large groups.
Suppose that we had a small group that repeated approval counts until
they reached some kind of equilibrium, i.e. until they stabilized. Is
there some way to use sincere range ballots to predict what this
equilibrium might be?
Of course not with 100 percent accuracy, because some voters are more
stubborn than others, etc.
But is there a reasonable way to predict an equilibrium?
Suppose that candidate X is the equilibrium winner, and that candidate
Y is ranked above X by 37 voters. Then Y we would expect that when the
approval votes stabilize with X winning, that Y would have an approval
of about 37, since there is no point in approving anybody you like less
than the first place candidate X.
For now let's don't worry about some of the voters rating X and Y
exactly equal.
Continuing onward ...
If X is the approval winner, then X must have approval greater than Y,
so X has approval greater than 37.
Similarly, if Z is ranked above X by 53 of the voters, then X must have
more than 53 approval votes to be the approval equilibrium.
Now suppose that this number 53 is the greatest pairwise opposition of
any candidate against X. Then an approval of 54 for X would be
sufficient to make X an approval equilibrium winner.
So each for each candidate X, if X can consistently get approval
greater than his greatest pairwise opposition, then X will be an
approval equilibrium candidate.
So the question becomes, "For which candidate X is it least difficult
to get the approval at the required level?"
I have some possible answers, but I would like to hear yours before I
prejudice your minds towards mine.
More information about the Election-Methods mailing list | {"url":"http://lists.electorama.com/pipermail/election-methods-electorama.com/2007-June/085723.html","timestamp":"2024-11-03T01:12:45Z","content_type":"text/html","content_length":"5086","record_id":"<urn:uuid:4959779e-15f9-4cef-82a0-af6fa51bdef0>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00780.warc.gz"} |
K.6 Numbers 0–20
Unit Goals
• Students answer “how many” questions and count out groups within 20. They understand that numbers 11 to 19 are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine
ones. They write numbers within 20.
Section A Goals
• Count groups of up to 20 objects.
Section B Goals
• Understand numbers 11-19 as 10 ones and some more ones.
Section C Goals
• Count groups of images up to 20.
• Represent quantities up to 20 with a written number.
Read More
Section A: Count Groups of 11-20 Objects
Problem 1
How many shapes do you see on the soccer ball?
Problem 2
How many soccer players are there in the picture?
How did you count them?
Section B: 10 Ones and Some More
Problem 1
Draw a line from each picture to the number that shows how many there are.
Problem 2
How many fingers are there?
Show how many fingers there are using a 10-frame.
Problem 3
Write a number to show how many dots there are.
Problem 4
Draw more dots to show each number.
Problem 5
Choose an expression that matches the dots.
\(10 + 5\)
\(10 +2\)
\(10 +7\)
Explain how the expression matches the dots.
Problem 6
Fill in the blanks to make each equation true.
1. \(10 + 7 = \underline{\hspace{1.4 cm}}\)
2. \(\underline{\hspace{1.4 cm}} + \underline{\hspace{1.4 cm}} = 18\)
3. \(\underline{\hspace{1.4 cm}} + \underline{\hspace{1.4 cm}} = 15\)
Problem 7
What are some different ways you know how to show 16?
What is your favorite way to show 16?
Share with a partner.
Problem 8
1. Arrange 18 dots in a way that helps you see there are 18.
2. Arrange 18 dots in a way that makes it hard to see how many there are.
3. Explain why you chose your arrangements.
Try again with other numbers up to 19.
Section C: Count Groups of 11–20 Images
Problem 1
1. How many triangles are there?
There are _______________ triangles.
2. Lin wrote the equation \(10 + 2 = 12\) to show the number of hexagons.
Color the hexagons to show Lin's equation.
Problem 2
1. How many squares are there?
There are _______________ squares.
2. How many circles are there?
There are _______________ circles.
Problem 3
Pick some of the flowers in the picture.
1. How many are there?
2. Share your answer with a partner.
Can you guess which flowers your partner counted?
Problem 4
Pick a shape in one of the designs and figure out how many there are. | {"url":"https://im.kendallhunt.com/K5/teachers/kindergarten/unit-6/practice.html","timestamp":"2024-11-07T06:01:13Z","content_type":"text/html","content_length":"152386","record_id":"<urn:uuid:7631a120-0639-467c-bf7d-39165a8d053e>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00522.warc.gz"} |
Graph log on calculator
graph log on calculator Related topics: Unit Circle Free Worksheet
free math answers problem solver
help solving exponential equations
solving quadratics by factoring worksheet
solving simultaneous equations ks3
math x and y
algebra math/116 how to solve linear equations using the equality properties
find worksheets on probability for a group of thirdgraders
printable prime and composite numbers worksheets
Second Order Differential Equations Online Graphing Calculator
freegrade 4 maths
factoring calculator quadratic
Author Message
Clga Posted: Thursday 26th of Dec 08:36
Hi math wizards! I am about halfway through the semester, and getting a bit worried about my course work. I just don’t seem to pick up the stuff I am learning, especially things to do
with graph log on calculator. Could somebody out there please explain to me with converting fractions, quadratic inequalities and factoring expressions. I can’t afford to hire a tutor,
but if anyone knows about other ways of getting better with topics like adding exponents or ratios effectively, please drop me a line Thanks
From: Igloo
Back to top
IlbendF Posted: Saturday 28th of Dec 12:18
What precisely are your troubles with graph log on calculator? Can you elaborate a little more. I remember that some time ago I too had to go through a similar time of unease. In my
instance, my anxious search led me to a coach in my locality . But he was so tied up that he simply did not have the time for me. He was the one who in fact pointed out that now-a-days
there is yet a new answer at hand. He initiated me to these wonderful software programs in math .
Back to top
Noddzj99 Posted: Sunday 29th of Dec 08:43
Yes I agree, Algebrator is a really useful tool. I bought it a few months back and I can say that it is the main reason I am passing my math class. I have recommended it to my friends
and they too find it very useful. I strongly recommend it to help you with your math homework.
From: the
Back to top
xmoltir Posted: Monday 30th of Dec 08:43
Is the software really that helpful? I’m just concerned because the software might not really help because it only solves the problem per ?e. I like to learn how a problem is solved and
not only know the answer. Nevertheless, could you give me a link for this software?
Back to top
alhatec16 Posted: Wednesday 01st of Jan 07:23
A extraordinary piece of math software is Algebrator. Even I faced similar problems while solving inverse matrices, side-side-side similarity and trigonometry. Just by typing in the
problem workbookand clicking on Solve – and step by step solution to my math homework would be ready. I have used it through several math classes - College Algebra, Pre Algebra and
Algebra 2. I highly recommend the program.
From: Notts,
Back to top
Admilal Posted: Thursday 02nd of Jan 15:36
`Leker Click on this: https://softmath.com/algebra-features.html. It has really helped me a lot , but just in case the software doesn’t do the trick for you then they also have an unconditional
money back guarantee which you can avail . So basically it’s a win-win situation for us as customers!
From: NW AR,
Back to top | {"url":"https://softmath.com/algebra-software-5/graph-log-on-calculator.html","timestamp":"2024-11-08T22:08:59Z","content_type":"text/html","content_length":"43599","record_id":"<urn:uuid:f8bb6bb0-17f3-4d40-bf70-3b494f17ec4f>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00850.warc.gz"} |
NCERT Maths Solutions Class 11th Chapter 7 Permutation and Combinations
NCERT Solutions Class 11 Maths Permutations and Combinations PDF
Download our free PDF of NCERT Solutions for Class 11 Maths Chapter 7 - Permutations and Combinations from www.manabadi.co.in. This resource includes all the questions from the NCERT books, prepared
by expert mathematics teachers in accordance with CBSE NCERT guidelines. Use this PDF to enhance your preparation for board and competitive exams and boost your scores.
NCERT Solutions for Class 11 Maths Chapter 7 Exercise.7.1
Ex 7.1 Class 11 Maths Question-1
How many 3 - digit numbers can be formed from the digit 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
There will be as many ways as there are ways of filling 3 vacant places
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 x 5 x 5 = 125
In this case, repetition of digts is not allowed. here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units
place of the three-digit number is 5.
Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaniing three digits.
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 x 4 x 3 = 60
Ex 7.1 Class 11 Maths Question-2
How many 3 - digit even numbers can be formed from the digit 1, 2, 3, 4, 5, 6 if the digits can be repeated?
There will be as many ways as there are ways of filling 3 vacant places
Therefore, by multiplication principle, the required number of three digit even numbers is 3 x 6 x 6 = 108
Ex 7.1 Class 11 Maths Question-3
How many 4- letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
There are as many codes as there are ways of filling 4 vacant places
The first place can be filled in 10 different ways by any of the first 10 letters of the English alphabet follewing which, the second place can be filled in by any of the remaining letters in 9
different ways. The third place can be filled in by any of the remaining 8 letters in 8 different ways and the fourth place can be filled in by any of the remaining 7 letters in 7 different ways.
Therefore, by multiplication principle, the required numbers of ways in which 4 vacant places can be filled is 10 x 9 x 8 7 = 5040
Hence, 5040 four-letter codes can be formed using the first 10 letters of the English alphabet, if no letter is repeated
NCERT Solutions for Class 11 Maths Chapter 7 Exercise.7.2
Ex 7.2 Class 11 Maths Question-1
(i) 8! (ii) 4! - 3!
(i) 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320
(ii) 4! = 1 x 2 x 3 x 4 = 24
3! = 1 x 2 x 3 = 6
Ex 7.2 Class 11 Maths Question-2
Is 3! + 4! = 7!?
3! = 1 x 2 x 3 = 6
4! = 1 x 2 x 3 x 4 = 24
7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040
Ex 7.2 Class 11 Maths Question-3
Ex 7.2 Class 11 Maths Question-4
NCERT Solutions for Class 11 Maths Chapter 7 Exercise.7.3
Ex 7.3 Class 11 Maths Question-1
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
3 - digit numbers have to be formed using the digit 1 to 9.
Here, the order of the digits matters.
Therefore, there will be as many 3 - digit numbers as there are permutations of 9 different digits taken 3 at a time.
Therefore, required number of 3 - digit numbers
Ex 7.3 Class 11 Maths Question-2
How many 4-digit numbers are there with no digit repeated?
The thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9 as the digit 0 cannot be included. Therefore, the number of ways in which thousands place can be filled is
The hundreds, tens, and units place can be filled by any of the digits from 0 to 9. However, the digits cannot be repeated in the 4-digit numbers and thousands place is already occupied with a digit.
The hundreds, tens, and units place is to be filled by the remaining 9 digits.
Therefore, there will be as many such 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.
Number of such 3-digit numbers
Thus, by multiplication principle, the required number of 3-digit numbers is 3 x 20 =60
NCERT Solutions for Class 11 Maths Chapter 7 Exercise.7.4
Ex 7.4 Class 11 Maths Question-1
Ex 7.4 Class 11 Maths Question-2
Ex 7.4 Class 11 Maths Question-3
NCERT Solutions for Class 11 maths Chapter Miscellaneous Solutions
Miscellaneous Question-1
How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
In the word DAUGHTER, there arre 3 vowels namely, A, U, and E, and 5 consonants namely D, G, H, T, and R.
Number of ways of selecting 2 vowels out of 3 vowels = ^3C[2] = 3
Numbers of ways of selecting 3 consonant out of 5 consonants ^5C[3] = 10
Therefore, numbers of combinations of 2 vowel and 3 consonants = 3 x 10 = 30
Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways.
Hence, required number of different words = 30 x 5! = 3600
Miscellaneous Question-2
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
In the word EQUATION, there arre 5 vowels namely, A, E, I, O, and U, and 3 consonants namely Q, T, and N.
Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. then, the permutations of these 2 objects taken all at a time are counted. this
number would be ^2P[2] = 2!
Corresponding to each of these permutations, these are 5! permurations of the five vowels taken all at time and 3! permutations of the 3 consonants taken all at a time.
Hence, by multiplication principle, required number of different words = 2! x 5! x 3! = 1440
Post your comments
Your mobile number will not be published.
View Comments
Very much composed answers
NCERT Solutions For Class 11 | {"url":"http://www.manabadi.co.in/cbse/ncert-maths-solutions-class-11th-chapter-7-permutation-and-combinations.aspx","timestamp":"2024-11-01T23:21:22Z","content_type":"application/xhtml+xml","content_length":"74990","record_id":"<urn:uuid:b9d3865a-3cf8-4f1e-82ce-d3d4d5596270>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00068.warc.gz"} |
[11:30am] Anand Sawant : School of Mathematics, TIFR
Algebraic Geometry seminar. Speaker: Anand Sawant. Affiliation: School of Mathematics, TIFR. Date and Time: Thursday 06 February, 11:30 am - 12:30 pm. Venue: Ramanujan Hall, Department of
Mathematics. Title: Central extensions of algebraic groups, II. Abstract: This talk will be a continuation of the Colloquium talk last week, where we will begin wth the work of
12:00pm Brylinski-Deligne.
[4:00pm] Mrinmoy Datta : Arctic University of Norway, Tromso
Geometry and Topology seminar. Speaker: Mrinmoy Datta. Affiliation: Arctic University of Norway, Tromso. Date and Time: Thursday 06 February, 04:00 pm - 05:00 pm. Venue: Ramanujan Hall,
Department of Mathematics. Title: Hermitian surfaces over finite fields and a conjecture by Sørensen. Abstract: Hermitian varieties, first studied by Bose and Chakravati in 1966, are a
4:00pm class of vastly studied objects in the area of finite geometry and coding theory. During 1991, in his PhD thesis, A. B. Sørensen proposed a conjecture on the maximum number of rational
points on the intersection of a Hermitian surface and a surface of degree d defined over the same field. Edoukou's work in 2006 towards proving the conjecture for d=2 marked the first
progress towards this conjecture. In 2018, in a joint work with Peter Beelen, we have shown that the conjecture is true for d=3. Finally, in a joint work with Peter Beelen and Masaaki
Homma, we have proved the conjecture completely. In this talk, we will give an account of these developments. | {"url":"https://www.math.iitb.ac.in/webcal/day.php?date=20200206","timestamp":"2024-11-11T04:12:02Z","content_type":"text/html","content_length":"21398","record_id":"<urn:uuid:21204097-f36d-49e3-8a57-8be8713ac2e7>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00464.warc.gz"} |
Constant Curvature Curve
Back to interactive Constant Curvature Curves
The only well known curves of constant curvature are planar circles and helices in R^3. One can put pieces of these curves together to make closed nonplanar curves of constant curvature, but these
curves cannot be 3 times differentiable because the torsion is a step function. The closed constant curvature curves on this page are analytic. Most of them are shown as anaglyphs, to be viewed with
red-green glasses.
Tangents of a curve are “best approximating” straight lines. “Osculating circles” of a curve are “best approximating circles”. Such circles have at the contact point the same first and second
derivative as the curve and therefore also the same curvature. Curves of constant curvature have osculating circles which all have the same radius. The shown curve has this property, it is a space
curve of constant curvature. The dotted lines are “symmetry normals”: 180 degree rotation around these normals is a symmetry of the curve.
One can roll a plane with a circle to a cylinder and adjust a deformation of the circle so that it continues to be a closed curve of constant curvature.
The same constant curvature curves on different cylinders as in the previous sequence.
A family of longitudinal constant curvature curves on a cylinder.
One of the previous curves as tube.
Similarly oscillating curves (as on cylinders) can be found on tori. Again, they are constant curvature space curves.
This curve looks as if it could be in a family of curves on a cylinder. But for this curve the four “circular” pieces are congruent, while on the cylinder only opposite pieces are congruent and
neighboring ones are not.
This curve with three upwards and three downwards almost circular pieces resembles the curve which we show on a torus. Both have the same symmetry group but very different constructions.
This curve has two parts where the torsion is small and the curve is close to a piece of a circle. And it has two parts where the torsion is large and the curve resembles a piece of a helix.
This curve has 6 circular pieces and six resembling pieces of a helix. This is not exactly true because the torsion is not piecewise constant but a smooth function.
A 5-2-knot as constant curvature space curve. The symmetry normals are a tool for their construction.
An 11-2-knot of constant curvature.
This is a family of non-closed constant curvature curves. The only difference between them is, that constants are added to the torsion function. As these constants increase, the symmetry normals are
spread apart to lie on steeper and steeper helicoid surfaces. The observable fact that the distance between adjacent symmetry normals increases roughly proportional to the added constants can be used
to construct closed curves.
In this family of constant curvature space curves a Fourier coefficient of the torsion (= a trigonometric polynomial) is changed and the mean value of the torsion is adjusted to keep the symmetry
normals passing through one point!! The possibility to do this relies on the observation of the previous sequence. - The animation shows how one can get further closed constant curvature space
curves, by starting from one and following the described deformation. The shown deformation passes through a 9-2-knot and almost reaches a curve with 4-fold symmetry.
A rotating 7-2-knot of constant curvature.
How to construct closed constant curvature space curves | {"url":"https://virtualmathmuseum.org/SpaceCurves/constant_curvature/constant_curvature.html","timestamp":"2024-11-02T04:25:05Z","content_type":"text/html","content_length":"8104","record_id":"<urn:uuid:db317d62-51f7-4c33-a9f0-a546aea279b3>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00079.warc.gz"} |
Introduction to Geometry
Basic Geometry Worksheet
Download this Basic Geometry Worksheet with 10 questions and answers as a practice test.
Geometry has allowed humanity to greatly expand our understanding of the objects around us, and it is used on a daily basis not only in mathematics but in many branches of science. Innumerable
industries rely on geometric principles, including construction, building design, product development, and city planning. Follow along as we give you an introduction to geometry, helping you see the
world from a new angle.
What is Geometry?
Geometry is a field of mathematics that relates to objects, or geometric shapes as they are referred to, and their sizes, shapes, positions, or spacial properties. This subject generally covers the
following areas:
• Distances
• Angles and Shapes
• Patterns
• Area and Perimeter
• Circumference and Volume
These, along with any other visually and spatially related concepts are considered to be a part of geometry. Within the vast world of geometry, there are two principal categories.
Euclidean Geometry
Geometry has been used since the time of the Ancient Greeks. The term "Geometry" even has Greek roots, from the Greek word “geo”, meaning earth, and “metrein” which translates to “to measure.” So it
comes as no surprise that the father of Geometry is the Greek mathematician, Euclid. He even has a branch of geometry named after him—Euclidean geometry.
This is generally the first category of geometry you learn in school. That's because it explains basic geometric principles. Euclidean geometry established the foundation for much of what we
understand about the field today. Academically, Euclidean geometry refers to the study of flat shapes and flat surfaces.
Non-Euclidean Geometry
There are two other major branches of geometry that are considered non-Euclidean. Non-Euclidean geometry is a rethinking of the properties of lines, points, and shapes. In other words, Euclidean
geometry deals with objects on a flat plane, whereas non-Euclidean geometry deals with our world (and non-flat surfaces).
The two main kinds of non-Euclidean geometry are below.
1. Spherical Geometry Explained
Spherical geometry is the study of geometry not on a plane, but rather on a sphere. In spherical geometry, like Euclidean geometry, lines are defined as the shortest distance between two points.
However, spherical geometry has no parallel lines. Triangles in spherical geometry also add up to more than 180 degrees. Spherical geometry also introduces new terms:
• A “line” between two points on a sphere is referred to as a “great circle”
• The shortest point between two points on the surface of a sphere is indicated by the arc of the great circle passing between them.
2. Hyperbolic Geometry Explained
A third major kind of geometry is hyperbolic. Hyperbolic geometry is defined as geometry on a curved surface. In hyperbolic geometry, one of the Euclidean postulates is replaced. In hyperbolic
geometry, parallel lines will become further and further apart. And triangles will have angles that are less than 180 degrees in total.
Quick Guide to Euclidean Geometry
Euclidean Principles
Euclidean geometry is also sometimes referred to as “plane geometry” because it concerns flat objects. Some key properties of Euclidean geometry are:
• The shortest distance between any two points is a straight line.
• Two parallel lines will be infinitely parallel and never intersect.
• The interior angles of a triangle will add up to 180 degrees.
• Right triangles have two perpendicular sides measuring 90 degrees forming a right angle within the triangle.
• Solid shapes have a size, position, and shape that can be moved from place to place.
• Euclidean geometry is the study of planes and solid objects.
Overall, this means Euclidean geometry will be concerned with angles, triangles, squares, lines, points, and more.
Euclid’s Postulates
Euclidean geometry is centered around five postulates. These important building blocks are as follows:
• A line extends indefinitely in both directions.
• All right angles are equal to one another.
• A line segment is drawn from any one point to another
• Any two straight lines that are an equal distance from each other at two points are parallel
• A circle can be described with any point and its radius.
Euclid’s Axioms
Euclid’s axioms are still an important part of geometry today. Though they may not be mathematically proven, they're widely accepted by modern mathematicians:
• Things that are equal to the same thing are equal to each other
• If equals and equals are added, the wholes are equal
• If equals are subtracted from equals, the remainders will be equal
• Things which coincide are equal to one another
• The whole is greater than the part
• Things that are double of the same thing are equal to each other
• Things that are halves of the same thing are equal to one another
5 Types of Euclidean Geometry
1. Plane Geometry
For the purpose of plane geometry, consider that lines, points, and angles are all placed on a two-dimensional plane that goes on infinitely in all directions (while remaining flat). Understanding
the concepts and study of plane geometry is essential in order to understand more complex forms of geometry. Some important terms to know are:
• A given point on a plane that is indicated with a set of coordinates
• Collinear points are points along the same line
• A line is infinite in two directions along the plane
• A line segment has a starting and ending point
• A ray has a starting point and then goes on infinitely in one direction
• Parallel means infinitely parallel. If two lines do not intersect anywhere on a plane, they are parallel
• Perpendicular lines are those that intersect at a right angle
Plane Geometry - Angles
If two lines are not parallel, they will intersect at some point. When they intersect, the two lines form an angle, typically measured in degrees. There are generally considered to be four kinds of
• A straight angle is exactly 180 degrees
• A right angle is exactly 90 degrees
• An obtuse angle is greater than 90 degrees, but less than 180
• An acute angle is less than 90 degrees
Angles can also be either complementary or supplementary. If the sum of two angles is 180 degrees, they are considered to be supplementary angles. If the sum is 90 degrees, they are instead
considered to be supplementary.
Plane Geometry - Shapes
Shapes in plane geometry are classified by their properties. These are all two-dimensional shapes and can be placed on a flat plane. In order to be considered a polygon, shapes must have more than
two lines and be closed. Some of the most commonly seen plane shapes include:
• Triangles (three sides, three vertices)
• Quadrilaterals (four sides, four vertices)
There are three kinds of triangles.
1. Equilateral triangles have three equal sides and angles.
2. A scalene triangle has three unequal sides and angles.
3. An isosceles triangle has only two equal sides.
When studying triangles in particular, there are a number of theorems to help identify their properties. These include the Pythagorean Theorem, the angle sum property, the exterior angle theorem, and
more. Triangles can be fully graphed with only two points and angles. For quadrilaterals, triangles, and all other polygons, it is also important to understand the concept of symmetry. Symmetry is
when an object can have a line drawn through it and be exactly the same on both sides. As an example, a circle has infinite lines of symmetry.
2. Solid Geometry
Solid geometry is concerned with objects and solid figures that are three-dimensional. This can include the following shapes, among others:
• Cylinder
• Cube
• Sphere
• Cone
• Prism
• Pyramid
• Cuboids
In order to be considered three-dimensional, an object needs to have length, width, and height. These three-dimensional shapes are classified by faces, edges, and vertices. Vertices are the corners
of the shapes, faces are the sides, and edges are the lines. For example, a cube has a total of 8 vertices, while a cone only has one single vertex.
3. Geometric Measurement
When measuring objects in geometry, there are many things to consider. Every geometric shape can be measured in perimeter, area, surface area, and volume.
• Perimeter is the distance around the outside of plane (two-dimensional) shapes
• Area is the amount of space occupied by a shape
• Volume is the area within a shape
• A solid's surface area is the area of all of the shape's faces
These measurements can be calculated in different ways, using specific formulas dependent on the nature of the shape.
4. Coordinate Geometry
Coordinate geometry, also referred to as Analytic Geometry, identifies a point on a plane with a set of ordered numbers, (X, Y). This set of numbers (coordinates) can be plotted on a graph, a
coordinate plane, along with other coordinates to indicate a particular shape. In coordinate geometry, the plane is divided into four quadrants. The top right quadrant will have two positive
coordinates, while the bottom left will have two negative coordinates.
Any coordinate to the left of the y-axis will be negative, while those on the right will be positive. Conversely, any coordinate above the x-axis will be positive, while those below it will be
negative. This is called a Cartesian coordinate system.
Coordinate geometry can show properties of geometric figures, such as lines, curves, ellipses, hyperbolas, circles, and more. Curves are represented with an algebraic equation. With the use of many
different formulas, including the distance formulas, the section formula, the midpoint formula, and more, these shapes can be completely graphed and identified. In other words, an algebraic equation
can indicate a specific curve.
Three-Dimensional Coordinate Geometry
Logically, as a two-dimensional plane shape has two-digit coordinates, three-dimensional objects have three. Instead of (X, Y), these coordinates are indicated with (X, Y, Z). It has all of the
concepts from two-dimensional coordinate geometry, and more.
The three-dimensional Cartesian coordinate system instead has three axes. The X, Y, and Z-axis are all perpendicular to one another. Since there are three axes instead of two, that means that instead
of quadrants, these graphs are divided into a total of eight sections (sometimes called octants). The coordinates are called:
• Abscissa is the first coordinate and indicates the distances from the origin along the x-axis
• Ordinate is the second coordinate, and is the perpendicular distance of the point from the x-axis. It is parallel to the y-axis
• Applicate is the third coordinate, the distance from the z-axis
Together, these three coordinates are like directions to find a given point in space. Any point in three-dimensional coordinate geometry can be indicated with three coordinates (X, Y, Z). This is one
of two ways to represent a point in three-dimensional geometry and the properties of space, referred to as the Cartesian form. The Cartesian form can be used to represent any three-dimensional shape
in geometry. The other way to represent a point, line of shape is called the vector form.
5. Differential Geometry
Differential geometry is the speciality of non-euclidean geometry that combines differential calculus with geometric principles. It’s used to study surfaces, manifolds, and curves within space.
Differential geometry can is used in numerous fields including:
• GPS
• Geometric controls in robotics
• Image Analytics
• Animation
• Medical Imaging
• Computer vision (AI)
• String Theory
Perhaps the easiest way to think about this is to imagine a screw. A screw has a solid shaft in the middle surrounded by the helical thread. The thread allows the screw to pull itself into another
material. Because the thread is not on a singular plane, you can’t get an accurate measurement for the radius without understanding how the curvature of curves and surfaces work.
Understanding these allows for the shortest path between two points on a surface to be measured.
Geometric Formulas
The distance formula helps find the shortest distance between two points. It is calculated by using the following formula. The distance equals the square root of (x1- x2) squared + (y2-y1) squared +
(z1-z1) squared). This formula can help you ascertain any missing coordinates.
The mid-point formula is a new point. The abscissa is the average of the x values of the given points, while the ordinate is the average of the y values of the two points that were given. The
midpoint is on the line joining two points as is located in the exact middle of the two points.
Section formula can help you find the coordinates of a point that divides the line segment. The point which divides the given two points is on the line adjoining the two points. It is available
either between the two points or on the line beyond the points indicated. The formula is used widely in Physics as well as Mathematics. In Physics, it is used to find the center of mass and points of
Geometric Terminology
Some of the key vocabulary words that are important for a geometry student to know and understand are below:
• Diameter - A straight line passing through the center of a circle (or sphere) and reaching the other end
• Ray - A line segment that goes on infinitely in one direction
• Parallel - Two lines that do not intersect
• Vertex (vertices) - the intersection point of two sides of a plane figure
• Face - Any of the individual flat surfaces of a solid object
• Edge - Any line segment joining one vertex to another
• Perpendicular - when two lines (or line segments) intersect at a 90-degree angle
• Vector - A line that shows magnitude and direction. Often used to represent distance or acceleration
• Transversal - A line intersecting two or more given lines in a plane at different points
• Complementary angles - when the sum of two angles is exactly 90 degrees
• Supplementary angles - when the sum of two angles is exactly 180 degrees
• Plane - A flat surface that extends infinitely in all directions (2-dimensional)
10 Basic Geometry Practice Problems
1. What is the volume of a pyramid whose base is a regular pentagon with an area of 60 m^2 and whose height is 10 m?
Volume of a Pyramid = ⅓ x [Base Area] x Height
Volume of a Pyramid = ⅓ x 60 m^2 x 10 m
Volume of a Pyramid = 200 m^3
2. What is the value of F + V - E for a prism whose cross-section is a heptagon?
Draw a heptagon prism. The prism has 9 Faces, 14 Vertices and 21 Edges.
F + V - E = 9 +41 - 21 = 2
3. A torus has the following values R = 15 m and r = 4 m. What is the volume of the torus?
Torus Volume = 2 x 𝞹^2 x R x r^2 m^3
Torus Volume = 2 x 𝞹^2 x 15 x 16 m^3
Torus Volume = 4,737.41 m^3
4. AB is an arc of length 14 in. on the circumference of a circle with center C. The size of angle ACB is 3.5 radians. Find the radius of the circle.
Arc Length Formula: L = θ × r
(Arc length) L = 14 in.
(Angle ACB) θ = 3.5 radians
(Radius) r = ?
14 = 3.5 × r
r = 14 ÷ 3.5
Radius = 4 in.
5. BC is an arc with a length of 72 feet on the circumference of a circle with the center D and a diameter of 24 feet. What is the size of the angle BDC in radians?
L = θ x r
72 = θ x (½ x 24)
72 = θ x 12
θ = 72 ÷ 12
θ = 6 radians
6. Convert 2𝞹/4 radians to degrees.
Degrees = 2𝞹/4 radians
Degrees = [(2 x 180°)/4]
Degrees = 360°/4
2𝞹/4 Radians = 90° Degrees
7. The following isosceles triangle has two equal sides, each with a length of 10.3. Angle A is a right angle. What is the size of the missing angle "B"?
A = 90°
A + B + B = 180°
90° + 2B = 180°
2B = 180° - 90°
2B = 90°
B = 90°/2
B = 45°
8. Find the side length of a parallelogram with perimeter of 22 in. and base of 8.3 in.?
Parallelogram Perimeter Formula:
P = 2(b + s)
(Perimeter) P = 22
(Base) b = 8.3
(Side) s = ?
22 = 2(8.3 + s)
11 = 8.3 + s
8.3 + s = 11
s = 2.7
Side length = 2.7 in.
9. In the following diagram, what is the size of angle BAC?
Exterior Angle Theorem: ∠BAC = ∠BCD − ∠ABC
BAC = 141° − 97°
BAC = 44°
10. What is the surface area for a regular tetrahedron when the total length of the edges is 48 mm? Round to the nearest 1/100th
One Edge Length = 48 mm ÷ 6 = 8 mm
Tetrahedron Surface Area = √3 x [Edge Length]^2
Tetrahedron Surface Area = √3 x 8^2 mm^2
Tetrahedron Surface Area = 1.732… x 64 mm^2
Tetrahedron Surface Area = 110.85 mm^2 | {"url":"https://www.intmath.com/functions-and-graphs/introduction-to-geometry.php","timestamp":"2024-11-15T03:41:57Z","content_type":"text/html","content_length":"119837","record_id":"<urn:uuid:929bde8b-cf2a-4dee-8697-aacb3fdccf8f>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00777.warc.gz"} |
Means of transport calculation preview
The submodule Means of Transport Calculation Preview can be found in the module Transportation Planning.
Here routes are evaluated according to the VDI norm 5586. The calculation of the VDI standard was extended here for a higher accuracy by empties and grouping transports, as well as an evaluation of
direct transports.
With 1 you can switch the view between "term based" and "formula based", so that you can also display the formula symbols here.
Above the graphs, the values used for the calculation are displayed.
Notes on the calculation:
Load units are treated like load carriers: If a load unit is larger than the standard load carrier, the NLT equivalent is 1, if a load unit is smaller than the standard load carrier, the NLT
equivalent is calculated.
Disposal chains are considered analogous to supply chains.
For grouping chains, the value of grouping transports per day is determined as follows:
Products per day / Number of compartments for the grouping box / Number of grouping boxes per grouping load unit
For the calculation of the NLT equivalent, the dimensions of the grouping load unit are always used here (regardless on whether grouping box or grouping load unit was selected)
Direct transports:
Via 2 you can switch between the routes and the direct transports.
All used combinations of means of transport are listed here, sorted in descending order according to the respective workload.
Notes on calculation:
The transportation means requirement between two stages is calculated as follows:
Transports per day per LOG ID = LC per day / division factor
Transports per day per LOG ID = LU per day / division factor
The division factor is always the minimum of the reorder quantity and the stacking factor for the respective transport (stored in the respectivesupply chain).
The transport duration is calculated from the distance and speed of the means of transport used.
The transports per day determined above are then multiplied by the transport duration: Result A
For the duration oft the processes, the loading and unloading times stored for the transport are added up.
The transports per day determined above are then multiplied by the process duration: Result B
Result A and result B are added and assigned to the respective transport means combination - this must be done for all stages where the respective transport means combination is used: Result C
Under 1 (see above) an empty trip percentage of 10% is stored by default. This value can be changed and confirmed with Enter so that the calculation is adjusted directly.
Result C is therefore multiplied by this precentage of empty trips and, if necessary, an allowance time (can be stored in the resource management) of the worker used.
This value is then divided by the net working time (in case of different shift schedules for workers and means of transport, the net working time of the means of transport applies) to obtain the
number of means of transport per shift. The number per day as well as the rounded number per day is calculated by counting the individual shifts.
Average workload = calculated number / rounded number
Grouping chains are not taken into account here.
0 comments
Please sign in to leave a comment. | {"url":"https://help.ipolog.ai/hc/en-us/articles/360016180260-Means-of-transport-calculation-preview","timestamp":"2024-11-13T01:33:05Z","content_type":"text/html","content_length":"26090","record_id":"<urn:uuid:0dd8b773-326f-4f3e-b698-8c6468024d61>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00504.warc.gz"} |
LM 35.7 Summary Collection
35.7 Summary by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
wavefunction — the numerical measure of an electron wave, or in general of the wave corresponding to any quantum mechanical particle
`?`— Planck's constant divided by `2pi` (used only in optional section 35.6)
`Psi` — the wavefunction of an electron
Light is both a particle and a wave. Matter is both a particle and a wave. The equations that connect the particle and wave properties are the same in all cases:
Unlike the electric and magnetic fields that make up a photon-wave, the electron wave function is not directly measurable. Only the square of the wave function, which relates to probability, has
direct physical significance.
A particle that is bound within a certain region of space is a standing wave in terms of quantum physics. The two equations above can then be applied to the standing wave to yield some important
general observations about bound particles:
1. The particle's energy is quantized (can only have certain values).
2. The particle has a minimum energy.
3. The smaller the space in which the particle is confined, the higher its kinetic energy must be.
These immediately resolve the difficulties that classical physics had encountered in explaining observations such as the discrete spectra of atoms, the fact that atoms don't collapse by radiating
away their energy, and the formation of chemical bonds.
A standing wave confined to a small space must have a short wavelength, which corresponds to a large momentum in quantum physics. Since a standing wave consists of a superposition of two traveling
waves moving in opposite directions, this large momentum should actually be interpreted as an equal mixture of two possible momenta: a large momentum to the left, or a large momentum to the right.
Thus it is not possible for a quantum wave-particle to be confined to a small space without making its momentum very uncertain. In general, the Heisenberg uncertainty principle states that it is not
possible to know the position and momentum of a particle simultaneously with perfect accuracy. The uncertainties in these two quantities must satisfy the approximate inequality
When an electron is subjected to electric forces, its wavelength cannot be constant. The “wavelength” to be used in the equation `p=h/lambda` should be thought of as the wavelength of the sine wave
that most closely approximates the curvature of the wavefunction at a specific point.
Infinite curvature is not physically possible, so realistic wavefunctions cannot have kinks in them, and cannot just cut off abruptly at the edge of a region where the particle's energy would be
insufficient to penetrate according to classical physics. Instead, the wavefunction “tails off” in the classically forbidden region, and as a consequence it is possible for particles to “tunnel”
through regions where according to classical physics they should not be able to penetrate. If this quantum tunneling effect did not exist, there would be no fusion reactions to power our sun, because
the energies of the nuclei would be insufficient to overcome the electrical repulsion between them.
Exploring further
The New World of Mr. Tompkins: George Gamow's Classic Mr. Tompkins in Paperback, George Gamow.
Mr. Tompkins finds himself in a world where the speed of light is only `30` miles per hour, making relativistic effects obvious. Later parts of the book play similar games with Planck's constant.
The First Three Minutes: A Modern View of the Origin of the Universe, Steven Weinberg.
Surprisingly simple ideas allow us to understand the infancy of the universe surprisingly well.
Three Roads to Quantum Gravity, Lee Smolin.
The greatest embarrassment of physics today is that we are unable to fully reconcile general relativity (the theory of gravity) with quantum mechanics. This book does a good job of introducing the
lay reader to a difficult, speculative subject, and showing that even though we don't have a full theory of quantum gravity, we do have a clear outline of what such a theory must look like.
Homework Problems
`sqrt` A computerized answer check is available online.
`int` A problem that requires calculus.
`***` A difficult problem.
1. In a television, suppose the electrons are accelerated from rest through a voltage difference of `10^4 V`. What is their final wavelength? `sqrt`
2. Use the Heisenberg uncertainty principle to estimate the minimum velocity of a proton or neutron in a ^208Pb nucleus, which has a diameter of about `13 fm` (`1 fm=10^(-15) m`). Assume that the
speed is nonrelativistic, and then check at the end whether this assumption was warranted. `sqrt`
3. A free electron that contributes to the current in an ohmic material typically has a speed of `10^5 m"/"s` (much greater than the drift velocity).
(a) Estimate its de Broglie wavelength, in nm. `sqrt`
(b) If a computer memory chip contains `10^8` electric circuits in a `1 cm^2` area, estimate the linear size, in nm, of one such circuit. `sqrt`
(c) Based on your answers from parts a and b, does an electrical engineer designing such a chip need to worry about wave effects such as diffraction?
(d) Estimate the maximum number of electric circuits that can fit on a `1 cm^2` computer chip before quantum-mechanical effects become important.
4. On page 970, I discussed the idea of hooking up a video camera to a visible-light microscope and recording the trajectory of an electron orbiting a nucleus. An electron in an atom typically has a
speed of about `1%` of the speed of light.
(a) Calculate the momentum of the electron. `sqrt`
(b) When we make images with photons, we can't resolve details that are smaller than the photons' wavelength. Suppose we wanted to map out the trajectory of the electron with an accuracy of `0.01`
nm. What part of the electromagnetic spectrum would we have to use?
(c) As found in homework problem 12 on page 789, the momentum of a photon is given by `p=E/c`. Estimate the momentum of a photon having the necessary wavelength. `sqrt`
(d) Comparing your answers from parts a and c, what would be the effect on the electron if the photon bounced off of it? What does this tell you about the possibility of mapping out an electron's
orbit around a nucleus?
5. Find the energy of a particle in a one-dimensional box of length `L`, expressing your result in terms of `L`, the particle's mass mm, the number of peaks and valleys nn in the wavefunction, and
fundamental constants. `sqrt`
6. The Heisenberg uncertainty principle, `Deltap Deltax?h`, can only be made into a strict inequality if we agree on a rigorous mathematical definition of `Deltax` and `Deltap`. Suppose we define the
deltas in terms of the full width at half maximum (FWHM), which we first encountered on p. 469 and revisited on page 925 of this book. Now consider the lowest-energy state of the one-dimensional
particle in a box. As argued on page 971, the momentum has equal probability of being `h"/"L` or `-h"/"L`, so the FWHM definition gives `Deltap=(2h)/lambda`.
(a) Find `Deltax` using the FWHM definition. Keep in mind that the probability distribution depends on the square of the wavefunction.
(b) Find `DeltaxDeltap`. `sqrt`
7. If `x` has an average value of zero, then the standard deviation of the probability distribution `D(x)` is defined by
where the integral ranges over all possible values of `x`.
Interpretation: if `x` only has a high probability of having values close to the average (i.e., small positive and negative values), the thing being integrated will always be small, because `x^2` is
always a small number; the standard deviation will therefore be small. Squaring `x` makes sure that either a number below the average (`x<0`) or a number above the average (`x>0`) will contribute a
positive amount to the standard deviation. We take the square root of the whole thing so that it will have the same units as `x`, rather than having units of `x^2`.
Redo problem 6 using the standard deviation rather than the FWHM.
Hints: (1) You need to determine the amplitude of the wave based on normalization. (2) You'll need the following definite integral:` int_(-pi"/"2)^(pi"/"2)u^2cos^2udu=(pi^3-6pi)"/"24`. `sqrt` `int`
8. In section 35.6 we derived an expression for the probability that a particle would tunnel through a rectangular potential barrier. Generalize this to a barrier of any shape. [Hints: First try
generalizing to two rectangular barriers in a row, and then use a series of rectangular barriers to approximate the actual curve of an arbitrary potential. Note that the width and height of the
barrier in the original equation occur in such a way that all that matters is the area under the `PE`-versus-`x` curve. Show that this is still true for a series of rectangular barriers, and
generalize using an integral.] If you had done this calculation in the 1930's you could have become a famous physicist. `int` `***`
9. The electron, proton, and neutron were discovered, respectively, in 1897, 1919, and 1932. The neutron was late to the party, and some physicists felt that it was unnecessary to consider it as
fundamental. Maybe it could be explained as simply a proton with an electron trapped inside it. The charges would cancel out, giving the composite particle the correct neutral charge, and the masses
at least approximately made sense (a neutron is heavier than a proton). (a) Given that the diameter of a proton is on the order of `10^(-15) m`, use the Heisenberg uncertainty principle to estimate
the trapped electron's minimum momentum. `sqrt`
(b) Find the electron's minimum kinetic energy. `sqrt`
(c) Show via `E=mc^2` that the proposed explanation fails, because the contribution to the neutron's mass from the electron's kinetic energy would be many orders of magnitude too large.
35.7 Summary by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license. | {"url":"https://www.vcalc.com/collection/?uuid=1f197b7c-f145-11e9-8682-bc764e2038f2","timestamp":"2024-11-13T08:33:14Z","content_type":"text/html","content_length":"71986","record_id":"<urn:uuid:341d5321-f668-43ab-b192-6adf183c4a0a>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00508.warc.gz"} |
3 batteries, 5 resistors using KCL & KVL
• Thread starter amy1983
• Start date
In summary: So, just keep plodding along. It will all come together eventually. If it helps, think of the sequence of steps like a dance routine. It's a series of steps in a sequence that you perform
to get the desired result, just like a dance routine. And like a dance routine, you practice it over and over until you've got it down pat. And like a dance routine, you can use your scratch paper to
write down the sequence of steps to help you remember them.In summary, the conversation discusses using Kirchhoff's Laws to find all the currents flowing in a circuit and determining the voltage
across each resistor to show compliance with KVL. The equations used are KVL and KCL
Homework Statement
1. Using Kirchhoff’s Laws, find all the currents flowing in the circuit.
2. Hence, determine the voltage across each resistor and show that all loops comply with KVL
Homework Equations
KVL and KCL
The Attempt at a Solution
Please see the pictures attached
amy1983 said:
Homework Statement
1. Using Kirchhoff’s Laws, find all the currents flowing in the circuit.
2. Hence, determine the voltage across each resistor and show that all loops comply with KVL
Homework Equations
KVL and KCL
The Attempt at a Solution
Please see the pictures attached
Hi amy1983, Welcome to Physics Forums.
For your Loop 2 equation be careful about the orientation of the voltage sources; as you go around the loop in the direction of mesh current I2, both give potential rises.
Take another look at your Loop 3 equation. Again, pay attention to the voltage source polarity with respect to the current loop direction. Also check that you haven't inadvertently counted some
voltage drops twice...
It's a bit confusing because you've taken clockwise mesh currents as negative. Also that loop 2 is not correct. As Gneill said, you've made an error with values of the voltage sources in that loop.
A good program to use for Nodal Analysis and Mesh Analysis is Pspice, for these kinds of circuits your answers can be checked very quickly.
Hi gneill / NewtonianAlch - thank you kindly for your welcome and for your suggestions, I will review this PM.
Hi g/neill / NewtonianAlch, I typed it all up and came up with the following according to your suggestions:
At node V_1,
⇒(15-V_1)/(1 k)+(10-V_1)/(2 k)=(V_1-V_2)/(1 k)
⇒(30-2V_1 )+(10-V_1 )=(2V_1-2V_2)
At node〖 V〗_2,
⇒(V_1-V_2)/(1 k)=(5-V_2)/(1 k)+V_2/(2 k)
⇒2(V_1-V_2 )=(10-2V_2 )+V_2
Solving the two equations;
V_1=-20V &V_2=30V
I_1=(15-V_1)/(1 k)
⇒I_1=(15-(-20))/(1 k)
⇒I_1=35/(1 k)
I_2=(10-V_1)/(2 k)
⇒I_2=(10-(-20))/(2 k)
⇒I_2=30/(2 k)
I_3=(V_1-V_2)/(1 k)
⇒I_3=((-20)-30)/(1 k)
⇒I_3=(-50)/(1 k)
I_4=(5-V_2)/(1 k)
⇒I_4=(5-30)/(1 k)
⇒I_4=(-25)/(1 k)
I_5=V_2/(2 k)
⇒I_5=30/(2 k)
Using KVL,
E_1-E_(R_1k )+E_(R_2k )-E_2=0
E_(R_1k )=I_1×R_1k
〖⇒E〗_(R_1k )=35×〖10〗^(-3)×1×〖10〗^3
〖⇒E〗_(R_1k )=35V
E_(R_2k )=I_2×R_2k
〖⇒E〗_(R_2k )=15×〖10〗^(-3)×2×〖10〗^3
〖⇒E〗_(R_2k )=30V
⇒E_1-E_(R_1k )+E_(R_2k )-E_2=0
amy1983 said:
Hi g/neill / NewtonianAlch, I typed it all up and came up with the following according to your suggestions:
At node V_1,
⇒(15-V_1)/(1 k)+(10-V_1)/(2 k)=(V_1-V_2)/(1 k)
⇒(30-2V_1 )+(10-V_1 )=(2V_1-2V_2)
Okay, looks good for Node V1.
At node〖 V〗_2,
⇒(V_1-V_2)/(1 k)=(5-V_2)/(1 k)+V_2/(2 k)
Oops. Check the polarity of the E3 source. Is it going to make the potential difference across R4 larger or smaller?
⇒2(V_1-V_2 )=(10-2V_2 )+V_2
At node V_1,
⇒(15-V_1)/(1 k)+(10-V_1)/(2 k)=(V_1-V_2)/(1 k)
⇒(30-2V_1 )+(10-V_1 )=(2V_1-2V_2)
At node〖 V〗_2,
⇒(V_1-V_2)/(1 k)=(5+V_2)/(1 k)+V_2/(2 k)
⇒2(V_1-V_2 )=(10+2V_2 )+V_2
Solving the two equations;
V_1=-20V &V_2=30V
I_1=(15-V_1)/(1 k)
⇒I_1=(15-(-20))/(1 k)
⇒I_1=35/(1 k)
I_2=(10-V_1)/(2 k)
⇒I_2=(10-(-20))/(2 k)
⇒I_2=30/(2 k)
I_3=(V_1-V_2)/(1 k)
⇒I_3=((-20)-30)/(1 k)
⇒I_3=(-50)/(1 k)
I_4=(5+V_2)/(1 k)
⇒I_4=(5+30)/(1 k)
⇒I_4=35/(1 k)
I_5=V_2/(2 k)
⇒I_5=30/(2 k)
Using KVL,
E_1-E_(R_1k )+E_(R_2k )-E_2=0
E_(R_1k )=I_1×R_1k
〖⇒E〗_(R_1k )=35×〖10〗^(-3)×1×〖10〗^3
〖⇒E〗_(R_1k )=35V
E_(R_2k )=I_2×R_2k
〖⇒E〗_(R_2k )=15×〖10〗^(-3)×2×〖10〗^3
〖⇒E〗_(R_2k )=30V
⇒E_1-E_(R_1k )+E_(R_2k )-E_2=0
Hi gneill, Thank you for your advice. Does the remainder of the equation look okay?
amy1983 said:
At node V_1,
⇒(15-V_1)/(1 k)+(10-V_1)/(2 k)=(V_1-V_2)/(1 k)
⇒(30-2V_1 )+(10-V_1 )=(2V_1-2V_2)
At node〖 V〗_2,
⇒(V_1-V_2)/(1 k)=(5+V_2)/(1 k)+V_2/(2 k)
⇒2(V_1-V_2 )=(10+2V_2 )+V_2
Oops again. I think you added the V_2's from the right hand side to the left hand side rather than subtracting them from both sides. But you're getting there...
⇒(V_1-V_2)/(1 k)=(5+V_2)/(1 k)+V_2/(2 k)
⇒2(V_1-V_2 )=(10+2V_2 )+V_2
I know it's bad, but I am finding this electrical element difficult.
:/ Worst case scenario if I put up what I have at least it shows I made the effort!
amy1983 said:
⇒(V_1-V_2)/(1 k)=(5+V_2)/(1 k)+V_2/(2 k)
⇒2(V_1-V_2 )=(10+2V_2 )+V_2
You seem to be having difficulty expanding and then collecting terms. On the LHS you've got: 2*V1 - 2*V2. On the RHS you've got: 10 + 2*V2 + V2 = 10 + 3*V2. So:
2*V1 - 2*V2 = 10 + 3*V2
Move the 3*V2 from the RHS to the LHS by subtracting 3*V2 from both sides. Then proceed.
I know it's bad, but I am finding this electrical element difficult.
:/ Worst case scenario if I put up what I have at least it shows I made the effort!
I know it can be tricky, and much of circuit analysis involves a good deal of finicky algebra. But you are indeed showing good effort and insight for what you need to do to get the result. As they
say, the devil is in the details
FAQ: 3 batteries, 5 resistors using KCL & KVL
1. What is KCL and KVL?
KCL (Kirchhoff's Current Law) states that the sum of currents entering a node in a circuit must equal the sum of currents leaving that node. KVL (Kirchhoff's Voltage Law) states that the sum of
voltage drops in a closed loop must equal the sum of voltage sources in that loop.
2. How do KCL and KVL apply to a circuit with 3 batteries and 5 resistors?
KCL and KVL can be used to analyze the flow of currents and voltages in a circuit with multiple components, such as 3 batteries and 5 resistors. By applying these laws, we can determine the currents
at different points in the circuit and the voltages across each component.
3. How can KCL and KVL help in solving circuit problems?
KCL and KVL provide a systematic approach to solving circuit problems by applying the principles of conservation of charge and energy. By using these laws, we can create equations that can be solved
to determine the unknown currents and voltages in a circuit.
4. Are there any limitations to using KCL and KVL?
KCL and KVL are based on ideal conditions and may not always accurately reflect the behavior of real circuits. They also assume that the circuit is in steady-state, meaning that the currents and
voltages are constant. In addition, KCL and KVL can only be applied to circuits that can be represented as a series or parallel combination of components.
5. How can I apply KCL and KVL in practical situations?
KCL and KVL are essential tools for circuit analysis and can be applied in various practical situations, such as designing and troubleshooting electronic circuits. By using these laws, we can
determine the optimal values for resistors and other components in a circuit and identify any potential problems that may arise. | {"url":"https://www.physicsforums.com/threads/3-batteries-5-resistors-using-kcl-kvl.588361/","timestamp":"2024-11-03T13:53:56Z","content_type":"text/html","content_length":"129196","record_id":"<urn:uuid:5aa8a1f6-2ecc-4a95-8601-d3a11edd4c46>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00036.warc.gz"} |
Evaluation of Transverse and Shear Strengths | Ansys Courses
This lesson covers the micro mechanics of lamina, focusing on the determination of longitudinal tensile strength, transverse tensile strength, and longitudinal compressive strength of a lamina. It
explains how the volume fraction influences the longitudinal tensile strength and how the strength of a composite or lamina is affected by factors such as fiber orientation, non-uniform strength of
fibers, discontinuous fibers, and the interface between the fiber and the matrix. The lesson also discusses the impact of residual stresses on the strength of a laminated composite. It provides a
detailed explanation of the failure modes under compression and the determination of transverse compressive strength.
Video Highlights
01:53 - Discussion on the critical volume fraction and how it affects stress strain curve
04:26 - Discussion on the factors influencing the strength of a composite or lamina
10:49 - Explanation of the determination of transverse tensile strength
33:47 - Discussion on the determination of longitudinal compressive strength
36:32 - Explanation of the determination of transverse compressive strength
41:42 - Discussion on the determination of in-plane shear strength
Key Takeaways
- The longitudinal tensile strength of a lamina is decided by the fiber failure only.
- The volume fraction plays a crucial role in deciding the longitudinal tensile strength of a lamina.
- The strength of a composite or lamina is influenced by factors such as fiber orientation, non-uniform strength of fibers, discontinuous fibers, and the interface between the fiber and the matrix.
- The transverse tensile strength of a lamina is influenced by factors like bond strength of the interface and presence of voids.
- The longitudinal compressive strength of a composite is complicated due to different failure modes under compression. | {"url":"https://innovationspace.ansys.com/courses/courses/micromechanics-of-lamina/lessons/evaluation-of-transverse-and-shear-strengths-lesson-4/","timestamp":"2024-11-09T10:27:06Z","content_type":"text/html","content_length":"175684","record_id":"<urn:uuid:8ac29034-4bf0-4815-962e-fa1d27752468>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00148.warc.gz"} |
Research on the controllability and energy saving of the pneumatic direct drive system
In order to analyze the motion characteristics and stability of the direct drive system of the small air compressor and achieve sound control results, it is necessary to model and analyze the system.
The modelling of the small air compressor system is performed in this section.
According to the law of conservation of energy and the ideal gas state equation, the pressure differential equation of the two cavities of the cylinder is obtained [18]:
$\frac{d P_{a}}{d t}=\frac{k}{V_{a}}\left(R T Q_{m a}-P_{a} \frac{d V_{a}}{d t}\right)$ (1)
$\frac{d P_{b}}{d t}=\frac{k}{V_{b}}\left(R T Q_{m b}+P_{b} \frac{d V_{b}}{d t}\right)$ (2)
P[a], P[b]—pressure of the air intake cavity and exhaust cavity, Pa;
k—ratio of specific heats;
V[a], V[b]—volume of the air intake cavity and exhaust cavity, m^3;
R—constant volume of the air, 287 J/(kg·K);
T—air temperature, K;
Q[ma], Q[mb]—mass flow rate of the air intake cavity and exhaust cavity,kg/s;
where, the volume of the two cavities of the cylinder is expressed as follows:
$V_{a}=A_{a}\left(x_{a 0}+y\right)$ (3)
$V_{b}=A_{b}\left(x_{b 0}+L-y\right)$ (4)
A[a], A[b]—the effective working area of the air in the air intake cavity and exhaust cavity, m^3;
x[a0], x[b0] —length of the dead zone in the air intake cavity and exhaust cavity, m;
L—total journey of the cylinder, m;
y—piston displacement, m
The heat insulation and air discharge in the exhaust cavity of the cylinder is treated as the isentropic process and the temperature change inside the square-cavity is [19]:
$T=T_{0}\left(\frac{P_{b}}{P_{b 0}}\right)^{k-1 / k}$ (5)
T[0]—initial temperature of the exhaust cavity, K;
P[b0]—initial pressure of the exhaust cavity, Pa.
The piston motion equation is obtained based on Newton's second law [20]:
$M \frac{d^{2} y}{d t^{2}}=\left(P_{a} A_{a}-P_{b} A_{b}\right)-F_{f}$ (6)
M—total mass of piston and moving parts, kg;
F[f]—frictional resistance, N.
The friction force during the cylinder movement is a variable, which is related to the cylinder structure, pressure of the two cavities, load size and other conditions. In this paper, the classical
empirical formula of the static friction force + coulomb friction + viscous friction force is adopted:
$F_{f}=\left\{\begin{array}{ll}{\left(F_{c}+F_{v}\right) \cdot \operatorname{sgn}(v),} & {|v|>0} \\ {F_{s}} & {,|v|<0}\end{array}\right.$ (7)
$F_{v}=B_{v} \cdot|v|$ (8)
F[c]—coulomb friction, N;
F[v]—viscous friction, N.
F[s]—static friction force, N;
B[v]—coefficient of viscous friction, N•s/m.
The flow of the gas through the valve port is determined by the effective circulation area of the valve port and the upstream and downstream pressure of the valve port. The flow characteristics of
the pneumatic system can be described by the sonic conductance and the critical pressure ratio. The mass flow equation of the system is shown in the following formula:
$Q_{m}=C_{v} \rho_{0} P_{u} \omega(\sigma, b)$ (9)
$\varphi(\sigma, b)=\left\{\begin{array}{ll}{1} & {\sigma=\frac{p_{d}}{p_{u}}<b} \\ {\sqrt{1-\left(\frac{\sigma-b}{1-b}\right)^{2}}} & {\sigma=\frac{p_{d}}{p_{u}} \geq b}\end{array}\right.$ (10)
C[v]—sonic conductance, m^3/s•MPa;
P[u], P[b]—upstream and downstream pressure of the component, MPa.
ρ[0]—density of air under standard conditions, kg/m^3;
b—critical pressure ratio, 0.2;
σ—pressure ratio.
Unlike the traditional valve-controlled cylinder system, the driving part of the small air compressor system is the air pump, and the input and output of the air pump directly affect the output of
the system. According to the vacuum system pumping equation, the continuity equation and the vacuum technology basic equation, the following formula can be obtained:
$V_{b} \frac{d P_{b}}{d t}=Q P_{b}$ (11)
$Q=Q_{s}\left(1-\frac{P_{s}}{P_{b}}\right)$ (12)
The relationship between the flow and pressure of the vacuum pump can be obtained through combining 11 and 12:
$V_{b} \frac{d P_{b}}{d t}=Q_{s}\left(1-\frac{P_{s}}{P_{b}}\right) P_{b}$ (13)
Q—the actual extraction flow of the air pump, L/min;
Q[s]—theoretical extraction flow of the air pump, L/min;
P[s]—extreme pressure of the air pump, MPa
Q[s] in formula 12 and 13 are the theoretical extraction flow of the air pump. It is related to the input voltage of the air compressor. The relationship between the input voltage U and the air pump
output flow is obtained by adopting the curve fitting method. The relational expression is as follows:
$f(x)=0.3342 x^{3}-7.569 x^{2}+51.82 x-13.11$ (14)
where, x is the input voltage of the air compressor and is the output flow of the air pump.
The system transfer function can be derived based on the above formula. The transfer function between the output displacement and the input voltage of the system is:
$G(x)=\frac{y(s)}{u(s)}=\frac{K_{v} \omega_{n}^{2}}{s\left(s^{2}+2 \xi \omega_{n} s+\omega_{n}^{2}\right)}$ (15)
K[v]—system speed gain, m^2/(s·V·K);
ω[n]—system inherent frequency, rad/s;
ξ—system damping ratio.
The parameters are expressed as follows:
$K_{v}=\frac{K_{q} T R\left(A_{a} V_{b}+A_{b} V_{a}\right)}{P_{a} A_{a}^{2} V_{b}+P_{b} A_{b}^{2} V_{a}}$ (16)
$\omega_{n}=\sqrt{\frac{k\left(P_{a} A_{a}^{2} V_{b}+P_{b} A_{b}^{2} V_{a}\right)}{M V_{a} V_{b}}}$ (17)
$\xi=\frac{B_{v}}{2} \sqrt{\frac{V_{a} V_{b}}{M k\left(P_{a} A_{a}^{2} V_{b}+P_{b} A_{b}^{2} V_{a}\right)}}$ (18) | {"url":"https://iieta.org/journals/ijht/paper/10.18280/ijht.350436","timestamp":"2024-11-10T17:35:46Z","content_type":"text/html","content_length":"95615","record_id":"<urn:uuid:1a25270e-98c9-4eba-a463-a4c90bcfa8d9>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00508.warc.gz"} |
Population Distributions: Probabilities for Discrete & Continuous Variables | Study notes Data Analysis & Statistical Methods | Docsity
Download Population Distributions: Probabilities for Discrete & Continuous Variables and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity! Topic (8) – POPULATION
DISTRIBUTIONS 8-1 Topic (8) – POPULATION DISTRIBUTIONS So far: We’ve seen some ways to summarize a set of data, including numerical summaries. We’ve heard a little about how to sample a population
effectively in order to get good estimates of the population quantities of interest (e.g. taking a good sample and calculating the sample mean as a way of estimating the true but unknown population
mean value) We’ve talked about the ideas of probability and independence. Now we need to start putting all this together in order to do Statistical Inference, the methods of analyzing data and
interpreting the results of those analyses with respect to the population(s) of interest. The Probability Distribution for a random variable can be a table or a graph or an equation. Topic (8) –
POPULATION DISTRIBUTIONS 8-2 Let’s start by reviewing the ideas of frequency distributions for populations using categorical variables. QUALITATIVE (NON-NUMERIC) VARIABLES For a random variable that
takes on values of categories, the Probability distribution is a table showing the likelihood of each value. EXAMPLE Tree species found in a boreal forest. For each possible species there would a
probability associated with it. E.g. suppose there are 4 species and three are very rare and one is very common. A probability table might look like: Species Probability 1 0.01 2 0.03 3 0.08 4 0.88
All 1.00 We interpret these values as the probability that a random selection would result in observing that species. We could also draw a bar chart but it would be fairly non- informative in this
instance since one value is so much larger than the others! An equation cannot be developed since the values that the variable takes on are non-numeric. Topic (8) – POPULATION DISTRIBUTIONS 8-5 Since
we have sampled the entire population (the set of counts for every quadrat in the region), this histogram represents the probability distribution of the random variable X = ”number of trees/quadrat”.
In general, the Poisson distribution is a common probability distribution for counts per unit time or unit area or unit volume. The graph can also be described using an equation known as the Poisson
Distribution Probability Mass Function. It gives the probability of observing a specific count (x) in any randomly selected quadrat as ! )Pr( x exX xµµ− == where )1)(2)(3)...(2)(1(! −−= xxxx and
,...2,1,0=x . In order for this distribution to be a valid probability distribution, we require that the total probability for all possible values equal 1 and that every possible value have a
probability associated with it. ∑∑ = − = === ,...2,1,0,...2,1,0 1 ! )Pr( X x X x exX µ µ Topic (8) – POPULATION DISTRIBUTIONS 8-6 and 0 ! )Pr( ≥== − x exX xµµ The mean of the Poisson distribution is
µ and the variance is µ as well. DISCRETE UNIFORM DISTRIBUTION: every discrete value that the random variable can take on has the same probability of occurring. For example, suppose a researcher is
interested in whether the number of setae on the first antennae of an insect is random or not. Further, the researcher believes that there must be at least 1 seta and at most 8. Then s/he is
postulating that every value between 1 and 8 are equally likely to be observed in a random draw of an insect from the population (or equivalently, that there are equal numbers of insects with 1, 2,
…, or 8 setae in the population). Such a distribution is known as the Discrete Uniform Distribution. Let K be the total number of distinct values that the random variable can take on (e.g. the set
{1, 2, …, 8} contains K = 8 distinct values). Then, K xX 1)Pr( == for x = 1, 2, …, 8 Topic (8) – POPULATION DISTRIBUTIONS 8-7 In addition, the mean for this particular discrete uniform is 5.4 8 36 ==
= ∑ K xµ and the variance is 25.5)5.4( 2 2 = − = ∑ K xσ . Also, it is easy to see that the probabilities sum to 1 as required. Finally, the graph of the distribution looks like a rectangle: 0 2 4 6 8
Topic (8) – POPULATION DISTRIBUTIONS 8-10 Fact 3: When the curve is describing frequency distribution of the population, every observation must fall within the limits of the distribution. Hence, 100%
of the observations are listed. Topic (8) – POPULATION DISTRIBUTIONS 8-11 When we combine these three facts, we get that the density curve describing the frequency distribution of values of a
quantitative variable 1) has a total area under the curve of 1 (analogous to 100%) and 2) the area over a range of values equals the relative frequency of that range in the population, i.e. the area
equals the probability of observing a value within that range Area in between these two lines is the probability that X falls between the values of 5 and 8. 5 8 There are many standard (common)
density curves: Topic (8) – POPULATION DISTRIBUTIONS 8-12 UNIFORM DISTRIBUTION – every subset interval of the same length is equal likely. For example, suppose we randomly selected a number from the
number line [0, 10]. Then the Probability distribution is given by LU abbXa − − =<< )Pr( for ],[ ULX ∈ and . 0, >UL Uniform 0 1 2 3 4 5 6 7 8 9 10 e.g. Pr(3<X<4) = The mean of a Uniform distribution
is 2 LU − =µ and the variance is Topic (8) – POPULATION DISTRIBUTIONS 8-15 Question: What do we do when the value of interest in the probability phrase does NOT fall exactly at the standard deviation
cutoffs? E.g. find Pr(IQ<110)? Answer: Convert the value to a Z-score and use it and a look up table (or a computer program) to calculate the probability. Recall the Z-SCORE for a value is the number
of standard deviations that value is from the mean: Z score z x− = = −* µ σ e.g. IQ of 110 ≡ z* .= − = − =110 110 100 15 0 667µ σ Topic (8) – POPULATION DISTRIBUTIONS 8-16 Defn: When X is normally
distributed, the Z-score has a STANDARD NORMAL DISTRIBUTION. The Standard normal distribution is a normal distribution with a mean of µ=0 and a standard deviation of σ=1. µ−3σ µ−1σ µ+1σ µ+3σ µ−2σ µ
µ+2σ Original IQ score 55 70 85 100 115 130 145 Equivalent Z-score -3 -2 -1 0 +1 +2 +3 Topic (8) – POPULATION DISTRIBUTIONS 8-17 So, the important point here is that we need to do the conversion )Pr
(Pr)Pr( zZaXaX <=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −< − =< σ µ σ µ in order to find probabilities of events under a normal distribution e.g. )667.0Pr( 15 100110 15 100Pr 110Pr)110Pr( <=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −< − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −< −
=< ZIQ IQIQ σ µ σ µ Next, look up the area (i.e. Probability) on a table: 7486.0)667.0Pr( =<Z , so approximately 75% of the population has an IQ less than 110. Topic (8) – POPULATION DISTRIBUTIONS
8-20 Some practice which also uses the rules for Probability that we learned earlier: 1. Find Pr(IQ>92) 2. Find Pr(70<IQ<120). Topic (8) – POPULATION DISTRIBUTIONS 8-21 Finding Quantiles for the
Normal Distribution Most often used to find extreme values in the very highest (or lowest) percentages EXAMPLE Suppose adult male heights are normally distributed with a mean of 69” and a standard
deviation of 3.5”. We have learned how to answer questions like: What proportion of the population are taller than 6’ (72”)? How do we answer a question like: Find the range of likely heights for the
shortest 5% of the male population, i.e. what height is the 5th percentile of the population? Here we are being asked to find the value of a that makes the following probability statement true: Pr
(Height < a) = 0.05 We know that Pr(Height < a) = Pr(Z < z*) So we’ll start by solving Pr(Z < z*)=0.05 Topic (8) – POPULATION DISTRIBUTIONS 8-22 for z*. Now, we’ll use the fact that z a* = − µ σ and
our knowledge of the values of µ and σ to solve for a. | {"url":"https://www.docsity.com/en/docs/population-distributions-lecture-notes-sta-6166/6918201/","timestamp":"2024-11-09T00:28:22Z","content_type":"text/html","content_length":"239616","record_id":"<urn:uuid:6bbcfda8-2418-47fa-aed8-dd183ce0ca74>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00194.warc.gz"} |
Optimal measure transportation with respect to non-traditional costs
We study optimal mass transport problems between two measures with respect to a non-traditional cost function, i.e. a cost c which can attain the value + ∞. We define the notion of c-compatibility
and strong c-compatibility of two measures, and prove that if there is a finite-cost plan between the measures then the measures must be c-compatible, and if in addition the two measures are strongly
c-compatible, then there is an optimal plan concentrated on a c-subgradient of a c-class function. This function is the so-called potential of the plan. We give two proofs of this theorem, under
slightly different assumptions. In the first we utilize the notion of c-path-boundedness, showing that strong c-compatibility implies a strong connectivity result for a directed graph associated with
an optimal map. Strong connectivity of the graph implies that the c-cyclic monotonicity of the support set (which follows from classical reasoning) guarantees its c-path-boundedness, implying, in
turn, the existence of a potential. We also give a constructive proof, in the case when one of the measures is discrete. This approach adopts a new notion of ‘Hall polytopes’, which we introduce and
study in depth, to which we apply a version of Brouwer’s fixed point theorem to prove the existence of a potential in this case.
All Science Journal Classification (ASJC) codes
• Analysis
• Applied Mathematics
Dive into the research topics of 'Optimal measure transportation with respect to non-traditional costs'. Together they form a unique fingerprint. | {"url":"https://cris.iucc.ac.il/en/publications/optimal-measure-transportation-with-respect-to-non-traditional-co","timestamp":"2024-11-06T22:13:05Z","content_type":"text/html","content_length":"50497","record_id":"<urn:uuid:e9d736f6-a185-4804-a80f-eaa2296bdf1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00722.warc.gz"} |
fading channel
MIMO Fading Channel
Filter input signal through MIMO multipath fading channel
Communications Toolbox / Channels
Communications Toolbox / MIMO
The MIMO Fading Channel block filters an input signal using a multi-input/multi-output (MIMO) multipath fading channel. This block models both Rayleigh and Rician fading and employs the Kronecker
model for modeling the spatial correlation between the links. For processing details, see the Algorithms section.
Signal Dimensions
The availability and dimensions of input and output port signals depends on:
Antenna Selection Signal Input (in Transmit Selection Input (Tx Sel Receive Selection Input (Rx Sel Initial Time Offset Input (Init Signal Output (Out1 Optional Channel Gain Output (Gain
Parameter ) ) ) Time) ) )
Off N[S]-by-N[T] N/A N/A N[S]-by-N[R]
Tx N[S]-by-N[ST] 1-by-N[T] N/A nonnegative scalar N[S]-by-N[R] N[S]-by-N[P]-by-N[T]-by-N[R]
Rx N[S]-by-N[T] N/A 1-by-N[R] N[S]-by-N[SR]
Tx and Rx N[S]-by-N[ST] 1-by-N[T] 1-by-N[R] N[S]-by-N[SR]
in — Input data signal
Input data signal, specified as an N[S]-by-N[T] or N[S]-by-N[ST] matrix.
• N[S] represents the number of samples in the input signal.
• N[T] represents the number of transmit antennas.
• N[ST] represents the number of selected transmit antennas.
Data Types: double | single
Complex Number Support: Yes
Tx Sel — Select active transmit antennas
binary vector
Select active transmit antennas, specified as a 1-by-N[T] binary vector. N[T] represents the number of transmit antennas. Elements set to 1 identify selected antenna indices and 0 identify
nonselected antenna indices.
To enable this port, on the Main tab, set Antenna selection to Tx or Tx and Rx.
Data Types: double
Rx Sel — Select active receive antennas
binary vector
Select active receive antennas, specified as a 1-by-N[R] binary vector. N[R] represents the number of receive antennas. Elements set to 1 identify selected antenna indices and 0 identify nonselected
antenna indices.
To enable this port, on the Main tab, set Antenna selection to Rx or Tx and Rx.
Data Types: double
Init Time — Initial time offset
nonnegative scalar
Initial time offset for the fading model in seconds, specified as a nonnegative scalar.
Init Time must be greater than the last frame end time. When Init Time is not a multiple of 1/Sample Rate (Hz), it is rounded up to the nearest sample position.
To enable this port, on the Realization tab, set Initial time source to Input port.
Data Types: double
Out1 — Output data signal for fading channel
Output data signal for the fading channel, returned as an N[S]-by-N[R] or N[S]-by-N[SR] matrix.
• N[S] represents the number of samples in the input signal.
• N[R] represents the number of receive antennas.
• N[SR] represents the number of selected receive antennas.
Gain — Discrete path gains
4-D array
Discrete path gains of the underlying fading process, returned as an N[S]-by-N[P]-by-N[T]-by-N[R] array.
• N[S] represents the number of samples in the input signal.
• N[P] represents the number of channel paths.
• N[T] represents the number of transmit antennas.
• N[R] represents the number of receive antennas.
Entries for nonselected paths are filled with NaN.
To enable this port, on the Realization tab, select Output channel path gains.
To edit block parameters interactively, use the Property Inspector. From the Simulink Toolstrip, on the Simulation tab, in the Prepare gallery, select Property Inspector.
Main Tab
Multipath parameters (frequency selectivity)
Inherit sample rate from input — Option to inherit the sample rate from input
on (default) | off
Select this parameter to use the sample rate of the input signal when processing. When you select Inherit sample rate from input, the sample rate is N[S]/T[S]. N[S] is the number of input samples,
and T[S] is the model sample time.
Sample rate (Hz) — Input signal sample rate
1 (default) | positive scalar
Input signal sample rate, specified in hertz as a positive scalar. To match the model settings, set the sample rate to N[S]/T[S], where N[S] is the number of input samples, and T[S] is the model
sample time.
This parameter appears when Inherit sample rate from input is not selected.
Discrete path delays (s) — Delays for each discrete path
0 (default) | nonnegative scalar | row vector
Delays for each discrete path in seconds, specified as a nonnegative scalar or row vector.
• When you set Discrete path delays (s) to a scalar, the MIMO channel is frequency flat.
• When you set Discrete path delays (s) to a vector, the MIMO channel is frequency selective.
Average path gains (dB) — Average gain for each discrete path
0 (default) | scalar | row vector
Average gain for each discrete path in decibels, specified as a scalar or row vector. Average path gains (dB) must have the same size as Discrete path delays (s).
Normalize average path gains to 0 dB — Option to normalize average path gains to 0 dB
on (default) | off
Select this parameter to normalize the fading processes so that the total power of the path gains, averaged over time, is 0 dB.
Fading distribution — Fading distribution of channel
Rayleigh (default) | Rician
Select the fading distribution of the channel, either Rayleigh or Rician.
K-factors — K-factor of Rician fading channel
3 (default) | positive scalar | row vector of nonnegative values
K-factor of a Rician fading channel, specified as a positive scalar or a 1-by-N[P] vector of nonnegative values. N[P] equals the value of the Discrete path delays (s) parameter.
• If you set K-factors to a scalar, the first discrete path is a Rician fading process with a Rician K-factor of K-factors. Any remaining discrete paths are independent Rayleigh fading processes.
• If you set K-factors to a row vector, the discrete path corresponding to a positive element of the K-factors vector is a Rician fading process with a Rician K-factor specified by that element.
The discrete path corresponding to any zero-valued elements of the K-factors vector are Rayleigh fading processes. At least one element value must be nonzero.
This parameter appears when you set Fading distribution to Rician.
LOS path Doppler shifts (Hz) — Doppler shifts for line-of-sight components
0 (default) | scalar | row vector
Doppler shifts for the line-of-sight components of the Rician fading channel in hertz, specified as a scalar or row vector. This parameter must have the same size as K-factors.
• If you set LOS path Doppler shifts (Hz) to a scalar, it represents the line-of-sight component Doppler shift of the first discrete path that is a Rician fading process.
• If you set LOS path Doppler shifts (Hz) to a row vector, the discrete path that is a Rician fading process has its line-of-sight component Doppler shift specified by the elements of LOS path
Doppler shifts (Hz) that correspond to positive elements in the K-factors vector.
This parameter appears when you set Fading distribution to Rician.
LOS path initial phases (rad) — Initial phases for line-of-sight components
0 (default) | scalar | row vector
Initial phases for the line-of-sight component of the Rician fading channel in radians, specified as a scalar or row vector. This parameter must have the same size as K-factors.
• If you set LOS path initial phases (rad) to a scalar, it is the line-of-sight component initial phase of the first discrete path that is a Rician fading process.
• If you set LOS path initial phases (rad) to a row vector, the discrete path that is a Rician fading process has its line-of-sight component initial phase specified by the elements of LOS path
initial phases (rad) that correspond to positive elements in the K-factors vector.
This parameter appears when you set Fading distribution to Rician.
Doppler parameters (time dispersion)
Maximum Doppler shift (Hz) — Maximum Doppler shift for all channel paths
0.001 (default) | nonnegative scalar
Maximum Doppler shift for all channel paths in hertz, specified as a nonnegative scalar.
Maximum Doppler shift (Hz) must be less than or equal to (Sample Rate (Hz)/10)/f[c] for each path, where f[c] is the cutoff frequency factor of the path. For more information, see Cutoff Frequency
Doppler spectrum — Doppler spectrum shape for all channel paths
doppler('Jakes') (default) | doppler('Flat') | doppler('Rounded', ...) | doppler('Bell', ...) | doppler('Asymmetric Jakes', ...) | doppler('Restricted Jakes', ...) | doppler('Gaussian', ...) |
doppler('BiGaussian', ...)
Doppler spectrum shape for all channel paths, specified as a single Doppler spectrum structure returned from the doppler function or a 1-by-N[P] cell array of such structures. The default value of
this parameter is the Jakes Doppler spectrum (doppler('Jakes')).
• If you assign a single call to doppler, all paths have the same specified Doppler spectrum.
• If you assign a 1-by-N[P] cell array of calls to doppler using any of the specified syntaxes, each path has the Doppler spectrum specified by the corresponding Doppler spectrum structure in the
array. In this case, N[P] equals the value of the Discrete path delays (s) parameter.
This parameter applies when Maximum Doppler shift (Hz) is greater than zero.
If you set Technique for generating fading samples to Sum of sinusoids, Doppler spectrum must be doppler('Jakes').
Antenna parameters (spatial dispersion)
Specify spatial correlation — Spatial correlation mode
None (default) | Separate Tx Rx | Combined
Select the spatial correlation mode: None, Separate Tx Rx, or Combined.
• Choose 'None' to specify the number of transmit and receive antennas.
• Choose 'Spatial Tx Rx' to specify the transmit and receive spatial correlation matrices separately. The number of transmit (N[T]) and receive (N[R]) antennas are derived from the dimensions of
the Transmit spatial correlation and Receive spatial correlation parameters, respectively.
• Choose 'Combined' to specify a single correlation matrix for the whole channel. The product of N[T] and N[R] is derived from the dimension of Combined spatial correlation.
Number of transmit antennas — Number of transmit antennas
2 (default) | positive integer
Number of transmit antennas, specified as a positive integer.
This parameter appears when you set Specify spatial correlation to None or Combined.
Number of receive antennas — Number of receive antennas
2 (default) | positive integer
Number of receive antennas, specified as a positive integer.
This parameter appears when you set Specify spatial correlation to None.
Transmit spatial correlation — Spatial correlation of transmitter
[1 0; 0 1] (default) | matrix | 3-D array
Specify the spatial correlation of the transmitter as an N[T]-by-N[T] matrix or N[T]-by-N[T]-by-N[P] array of complex or real values. N[T] is the number of transmit antennas, and N[P] equals the
value of the Discrete path delays (s) parameter.
• If you set Discrete path delays (s) to a scalar, the channel is frequency flat, and Transmit spatial correlation is an N[T]-by-N[T] Hermitian matrix. The magnitude of any off-diagonal element
must be no larger than the geometric mean of the two corresponding diagonal elements.
• If you set Discrete path delays (s) to a vector, the channel is frequency selective, and you can specify Transmit spatial correlation as a matrix. Each path has the same transmit spatial
correlation matrix.
• Alternatively, you can specify Transmit spatial correlation as an N[T]-by-N[T]-by-N[P] array, where each path can have its own different transmit spatial correlation matrix.
This parameter appears when you set Specify spatial correlation to Separate Tx Rx.
Receive spatial correlation — Spatial correlation of receiver
[1 0; 0 1] (default) | matrix | 3-D array
Specify the spatial correlation of the receiver as an N[R]-by-N[R] matrix or N[R]-by-N[R]-by-N[P] array of complex or real values. N[R] is the number of receive antennas, and N[P] equals the value of
the Discrete path delays (s) parameter.
• If you set Discrete path delays (s) to a scalar, the channel is frequency flat, and Receive spatial correlation is an N[R]-by-N[R] Hermitian matrix. The magnitude of any off-diagonal element must
be no larger than the geometric mean of the two corresponding diagonal elements.
• If you set Discrete path delays (s) to a vector, the channel is frequency selective, and you can specify Receive spatial correlation as a matrix. Each path has the same receive spatial
correlation matrix.
• Alternatively, you can specify Receive spatial correlation as an N[R]-by-N[R]-by-N[P] array, where each path can have its own different receive spatial correlation matrix.
This parameter appears when you set Specify spatial correlation to Separate Tx Rx.
Combined spatial correlation — Combined spatial correlation matrix
[1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1] (default) | matrix | 3-D array
Specify the combined spatial correlation matrix as an N[TR]-by-N[TR] matrix or N[TR]-by-N[TR]-by-N[P] array of complex or real values. N[TR] = (N[T] ✕ N[R]), and N[P] equals the number of delay paths
specified by the Discrete path delays (s) parameter.
• If you set Discrete path delays (s) to a scalar, the channel is frequency flat, and Combined spatial correlation is an N[TR]-by-N[TR] Hermitian matrix. The magnitude of any off-diagonal element
must be no larger than the geometric mean of the two corresponding diagonal elements.
• If you set Discrete path delays (s) to a vector, the channel is frequency selective, and you can specify Combined spatial correlation as a matrix. Each path has the same spatial correlation
• Alternatively, you can specify Combined spatial correlation as an N[TR]-by-N[TR]-by-N[P] array, where each path can have its own different combined spatial correlation matrix.
This parameter appears when you set Specify spatial correlation to Combined.
Normalize outputs by number of receive antennas — Normalize channel output
on (default) | off
Select this parameter to normalize the channel outputs by the number of receive antennas.
Simulate using — Compilation type
Interpreted execution (default) | Code generation
Compilation type, specified as Interpreted execution or Code generation.
Antenna selection — Antenna mode
Off (default) | Tx | Rx | Tx and Rx
The antenna mode you select corresponds to additional input ports on the block.
Antenna selection Setting Input Ports Added
Off None
Tx Tx Sel
Rx Rx Sel
Tx and Rx Tx Sel, Rx Sel
Realization Tab
Technique for generating fading samples — Channel modeling technique
Filtered Gaussian noise (default) | Sum of sinusoids
Select the channel modeling technique, either Filtered Gaussian noise or Sum of sinusoids.
Number of sinusoids — Number of sinusoids used
48 (default) | positive integer
Number of sinusoids used to model the fading process, specified as a positive integer.
This parameter appears when you set Technique for generating fading samples to Sum of sinusoids.
Initial time source — Source of initial time offset
Property (default) | Input port
Indicate the source of the initial time offset for the fading model, either Property or Input port.
• When you set Initial time source to Property, use Initial time (s) to set the initial time offset.
• When you set Initial time source to Input port, use the input port Init Time to set the initial time offset.
This parameter appears when you set Technique for generating fading samples to Sum of sinusoids.
Initial time (s) — Initial time offset
0 (default) | nonnegative scalar
Initial time offset for the fading model, specified as a nonnegative scalar.
When Initial time (s) is not a multiple of 1/Sample Rate (Hz), it is rounded up to the nearest sample position.
This parameter appears when you set Technique for generating fading samples to Sum of sinusoids and Initial time source to Property.
Initial seed — Random number generator initial seed
73 (default) | nonnegative integer
Random number generator initial seed for this block, specified as a nonnegative integer.
Output channel path gains — Option to output channel path gains
off (default) | on
Select this parameter to add the Gain output port to the block and output the channel path gains of the underlying fading process.
Visualization Tab
Channel visualization — Select the channel visualization
Off (default) | Impulse response | Frequency response | Doppler spectrum | Impulse and frequency responses
Select the channel visualization: Off, Impulse response, Frequency response, Doppler spectrum, or Impulse and frequency responses. When visualization is on, the selected channel characteristics, such
as impulse response or Doppler spectrum, display in a separate window. For more information, see Channel Visualization.
Antenna pair to display — Transmit-receive antenna pair to display
[1,1] (default) | vector
Transmit-receive antenna pair to display, specified as a 1-by-2 vector, where the first element corresponds to the desired transmit antenna and the second corresponds to the desired receive antenna.
At this time, only a single pair can be displayed.
This parameter appears when Channel visualization is not Off.
Percentage of samples to display — Percentage of samples to display
25% (default) | 10% | 50% | 100%
Select the percentage of samples to display: 10%, 25%, 50%, or 100%. Increasing the percentage improves display accuracy at the expense of simulation speed.
This parameter appears when you set Channel visualization to Impulse response, Frequency response, or Impulse and frequency responses.
Path for Doppler spectrum display — Path for which Doppler spectrum is displayed
1 (default) | positive integer
Path for which the Doppler spectrum is displayed, specified as a positive integer from 1 to N[P], where N[P] equals the value of the Discrete path delays (s) parameter.
This parameter appears when you set Channel visualization to Doppler spectrum.
Block Characteristics
Data Types double | single
Multidimensional Signals yes
Variable-Size Signals yes
The fading processing per link is described in Methodology for Simulating Multipath Fading Channels and assumes the same parameters for all (N[T] × N[R]) links of the MIMO channel. Each link
comprises all multipaths for that link.
The Kronecker Model
The Kronecker model assumes that the spatial correlations at the transmit and receive sides are separable. Equivalently, the direction of departure (DoD) and directions of arrival (DoA) spectra are
assumed to be separable. The full correlation matrix is:
${R}_{H}=E\left[{R}_{t}\otimes {R}_{r}\right]$
• The ⊗ symbol represents the Kronecker product.
• R[t] is the correlation matrix at the transmit side, ${R}_{t}=E\left[{H}^{H}H\right]$, and is of size N[T]-by-N[T].
• R[r] is the correlation matrix at the receive side, ${R}_{r}=E\left[H{H}^{H}\right]$, and is of size N[R]-by-N[R].
You can obtain a realization of the MIMO channel matrix as:
A is an N[R]-by-N[T] matrix of independent identically distributed complex Gaussian variables with zero mean and unit variance.
Cutoff Frequency Factor
The cutoff frequency factor, f[c], is dependent on the type of Doppler spectrum.
• For any Doppler spectrum type other than Gaussian and bi-Gaussian, f[c] equals 1.
• For a doppler('Gaussian') spectrum type, f[c] equals NormalizedStandardDeviation$\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\sqrt{2\mathrm{log}2}$.
• For a doppler('BiGaussian') spectrum type:
□ If the PowerGains(1) and NormalizedCenterFrequencies(2) field values are both 0, then f[c] equals NormalizedStandardDeviation(1)$\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\sqrt{2\mathrm
□ If the PowerGains(2) and NormalizedCenterFrequencies(1) field values are both 0, then f[c] equals NormalizedStandardDeviation(2)$\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\sqrt{2\mathrm
□ If the NormalizedCenterFrequencies field value is [0,0] and the NormalizedStandardDeviation field has two identical elements, then f[c] equals NormalizedStandardDeviation(1)$\text{\hspace
□ In all other cases, f[c] equals 1.
Antenna Selection
When the object is in antenna-selection mode, it uses these algorithms to process an input signal.
• All random path gains are always generated and keep evolving for each link, whether or not a given link is selected. The path gain values output for the nonselected links are populated with NaN.
• The spatial correlation applies to only the selected transmit and receive antennas, and the correlation coefficients are the corresponding entries in the transmit, receive, or combined
correlation matrices. That is, the spatial correlation matrix for the selected transmit or receive antennas is a submatrix of the transmit, receive, or combined spatial correlation matrix
property value.
• For signal paths that are associated with nonactive antennas, a signal with zero power is transmitted to the channel filter.
• Channel output normalization happens over the number of selected receive antennas.
[1] Oestges, C., and B. Clerckx. MIMO Wireless Communications: From Real-World Propagation to Space-Time Code Design. Academic Press, 2007.
[2] Correira, L. M. Mobile Broadband Multimedia Networks: Techniques, Models and Tools for 4G. Academic Press, 2006.
[3] Kermoal, J. P., L. Schumacher, K. I. Pedersen, P. E. Mogensen, and F. Frederiksen. "A stochastic MIMO radio channel model with experimental validation." IEEE Journal on Selected Areas of
Communications. Vol. 20, Number 6, 2002, pp. 1211–1226.
[4] Jeruchim, M., P. Balaban, and K. S. Shanmugan. Simulation of Communication Systems. Second Edition. New York: Kluwer Academic/Plenum, 2000.
[5] Pätzold, Matthias, Cheng-Xiang Wang, and Bjorn Olav Hogstand. "Two New Sum-of-Sinusoids-Based Methods for the Efficient Generation of Multiple Uncorrelated Rayleigh Fading Waveforms." IEEE
Transactions on Wireless Communications. Vol. 8, Number 6, 2009, pp. 3122–3131.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using Simulink® Coder™.
Version History
Introduced in R2013b
R2022b: Updates to channel visualization display
The channel visualization feature now presents:
• Configuration settings in the bottom toolbar on the plot window.
• Plots side-by-side in one window when you select the Impulse and frequency response channel visualization option.
See Also | {"url":"https://de.mathworks.com/help/comm/ref/mimofadingchannel.html","timestamp":"2024-11-05T07:32:07Z","content_type":"text/html","content_length":"175751","record_id":"<urn:uuid:f2594537-de12-4e4b-aeac-7bf6c0e531b9>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00493.warc.gz"} |
Production: Laser Light Show
I originally wanted the lines drawn from random point to random point on the edge to generate with each click of the mouse. However, on my first attempt, I could only get the lines to draw in
completely random areas with random lengths. In other words they didn't reach from edge to edge.
Within Class Line, I had already defined:
With these variables I could constrain the starting points of the lines to the x and y edges of the artboard. I started by defined the choice function to be random between 0 and 1, or otherwise
choose whether the starting point would be on the x or the y edge. With the the choice function in the if statement:
// the starting point is anywhere from point 0 to edge of canvas in the Y direction
startY = random (0,height);
// y direction is randomized
// the starting point is anywhere from point 0 to edge of canvas in the X direction
startX = random (0,width);
// x direction is randomized
This if statement constrained the starting points, but I also wanted the end points to be constrained to the edges so that the lines drew all the way across the screens. So, I had to create two more
variables (x1 and y1) to define the placements of the endpoints. I first defined the new variables alongside the original ones in mousepressed:
Then, in Class Line(), I added x1 and y1 to the constructor and changed the endpoints of the line from random(0,400) to this.x1 and this.y1.
constructor(x,y,dx,dy,x1,y1) {
randomColor = color(random(255), random(255), random(255));
line(this.x, this.y, this.x1, this.y1); | {"url":"https://www.isabellerieken.com/post/production-laser-light-show","timestamp":"2024-11-14T11:54:20Z","content_type":"text/html","content_length":"1050484","record_id":"<urn:uuid:6826e666-5456-4105-af88-58c68fc415c0>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00743.warc.gz"} |
Counter Log only shows and stores partial value (incremental) / ComfortClick / ComfortClick
Paul G 4 years ago
Devices / KNX
updated 4 years ago • 2
BoS: 4.8.14
KNX data point value
Counter log pointed to KNX data point value above
Why does the counter log only records the partial values and graphically displays the partial values as opposed to the real value of the KNX data point?
The counter log only saves the difference of the values in a 1h interval.
The best way to explain is to check one previous post that I've posted, where I was having some problems with the counter logs with the power consumption value.
Check the reply of Matic Godejsa from ComfortClick on that post, and you'll understand why the counter log doesn't save the exact real time value shown at the moment, but rather the diferences of the
value in a 1h interval section.
Counter Log - Help with kWh calculation / ComfortClick / ComfortClick
A short piece of the reply:
"When we first link the input value to Counter log, it remembers the value. Let's say that when we link our electrical energy meter to counter log it's current value is 100 kWh and counter logs
starting value will be 100. However this value is not saved in database, because counter log stores only difference in values in interval of 1 hour so it will actually start with 0.
Let say that in next hour our energy meter value will be 110 kWh, that means that consumption in that hour was 10 kWh and Counter log will save value 10 in the database. Then after another hour the
energy meter value would be 115 kWh and counter log will add 5 to previous database entry and the new entry would be 15 and so on..." | {"url":"https://support.comfortclick.com/communities/4/topics/5896-counter-log-only-shows-and-stores-partial-value-incremental","timestamp":"2024-11-12T05:51:25Z","content_type":"text/html","content_length":"38398","record_id":"<urn:uuid:a6fad65a-73ad-495c-8bc7-9722e7b33153>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00549.warc.gz"} |
compass point to full turn
angle units conversion
Amount: 1 compass point (point) of angle
Equals: 0.031 full turns (turn) in angle
Converting compass point to full turns value in the angle units scale.
TOGGLE : from full turns into compass points in the other way around.
CONVERT : between other angle measuring units - complete list.
How many full turns are in 1 compass point? The answer is: 1 point equals 0.031 turn
0.031 turn is converted to 1 of what?
The full turns unit number 0.031 turn converts to 1 point, one compass point. It is the EQUAL angle value of 1 compass point but in the full turns angle unit alternative.
point/turn angle conversion result
From Symbol Equals Result Symbol
1 point = 0.031 turn
Conversion chart - compass points to full turns
1 compass point to full turns = 0.031 turn
2 compass points to full turns = 0.063 turn
3 compass points to full turns = 0.094 turn
4 compass points to full turns = 0.13 turn
5 compass points to full turns = 0.16 turn
6 compass points to full turns = 0.19 turn
7 compass points to full turns = 0.22 turn
8 compass points to full turns = 0.25 turn
9 compass points to full turns = 0.28 turn
10 compass points to full turns = 0.31 turn
11 compass points to full turns = 0.34 turn
12 compass points to full turns = 0.38 turn
13 compass points to full turns = 0.41 turn
14 compass points to full turns = 0.44 turn
15 compass points to full turns = 0.47 turn
Category: main menu • angle menu • Compass points
Convert angle of compass point (point) and full turns (turn) units in reverse from full turns into compass points.
This calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation
measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which
has built in all angle unit-variations. Page with individual angle units.
Converter type: angle units
First unit: compass point (point) is used for measuring angle.
Second: full turn (turn) is unit of angle.
15 point = ? turn
15 point = 0.47 turn
Abbreviation, or prefix, for compass point is:
Abbreviation for full turn is:
Other applications for this angle calculator ...
With the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool:
1. in practicing compass points and full turns ( point vs. turn ) measures exchange.
2. for conversion factors between unit pairs.
3. work with angle's values and properties. | {"url":"https://www.traditionaloven.com/tutorials/angle/convert-compass-navigation-point-to-full-turn-unit.html","timestamp":"2024-11-05T20:35:18Z","content_type":"text/html","content_length":"45755","record_id":"<urn:uuid:6f87a62b-290d-4fde-b192-a0e7705d01ed>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00052.warc.gz"} |
Fencing Fields -
A farmer has two fields, one for crops and the other for livestock. The field used for crops is 8 times larger in area than the field used for livestock. Surprisingly, the fencing required to enclose
the smaller field (i.e. its perimeter) is twice the fencing required to enclose the larger field. Both fields are rectangular and the fence around each is continuous (i.e. no gaps). The fencing is
available only in 1 meter units. How many meters of fencing is required to enclose the two fields?
This puzzle has multiple solutions. Logically, we know that small field (area-wise) will be a thin/skinny rectangle given it has twice the perimeter of the larger field despite being so much smaller
in terms of area. Here are a few solutions:
Large Field: 17 x 264, Small Field: 1 x 561 (total fencing: 1686 m)
Large Field: 18 x 140, Small Field: 1 x 315 (total fencing: 948 m)
Large Field: 20 x 78, Small Field: 1 x 195 (total fencing: 588 m)
Large Field: 24 x 47, Small Field: 1 x 141 (total fencing: 426 m)
Large Field: 33 x 1024, Small Field: 2 x 2112 (total fencing: 1686 m)
Large Field: 63 x 64, Small Field: 2 x 252 (total fencing: 762 m)
Large Field: 49 x 2280, Small Field: 3 x 4655 (total fencing: 13974 m)
Large Field: 50 x 1164, Small Field: 3 x 2425 (total fencing: 7284 m)
Large Field: 52 x 606, Small Field: 3 x 1313 (total fencing: 3948 m)
Large Field: 56 x 327, Small Field: 3 x 763 (total fencing: 2298 m)
Large Field: 57 x 296, Small Field: 3 x 703 (total fencing: 2118 m)
Large Field: 79 x 120, Small Field: 3 x 395 (total fencing: 1194 m)
Large Field: 84 x 110, Small Field: 3 x 385 (total fencing: 1164 m) EXPLANATION
: This is from an old book of Mensa puzzles. The book listed a single solution (the example with 762 m of fencing) without any explanation, but there are in fact multiple (infinite) solutions. There
are 2 equations:
ab = 8cd
(2a + 2b) × 2 = 2c + 2d
. The next step requires assigning a value to one of the 4 variables, for example, we can consider a solution where C (the height of the thin/skinny field) = 1. To continue, an understanding of
derivatives is required. You can find an explanation of the solution in the answer posted
Graphically, we're looking at pairs of rectangles where the second has twice the perimeter of the first and then looking at all the different possible areas for each rectangle and trying to find an
area in the first rectangle that is eight times an area in the second rectangle. For example, in the 4th solution listed above:
Do you have a
for this puzzle (e.g. something that should be mentioned/clarified in the question or solution, bug, typo, etc.)? | {"url":"https://puzzlesandriddles.com/ForGeeksAndBrainiacs35.html","timestamp":"2024-11-05T18:33:07Z","content_type":"text/html","content_length":"14653","record_id":"<urn:uuid:5a107018-c121-4b46-ac0e-932768d32f99>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00091.warc.gz"} |
Data Handling - Definition, Types, Examples | Graphical Representation of Data
Data Handling – Definition, Types, Examples | Graphical Representation of Data
Our life is all surrounded by numbers likewise marks scored, making a note of the height, weight, runs made, etc. All these are nothing but data and this article explains all about what is meant by
data handling, how to organize data in the form of graphs, charts, types of data handling, calculation of mean, median, mode, etc. Practice the different questions available for representing various
data in the form of bar graphs, pictographs, histograms, line graphs, etc., and get a good hold of the topics.
List of Data Handling Topics
Simply click on the links available below to learn in-depth about the respective topic.
What is Data Handling?
Data Handling is the process of collecting data and representing it in different forms. It is Securing the Research Data Gathered, Archived, Disposed of in a safe way once the analysis process is
completed. You can use the data for comparing, taking out mean, median, mode, etc. The information which is collected initially is known as raw data and it can be in any form be it words,
measurements, numbers, or descriptions.
Types of Data
Data Handling can be performed depending on the type of data. In general, the data is classified into two types namely
• Qualitative Data
• Quantitative Data
Qualitative Data: It gives descriptive information about a certain thing.
Quantitative Data: It provides numerical information and is further classified into two types namely discrete data, continuous data. Discrete Data takes values like whole numbers whereas continuous
data takes a certain range.
How to Represent Data Handling?
Data can be denoted in the following ways and they are as such
• Bar Graph
• Pictograph
• Line Graph
• Stem and Leaf Plots
• Histogram
• Dot Plots
• Cumulative Tables and graphs
• Frequency Distribution
Steps Involved in Data Handling
Steps Explanation
Purpose Firstly identify the purpose
Collection of Data Collect the data relevant to the purpose
Presentation of Data Represent the collected data in the form of a meaningful and easy-to-understand way be it in the form of a table, tally marks, etc.
Graphical Representation of Data Visual Representation of Data can be quite easier for analysis as well as for understanding and has a greater impact.
Analyzing Data Look for ways to derive useful information so that you can proceed further
Conclusion Based on the analysis we did draw a conclusion to the given problem
Problems on Data Handling
Example 1.
Below is the table that tells us the number of chocolates present in different boxes. Draw a Pictograph for the given information?
Boxes Number of chocolates
Box 1 21
Box 2 14
Box 3 42
Box 4 7
We have considered each chocolate symbol as 7 chocolates and interpreted the above data in pictorial way for better visualization.
Boxes Number of chocolates
Box 1
Box 2
Box 3
Box 4
Example 2.
Students are given a pictograph describing the favorite ice cream of children. Based on the information given answer the questions asked?
1. How many students voted for the vanilla?
2. What flavor did the students like the most?
3. What flavor did students like the least?
4. How many students voted for chocolate than vanila?
1. No. of votes students cast for vanilla = 7
2. The flavor students liked the most is chocolate chips
3. The flavor students liked the least is strawberry
4. No. of students voted for chocolate than vanilla is 1.
Example 3.
The Sales of 3 Laptop company’s in the three consecutive months is represented by the table. Draw a Bar Graph.
Lenovo Hp Dell
Jan 200 300 400
Feb 500 400 600
Mar 600 500 700
Apr 800 700 900
Example 4.
The below pie chart describes how often 100 people use different modes of transport. Answer the Questions on the below diagram?
i) What percentage of people use the train most often?
ii) How many people use trains most often?
iii) How many people use buses most often?
i) 10% of people use the trains most often
ii) We need to find how many people use cars
No. of People given = 100
40% of people use cars
thus \(\frac { 40 }{ 100 } \)*100
= 40
Therefore, 40 people use cars most often
iii) No. of people = 100
Perecentage of People using buses = 20%
thus \(\frac { 20 }{ 100 } \)*100 = 20
Therefore, 20 people uses buses often.
Example 5.
The numbers of newspapers sold at a local shop over the 6 days are: 24, 20, 16, 16, 28, 32. State the frequency?
Given that,
The total number of newspapers sold at the local shop past 6 days are 24, 20, 16, 16, 28, 32
By arranging the newspapers numbers in ascending order we get 16, 16, 20, 24, 28, 32
The frequency table for the papers sold is given here:
┃PAPER SOLD │FREQUENCY ┃
┃16 │2 ┃
┃20 │1 ┃
┃24 │1 ┃
┃28 │1 ┃
┃32 │1 ┃
┃Total │6 ┃
Example 6.
Find the mean, median, mode, and range for the following list of values 5, 11, 7, 7, 1, 8, 2, 6, and 13?
Mean= 5+11+7+7+1+8+2+6+13=66 ; 60/8= 7.33
Median= 7; the middle number when we arrange numbers in Ascending or Descending : 1,2,5,6,7,7,8,11,13
Mode=7; the number that occurs most often
Range=13-1=12; the difference between the highest and the lowest values in the data set.
FAQs on Data Handling
1. What is Data Handling in Simple Words?
Data Handling is the process of collecting data and representing it in different forms. It is sometimes referred to as Statistics.
2. What are the two types of data handling?
There are two types of data handling namely qualitative data, quantitative data.
3. How to find Mean in Data Handling?
Mean is the average of a set of data. We can find the mean by simply adding all the numbers in the data set and then dividing the sum by the number of values in the data set.
4. What is meant by Class Size in Data Handling?
Class Size refers to the range i.e. the difference between the upper limit and lower limit.
5. How is Data Represented Graphically?
We can represent data graphically in various forms such as
• Bar Graph
• Scatter Plot
• Line Graph
• Area Plot
• Pie Chart/ Circle Chart
• Picture Graph | {"url":"https://ccssanswers.com/data-handling/","timestamp":"2024-11-04T04:06:49Z","content_type":"text/html","content_length":"163730","record_id":"<urn:uuid:8bdd449c-801c-42be-986e-766ed1a8230c>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00790.warc.gz"} |
Worksheet Solutions: Use of Statistical Tools | Economics Class 11 - Commerce PDF Download
Multiple Choice Questions
Q1: Which statistical tool is used to measure the average of a set of data?
(a) Histogram
(b) Mean
(c) Range
(d) Median
Q2: In statistics, the measure of central tendency that is least affected by outliers is:
(a) Mode
(b) Median
(c) Standard Deviation
(d) Variance
Q3: What does the interquartile range represent?
(a) The spread of data
(b) The middle 50% of data
(c) The maximum value in the dataset
(d) The minimum value in the dataset
Q4: Which of the following is a measure of dispersion?
(a) Mean
(b) Mode
(c) Standard Deviation
(d) Median
Q5: If the coefficient of variation is 10%, what does it imply?
(a) Data is highly variable
(b) Data is not variable
(c) Data is moderately variable
(d) Data is normally distributed
True or False
Q1: The mean and median will always be the same in any dataset.
Q2: A high standard deviation indicates that the data points are closely clustered around the mean.
Q3: The range of a dataset is the difference between the highest and lowest values.
Q4: The coefficient of variation is expressed as a percentage.
Q5: Skewness measures the symmetry of a dataset.
Very Short Answers
Q1: Define 'Statistics.'
Ans: Statistics is the science of collecting, organizing, analyzing, interpreting, and presenting data to make informed decisions.
Q2: Explain the term 'Central Tendency.'
Ans: Central tendency refers to the measure that represents the center or average of a dataset. It includes mean, median, and mode.
Q3: What is the primary purpose of a histogram?
Ans: The primary purpose of a histogram is to represent the frequency distribution of data in graphical form.
Q4: Define 'Variance.'
Ans: Variance is a measure of how spread out the data points are from the mean in a dataset. It quantifies the dispersion or variability of data.
Q5: What is the significance of the coefficient of variation (CV)?
Ans: The coefficient of variation (CV) is significant because it measures the relative variability of data, allowing for comparisons between datasets with different units or scales.
Short Answers
Q1: Explain the concept of 'Measures of Dispersion' and provide examples.
Ans: Measures of dispersion quantify how data points vary or spread out in a dataset. Examples include the range, variance, and standard deviation. The range is the difference between the highest and
lowest values. Variance measures the average squared deviation from the mean, and standard deviation is the square root of variance.
Q2: Describe the 'Coefficient of Variation' (CV) and its importance.
Ans: The coefficient of variation (CV) is the ratio of the standard deviation to the mean, expressed as a percentage. It is important because it helps compare the relative variability of datasets
with different units or scales. A lower CV indicates less relative variability, while a higher CV suggests greater relative variability.
Q3: Explain the concept of 'Skewness' in statistics.
Ans: Skewness measures the asymmetry of the probability distribution of a dataset. A positive skew indicates that the tail on the right side is longer or fatter, while a negative skew means the left
tail is longer or fatter. A perfectly symmetrical distribution has zero skewness.
Q4: Differentiate between 'Mean' and 'Median' as measures of central tendency.
Ans: Mean is the arithmetic average of a dataset and is sensitive to outliers. Median is the middle value when data is arranged in ascending or descending order and is less affected by outliers. Mean
provides a balance point, while the median represents the middle value.
Q5: Explain the construction and interpretation of a 'Box-and-Whisker Plot.'
Ans: A Box-and-Whisker Plot is a graphical representation of the five-number summary of a dataset: minimum, first quartile (Q1), median, third quartile (Q3), and maximum. It helps visualize the
spread and skewness of data. The box represents the interquartile range (Q3-Q1), and the whiskers extend to the minimum and maximum values. Outliers may be shown as individual points. | {"url":"https://edurev.in/studytube/Worksheet-Solutions-Use-of-Statistical-Tools/a1931858-793f-4e6f-9a00-5adca5f2128a_t","timestamp":"2024-11-10T06:20:18Z","content_type":"text/html","content_length":"279146","record_id":"<urn:uuid:d2c0311a-23f1-4c9b-87e9-84ea15f4e3c8>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00887.warc.gz"} |
Tumu’s class was given an assignment to feature a scientist that contributed. - WorkSheets Buddy
Tumu’s class was given an assignment to feature a scientist that contributed.
Tumu’s class was given an assignment to feature a scientist that contributed to the development of the cell theory. The class decided to use images to feature the scientist of their choice.
Which image would Tumu most likely use in his assignment to feature Rudolf Virchow?
the magnified shapes of cells in a piece of cork
the different cells that make up a plant leaf
the types of cells at work in the human body
the way root cells reproduce to increase root length
the way root cells reproduce to increase root length.
The image that Tumu would most likely use in his assignment to feature Rudolf Virchow is “the types of cells at work in the human body.”Rudolf Virchow contributed to the development of the cell
theory by adding the idea that all cells arise from pre-existing cells. He studied and made significant contributions to the understanding of human cells, therefore, the types of cells at work in the
human body would be the most appropriate image to use in Tumu’s assignment to feature Rudolf Virchow.The magnified shapes of cells in a piece of cork would be a better fit to illustrate Robert
Hooke’s work with cork cells while Antonie van Leeuwenhoek was the first to observe living cells under a microscope. The different cells that make up a plant leaf or the way root cells reproduce to
increase root length are more related to the plant cell theory and not human cell theory.
Virchow discovered how cells reproduce so this would be the most reasonable answer.
More Answers:
Leave a Comment | {"url":"https://www.worksheetsbuddy.com/tumus-class-was-given-an-assignment-to-feature-a-scientist-that-contributed/","timestamp":"2024-11-12T23:38:51Z","content_type":"text/html","content_length":"131523","record_id":"<urn:uuid:ae1f0093-243b-4bff-919e-873c28285564>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00026.warc.gz"} |
Methods and Applications of Singular Perturbations: Boundary Layers and Multiple Timescale Dynamics PDF
50 Texts in Applied Mathematics Editors J.E. Marsden L. Sirovich S.S. Antman Advisors G. Iooss P. Holmes D. Barkley M. Dellnitz P. Newton Texts in Applied Mathematics 1.
Sirovich:IntroductiontoAppliedMathematics. 2. Wiggins:IntroductiontoAppliedNonlinearDynamicalSystemsandChaos. 3. Hale/Ko¸cak:DynamicsandBifurcations. 4. Chorin/
Marsden:AMathematicalIntroductiontoFluidMechanics,3rded. 5. Hubbard/Weist:DifferentialEquations:ADynamicalSystemsApproach: OrdinaryDifferentialEquations. 6.
Sontag:MathematicalControlTheory:DeterministicFiniteDimensional Systems,2nded. 7. Perko:DifferentialEquationsandDynamicalSystems,3rded. 8. Seaborn:HypergeometricFunctionsandTheirApplications. 9.
Pipkin:ACourseonIntegralEquations. 10. Hoppensteadt/Peskin:ModelingandSimulationinMedicineandtheLife Sciences,2nded. 11. Braun:DifferentialEquationsandTheirApplications,4thed. 12. Stoer/
Bulirsch:IntroductiontoNumericalAnalysis,3rded. 13. Renardy/Rogers:AnIntroductiontoPartialDifferentialEquations. 14. Banks:GrowthandDiffusionPhenomena:MathematicalFrameworksand Applications. 15.
Brenner/Scott:TheMathematicalTheoryofFiniteElementMethods,2nded. 16. VandeVelde:ConcurrentScientificComputing. 17. Marsden/Ratiu:IntroductiontoMechanicsandSymmetry,2nded. 18. Hubbard/
West:DifferentialEquations:ADynamicalSystemsApproach: Higher-DimensionalSystems. 19. Kaplan/Glass:UnderstandingNonlinearDynamics. 20. Holmes:IntroductiontoPerturbationMethods. 21. Curtain/
Zwart:AnIntroductiontoInfinite-DimensionalLinearSystems Theory. 22. Thomas:NumericalPartialDifferentialEquations:FiniteDifferenceMethods. 23. Taylor:PartialDifferentialEquations:BasicTheory. 24.
Merkin:IntroductiontotheTheoryofStabilityofMotion. 25. Naber:Topology,Geometry,andGaugeFields:Foundations. 26. Polderman/Willems:IntroductiontoMathematicalSystemsTheory: ABehavioralApproach. 27.
Reddy:IntroductoryFunctionalAnalysiswithApplicationstoBoundary- ValueProblemsandFiniteElements. 28. Gustafson/Wilcox:AnalyticalandComputationalMethodsofAdvanced EngineeringMathematics. 29. Tveito/
Winther:IntroductiontoPartialDifferentialEquations: AComputationalApproach. 30. Gasquet/Witomski:FourierAnalysisandApplications:Filtering,Numerical Computation,Wavelets. (continuedafterindex)
Ferdinand Verhulst Methods and Applications of Singular Perturbations Boundary Layers and Multiple Timescale Dynamics With 26 Illustrations FerdinandVerhulst UniversityofUtrecht Utrecht3584CD
TheNetherlands SeriesEditors J.E.Marsden L.Sirovich ControlandDynamicalSystems,107–81 DivisionofAppliedMathematics CaliforniaInstituteofTechnology BrownUniversity Pasadena,CA91125 Providence,RI02912
USA USA marsden@cds.caltech.edu chico@camelot.mssm.edu S.S.Antman DepartmentofMathematics and InstituteforPhysicalScience andTechnology UniversityofMaryland CollegePark,MD20742-4015 USA
ssa@math.umd.edu MathematicsSubjectClassification(2000):12147 LibraryofCongressCataloging-in-PublicationData Verhulst,F.(Ferdinand),1939–
Methodsandapplicationsofsingularperturbations:boundarylayersandmultiple timescaledynamics/FerdinandVerhulst. p.cm. Includesbibliographicalreferencesandindex. ISBN0-387-22966-3
1.Boundaryvalueproblems—Numericalsolutions. 2.Singularperturbations (Mathematics) I.Title. QA379.V47 2005 515′.35—dc22 2005042479 ISBN-10:0-387-22966-3 Printedonacid-freepaper.
ISBN-13:978-0387-22966-9 ©2005SpringerScience+BusinessMedia,Inc. Allrightsreserved.Thisworkmaynotbetranslatedorcopiedinwholeorinpartwithoutthewritten permissionofthepublisher
(SpringerScience+BusinessMedia,Inc.,233SpringStreet,NewYork,NY 10013,USA),exceptforbriefexcerptsinconnectionwithreviewsorscholarlyanalysis.Useinconnection
withanyformofinformationstorageandretrieval,electronicadaptation,computersoftware,orbysimi- larordissimilarmethodologynowknownorhereafterdevelopedisforbidden.
Theuseinthispublicationoftradenames,trademarks,servicemarks,andsimilarterms,eveniftheyare notidentifiedassuch,isnottobetakenasanexpressionofopinionastowhetherornottheyaresubject toproprietaryrights.
PrintedintheUnitedStatesofAmerica. (EB) 9 8 7 6 5 4 3 2 1 SPIN11005988 springeronline.com Texts in Applied Mathematics (continuedfrompageii) 31.
Br´emaud:MarkovChains:GibbsFields,MonteCarloSimulation, andQueues. 32. Durran:NumericalMethodsforWaveEquationsinGeophysicalFluids Dynamics. 33.
Thomas:NumericalPartialDifferentialEquations:ConservationLawsand EllipticEquations. 34. Chicone:OrdinaryDifferentialEquationswithApplications. 35.
Kevorkian:PartialDifferentialEquations:AnalyticalSolutionTechniques, 2nded. 36. Dullerud/Paganini:ACourseinRobustControlTheory:AConvex Approach. 37. Quarteroni/Sacco/Saleri:NumericalMathematics. 38.
Gallier:GeometricMethodsandApplications:ForComputerScienceand Engineering. 39. Atkinson/Han:TheoreticalNumericalAnalysis:AFunctionalAnalysis Framework,2nded. 40. Brauer/
Castillo-Cha´vez:MathematicalModelsinPopulationBiologyand Epidemiology. 41. Davies:IntegralTransformsandTheirApplications,3rded. 42. Deuflhard/Bornemann:ScientificComputingwithOrdinaryDifferential
Equations. 43. Deuflhard/Hohmann:NumericalAnalysisinModernScientificComputing: AnIntroduction,2nded. 44. Knabner/Angermann:NumericalMethodsforEllipticandParabolicPartial DifferentialEquations. 45.
Larsson/Thom´ee:PartialDifferentialEquationswithNumericalMethods. 46. Pedregal:IntroductiontoOptimization. 47. Ockendon/Ockendon:WavesandCompressibleFlow. 48. Hinrichsen:MathematicalSystemsTheoryI.
49. Bullo/Lewis:GeometricControlofMechanicalSystems;Modeling,Analysis, andDesignforSimpleMechanicalControlSystems. 50. Verhulst:MethodsandApplicationsofSingularPerturbations:Boundary
LayersandMultipleTimescaleDynamics. Series Preface Mathematicsisplayinganevermoreimportantroleinthephysicalandbiolog- ical sciences, provoking a blurring of boundaries between scientific disciplines
and a resurgence of interest in the modern as well as the classical techniques of applied mathematics. This renewal of interest, both in research and teach- ing, has led to the establishment of the
series Texts in Applied Mathematics (TAM). The development of new courses is a natural consequence of a high level of excitement on the research frontier as newer techniques, such as numerical
andsymboliccomputersystems,dynamicalsystems,andchaos,mixwithand reinforce the traditional methods of applied mathematics. Thus, the purpose
ofthistextbookseriesistomeetthecurrentandfutureneedsoftheseadvances and to encourage the teaching of new courses. TAM will publish textbooks suitable for use in advanced undergraduate and beginning
graduate courses, and will complement the Applied Mathe- matical Sciences (AMS) series, which will focus on advanced textbooks and research-level monographs. Pasadena, California J.E. Marsden
Providence, Rhode Island L. Sirovich Houston, Texas M. Golubitsky College Park, Maryland S.S. Antman Preface Mathematics is more an activity than a theory (Mathematik is mehr ein Tun als eine Lehre)
Hermann Weyl, after L.E.J. Brouwer Perturbation theory is a fundamental topic in mathematics and its appli- cations to the natural and engineering sciences. The obvious reason is that hardly any
problem can be solved exactly and that the best we can hope for is the solution of a “neighbouring” problem. The original problem is then a perturbation of the solvable problem, and what we would
like is to establish the relation between the solvable and the perturbation problems. Whatisasingularperturbation?Thetraditionalideaisadifferentialequa- tion (plus other conditions) having a small
parameter that is multiplying the highest derivatives. This covers a lot of cases but certainly not everything. It refers to boundary layer problems only.
Themodernviewistoconsideraproblemwithasmallparameterεandso- lution x(t,ε). Also defined is an “unperturbed” (neighbouring) problem with solution x(t,0). If, in an appropriate norm, the difference
(cid:2)x(t,ε)−x(t,0)(cid:2) does not tend to zero when ε tends to zero, this is called a singular per- turbation problem. The problems in Chapters 1–9 are covered by both the old(fashioned)
definitionandthenewone.Slow-timeproblems(multipletime dynamics),aswillbediscussedinlaterchapters,fallunderthenewdefinition. Actually, most perturbation problems in this book are singular by this defi-
nition; only in Chapter 10 shall we consider problems where “simple” contin- uation makes sense. This book starts each chapter with studying explicit examples and intro- ducing methods without proof.
After many years of teaching the subject of singular perturbations, I have found that this is the best way to introduce thisparticularsubject.Ittendstobesotechnical,bothincalculationsandin theory,
that knowledge of basic examples is a must for the student. This view VIII Preface isnotonlyconfirmedbymylecturesinUtrechtandelsewherebutalsobylec- turers who used parts of my text in various places.
In this respect, Hermann Weyl’s quotation which is concerned with the fundamentals of mathematics, gives us the right perspective. I have stressed that the proposed workbook format is very suitable
for singular perturbation problems, but I hope that the added flavour of precise estimates and excursions into the theoretical background makes the book of interest both for people working in the
applied sciences and for more theoret- ically oriented mathematicians. Let me mention one more important subject of the forthcoming chapters. There will be an extensive discussion of timescales and a
priori knowledge of thepresenceofcertaintimescales.Thisisoneofthemostwidelyusedconcepts inslow-timedynamics,andthereisalotofconfusionintheliterature.Ihope to have settled some of the questions
arising in choosing timescales. Whatabouttheory and proofsonemay ask. Tolimitthesizeof thebook, thosemathematicalproofsthatareeasytoobtainfromtheliteraturearelisted at the end of each chapter in a
section “Guide to the Literature”. If they are readily accessible, it usually makes no sense to reproduce them. Exceptions are sometimes cases where the proof contains actual constructions or where a
line of reasoning is so prominent that it has to be included. In all cases discussed in this book - except in Chapter 14 - proofs of asymptotic validity are available. Under “Guide to the Literature”
one also finds other relevant and recent references. In a final chapter I collected pieces of theory that are difficult to find in the literature or a summary such as the one on perturbations of matrices
or a typical and important type of proof such as the application of maximum principlesforellipticequations.Also,intheepilogueIreturntothediscussion about “proving and doing”. To give a general
introduction to singular perturbations, I have tried to cover as many topics as possible, but of course there are subjects omitted. The first seven chapters contain standard topics from ordinary
differential equations and partial differential equations, boundary value problems and problems with initial values within a mathematical framework that is more rigorous in formulation than is usual in
perturbation theory. This improves the connection with theory-proof approaches. Also, we use important, but nearly forgotten theorems such as the du Bois-Reymond theorem. Some topics are missing
(such as the homogenisation method) or get a sketchy treatment (such as the WKBJ method). I did not include relaxation oscillations, as an elementary treatment can be found in my book Nonlinear
Differential Equations and Dynamical Systems.Alsotherearebooksavailable on this topic, such as Asymptotic Methods for Relaxation Oscillations and Applications by Johan Grasman. Perturbation theory is
a fascinating topic, not only because of its appli- cations but also because of its many unexpected results. A long time ago, Wiktor Eckhaus taught me the basics of singular perturbation theory, and
at Preface IX aboutthesametimeBobO’MalleyintroducedmetoTikhonov’stheoremand multiple scales. Many colleagues and students made remarks and gave suggestions. I mention Abadi, Taoufik Bakri, Arjen
Doelman, Hans Duistermaat, Wiktor Eckhaus, Johan Grasman, Richard Haberman, Michiel Hochstenbach, James Murdock, Bob O’Malley, Richard Rand, Bob Rink, Thijs Ruijgrok, Theo Tuwankotta, Adriaan van der
Burgh. I got most of section 15.5 from Van Harten’s (1975) thesis, section 15.9 is based on Buitelaar’s (1993) thesis. The figures in the first nine chapters were produced by Theo Tuwankotta; other
figures were obtained from Abadi, Taoufik Bakri and Hartono. Copyed- itor Hal Henglein of Springer proposed the addition of thousands of comma’s and many layout improvements. I am grateful to all of
them. Corrections and additions will be posted on http://www.math.uu.nl/people/verhulst Ferdinand Verhulst, University of Utrecht Contents Preface
........................................................VII 1 Introduction............................................... 1 2 Basic Material............................................. 9 2.1
Estimates and Order Symbols. ............................ 9 2.2 Asymptotic Sequences and Series .......................... 12 2.3 Asymptotic Expansions with more Variables ................ 13 2.4
Discussion .............................................. 16 2.4.1 The Question of Convergence........................ 17 2.4.2 Practical Aspects of the du Bois-Reymond Theorem.... 17 2.5 The
Boundary of a Laser-Sustained Plasma................. 18 2.6 Guide to the Literature .................................. 20 2.7 Exercises ............................................... 21 3
Approximation of Integrals ................................ 25 3.1 Partial Integration and the Laplace Integral................. 25 3.2 Expansion of the Fourier Integral. ......................... 27
3.3 The Method of Stationary Phase .......................... 27 3.4 Exercises ............................................... 28 4 Boundary Layer Behaviour ................................ 31 4.1
Regular Expansions and Boundary Layers .................. 31 4.1.1 The Concept of a Boundary Layer................... 33 4.2 A Two-Point Boundary Value Problem..................... 35 4.3 Limits of
Equations and Operators ........................ 38 4.4 Guide to the Literature .................................. 41 4.5 Exercises ............................................... 42
See more | {"url":"https://www.zlibrary.to/dl/methods-and-applications-of-singular-perturbations-boundary-layers-and-multiple-timescale-dynamics-0","timestamp":"2024-11-15T00:36:01Z","content_type":"text/html","content_length":"127114","record_id":"<urn:uuid:799f440c-9439-479e-a258-476a40632316>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00393.warc.gz"} |
Top 17 Sensational 17 Stats by Hampshire Mathematics Professor David Kelly
Comment: 0
1. The ancient Greeks knew how to construct squares, equilateral triangles, and regular pentagons with straightedges and compasses; in 1796 Carl Friedrich Gauss proved that the euclidean tools
sufficed also for the construction of a regular 17-sided polygon.2. Any sequence of fewer than 17 consecutive positive integers containsat least one number which has no divisor in common with any of
the other numbers; the 17-term sequence {2184, 2185, …, 2200} contains no such number.3. You can color all the (136) edges joining pairs of 16 points with 3 colors without having 3 of the edges
forming a triangle be all the same color; with 17 (or more) points, there must be a monochromatic triangle.4. A cube can be cut along 7 edges and unfolded to make a cross; a tesseract or
4-dimensional hypercube can be cut along 17 2-dimsional faces and unfolded into a three-dimensional cross.
5. There are 17 muscles in a horse’s ear.
6. Roman historian Plutarch records “The Pythagoreans also have a horror for the number 17, for 17 lies exactly halfway between 16, which is a square, and the number 18, which is the double of a
square, these two, 16 and 18, being the only two numbers representing areas for which the perimeter equals the area.”
7. There are 17 columns on the long side of the Parthenon in Greece.
8. There are 17 mathematically distinct wallpaper patterns. Tilings in the Alhambra, a Moorish castle in Spain, and M.C.Escher’s tessellations exhibit all the 17 different combinations of
translations, rotations, and reflections.
9. Horses have recently been reported to distinguish among 17 facial expressions.
10. The 17th century was a great century for Japanese Haiku which has 17 syllables (comfortably spoken with a single breath).
11. 17 is the smallest number of clues that can be provided for a 9×9 Sudoku puzzle to have a unique solution.
12. There are 17 ways to write 17 as a sum of primes.
13. 1/17 is the 1st reciprocal of a positive integer whose periodic decimal expansion contains all 10 digits. (1/7 = 0.142857 142857 … ; 1/11 = .090909…; 1/17 = 0.0588235294117647 0588235294117647…)
14. To calculate a 17% tip, divide the bill by 6 (and round up by 0.33333…%).
15. Morris the Cat died at age 17, the average lifespan of a goldfish.
16. The 17-year locust.
17. There are almost 17 ounces in a pound.
– See more here.
Leave a Reply Cancel reply | {"url":"https://sites.hampshire.edu/educatingforchange/top-17-sensational-17-stats-by-hampshire-mathematics-professor-david-kelly/","timestamp":"2024-11-12T17:06:31Z","content_type":"text/html","content_length":"48781","record_id":"<urn:uuid:185546f5-f560-4aa7-9add-0a6f4ee84f54>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00130.warc.gz"} |
1 year cmt rate
Constant Maturity Treasury (CMT) rates are the interpolated yields based on the yields of the recently auctioned treasury bills, notes, and bonds. For example, 1 Year CMT rate is the yield on
treasury securities having a 1 year term. CMT rates are also known as the Treasury Yield Curve rates. View a 1-year yield estimated from the average yields of a variety of Treasury securities with
different maturities derived from the Treasury yield curve. 1-Year Treasury Constant Maturity Rate Skip to main content
Treasury discontinued the 20-year constant maturity series at the end of calendar year 1986 and reinstated that series on October 1, 1993. The 20-year constant maturity rate for the time period from
January 2, 1990 through September 30, 1993 is the arithmetic average of the 10-year and 30-year constant maturity rates. The CMT yield values are read from the yield curve at fixed maturities,
currently 1, 2, 3 and 6 months and 1, 2, 3, 5, 7, 10, 20, and 30 years. This method provides a yield for a 10 year maturity, for example, even if no outstanding security has exactly 10 years
remaining to maturity. Treasury Securities ("T-Secs", also known as TCM, or CMT, or CMT, or T-Sec) values are calculated by the Treasury Department and reported by the Federal Reserve in Publication
H.15.On this page, you will find current and historical weekly yields for 3 month, 6 month Treasuries, as well as values for 1-, 2-, 3-, 5-, 7-, 10-, 20-, and 30 year treasuries. The 1 year treasury
yield is included on the shorter end of the yield curve and is important when looking at the overall US economy. Historically, the 1 year treasury yield reached upwards of 17.31% in 1981 and nearly
reached 0 in the 2010s after the Great Recession. 1 Year Treasury Rate is at 1.59%,
London Interbank Offer Rate (LIBOR); FHFA Monthly Interest Rate Survey (MIRS) ; 12-month Treasury Average Index (MTA); Constant Maturity Treasury (CMT)
One-Year Constant Maturity Treasury - 1-Year CMT: The interpolated one-year yield of the most recently auctioned four-, 13- and 26-week U.S. Treasury bills , plus the most recently auctioned 2-, 3
Rate Comparisons of Adjustable Rate Loan Indexes Comparison Charts: All on One Page 1 Year LIBOR-Prime Rate-CMT | LIBOR-1 Year-6 Month-3 Month-1 Month 1 Year LIBOR-12MTA-CMT | Fed Funds-Prime Rate
COFI-CMT-1 Year LIBOR | COFI-COSI-CODI. Reasonable efforts are made to maintain accurate information. The CMT yield values are read from the yield curve at fixed maturities, currently 1, 2, 3 and 6
months and 1, 2, 3, 5, 7, 10, 20, and 30 years. This method provides a yield for a 10 year maturity, for example, even if no outstanding security has exactly 10 years remaining to maturity. 1-Year
Constant Maturity Treasury index (1 Yr CMT) This is the most widely used index. Roughly half of all ARMs are based on this index. It's used on ARMs with annual rate adjustments. It is also referred
to as the 1-Year Treasury Bill (1Yr T-Bill) [see note], the 1-Year Treasury Security (1Yr T-Sec), or the 1-Year Treasury Spot index.
The CMT yield values are read from the yield curve at fixed maturities, currently 1, 2, 3 and 6 months and 1, 2, 3, 5, 7, 10, 20, and 30 years. This method provides a yield for a 10 year maturity,
for example, even if no outstanding security has exactly 10 years remaining to maturity.
One-Year Constant Maturity Treasury - 1-Year CMT: The interpolated one-year yield of the most recently auctioned four-, 13- and 26-week U.S. Treasury bills , plus the most recently auctioned 2-, 3
Rate Comparisons of Adjustable Rate Loan Indexes Comparison Charts: All on One Page 1 Year LIBOR-Prime Rate-CMT | LIBOR-1 Year-6 Month-3 Month-1 Month 1 Year LIBOR-12MTA-CMT | Fed Funds-Prime Rate
COFI-CMT-1 Year LIBOR | COFI-COSI-CODI. Reasonable efforts are made to maintain accurate information. The CMT yield values are read from the yield curve at fixed maturities, currently 1, 2, 3 and 6
months and 1, 2, 3, 5, 7, 10, 20, and 30 years. This method provides a yield for a 10 year maturity, for example, even if no outstanding security has exactly 10 years remaining to maturity. 1-Year
Constant Maturity Treasury index (1 Yr CMT) This is the most widely used index. Roughly half of all ARMs are based on this index. It's used on ARMs with annual rate adjustments. It is also referred
to as the 1-Year Treasury Bill (1Yr T-Bill) [see note], the 1-Year Treasury Security (1Yr T-Sec), or the 1-Year Treasury Spot index. View a 1-year yield estimated from the average yields of a variety
of Treasury securities with different maturities derived from the Treasury yield curve. 1-Year Treasury Constant Maturity Rate Skip to main content 1 Year Treasury Rate - 54 Year Historical Chart.
Interactive chart showing the daily 1 year treasury yield back to 1962. The values shown are daily data published by the Federal Reserve Board based on the average yield of a range of Treasury
securities, all adjusted to the equivalent of a one-year maturity. The current 1 year treasury yield as of October 17, 2019 is 1.59%.
The 1 year treasury yield is included on the shorter end of the yield curve and is important when looking at the overall US economy. Historically, the 1 year treasury yield reached upwards of 17.31%
in 1981 and nearly reached 0 in the 2010s after the Great Recession. 1 Year Treasury Rate is at 1.59%,
Among the most common indices are the rates on 1-year constant-maturity Treasury (CMT) securities, the cost of funds index (COFI), and the London Interbank Bankrate.com provides today's current 1
year CMT treasury note constant maturity rate and index rates. As a result, there are no 20-year rates available for the time period January 1, 1987 Negative Yields and Nominal Constant Maturity
Treasury Series Rates View a 1-year yield estimated from the average yields of a variety of Treasury securities with different maturities derived from the Treasury yield curve. 21 Feb 2020 The
monthly one-year CMT value is a popular mortgage index to which many adjustable-rate mortgages (ARMs) are tied. CMTs and Mortgage
most common indexes are the rates on 1-year constant-maturity. Treasury (CMT) securities, the Cost of Funds Index (COFI), and the London Interbank Offered
is the spot or zero-coupon yield on a bond with t years to maturity. Dt ≡ 1/(1 + rst) t. = the corresponding discount factor. In 4.1, rs1 is the current one-year spot London Interbank Offer Rate
(LIBOR); FHFA Monthly Interest Rate Survey (MIRS) ; 12-month Treasury Average Index (MTA); Constant Maturity Treasury (CMT)
The following CMT indexes are the most often used for ARMs: 1-Year Constant Maturity Treasury index (1 Yr CMT) This is the most widely used index. Roughly 6 Mar 2020 Rate for conventional loans and
the Constant Maturity Treasury for A 5/1 ARM has a fixed rate of interest for the first 5 years of the loan. 1 Year CMT. Enter Margin for 1 Year CMT %. 5 Year CMT. Enter Margin for 5 Year CMT %. 10
Year CMT. Enter Margin for 10 Year CMT %. 12 MAT. Product, Term, Mortgage Rate, APR1 as low as Print Mortgages day every 12th month thereafter based on the one-year Constant Maturity Treasury (CMT).
Current and historical US treasury yields, swap rates, LIBOR, SOFR, SIFMA, Fed Funds, Prime, and other interest rate risk benchmarks for Ago, 1 Year Ago is the spot or zero-coupon yield on a bond
with t years to maturity. Dt ≡ 1/(1 + rst) t. = the corresponding discount factor. In 4.1, rs1 is the current one-year spot | {"url":"https://optionsephidv.netlify.app/worthy28656riji/1-year-cmt-rate-80.html","timestamp":"2024-11-05T06:39:53Z","content_type":"text/html","content_length":"36932","record_id":"<urn:uuid:8bab6dee-4446-4001-b667-5b7f78b4940c>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00215.warc.gz"} |
How to Lead a More Rational Life with Bayes' Theorem
Meet Steve. He's a shy, withdrawn American who's helpful when asked, but otherwise not interested in people. Steve likes detail, order, and putting things in the right places. Is he a farmer or a
As the psychologists Amos Tversky and Daniel Kahneman showed in their paper Judgment under Uncertainty, most people guess that Steve is a librarian. But he's probably not. He's almost certainly a
farmer. Sounds counter-intuitive? It's not. It's Bayes' theorem at work.
This article will explain what Bayes' theorem is (including why Steve is a farmer) and how you can use it to lead a more rational life where you make smarter decisions and fewer mistakes.
What is Bayes' Theorem?
Bayes' theorem is a formula for calculating conditional probabilities. In plain words, it's a way to figure out how likely something is under certain conditions. Let's explain the theorem with an
intuitive example first and a mathematical example next.
Example 1: The Problem with Medical Tests
Bayes' theorem saves lives every day
You're at the doctor's office for a routine checkup. She tests you for a rare disease that occurs in 0.1% of the population. The disease has debilitating effects and is impossible to cure.
You test positive.
The test correctly identifies 99% of the people who have the disease. It incorrectly identifies 1% of the people as having the disease even though they don't—the so-called false positives.
If the test is 99% accurate, does this mean there's a 99% chance you have the disease when you test positive? It doesn't. The chances that you have the disease are much lower. To understand why,
let's flip around the scenario.
Consider 1,000 randomly sampled people in a big room. One person in the room will have the disease, because its prevalence is 0.1%. However, because the test has a false positive rate of 1%, ten
people will test positive even though they don't have the disease.
So there will be eleven people who test positive, but only one will have the disease. That's why, once you've tested positive, your chances of having the disease aren't 99%. They're one in eleven, or
Example 2: Let's Talk About Steve
Most people think Steve is a librarian because of his description. He's shy, withdrawn, likes order, etc. All characteristics that seem to fit a librarian. We place so much emphasis on his
description that we forget to incorporate information about the total number of farmers versus librarians.
According to the American Library Association, there are 166,194 librarians in the United States today. Meanwhile, according to the US Department of Agriculture, there are at least 2.6 million
farmers. So the ratio is approximately one librarian for every fifteen farmers.
For the sake of explaining Bayes' theorem, let's assume that Steve's description fits around 70% of all librarians and 30% of all farmers. We now have enough information to dig into Bayes' formula.
We're looking for the Probability that Steve is a Librarian given the Description. In the formula, we write this down as P(L|D). All the other parts of the formula are:
• P(L) = the probability that Steve is a librarian, 1/15 or 0.067.
• P(F) = the probability that Steve is a farmer, 0.933.
• P(D|L) = the probability that the description fits a librarian, 0.7.
• P(D|F) = the probability that the description fits a farmer, 0.3.
Let's now build Bayes' formula. For the numerator, we need to multiply the probability that the description fits a librarian, P(D|L), with the probability that Steve is a librarian before we'd read
his description, P(L).
Next, we need to divide the numerator with the probability that Steve is a librarian, P(L), times the probability that the description fits a librarian, P(D|L), plus the probability that Steve is a
farmer, P(F), times the probability that the description fits a farmer, P(D|F).
Time to plug in the numbers:
P(L|D) = (0.7 x 0.067) / (0.067 x 0.7 + 0.933 x 0.3)
P(L|D) = 0.0469 / (0.0469 + 0.2799)
P(L|D) = 0.0469 / 0.3268
P(L|D) = 0.14
There you have it. The chances that Steve is a librarian are only 14%.
This is what makes Bayes' theorem so powerful. It allows you to quantify probabilities, which is why it's heavily used in medicine, statistics, machine learning, risk analysis, and other math-heavy
fields full of probabilities.
But it's also a powerful tool to think more rationally as an individual.
How to Apply Bayesian Thinking to Your Life
The point of this article isn't for you to memorize Bayes' formula and calculate probabilities wherever you go. That's entirely impractical. The power of Bayes' theorem for the individual lies in the
three implicit lessons that come with it. Let's go over each one.
Be Careful with New Evidence
Had you been asked if Steve were a librarian or a farmer before you'd read his description, what would you have said? You wouldn't have known anything about Steve, so you might have said he could be
either of the two jobs. You might even have considered the ratio between farmers and librarians, and said he was probably a farmer.
But when you're given a small piece of evidence, Steve immediately becomes a librarian. All of us are guilty of placing far too much emphasis on new evidence, to the point where we sometimes
flip-flop between opinions because of evidence and counter-evidence. For example:
• UFOs didn't exist until the Pentagon UFO report and now it's beyond a shadow of a doubt that UFOs exist.
• Nuclear power was a great way to generate clean energy until the Fukushima Daiichi accident and now we must shut down all nuclear plants.
This isn't rational thinking, because it's too volatile, too fast. Bayes' theorem is a reminder to update our beliefs incrementally, according to the strength of the evidence. If the evidence isn't
empirically convincing and consistently replicable, don't take it at face value. Instead, evaluate carefully and move your beliefs appropriately.
Accept Other Opinions
Bayes' theorem explains why two people might see the same evidence and come up with entirely different conclusions. They simply have different priors. A prior, or prior probability, is what you
believed before you encountered new evidence. In our example with Steve, that was the probability that he was a librarian before we'd read his description, P(L).
When you're talking about beliefs, priors aren't easily quantifiable. You have to look at them relatively. For example, if you grew up in a neighborhood full of crime, your belief that crime is part
of life might be 75% when you compare it to the person who grew up in a neighborhood without any crime, for who it might be 5%.
If you both see a crime, your beliefs that crime is a part of life should increase incrementally, but for you it might move to 80% while for the other it might move to only 10%.
Bayes' theorem makes it easier to understand why people see the world in different ways. It encourages us to empathize with others. They might be thinking rationally; they're just coming at it from
different backgrounds.
Prove Yourself Wrong
Bayes' theorem is a tool for thinking rationally, but only if you're willing to search for evidence that proves you wrong. Do not ignore such evidence and do not disregard it without proper
diligence. Instead, seek it out.
If you don't, you'll end up in an alternate reality where your beliefs are constantly proven right despite possibly overwhelming evidence to the contrary. It's the world of conspiracy theories, home
to those who are willingly blind to the truth:
• Climate change deniers;
• Anti-vaxxers;
• Flat-Earthers;
• Holocaust deniers.
Finding evidence that proves you wrong is harder than it seems, in large part because the Internet is set up in such a way it often encourages what you already believe without showing you the flip
side of the coin.
But it's exactly because it's hard that it's important. You have to search for the evidence you don't like, so your beliefs can inch ever closer to the truth.
In Conclusion
This article has explained that Bayes' theorem is a mathematical way to calculate conditional probabilities. It did so with an intuitive example and a mathematical example.
Next, it spoke about the three ways you can apply Bayesian thinking to your life. Use Bayes' theorem to evaluate new evidence, accept other opinions, and seek evidence that proves you wrong. Do this
consistently, and you'll be able to live a life of rational integrity.
Top comments (0)
For further actions, you may consider blocking this person and/or reporting abuse | {"url":"https://practicaldev-herokuapp-com.global.ssl.fastly.net/x-team/how-to-lead-a-more-rational-life-with-bayes-theorem-2d27","timestamp":"2024-11-02T11:41:28Z","content_type":"text/html","content_length":"81987","record_id":"<urn:uuid:bfe67090-990a-48a8-bc22-ecc0ad983b47>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00751.warc.gz"} |
WEDNESDAY, August 5, 2009 — Gary Steinmehl
Theme: More fun than a barrel of monkeys
— The last word of each theme answer comes after the word
in a familiar phrase.
Theme answers:
• 20A: Stage, screen, etc. (SHOW BUSINESS).
• 25A: They have latte charges (COFFEE BARS).
• 38A: Action in court (LAWSUIT).
• 47A: Oxford brighteners (SHOESHINES).
• 53A: Spend time idly (and a hint to what can precede the last word of 20-, 25-, 38- or 47-Across) (MONKEY AROUND).
Crosswordese 101:
There are three famous women named
that you need to know.
63D: Peeples of "Fame,"
who appears in today's puzzle, is known for both acting and singing. Actress NIA Long is typically clued with a reference to "Third Watch" or "Boyz N the Hood." Finally, NIA Vardalos rocketed into
the fourth dimension of Eternal Crossword Fame when she starred in the 2002 sleeper hit "My Big Fat Greek Wedding."
I know you were expecting Orange today but she's off gallivanting around our nation's capital with her family for a few days, so you're stuck with me. I'm also covering
her other blog
if you're interested in my delightful musings on the day's other puzzles.
Got through this one pretty quickly. Funky looking grid! Only real hang-up was right in the middle and I'm going to guess a lot of you were hung up too. Of course I'm talking about the cross of CAVA
44A: Vena __
) and ANSCO (
29D: Inexpensive former camera brand
). My gut reaction is that this crossing is unfair, possibly even a
I'd like to hear what you all think before I call it for sure.
Quickly now:
• 16A: Shire of "Rocky" (TALIA). Adriaaaaaaan!
• 22A: __ Miguel: Azores island (SAO). Popped in San without thinking about it. That one gets me just about every time.
• 30A: It follows that (ERGO). 10-year-old PuzzleSon used this word the other day. Pretty sure he doesn't have any idea what it means.
• 37A: Tony winner Hagen (UTA). I talked a little about Uta back in this blog's infancy. Good times.
• 66A: New parent's lack? (SLEEP). Not sure what the question mark is about here. I'm totally sure that the answer is right though. Advice to women pregnant for the first time: After your baby is
born you will be extremely sleep deprived. This will cause you to feel as if you're going completely insane. That's because sleep deprivation causes insanity. Which is why it's used as torture.
• 67A: "The Company" (CIA). This is how dumb I am. I'm going, "IBM?"
• 1D: Skipper, to Barbie (SIS). I was not allowed to have Barbie dolls when I was a kid. *cue violin*
• 54D: Lionel layout, maybe (OVAL). Lionel Corporation, previously, and Lionel LLC, currently, produce toy trains and model railroads.
• 55D: 15th century caravel (NINA). I do not know what this means. Oh wait, is a caravel a boat? Okay, I get it now.
Follow PuzzleGirl
on Twitter.]
Everything Else — 1A: Wire wearer (SPY); 4A: Low men (BASSI); 9A: Highly capable (ADEPT); 14A: Witness stand oath (I DO); 15A: __ Park, Colorado (ESTES); 17A: Occupants of abandoned buildings, e.g.
(SQUATTERS); 19A: Measuring aid (RULER); 24A: Q.E.D. part (ERAT); 34A: Sock-in-the-gut grunt (OOF); 35A: Scientology's __ Hubbard (L. RON); 36A: Big name in Arizona political history (UDALL); 41A:
Kilmer of "The Saint" (VAL); 42A: Snacker's bagful (CHIPS); 45A: Nest egg initials (IRA); 46A: War honoree (HERO); 50A: Bake sale item (CAKE); 52A: Homer Simpson's neighbor (NED); 60A: Prestigious
university octet (IVIES); 61A: Using a DVR, say (REPLAYING); 65A: __ Domingo (SANTO); 68A: Purse closer (CLASP); 69A: Color qualities (TONES); 70A: Big fat mouth (YAP); 2D: Brief "At once!" (PDQ);
3D: Uncle Sam poster word (YOU); 4D: LPGA Hall of Famer Daniel (BETH); 5D: Regarding (AS TO); 6D: Ladled dish (STEW); 7D: Balkan native (SERB); 8D: Bank named on a credit card (ISSUER); 9D: Skylit
areas (ATRIA); 10D: Discouraged (DAUNTED); 11D: Model Macpherson (ELLE); 12D: Slapstick ammo (PIES); 13D: Blackens, in a way (TARS); 18D: Beginning on (AS OF); 21D: Yearbook sect. (SRS); 22D: Bring
relief to (SOOTHE); 23D: Catered event (AFFAIR); 25D: Potato's place? (COUCH); 26D: Plumbing joints (ELLS); 27D: Procter & Gamble detergent (ERA); 28D: Respond to an ovation (BOW); 31D: Formation
from stream erosion (RAVINE); 32D: Stared angrily (GLARED); 33D: Southwestern crocks (OLLAS); 36D: Six-sided state (UTAH); 39D: Dubaiís federation: Abbr. (UAE); 40D: ICU drips (IVS); 43D: Cargo pants
features (POCKETS); 47D: Limit, in a saying (SKY); 48D: Publisher who was the inspiration for "Citizen Kane" (HEARST); 49D: Race since 1911, informally (INDY); 51D: "The Lion and the Mouse" fabulist
(AESOP); 53D: Junk drawer label (MISC); 56D: Move, in Realtor-speak (RELO); 57D: Receptive (OPEN); 58D: Beekeeper in a 1997 movie (ULEE); 59D: Is off guard (NAPS); 62D: Hardly friendly (ICY); 64D:
Dental problem (GAP).
29 comments:
Being on Paris time means I get to leave the first comment, yay! Thanks for the write-up!
I definitely agree with the Natick comment. Had no clue whatsoever about those two, haven't even heard of them.
Did the same thing with San vs. SAO.
On the subject of the theme...I don't believe I've ever heard the expression "monkey shines"--what's that about?
Wow, early writeup today. I had to smile at 66A: New parent's lack? (SLEEP), because the only reason I'm posting at 1am is that I'm waiting to do our 5-week-old's next feeding. My wife's pretty
fanatical about breastfeeding, but if I don't get her a stretch of sleep longer than 2 or 3 hours once in a while (i.e., long enough to get through a full sleep cycle), that's when the insanity
sets in.
I can't give an unbiased opinion on 44A: Vena ___ because I'm an MD, but it sure seems like it shows up in crosswords a lot, along with its anatomical neighbor, the AORTA. Didn't know ANSCO, but
I liked the CAVA/IVS cross.
@Soozy: Monkeyshines are horseplay, shenanigans, etc. Basically what my 3-year-old twin boys are up to most of the time (add them to the newborn... yeesh!)
@KJ Thanks for the monkeyshines explanation! Good luck with the lil' ones. :)
Never heard of ANSCO, but vena CAVA was familiar to me from various sources, including high school biology and a (brief) stint as a paramedic. Can't say I see the word too often in puzzles,
Vena CAVA was no problem as I have a medical background, but I've never heard of an ANSCO camera. I got all of ANSCO through crosses and didn't even read that clue (oops).
I had SOUP before STEW for the ladled dish. My mom reminisces about vacationing in ESTES Park, CO as a child -- haven't been there myself. I got NINA from the crosses but looked up caravel
afterward to understand the clue/answer.
COUCH for potato's place and PIE for slapstick ammo both gave me a chuckle. Had a good time MONKEYingAROUND with this puzzle -- thanks, Gary.
ANSCO/VERA is a horrible crossing.
Sorry, not a Natick for me, and I do not have a medical background. I did not "know" CAVA, but it was certainly at the "confident recall" level once I got some of its fill. I assume I have seen
it puzzles before, but perhaps I gleaned it from some other source along the way.
SAO is the one that got me.
Like many others, I didn't know ANSCO but quickly filled in around it, so the center was no problem. Knew vena CAVA from long ago college anatomy --veins carrying blood from the heart
(superior-to the upper body; inferior-to the lower body). Saw the great actress UTA Hagen on Broadway in 1962 in Edward Albee's "Who's Afraid of Virginia Woolfe?", in the role played
magnificently by Elizabeth Taylor in the movie. No trouble getting the theme so the rest was easy. This was fun!
Had no trouble with Vena Cava which is the only reason I got the Ansco camera.
Didn't care for both AS TO and AS OF in the same NW corner.
Had some problems with the whole top of the puzzle - never heard of Talia - got it from the crosses; but, any day I finally get the whole puzzle with NO Googles is a good day for me!
I like your mother! I didn't give my daughters Barbies either. Nothing like making little girls start to think like fashionistas and think their bodies should look like hers as they are growing
up! I think her body is anatomically impossible!
@anonymous 7:04 - I think you meant to say the vena cava veins carry blood TO the heart, not FROM the heart.
@BEV - Right! Just typed it in wrong. Thanks!
This constructor (of course in conjunction with the fine ed. staff at LAT) has a knack for quality, entertaining cluing.
Minor tsk-tsk for the non-plural LAWSUIT, which of course can;t be centered with another letter. This is made up for by Elle Macpherson reference ...
Hey PG, what is a Natick?
I too found the CAVA/ANSCO part troubling; I also monkeyed around too long w the top center..BASSI? ESTES? ASTP and ASOF so close by as criticized by another poster...also wary of DAUNTED in this
usage...Alas, I know none of these NIAs
Enjoyed the theme; enjoyed UDALL, RAVINE, SQUATTERS
But on the whole, on finishing, I did not have feeling of humorous and satisfied accomplishment. Puzzle left me flat sorry to say
Any puzzle with monkeys as the theme is good with me. The vena CAVA fill has been seen a bunch of times maybe Crosswordese 301? I also put SAN Miguel, I've seen that one before and should
remember it.
As much fun as a barrel of monkeys.
ANSCO is the kind of name that was around in the 50s -- combine part of the word with "co" for company, but I knew all the crosses. Never really heard of ANSCO.
I have four daughters -- the older two were not allowed Barbies, but the first time they went to a yard sale with "their own $" they bought a couple of very used ones. They had so much very
creative fun with those darned dolls.
#3 daughter was never interested in dolls at all! #4? Who can remember?
CAVA posed no problems; ANSCO seems obscure. but overall this puzzle was easier than yesterday's for me. in fact, this seems to be my fastest-ever LAT wednesday.
here's my SÃO miguel hint: azores =
portugal. SAN is spanish.
Fun puzzle. Knew CAVA, and there's a faint glimmer of remembering ANSCO cameras. Crosses confirmed.
"Monkey Business" is, of course, a favorite movie of mine!
Liked "Potato's place" very much.
There's a blast from the past. Ansco cameras, so no natick for me. Is natick supposed to be capitalized? golfballman
Took some anatomy courses in college (thought I wanted to be a doctor!) and am now a photographer, so VENA CAVA and ANSCO were gimmes for me. Interesting article on Ansco here: http://
I see that Ansco was headquartered in Binghamton, NY, Rex's hometown!
Ok, so I had a good joke about PERRYWHITE being too cheap to buy Jimmy Olsen a real camera, so ANSCO was a tie in to the theme, not anachronistic, etc. Then I realized I was conflating my
puzzles, NY & LA Times. Made it much less funny.
@golfballman - Natick (Mass.), should indeed be capitalized when refering to the town. When used refering to evil crossword crossings, I think the small n is okay IMOO.
I was born in Binghamton and my godfather actually worked for Ansco, so no problem there.
Nice write-up, as always - including the funky grid comment - looked a bit unusual to me too, but a very nice Wednesday puzzle.
Thanks Mr. Steinmehl
some how I knew cava so then I got ansco which I don't remember from my youth.
33D southwestern crocks. I was trying to think of an animal - gila-. An olla is not a crock IMO. You cook in an olla. You store things like pickles in a crock
@chefbea I disagree--I recall my mother having a crockpot, in which she'd let beans or soup or meat stew for hours, so it'd be ready at dinnertime or for the next day's lunch.
Had no problem with the puzzle other than the hole where the C needed to go in CAVA ANSCO. Didn't know either.
I love sitting on the COUCH eating CHIPS with dip. mmmmmm!
There are a few bio words crossworders should commit to memory.
Would the constructors have the nerve to include vas deferens?
"Quit those monkeyshines," another Daddy quotes from my '50s childhood.
Speaking of that, I was too old when Barbie made the scene. I had Jill, the older sister of Ginny. Much more normal looking proportion-wise except for a big head. Had pierced ears.
I now have a Munster set of Barbie and Ken - green.
Check out Dave Barry setting things on fire with Rollerblade Barbie
@Sfingi - Looked at the Dave Barry piece and thought it was very funny. Thanks!
Fun info-ANSCO, founded in 1842; pre-dated Kodak, orig. name: E. and H.T. Anthony & Co. Merged in 1928 with German photo co. Agfa to form Agfa-Ansco.Eventually became GAF. Last camera produced in
1990's acc. to Wiki....Wanted to put Barry in Arizona Pol family. Worked on the Hill many moons ago and actually saw Mr. Udall a few times! Oxfords-kept thinking shirts with the 27d Era, wanted
some P & G detergent in there...lol
A nice, smooth run with today's puzzle, except for 33D. I misread the clued word "crocks" for "crocs," thinking "gator" could be the only answer. Couldn't imagine what kind of reptiles "ollas"
Why did you enter PCS for 21 across instead of MCS? And what is a tyro? | {"url":"https://latcrossword.blogspot.com/2009/08/wednesday-august-5-2009-gary-steinmehl.html","timestamp":"2024-11-11T03:32:44Z","content_type":"application/xhtml+xml","content_length":"121085","record_id":"<urn:uuid:57d597a2-b311-4799-af1b-1d3d6b3b0172>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00421.warc.gz"} |
Journal of the Korean Mathematical Society
Jangwon Ju
Abstract : For an arbitrary integer $x$, an integer of the form $T(x)\!=\!\frac{x^2+x}{2}$ is called a triangular number. Let $\alpha_1,\dots,\alpha_k$ be positive integers. A sum $\Delta_{\alpha_1,\
dots,\alpha_k}(x_1,\dots,x_k)=\alpha_1 T(x_1)+\cdots+\alpha_k T(x_k)$ of triangular numbers is said to be {\it almost universal with one exception} if the Diophantine equation $\Delta_{\alpha_1,\
dots,\alpha_k}(x_1,\dots,x_k)=n$ has an integer solution $(x_1,\dots,x_k)\in\mathbb{Z}^k$ for any nonnegative integer $n$ except a single one. In this article, we classify all almost universal sums
of triangular numbers with one exception. Furthermore, we provide an effective criterion on almost universality with one exception of an arbitrary sum of triangular numbers, which is a generalization
of ``15-theorem" of Conway, Miller, and Schneeberger. | {"url":"https://jkms.kms.or.kr/journal/list.html?pn=mostread","timestamp":"2024-11-09T00:28:50Z","content_type":"text/html","content_length":"44938","record_id":"<urn:uuid:a15d5aa2-1414-4437-a615-013fc65b74d0>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00887.warc.gz"} |
Improved N-Best Extraction with an Evaluation on Language Data
We show that a previously proposed algorithm for the N-best trees problem can be made more efficient by changing how it arranges and explores the search space. Given an integer N and a weighted tree
automaton (wta) M over the tropical semiring, the algorithm computes N trees of minimal weight with respect to M. Compared with the original algorithm, the modifications increase the laziness of the
evaluation strategy, which makes the new algorithm asymptotically more efficient than its predecessor. The algorithm is implemented in the software Betty, and compared to the state-of-the-art
algorithm for extracting the N best runs, implemented in the software toolkit Tiburon. The data sets used in the experiments are wtas resulting from real-world natural language processing tasks, as
well as artificially created wtas with varying degrees of nondeterminism. We find that Betty outperforms Tiburon on all tested data sets with respect to running time, while Tiburon seems to be the
more memory-efficient choice.
Trees are standard in natural language processing (NLP) to represent linguistic analyses of sentences. Similarly, tree automata provide a compact representation for a set of such analyses. Bottom–up
tree automata act as recognizing devices and process their input trees in a step-wise fashion, working upwards from the leaf nodes towards the root of the tree. For memory, they have a finite set of
states, some of which are said to be accepting, and their internal logic is represented as a finite set of transition rules. A run of a tree automaton M on an input tree t is a mapping from the nodes
of t to the states, which is compatible with the transition rules. In general, M can have several distinct runs on t, and acceptst if one of these runs maps the root of t to an accepting state. By
equipping the transition rules with weights, M can be made to associate t with a likelihood or score: The weight of a run of M on t is the product of the weights of the transition rules used in the
run, and the weight of t is the sum of the weights of all runs on t. This type of automaton is called a weighted-tree automaton (wta) and is popular in, for example, dependency parsing and machine
A central task is the extraction of the highest ranking trees with respect to a weighted tree automaton. For example, when the automaton represents a large set of intermediate solutions, it may be
desirable to prune these down to a more manageable number before continuing the computation. This problem is known as the best trees problem. It is related to the best runs problem that asks for
highest-ranking runs of the automaton on not necessarily distinct trees. The computational difficulty of the N-best trees problem depends on the algebraic domain from which weights are taken. Here we
assume this domain to be the tropical semiring. Hence, the weight of a tree t is the minimal weight of any run on t, and the weight of a run is the sum of the weights of the transitions used in the
run, which are non-negative real numbers. The tropical semiring is particularly common in speech and text processing (Benesty, Sondhi, and Huang 2008), since probabilistic devices can be modeled
using negative log likelihoods. Moreover, the semiring has the advantage of being extremal: The sum of two elements a and b always equals one of a and b. As a consequence, it is not necessary to
consider all runs of an automaton on an input tree to find the weight of the tree, as its weight is equal to the weight of the optimal run. (In the non-extremal case, the problem is NP-complete even
for strings (Lyngsø and Pedersen 2002).) The N-best trees problem for a given wta M over an extremal semiring can be solved indirectly by computing a list of N′ best runs for M, for a sufficiently
large number N′, and outputting the corresponding trees while discarding previously outputted trees. A complicating factor with this approach, however, is that M can have exponentially many runs on a
single tree, so N′ may have to be very large to guarantee that the output contains N distinct trees.
Best trees extraction is useful in any application that includes some type of re-ranking of hypotheses. One example that makes use of best trees extraction is the work by Socher et al. (2013) on
syntactical language analysis. The team of authors improve the Stanford parser by composing a probabilistic context-free grammar (PCFG) with a recurrent neural network (RNN) that learns vector
representations. Intuitively, each nonterminal in the PCFG is associated with a continuous vector space. The vector space induces an unbounded refinement of the category represented by the
nonterminal into subcategories, and the RNN computes transitions between such vectors. For efficiency reasons, the device is not applied to the input sentence directly. Instead, the N = 200
highest-scoring parse trees with respect to the PCFG are computed, whereupon the RNN is used to rerank these to find the best parse tree. The work has raised interest in hybrid finite-state
continuous-state approaches (see, e.g., the work by Zhao, Zhang, and Tu (2018)), and underlines the value of the N-best problem in language processing.
In previous work, Björklund, Drewes, and Zechner (2019) generalized an N-best algorithm by Mohri and Riley (2002) from strings to trees, resulting in the algorithm Best Trees v.1. Intuitively, the
algorithm performs a lazy implicit determinization and uses a priority queue to output N best trees in the right order. The running time of Best Trees v.1 was shown to be in $O(max(Nmn⋅
(Nr+rlogr+NlogN),N2n3,mr2))$, where m and n are the numbers of transitions and states of M, respectively, and r is the maximum number of children (the rank) of symbols in the input alphabet. Best
Trees v.1 was evaluated empirically in Björklund, Drewes, and Jonsson (2018) against the N best runs algorithm by Huang and Chiang (2005), which represents the state of the art. Although Büchse et
al. (2010) proved that the algorithm by Huang and Chiang works for cyclic input wtas and generalized it by extending it to structured weight domains, the core idea of the algorithm remains the same.
The algorithm by Huang and Chiang (2005) is implemented in the widely referenced Tiburon toolkit (May and Knight 2006). From here on, we refer to this implementation simply as Tiburon, even though
the best runs procedure is only one out of many that the toolkit has to offer. The conclusion of Björklund, Drewes, and Jonsson (2018) was that Best Trees v.1 is faster if the input wtas exhibit a
high degree of nondeterminism, whereas Tiburon is the better option when the input wtas are large but essentially deterministic.
We now improve Best Trees v.1 by exploring the search space in a more structured way, resulting in the algorithm Best Trees. In Best Trees v.1, all assembled trees were kept in a single queue. In
this work, we split the queue into as many queues as there are transitions in the input automaton. The queue K[τ] of transition τ contains trees that are instantiations of τ, that is, trees with a
run that applies τ at the root. This makes it possible to improve the strategy to prune the queue that was used by Björklund, Drewes, and Zechner (2019), and avoid pruning altogether. The intuition
is simple: To assemble N distinct output trees, at most N instantiations of any one transition may be needed. We furthermore assemble the instantiations of τ in a lazy fashion, constructing an
instantiation explicitly only when it is dequeued from K[τ]. We formally prove the correctness of Best Trees and derive an upper bound on its running time, namely, $O(Nm(log(m)+r2+rlog(Nr)))$.
In addition to solving the best trees problem, Best Trees can also solve the best runs problem by removing the control structure that makes it discard duplicate trees (see Section 5). In this
article, we make use of this possibility to compare this algorithm, implemented as Betty, with Tiburon on the home turf of the latter, that is, with respect to the computation of best runs rather
than best trees. For our experiments, we use both largely deterministic wtas from a machine translation project and from Grammatical Framework (Ranta 2011), and more nondeterministic wtas that were
artificially created to expose the algorithms to challenging instances. Our results show that Betty is generally more time efficient than Tiburon, despite the fact that the former is more general as
it can also compute best trees (with almost the same efficiency as it computes best runs). Moreover, we perform a limited set of experiments measuring the memory usage of the applications, and can
conclude that overall, the memory efficiency of Tiburon is slightly better than that of Best Trees.
1.1Related Work
The proposed algorithm adds to a line of research that spans two decades. It originates with an algorithm by Eppstein (1998) that finds the N best paths from one source node to the remaining nodes in
a weighted directed graph. When applied to graphs representing weighted string automata, the list returned by Eppstein’s algorithm may in case of nondeterminism contain several paths that carry the
same string, that is, the list is not guaranteed to be free from duplicate strings. Four years later, Mohri and Riley (2002) presented an algorithm that computes the N best strings with respect to a
weighted string automaton and thereby creates duplicate-free lists. To reduce the amount of redundant computation, consisting in the exploration of alternative runs on one and the same substring,
they incorporate the N shortest paths algorithm by Dijkstra (1959). Moreover, Mohri and Riley work with on-the-fly determinization of the input automaton M, which avoids the problem that the
determinized automaton can be exponentially larger than M, but has at most one run on each input string. Jiménez and Marzal (2000) lift the problem to the tree domain by finding the N best parse
trees with respect to a context-free grammar in Chomsky normal form for a given string. Independently, Huang and Chiang (2005) published an algorithm that computes the best runs for weighted
hypergraphs (which is equivalent to weighted tree automata and weighted regular tree grammars), and that is a generalization of the algorithm by Jiménez and Marzal in that it does not require the
input to be in normal form. Huang and Chiang combine dynamic programming and lazy evaluation to keep the number of intermediate computations small, and derive a lower bound on the worst-case running
time of their algorithm than Jiménez and Marzal do.
As previously mentioned, the algorithm by Huang and Chiang (2005) is implemented in the Tiburon toolkit by May and Knight (2006). Its initial usage was as part of a machine-translation pipeline, to
extract the N best trees from a weighted tree automaton. In connection with this work, Knight and Graehl (2005) noted that there was no known efficient algorithm to solve this problem directly. As an
alternative way forward, Knight and Graehl thus enumerate the best runs and discard duplicate trees, until sufficiently many unique trees have been found. Since Tiburon is highly optimized and the
current state-of-the-art tool for best runs extraction, it is a natural choice of reference implementation for an empirical evaluation of our solution. The algorithm by Huang and Chiang (2005) was
later generalized by Büchse et al. (2010) to allow a linear pre-order on the weights, as opposed to a total order. Büchse et al. (2010) also prove that the algorithm is correct on cyclic input
hypergraphs, provided that the Viterbi algorithm for finding an optimal run (Jurafsky and Martin 2009) is replaced by Knuth’s algorithm (Knuth 1974).^^1
Also, Finkel, Manning, and Ng (2006) remark on the lack of sub-exponential algorithms for the best trees problem. They propose an algorithm that approximates the solution when the automaton is
expressed as a cascade of probabilistic tree transducers. The authors model the cascade as a Bayesian network and consider every step as a variable. This allows them to sample a set of alternative
labels from each prior step, to propagate onward in the current step. The approximation algorithm runs in polynomial time in the size of the input device and the number of samples, but the
convergence rate to the exact solution is not analyzed. To explain why their approach is preferable to finding N best runs, they extract the N = 50 best runs from the Stanford parser and observe that
about half of the output trees are actually duplicates—enough to affect the outcome of the processing pipeline. Thus, they argue, extracting the highest ranking trees rather than the highest ranking
runs is not only theoretically better, but is also of practical significance.
Finally, we note that the best trees problem also has applications outside of NLP. In fact, whenever we are considering a set of objects, each of which can be expressed uniquely by an expression in
some particular algebra, and the set of expressions is the language of a wta, then the best trees algorithm can be used to produce the best objects. For instance, Björklund, Drewes, and Ericson (2016
) propose a restricted class of hypergraphs that are uniquely described by expressions in a certain graph algebra, so the best trees algorithm makes it possible to find the optimal such graphs with
respect to a wta. The reason why the representation needs to be unique is that otherwise we will have to check equivalence between objects as an added step. In the case of graphs, this would mean
deciding graph isomorphism, which is not known to be tractable in general.
We write ℕ for the set of nonnegative integers, ℕ[ +] for ℕ ∖{0}, and ℝ[ +] for the set of non-negative reals; $N∞$ and $R+∞$ denote $N∪{∞}$ and $R+∪{∞}$, respectively. For n ∈ℕ, [n] = {i ∈ℕ∣1 ≤ i ≤
n}. Thus, in particular, [0] = ∅ and $[∞]=N$. The cardinality of a (countable) set S is written |S|. The n-fold Cartesian product of a set S with itself is denoted by S^n.
The set of all finite sequences over S is denoted by S^*, and the empty sequence by λ. A sequence of l copies of a symbol s is denoted by s^l. Given a sequence σ = s[1]⋯s[n] of n elements s[i] ∈ S,
we denote its length n by |σ|. Given an integer i ∈ [n], we write σ[i] for the i-th element s[i] of σ. For notational simplicity, we occasionally use sequences as if they were sets, for example,
writing s ∈ σ to express that s occurs in σ, or S ∖ σ to denote the set of all elements of a set S that do not occur in the sequence σ.
A (commutative) semiring is a structure $(D,⊕,⊗,0,1)$ such that both $(D,⊕,0)$ and $(D,⊗,1)$ are commutative monoids, the semiring multiplication ⊗ distributes over the semiring addition ⊕ from both
left and right, and 0 is an annihilator for ⊗, that is, 0 ⊗ d = 0 = d ⊗ 0 for all $d∈D$. In this article, we will exclusively consider the tropical semiring. Its domain is $R+∞$, with $min$ serving
as semiring addition and ordinary plus as semiring multiplication.
For a set
, an
is a partial function
whose domain dom(
) is a finite non-empty set that is closed to the left and under taking prefixes; whenever
) for some
[ +]
, it holds that
) for all 1 ≤
(closedness to the left) and
) (prefix-closedness). The size of
is |
| = |dom(
)|. An element
of dom(
) is called a
, and |{
[ +]
)}| is the
. The
rooted at
is the tree
defined by
) =
) for every
. If
(λ) =
for all
∈ [
], where
is the rank of λ in
, then we denote
], which may be simplified to
= 0.
A ranked alphabet is a disjoint union of finite sets of symbols, $Σ=⋃k∈NΣ(k)$. For f ∈ Σ, the k ∈ℕ such that f ∈ Σ[(k)] is the rank of f, denoted by rank(f). The set T[Σ] of ranked trees over Σ
consists of all Σ-labeled trees t in which the rank of every node v ∈dom(t) equals the rank of t(v). For a set T of trees we denote by Σ(T) the set of trees which have a symbol from Σ at their root,
with direct subtrees in T, more precisely, {f[t[1],…,t[k]]∣k ∈ℕ,f ∈ Σ[(k)],and t[1],…,t[k] ∈ T}.
In the following, let
be a special symbol of rank 0. The set of
contexts over Σ
is the set
of trees
containing exactly one node
) with
. We define the
to be
) = |
|, i.e., the depth of
is the distance of
from the root of
. The
of another tree
results in the tree
given by
and, for all
$c⟦t⟧(w)=llc(w)ifw∈dom(c)∖{v}t(u)ifw=vufor someu∈dom(t).$
A weighted tree language over the tropical semiring is a mapping $L:TΣ→R+∞$, where Σ is a ranked alphabet. Weighted tree languages can be specified in a number of equivalent ways. Three of the
standard ones, mirroring the ways in which regular string languages are traditionally specified, are weighted regular tree grammars, weighted tree automata, and weighted finite-state diagrams
formalized as hypergraphs. The equivalence of the second and the third is shown explicitly in Jonsson (2021). All three have been used in the context of N-best problems: weighted regular tree
grammars by May and Knight (2006), weighted tree automata by Björklund, Drewes, and Zechner (2019), and hypergraphs by Huang and Chiang (2005) and Büchse et al. (2010). In this article, we use
weighted tree automata.
A weighted tree automaton (wta) over the tropical semiring is a system M = (Q, Σ, R, ω, q[f]) consisting of:
• a finite set Q of symbols of rank 0 called states;
• a ranked alphabet Σ of input symbols disjoint with Q;
• a finite set $R⊆⋃k∈NQk×Σ(k)×Q$ of transition rules;
• a mapping ω: R →ℝ[ +]; and
• a final state q[f] ∈ Q.
From here on, we write $f[q1,…,qk]→wq$ to denote that τ = (q[1],…,q[k],f,q) ∈ R and ω(τ) = w, and consider R to be the set of these weighted rules, thus dropping the component ω from the definition
of M.
A transition rule $τ:f[q1,…,qk]→wq$ will also be viewed as a symbol of rank k, turning R into a ranked alphabet. We let tar(τ) denote the target state q of τ, src(τ) denotes the sequence of source
states q[1]…q[k], and rank(τ) = rank( f). In addition, we view every state q ∈ Q as a symbol of rank 0.
We define the set runs[M] ⊆ T[R∪Q] of runs ρ ofM, their input trees input[M](ρ), their intrinsic weightswt[M](ρ), and their target statetar(ρ) inductively, as follows:
1. For every q ∈ Q, we have that q ∈runs[M] with input[M](q) = q, wt[M](q) = 0, and tar(q) = q.
2. For every transition rule
and all runs ρ
such that
) =
for all
∈ [
], we let ρ = τ[ρ
] ∈
$inputM(ρ)=f[inputM(ρ1),…,inputM(ρk)]wtM(ρ)=w+∑i∈[k]wtM(ρi), andtar(ρ)=q$
of a run ρ ∈
Now, the weighted tree language
$M:TΣ→R+∞$recognized by M
is given by
for all
(where, by convention,
). In other words,
) is the minimal weight of any run resulting in
– which is the sum of all weights of
in the tropical semiring. Note that we, by a slight abuse of notation, denote by
both the wta and the weight assignments to runs and trees it computes. Moreover, for
we define the mapping
for every
Throughout the rest of the article, we will generally drop the subscript M in runs[M], input[M], and wt[M], because the wta in question will always be clear from the context.
Given as input a wta M and an integer N ∈ℕ, the N-bestruns problem consists in computing a sequence of N runs of minimal weight according to M. More precisely, an algorithm solving the problem will
output a sequence ρ[1],ρ[2],… of N pairwise distinct runs such that there do not exist i ∈ [N] and ρ ∈ runs ∖{ρ[1],…,ρ[i]} with M(ρ) < M(ρ[i]).
General Assumption. To make sure that the N-best runs problem always possesses a solution, and to simplify the presentation of our algorithms, we assume from now on that all considered wtas M have
infinitely many runs ρ such that tar(ρ) = q[f]. In particular, T[Σ] is assumed to be infinite. Apart from simplifying some technical details, this assumption does not affect any of the reasonings in
the paper.
Similarly to the N-best runs problem, the N-besttrees problem for the wta M consists in computing a sequence of pairwise distinct trees t[1],t[2],… in T[Σ] of minimal weight. In other words, we seek
a sequence of trees such that there do not exist i ∈ [N] and t ∈ T[Σ] ∖{t[1],…,t[i]} with M(t) < M(t[i]). Note that the N-best trees problem always has a solution because we assume that T[Σ] is
Figure 1
shows an example wta, and
Table 1
contains a side-by-side comparison of the input trees of the 10 best runs and the 10 best trees of the automaton. Because the weight of every transition rules is 1, the weight of every run on an
input tree
is its size |
|. Looking at the rules, we obtain the following recursive equations for the number
) of runs ρ on a tree
ending in state
(ρ) =
Hence, every tree
will occur
) times in an
-best list based on best runs (provided that
is large enough).
Table 1
Best runs . Best trees .
Input tree . Weight . Tree . Weight .
a 1 a 1
f[a,a] 3 f[a,a] 3
f[a,a] 3 f[f[a,a],a] 5
f[a,a] 3 f[a,f[a,a]] 5
f[a,f[a,a]] 5 f[f[a,a],f[a,a]] 7
f[f[a,a],a] 5 f[f[f[a,a],a],a] 7
f[a,f[a,a]] 5 f[f[a,f[a,a]],a] 7
f[f[a,a],a] 5 f[a,f[f[a,a],a]] 7
f[a,f[a,a]] 5 f[a,f[a,f[a,a]]] 7
f[f[a,a],a] 5 f[f[a,f[a,a]],f[a,a]] 9
Best runs . Best trees .
Input tree . Weight . Tree . Weight .
a 1 a 1
f[a,a] 3 f[a,a] 3
f[a,a] 3 f[f[a,a],a] 5
f[a,a] 3 f[a,f[a,a]] 5
f[a,f[a,a]] 5 f[f[a,a],f[a,a]] 7
f[f[a,a],a] 5 f[f[f[a,a],a],a] 7
f[a,f[a,a]] 5 f[f[a,f[a,a]],a] 7
f[f[a,a],a] 5 f[a,f[f[a,a],a]] 7
f[a,f[a,a]] 5 f[a,f[a,f[a,a]]] 7
f[f[a,a],a] 5 f[f[a,f[a,a]],f[a,a]] 9
We end this section by discussing the choice of our particular weight structure, the tropical semiring. In the literature on wta, Definitions 1 and 2 are generalized to wta over arbitrary commutative
semirings, and their resulting weighted tree languages, simply by replacing + and $min$ by ⊕ and ⊗, respectively. The tropical semiring $(R+∞,min,+,∞,0)$ used in Definitions 1 and 2 is frequently
used in natural language processing. Equally popular is the Viterbi semiring$([0,1],max,⋅,0,1)$ that acts on the unit interval of probabilities, with maximum and standard multiplication as
operations. In the setting discussed here, both semirings are equivalent. To see this, transform a wta M over the Viterbi semiring to a wta M′ over the tropical semiring by simply mapping every
weight p of a transition rule of M to $−lnp$, that is, taking negative logarithms everywhere. Since $−lnp+−lnp′=−ln(p⋅p′)$ and $−lnp<−lnp′⇔p>p′$, it holds that M(t) (now calculated using the
Viterbi semiring operations) is equal to $exp(−M′(t))$ for all trees t. It follows that the trees t[1],…,t[N] form an N-best list according to M (now looking for trees with maximal weights) if and
only if they form an N-best list according to M′ in the sense defined above.
3The Improved Best Trees Algorithm
In this section, we explain how the algorithm in Björklund, Drewes, and Zechner (2019) can be made lazier, and hence more efficient, by exploring the search space with respect to transitions rather
than states. From here on, let M = (Q,Σ,R,q[f]) be a wta with m transition rules, n states, and a maximum rank of r among the symbols in Σ.
3.1Best Trees v.1
We first summarize the approach of Björklund, Drewes, and Zechner (
). The algorithm maintains two data structures: an initially empty set
that collects all processed trees, and a priority queue
of trees in Σ(
) that will be examined next. The priority of a tree
is determined by the minimal value in the set of all
, where
ranges over all possible contexts. Let
denote the wta obtained from
by making
its final state. Then, for every context
and all trees
) is independent of
, it is possible to compute in advance a
best context
that minimizes it. The technique will be discussed in
Section 3.2
best context
of a state
is a context 𝕔
The value
) is denoted by 𝕨
The algorithm uses the best contexts to compute an
optimal stateopt
) for each tree
that it encounters:
. Throughout the article, we shall make use of the following weight functions derived from
, where
Note that
The priority queue
is initialized with the trees in Σ
. Its priority order <
is defined as follows, for all trees
Here, <
is any lexical order that orders trees first by size and then by viewing them as strings to be compared alphabetically from left to right.
We can now reproduce the pseudocode of the base algorithm from Björklund, Drewes, and Zechner (2019) in Algorithm 1. Given a wta M and N ∈ℕ, it solves the N-best trees problem. After outputting i ∈ [
N] trees, the set of trees enqueued in line 23 is pruned so that for every q ∈ Q, at most N − i trees are kept for which q is an optimal state. The function expand(T,t), which computes the trees to
be enqueued in each step, returns the set of all trees in Σ(T) such that the “new” tree t occurs at least once among the direct subtrees of the root.
The correctness of this approach is formally proved in Björklund, Drewes, and Zechner (2019).
3.2Computation of Best Contexts
We now recall the computation of best contexts 𝕔[q], q ∈ Q, to the extent needed to understand the improved algorithm. This computation consists of two phases: first, a 1-best tree $tqbest$ is
computed for each state q ∈ Q. The desired property of $tqbest$ is that it is a 1-best tree of M^q, that is, it is a tree t ∈ T[Σ] that minimizes M^q(t). After that, the second phase computes the
actual best context 𝕔[q] for every q ∈ Q, in other words, a context c ∈ C[Σ] with M[q](c) =𝕨[q].
The first phase can be accomplished using a dynamic programming algorithm by Knuth (1977) that computes 1-best runs. (Note that if ρ is a 1-best run, then input(ρ) is a 1-best tree.) The algorithm
maintains a min-priority queue of all transition rules and collects, in |R| iterations, the desired best runs ρ[q]. Initially, ρ[q] is undefined for every q ∈ Q. The value determining the priority of
a transition rule $τ:f[q1,…,qk]→wq$ is $∞$ if any $ρqi$ (i ∈ [k]) is still undefined. Otherwise, it is $w+∑i∈[k]wt(ρqi)$, that is, the weight of the run $τ[ρq1,…,ρqk]$. The algorithm repeatedly
dequeues the highest priority element $τ:f[q1,…,qk]→wq$ from the queue. If ρ[q] is still undefined, it sets $ρq=τ[ρq1,…,ρqk]$ and $tqbest=f[tq1best,…,tqkbest]$. It then updates the priorities of
transition rules having q among the states in their right-hand sides and repeats. We note that the trees $tqbest$ are discovered by the algorithm in the order of ascending weight. This observation
will soon become important for the initialization phase of Algorithm 2.
As a side remark, we note that the set of 1-best runs and 1-best trees determined by this algorithm are subtree closed, meaning that every subtree of ρ[q] and $tqbest$ is itself one of the trees ρ[q′
] and $tq′best$, respectively. It follows that the entire set of these trees can be stored as a maximally shared directed acyclic graph with |Q| nodes.
The second phase applies Dijkstra’s shortest paths algorithm (Dijkstra
) to
, where
is viewed as a weighted edge-labeled graph. More precisely, consider the graph with node set
such that, for every transition rule
and every
}, there is an edge
= (
) from
with label τ. The weight of such an edge is given by
The intuition behind this weight assignment is that the edge represents the possibility that
), in which case
) is the weight that needs to be added to 𝕨
to obtain 𝕨
(under the assumption that, indeed,
Now, having computed the paths of minimal weight in this graph using Dijkstra’s algorithm, consider the edge sequence π of least weight from q[f] to a state q′. Then 𝕔[q′] = 𝕔(π), where 𝕔(π) is
defined recursively as follows:
1. If π = λ then q′ = q[f] and we set $c(π)=□$.
2. If π =
= (
, τ,
) for some transition rule
, we choose some
∈ [
] with
, and set
We note here that this way of computing best contexts results in contexts of a very peculiar kind: Every such context 𝕔[q] consists of a “spine” leading to the node v such that $cq(v)=□$, and all
subtrees that branch out from this spine are of the form $tq′best$ for the required states q′.
3.3Transition-Based Best Trees Computation
The improved algorithm for computing N best trees hinges on the observation that to generate N best trees, at most N distinct instantiations of each transition rule are needed. Here, an instantiation
of a rule $τ=f[q1,…,qk]→wq$ is a tree f[t[1],…,t[k]]. Furthermore, the algorithm creates these instantiations in a lazy fashion.
To this end, we build sequences T[q] of N′ ≤ N best trees for each state q. Most importantly, a separate priority queue K[τ] is kept for each transition rule $τ=f[q1,…,qk]→wq$ in R. Every tree in
this queue is of the form $f[(Tq1)i1,…,(Tqk)ik]$ and is hence uniquely determined by the tuple (i[1],…,i[k]), each i[j] working as a pointer into the sequence $Tqj$.^^3 Hence, each such tuple can be
understood as an abstract instruction of how to instantiate τ by previously dequeued trees. As outlined in Section 3.5, an efficient implementation of the algorithm can make this assembly “just in
time,” so as to avoid unnecessary work.
In each iteration, the algorithm chooses the highest-priority element across all of the queues K[τ], where the priority order (to be described later) is similar to <[K], but improved by replacing the
use of the lexical order by a more goal-oriented component. To efficiently pick the highest-priority element across all queues, we organize the queues in a meta-queue K′, where the priority of every
K[τ] in K′ is given by the priority of its highest-priority element. This organization is schematically illustrated in Figure 2.
The algorithm itself is outlined in Algorithm 2. The sequences T[q] mentioned above are the previously dequeued best instantiations of rules τ with tar(τ) = q. Thus, as mentioned before, T[q] is a
prefix of a solution of the N-best problem for the wta M^q.
For every $τ:f[q1,…,qk]→wq$ and every index tuple u = (i[1],…,i[k]) ∈ℕ^k, the instantiated transition rule is denoted by τ[u]. As previously mentioned, it is defined as the tree $f[(Tq1)i1,…,(Tqk)ik]
$, but since this is well defined only if $ij≤|Tqj|$ for every j ∈ [k], we let τ[u] be undefined otherwise.
The function
returns, for every tuple
= (
) ∈ℕ
∈ℕ), the set of all its successors obtained by increasing exactly one of
∈ [
], by 1. Formally, for every tuple
= (
) ∈ℕ
Each queue K[τ] in Algorithm 2 is a min-priority queue in which the least priority is assigned to elements u such that τ[u] is undefined. This reflects that we do not yet know the weight of the
instantiation of τ with respect to u, and that every instantiation for which we do know the weight of the resulting tree can be shown to be preferable.
To define the priority used in
, we follow the previous approach, using a variant of <
, but in addition to the necessary adaptations, we shall replace the lexical component by a more goal-oriented one. The priority of every element
is primarily determined by a variant of Δ, denoted by Δ
). Recall that, for
, Δ
) is the least weight of all runs on trees of the form
, where only runs are considered whose target state at the root of the subtree
. Thus, Δ
) ≥ Δ(
), where equality holds for
). Now, Δ
specializes this further by assuming that the specific transition rule applied at the root of the subtree
is τ. Moreover, because we only apply this weight function to trees of the form τ[
], we let its argument be
rather than τ[
]. Formally, consider a transition rule
. If τ[
] is undefined, we simply put
. If τ[
] is defined, we define
This expression may look a bit complicated, but the intuition behind it is actually straightforward. As we shall see,
is an optimal state for
∈ [
]. Thus, if ρ
is the lowest-weight run on
) =
, and we set ρ = τ[ρ
], then the expression
is simply the definition of
(ρ), which is equal to
]). Thus, adding 𝕨
, Δ
) turns out to be the minimum of all
(ρ), where ρ ranges over all runs with ρ(
) = τ and
) = τ[
] for some
∈domρ. Thus, Δ
) ≥ Δ
]) ≥ Δ(τ[
Finally, to complete the definition of the priority used in
(τ ∈
) and in
, let
be given by
where δ
)) for every
. We order
as usual, namely, (
) < (
) if
Now, let q ∈ Q and $(τ:f[q1,…,qk]→wq)∈R$. Having computed best contexts (and, in the process, also best trees) for all states, T[q] is initialized to contain only the best tree $tqbest$ for q. The
queue K[τ] is initialized to contain only 1^k if τ≠ρ[q](λ), since the latter means that τ was not used to build $tqbest$, which means that the first instantiation of τ is still waiting to be added
to T[q] at a suitable position. Otherwise, K[τ] is initialized to contain all successors of 1^k because, by the way in which $tqbest$ was constructed and the definition of τ[u], we have $(Tq)1=tqbest
=τ[1k]$ and are now looking for the next best instantiation of τ.
After initializing T[q] and K[τ], the algorithm enters the main loop. This loop is executed until N trees have been outputted. The first step in the loop is to extract a minimal u from the set of all
queues K[τ], τ ∈ R (by first dequeuing K[τ] from K′ and then u from K[τ]). Thanks to our assumption that M possesses infinitely many runs ending in q[f], it can be shown that Δ[τ](u) ∈ℕ, that is, the
tree τ[u] is defined (see the next section). The tree is appended at the end of the list of the best (smallest-weighted) trees that reach the target state tar(τ) of τ. If tar(τ) = q[f] and τ[u] is
“new,” that is, has not been outputted before, then τ[u] is outputted now, and the counter c tracking the number of output trees is increased. Note that, in contrast to Algorithm 1, we have to check
whether τ[u] was outputted before, because some τ′[u′] outputted earlier may actually have been equal to τ[u]. However, this can only be the case if τ≠τ′, and can thus only happen m times for every
Let us now show that Algorithm 2 is correct. To simplify the reasoning, we shall first consider a variant of the algorithm, referred to as Algorithm 2′, obtained by removing the inequality |T[tar(τ)]
| < N from the condition on line 14. In the following lemma, we say that T[q] = t[1]⋯t[m] is appropriate if t[1],…,t[m] is a solution of the m-best trees problem for M^q.
Consider a run of Algorithm 2′. During every execution of the main loop the following statements hold:
• (1)
When the loop is entered, each T[q] (q ∈ Q) is appropriate.
• (2)
When line 13 has been executed with tar(τ) = q, there do not exist any q′ ∈ Q and t ∈ T[Σ] ∖ T[q′] such that Δ[q′](t) < Δ[τ](u) (where τ and u denote the values of the corresponding variables in
Algorithm 2′ at that point).
We use (1) as a loop invariant. Because it holds before the first execution of the loop (as each T[q] consists of a single best tree with respect to M^q), we need to show that, under the condition
that (1) holds when the loop is entered, statement (2) holds as well and, when line 21 has been executed, (1) still holds.
We first show (2). Assume for a contradiction that
did actually exist and let
, where
, that is,
is the node such that
. By the definition of Δ
, there is a run ρ with
(ρ) =
) =
, and
(ρ) = Δ
). For every
), let
) be the state the run is in after having processed
. There is at least one node
) such that
). Now, choose
) with
in such a way that |
| is minimal, and suppose that
] and
. (Thus,
for all
∈ [
].) By the minimality of |
, and thus there is
= (
) ∈ℕ
such that
for all
∈ [
]. With
) it follows that
Because $s/v∉Tqv$, the tuple u′ has not yet been dequeued from K[τ′]. Thus, while u′ itself may not yet be in K[τ′], K[τ′] must contain some u″ = (j[1]′,…,j[ℓ]′) with j[i]′ ≤ j[i] for all i ∈ [ℓ].
This is because when an element is dequeued on line 13 then all of its direct successors will be enqueued on line 21. In particular, K[τ′] cannot be empty at the start of an iteration, as long as
there is some u′ ∈ℕ^ℓ that has not yet been dequeued from K[τ′] (which will always be the case if ℓ > 0).
Note that j[i]′ ≤ j[i] for all i ∈ [ℓ] implies that $Δqv(u″)≤Δqv(u′)$ since $Tp1,…,Tpℓ$ are appropriate. Hence, the inequality Δ[τ′](u″) ≤ Δ[τ′](u′) < Δ[τ](u) contradicts the assumption that u was
dequeued on line 13.
We have thus proved (2). Because τ[
] is appended to
on line 15, it remains to be shown that
]) ≤
]) for all trees τ[
] that have been appended to
during earlier iterations. Choosing
, τ[
] as
, and
in (2) we get
where the first inequality holds by the appropriateness of
and the second holds because
(τ) =
. Furthermore, by (2) there is no tree
such that
) <
]). Thus,
is still appropriate when τ[
] has been appended to it.
Algorithm 2′ computes a solution to the N-best trees problem.
We first observe that, due to the output condition on line 17 of Algorithm 2′, the sequence of trees outputted by the algorithm does not contain repetitions.
Next, whenever a tree s = τ[u] is outputted on line 18, we show that M(t) ≥ M(s) for all trees t ∈ T[Σ] that have not yet been outputted. Let $τ:f[q1,…,qk]→wq$. Then q = q[f] and thus $cq=□$, 𝕨[q]
= 0, and M(s) = Δ[τ](u). Now, consider any t ∈ T[Σ] that has not yet been outputted. Then we have $t∈TΣ∖Tqf$. Consequently, $M(t)=Δqf(t)≥Δτ(u)$ by Lemma 1, as required.
To complete the proof, we have to show that there cannot be an infinite number of iterations without any tree being outputted. We show the following, stronger statement:
Claim 1. Let $δ=maxq∈Qδq$, where δ[q] is defined as in Equation 2 for all q ∈ Q. At any point in time during the execution of Algorithm 2′, it takes at most δ iterations until the selected
transition rule τ satisfies tar(τ) = q[f].
To see this, let
be dequeued on line 12 and let
(τ). If
, the statement holds. Otherwise, the context 𝕔
has the form
for a transition rule
. The tree
= τ[
] is appended to
, say at position
. It follows that
] with
becomes defined, where
. We also know that
1. there is no u″ in any of the queues K[τ″] with Δ[τ″](u″) < Δ[τ](u) (Lemma 1(2)),
2. $Δτ′(u′)=M(cq⟦t⟧)=M(cq′⟦t′⟧)=Δτ(u)$, and
3. δ[q′] = δ[q] − 1.
It follows that the queue K[τ″] from which a tuple is dequeued on line 12 at the start of the next iteration satisfies δ[tar(τ″)] = δ[q] − 1, thus bounding the number of iterations until this
quantity reaches 0 from above by δ.
Now, to finish the proof, note that τ[u]≠τ[u′] whenever u≠u′ because the sequences T[q], q ∈ Q, do not contain repetitions. Because, furthermore, no tuple u is enqueued twice in K[τ], line 18 will be
reached after at most δ iterations.
We finally show that Algorithm 2 is correct as well.
Algorithm 2 computes a solution to the N-best trees problem.
Consider an execution of Algorithm 2′ and assume that we assign every tree a color, red or black, where black is the default color. (The color attribute of a subtree may differ from that of the tree
itself.) Suppose that, in an iteration of the main loop, we have $τ:f[q1,…,qk]→wq$ and u = (i[1],…,i[k]), and thus τ[u] = f(t[1],…,t[k]) where $tj=(Tqj)ij$ for all j ∈ [k]. If |T[q]|≥ N on line 14
and we append τ[u] to it on line 15, we color τ[u] red while its subtrees t[1],…,t[k] keep their colors as given by their positions in $Tq1,…,Tqk$.
Now, assume that some T[q] contains a tree t = f[t[1],…,t[k]] such that t[j] is red for some j ∈ [k]. We show that this implies that t is red. We know that t was appended to T[q] as a tree of the
form τ[u] for some u = (i[1],…,i[k]) with i[j] > N (because of the assumption that t[j] is red). We know also that earlier iterations have dequeued all tuples from K[τ] of the form u^i = (i[1],…,i[j
−1],i,i[j +1],…,i[k]) for i = 1,…,N. (This is an immediate consequence of Lemma 2 because, by Lemma 1(1), Δ[τ](u^i) < Δ[τ](u) for all i ∈ [k].) Since the τ[u^i] are pairwise distinct (as they differ
in the j-th direct subtree) this means that |T[q]|≥ N before τ[u] was enqueued, thus proving that t is red, as claimed.
Because Algorithm 2′ terminates when $|Tqf|=N$, none of the trees in $Tqf$ will ever be red. By the above, this implies that all subtrees of trees in $Tqf$ are black as well. As subtrees inherit
their color from the T[q] they are taken from, this shows that no red tree occurring in an execution of Algorithm 2′ can ever have an effect on the output of the algorithm. Hence, the output of
Algorithm 2 is the same as that of Algorithm 2′ and the result follows from Lemma 2.
3.5Time Complexity
Recall that the input M = (Q,Σ,R,q[f]) is assumed to be a wta with m transition rules, n states, and a maximum rank of r among its symbols. In the complexity analysis, we consider an efficient
implementation of Algorithm 2 along the lines illustrated in Figure 2, with priority queues based on heaps (Cormen et al. 2009). This enables us to implement the following details efficiently:
• Consider a transition $τ:f[q1,…,qk]→wq$. At a given stage of the algorithm, some of the elements u = (u[1],…,u[k]) in K[τ] may contain elements u[i] such that $|Tqi|<ui$, that is, τ[u] is still
undefined and hence $Δτ(u)=∞$. Thus, Δ[τ](u) must be decreased from $∞$ to its final value in $R+∞$ when $|Tqi|$ has reached u[i] for all i ∈ [k]. For this, we record for every p ∈ Q and for |T[
p]| < j < |N| a list of all u ∈ K[τ] such that τ is as above, q[i] = p for some i ∈ [k], and u[i] = j. When |T[p]| reaches the value j, this list is used to adjust the priority of each u on that
list to the new value of Δ[τ](u) (which is either still $∞$ or has reached its final value).
• The queue K′ that contains the individual queues K[τ] as elements is implemented in the straightforward way, also using priority queues based on heaps. As described earlier, the priority between
K[τ] and K[τ′] is given by comparing their top-priority elements: if these elements are u and u′, respectively, then K[τ] takes priority over K[τ′] if (Δ[τ](u),δ[tar(τ)]) <(Δ[τ′](u′),δ[tar(τ′)]).
If one of the queues K[τ] runs empty (which can only happen if rank(τ) = 0), then K[τ] is removed from K.
We now establish an upper bound on the running time of the algorithm.
Algorithm 2 runs in time
For the proof, we look at the maximum number of instantiations that we encounter during a run of the algorithm.
Because K′ contains (at most) the m queues K[τ], enqueuing into K′ is in $O(log(m))$. Furthermore, each rule is limited to N instantiations, due to the fact that the sequences T[q] do not contain
repetitions and thus τ[u]≠τ[u′] for u≠u′. This implies that the maximum number of iterations of the main loop is Nm, yielding an upper bound of $O(Nmlog(m))$ for the management of K′.
Next, we have the rule-specific queues K[τ]. Each time a tuple u ∈ℕ^k is dequeued from K[τ], at most |inc(u)| = k ≤ r new tuples are enqueued. In total, the creation of these k tuples of size k each
takes k^2 ≤ r^2 operations. Thus, there will be at most Nr elements in K[τ] for any τ, which gives us a time bound of $O(log(Nr))$ per queue operation, a total time of $O(N(r2+rlog(Nr)))$ for the
management of K[τ], and thus a total of $O(Nm(r2+rlog(Nr)))$ for the m queues K[τ] altogether.
Summing up the upper bounds for the management of the two queue types yields
as claimed.
To complete the analysis, we have to argue that the time that needs to be spent to check whether τ[u] has been outputted before, can be made negligible. We do this by implementing the forest of
outputted trees in such a way that equal subtrees are shared. Hence, trees are equal if (and only if) they have the same address in memory. Assuming a good hashing function, the construction of τ[u]
from the (already previously constructed) trees referred to by u, can essentially be done in constant time. Now, if we maintain with every previously constructed tree a flag $✓$ indicating whether
that tree had already been outputted once, the test boils down to constructing τ[u] (which would return the already existing tree if it did exist) and checking the flag $✓$.
The running time of Algorithm 2 should be contrasted with the running time
of Algorithm 1 (Björklund, Drewes, and Zechner
). By handling instantiations of transition rules rather than states, we reduce the running times of the algorithm roughly by a factor of
We have not performed a detailed space complexity analysis, but because we know that the logarithmic factors are due to heap operations, we can conclude that the memory space consumption of the
algorithm is in O(Nm).
In practical applications, we expect to see Algorithm 2 used in two ways. The first is, as discussed in the Introduction, the situation in which N best trees are computed for a relatively small value
of N, for example, N = 200 as suggested by Socher et al. (2013). Here, based on a comparison of the upper bounds on the running time, and assuming that they are reasonably tight, Algorithm 2 will
outperform Algorithm 1 if N is larger than $m$, which we believe is the common case. In the second scenario, Algorithm 2 is invoked with a very large N to enumerate the trees recognized by the input
automaton, outputting them in ascending order by weight. In this scenario, our exploration by transition rule is even more valuable, as it saves redundant computation.
4The Algorithm Best Runs
We now recall the algorithm by Huang and Chiang (2005), which we henceforth will refer to as Best Runs. To facilitate comparison, we express Best Runs in terms of wta.
The type of wta used as input to Best Runs differs slightly from the one used in Definition 1 in that the admissible weight structures are not restricted to the tropical semiring. Instead, each
transition rule τ: f[q[1],…,q[k]] → q is equipped with a weight function $wtτ:R+k→R+$. The definition of the weight of a run ρ = τ[ρ[1],…,ρ[k]] is then changed to wt(ρ) =wt[τ](wt(ρ[1]),…,wt(ρ[k])).
For the algorithm to work, these weight functions wt[τ] are required to be monotonic: wt[τ](w[1],…,w[k]) ≥wt[τ](w[1]′,…,w[k]′) whenever w[i] ≥ w[i]′ for all i ∈ [k]. Definition 1, which Best Trees is
based on, corresponds to the special case where each of these weight functions is of the form $wtτ(w1,…,wk)=w+∑i∈[k]wi$ for a constant w. In other words, the weight functions Best Trees can work with
are a restriction of those Best Runs works on. This will be discussed in Section 5.
The input to Best Runs is a pair (M,N), where M is a wta with m transition rules and n states, and N ∈ℕ. The algorithm is outlined in Algorithm 3. Line 2 is a preprocessing step that can be performed
in O(m) time using the Viterbi algorithm, given that the rank of the alphabet used is considered a constant. A list input[q] is used to store, for each state q ∈ Q, the at most N discovered best runs
arriving at q. The search space of candidate runs is represented by an array of heaps, here denoted cands. For each state q, cands[q] holds a heap storing the (at most N) best, so far unexploited,
runs arriving at q. That is, if we have already picked the N′ best runs arriving at a node, the heap lets us pick the next best unpicked candidate efficiently when so requested by the recursive call.
To expand the search space, the N′-th best run ρ is used as follows: if τ = ([q[1],…,q[k]] → q), we obtain a new candidate by replacing the i-th direct subtree of ρ with the next (and thereby
minimally worse) run in input arriving at q[i]. Note that this is equivalent to the increment method used in Best Trees.
As shown by Huang and Chiang (2005), the worst case running time of Best Runs is $O(mlog|V|+smaxNlogN)$ where $smax$ is the size of the largest run among the N results.^^4 Using similar reasoning for
the memory complexity as for Best Trees, Best Runs achieves a $O(m+smaxN)$ memory complexity bound.
A major difference between Best Runs and Best Trees is that the former solves the N-best runs problem whereas the latter solves the N-best trees problem: On lines 14 and 17 of Algorithm 2, duplicate
trees are discarded. If these conditions are removed, Best Trees solves the best runs problem. (Provided, of course, that the objects outputted are changed to being runs rather than trees.)
For the sake of comparison, we adopt the view of Huang and Chiang (2005) that the ranked alphabet Σ can be considered fixed. This yields that the running time of Best Trees is $O(Nm(logm+logN))$.
Since m ∈ O(n^r +1), where r is the (now fixed) maximal rank of symbols in Σ, it follows that $logm∈O(logn)$, so the second expression simplifies to $Nm(logn+logN)≤mlogn⋅NlogN$. Moreover, recall
that the worst case running time of Best Runs is $O(mlogn+smaxNlogN)$. If N is large enough to make the second term the dominating one, the difference between both running times is thus a factor of
$(mlogn)/smax$ (assuming for the sake of the comparison that the given bounds are reasonably tight). A further comparison between the running times does not appear to be all that meaningful because
$smax$ depends on both N and the structure of the input wta, and the algorithms are specialized for different problems.
A conceptual comparison of the two algorithms may be more insightful. The algorithms differ mainly in two ways. The first difference is that, while Best Runs enumerates candidates of best runs using
one priority queue per node of G, Best Trees uses a more fine-grained approach, maintaining one priority queue per hyperedge. Since the upper bound on the length of queues is O(N) in both cases, in
total Best Trees may need to handle O(Nm) candidates whereas Best Runs needs only O(Nn). This ostensible disadvantage of Best Trees is not a real one as it occurs only when solving the N-best trees
problem. The difference vanishes if the algorithm is used to compute best runs (by changing lines 14 and 17). To see this, consider a given state q ∈ Q. Each time a queue element (i[1], …, i[k]) is
dequeued from a queue K[τ] with τ: f[q[1],…,q[k]] → q, the corresponding run is appended to the list T[q] on line 15, and k queue elements are inserted on line 21. As this can only happen at most N
times per state q, in total only O(Nn) queue elements are ever created in the worst case. In other words, if Best Trees is set to solve the N-best runs problem, the splitting of queues does not
result in a disadvantage compared to Best Runs.
The second conceptual difference between the algorithms is that
Best Trees
adopts an optimization technique known from the
-best strings algorithm by Mohri and Riley (
). It precomputes and uses (the weight and depth of) a best context
for every state
in order to explore the search space in a more goal-oriented fashion. The possibility of using this optimization depends on the use of the tropical semiring. As mentioned in the Introduction, Büchse
et al. (
) extend the algorithm of Huang and Chiang to structured weight domains. This does not seem to be possible for Algorithm 2 (nor for Algorithm 1). The reason is that the computation and use of best
contexts requires that the semiring is extremal, which is the case for the tropical semiring, but not for structured weight domains in general. Let
be the maximum of the distances of states in
, that is,
) denotes the length of the shortest sequence of transition rules τ
such that
) ∈
) for all
∈ [
], and
). The argument used to show Claim 1 in the proof of
Lemma 2
yields that, at every point in time at most
further loop executions are made before the next best run is outputted. To see this, consider an execution of the main loop, in which a run ρ = τ[⋯ ] arriving at a state
is constructed. The priority of the corresponding queue element on line 13 is given by (wt(ρ) +
), where
are the weight and depth of
. If
, then
, and thus it is of the form
, where the ρ
are the 1-best runs arriving at their respective nodes (see
Figure 3
It follows that line 21 inserts the element into K[τ] that represents the run ρ′ = τ′[ρ[1],…,ρ[i−1],ρ,ρ[i +1],…,ρ[k]]. Because each of the runs ρ[i], arriving at some state q[i], is already in the
respective list $Tqi$, the run ρ′ immediately becomes a current candidate arriving at q = tar(τ). The corresponding pair (w[q],d[q]) = (wt(c[q]),depth(c[q])) satisfies w[q] + wt(ρ′) = w + wt(ρ) and
d[q] = d − 1. Hence, ρ′ (or another run with the same priority) will be picked in the next loop execution, meaning that after at most ℓ steps the node arrived at by the constructed run will be q[f],
that is, the run will be outputted. This also shows that queues the elements of which are not needed for generating output trees will never become filled beyond their initial element.
We end the comparison by looking at Table 2, which summarizes the discussion above and provides comparison data for the other N-best algorithms discussed in this article. Keep in mind that N must be
interpreted differently depending on the problem at hand. For example, the output of Best Runs is not equivalent to the output of Best Trees (unless the input is deterministic), which is why we
cannot directly compare the time complexities. This output inequivalence should also be considered in Section 7 where, for simplicity, we plot data for Best Runs and Best Trees side by side.
Table 2
Algorithm . Objects . Time complexity . Best contexts . Search-space expansion .
Eppstein (1998) Paths $O(nlogn+Nn+m)$ No Adding sidetracks to implicit heap representations of paths
Mohri and Riley (2002) Strings No formal analysis provided Yes On-the-fly determinization
Huang and Chiang (2005) Runs $O(mlogn+smaxNlogN)$^^5 No Increment
Best Trees v.1 Trees $O(N2n(n2+mlogN))$ Yes Eppstein’s algorithm
Best Trees Trees $O(Nm(logm+logN))$ Yes Increment
Runs $O(N(logm+logN))$^^6 Yes Increment
Algorithm . Objects . Time complexity . Best contexts . Search-space expansion .
Eppstein (1998) Paths $O(nlogn+Nn+m)$ No Adding sidetracks to implicit heap representations of paths
Mohri and Riley (2002) Strings No formal analysis provided Yes On-the-fly determinization
Huang and Chiang (2005) Runs $O(mlogn+smaxNlogN)$^^5 No Increment
Best Trees v.1 Trees $O(N2n(n2+mlogN))$ Yes Eppstein’s algorithm
Best Trees Trees $O(Nm(logm+logN))$ Yes Increment
Runs $O(N(logm+logN))$^^6 Yes Increment
6Implementation Details
In the upcoming section, we experimentally compare Best Runs and Best Trees. In preparation of that, we want to make a few comments on the implementation of Best Trees. As previously mentioned, Best
Runs is implemented in the Java toolkit Tiburon by May and Knight, and this is the implementation we use in our experiments. Therefore, we simply refer to the Tiburon GitHub page^^7 for
implementation details.
We have extended our code repository Betty,^^8 which originally provided an implementation of Best Trees v.1, to additionally implement the improved Best Trees as its standard choice of algorithm. A
flag -runs can be passed on as an argument to compute best runs instead of best trees. Below follow a number of central facts about the Betty implementation.
First, recall Algorithm 2, and in particular that the best tree $tqbest$ is inserted into T[q] for all q ∈ Q prior to the start of the main loop rather than letting T[q] be empty and initializing K
[τ] to 1^rankτ for all q ∈ Q and τ ∈ R. The reason is that the latter would not guarantee that at most ℓ iterations are made until a run is outputted because some ρ[i] may still not be in T[q]. If
this happens, transition rules adding the weight 0 may repeatedly be picked because some other τ′ is not yet enabled. However, with a trick the initialization can nevertheless be simplified as
indicated. The idea is to delay the execution of line 22 for every queue K[τ] until τ has actually appeared in an output tree. Thus, until this has happened, K[τ] is disabled from contributing
another tree to T[tar(q)]. This variant turned out to have efficiency advantages in practice and is therefore the variant implemented in Betty. Another advantage is that it allows Betty to handle a
set of final states rather than a single one. This is not possible with the original initialization of Algorithm 2, because line 3 would have to be generalized to outputting the best trees of all
final states, which cannot be done while maintaining the correctness of the algorithm, since the second best tree for a state q may have a lesser weight than the best tree for another state q′.
In lines 14 and 17 of Algorithm 2, trees are checked for equivalence. To perform the comparisons efficiently, we make use of hash tables. We use immutable trees, which need to be hashed only once
when they are created; we then save the hash code together with the tree to make the former accessible in constant time for each tree. To compare trees for inequality, their hash codes are compared.
If equal (which seldom happens if the trees are not equal), the comparison is continued recursively on the direct subtrees. This is theoretically less efficient than the method of representing trees
uniquely in memory (as described in the last paragraph of the proof of Theorem 2), but practically sufficient and much easier to implement.
When creating new tuples for a transition rule τ as given by line 21, we add the tuples that can be instantiated directly to the corresponding queue K[τ]. The tuples that cannot be instantiated must,
however, be stored until they can. We want to be able to efficiently access the tuples that are affected when adding a tree t to T[q′] for some q′ ∈ Q (line 15). Therefore, we connect each tuple to
the memory locations that will contain the data needed by the tuple. In more detail: Let τ = ( f[q[1]⋯q[k]] → q) be any transition rule in R and let (i[1],…,i[k]) be a tuple originating from τ. For
every i[j] ∈{i[1],…,i[k]}, the tuple is saved in a list of tuples affected by $Tqj(ij)$ and marked with a counter that shows how many trees remain until it can be instantiated. Thus, when t is added
to T[q′](i[j]) for i[j] ∈{i[1],…,i[k]} and q′ = q[j], we can immediately access all tuples that can possibly be instantiated. If a tuple cannot be instantiated, its counter is decreased
Let us now describe our experiments. For each problem instance (i.e., combination of input file and value of N), we perform a number of test runs, measure the elapsed time, and compute the average
over the test runs. To avoid noise in our data caused by, for example, garbage collection, we only measure the time consumed by the thread the actual application runs in. Also, we disregard the time
it takes to read the input files.
The number of test runs that are performed per problem instance is decided by the relationship between the mean μ and the standard deviation σ of the recorded times—these values are computed every
fifth test run, and to finish the testing of the current problem instance, we require that σ < 0.01μ. However, five test runs per problem instance has turned out to be sufficient for fulfilling the
requirement in practically all instances seen.
The memory usage is measured in terms of the maximum resident set size of the application process by using the Linux time command. Moreover, the strategy described above for computing the average
over several test runs is used here as well.
All test scripts have been written in Python, in contrast to the tested implementations, which use Java. We run the experiments on a computer with a 3.60 Hz Intel Core i7-4790 processor.
The corpora that were used in our experiments (see the list below) contain both real-world and synthetic data. The first corpus is derived from an actual machine-translation system and is thus
representative for real-world usage. The second corpus consists of manually engineered grammars for a set of natural languages, used in a range of research and industry applications. The last two
corpora are artificially created for the purpose of investigating the effect of increasing degrees of nondeterminism. Let us now present each in closer detail.
• MT-data
This data set consists of tree automata resulting from an English-to-German machine-translation task, described in the doctoral thesis of Quernheim (2017). The data consists of 927 files, each
file containing a wta corresponding to one sentence; the files are indexed from 0 to 926. The smallest file has 24 lines and the largest one has 338, 937 lines; all lines but the first hold a
transition rule. Moreover, these wtas have a large number of states and are essentially deterministic in the sense that each state corresponds to a particular part-of-sentence structure and input
• GF-data
Unweighted context-free grammars from the Grammatical Framework (GF) by Ranta (2011) provided the basis for this corpus. GF is, among other things, a programming language and processing platform
for multilingual grammar applications. It provides combinatory categorial grammars (Steedman 1987) for more than 60 natural languages, out of which we export a subset to context-free grammars in
Backus–Naur form with the help of the built-in export tool, and assign every grammar rule the weight 1. The number of production rules varies between languages: The Latin grammar has, for
example, 1.6 million productions, whereas the Italian grammar has 5.8 million. (These differences are due to the level of coverage chosen by the grammar designers, and do not necessarily reflect
inherent complexities of the languages.)
• PolyNonDet
This is a family of automata of increasing size indexed by i. The states of member i are q[0],…,q[i], where q[i] is the final state and the rules are:
□ $a→0qj$ for j = 0,…,i
□ $f[qj,qj]→1qj$ for j = 0,…,i
□ $f[qj,qj−1]→1qj−1$ for j = 1,…,i
There are then Θ(n^i) runs for a tree of size n (and the number of rules grows linearly). Therefore, we call these polynomially nondeterministic.
• ExpNonDet
Finally, we use another family of automata of increasing size, also indexed by i and with states q[0],…,q[i] and a final state q[f]. The rules for member i of the family for j, k = 0,…,i and j≠k
□ $a→0qj$
□ $f[qj,qk]→1qj$
□ $f[qk,qj]→1qj$
□ $qj→0qf$
Thus, the number of rules grows quadratically in i and the number of runs for a tree of size n is Ω((i +1)^n); we say that these are exponentially nondeterministic.
The variables in all result-displaying plots in this article are either N (the number of trees or runs to output), m (the number of transition rules, i.e., lines in the input file), or both. For the
artificially created corpora PolyNonDet and ExpNonDet, we exchange m for the variable i with which they are indexed. Note that for the former, m and i are interchangeable, but for the latter, m grows
quadratically with increasing i.
7.2Comparison with Best Trees v.1
First, we verify that the algorithm Best Trees is at least as efficient as its predecessor Best Trees v.1. Therefore, we run experiments on the ExpNonDet data set for both of the algorithms—both
solving the best trees problem. The ExpNonDet data set was chosen because it has the largest degree of nondeterminism achievable, which should challenge both of the algorithms maximally. The results
are presented in figures 4 and 5 for varying N and i, respectively. Note that both of these plots have logarithmic y axes. We see that Best Trees outperforms Best Trees v.1 considerably for both
increasing N and increasing i.
More interestingly, in Figure 4, Best Trees displays a step-like behavior at N ≈ 100, 200, and 600. The nature of the ExpNonDet corpus is the explanation of this: When we have found all of the
distinct trees of size s (all of the same weight), then the algorithm has to discard all of the duplicates of size s that are still in the queue, before arriving at a tree of size s + 1 with higher
7.3Comparison with Tiburon
We have presented an improved version of the algorithm by Björklund, Drewes, and Zechner (2019) that solves the N-best trees problem for weighted tree automata over the tropical semiring. The main
novelty lies in the exploration of the search space with a focus on instantiations of transition rules rather than on states, and the lazy assembly of these instantiations. We have proved the new
algorithm to be correct and derived an upper bound on its running time—a bound that is smaller than that of the previous algorithm. We believe that this speed-up makes it superior for usage in
typical language-processing applications.
Moreover, we have complemented the theoretical work with an experimental evaluation. Because Best Trees can be easily modified to produce the best runs instead of the best trees, we considered both
tasks in our evaluation. To achieve a large coverage, we used two types of data: data from real-world language processing tasks and artificially created data. The real-world data consist of machine
translation output and corpus-based rule sets for natural languages; these corpora are meant to display the kind of behavior one can expect when applying our algorithm to language processing tasks.
The artificial corpora were designed to expose Best Trees to its worst-case scenario for the best trees task: the case when we have an exponential number of duplicate trees in a best runs list. We
also covered the more moderate case where the nondeterminism only gives rise to a polynomial number of duplicates.
In the experiments focusing on running time, we first used the exponentially nondeterministic data to show that Best Trees is better at the best trees task than its predecessor Best Trees v.1 that
uses a less efficient pruning scheme. Then, we compared Best Trees with the state-of-the-art best-runs algorithm of Huang and Chiang (2005), implemented in Tiburon by May and Knight (2006) for both
tasks on all data sets. The results made it clear that Best Trees is preferable when extracting N-best lists of both runs and trees: Betty outperforms Tiburon for the best runs task on all 2,269
input wtas except one. The single exception revealed a corner case where the Huang and Chiang algorithm is faster, namely, when there are millions of rules and only a small percentage of them are
used to produce N very small (height 0 or 1) runs. Additionally, we performed a smaller number of experiments to measure the memory usage of the three applications, and, while no final conclusion
could be made, it seemed as though Tiburon had an advantage over Best Trees with respect to memory usage in total.
Prior to the experiments, we compared the two algorithms at a conceptual level and discussed the expected effects on their running time. The allocation of queues to transition rules instead of states
mainly serves to structure the implementation. As we saw, it does not have a disadvantage with respect to running time. The use of best contexts, generalizing the idea of Mohri and Riley (2002) to
trees, has a positive effect: it ensures that the maximum distance of a state to the final state is an upper bound on the maximum number of main loop iterations before the next run is outputted. This
is because the best contexts guide the algorithm to take the shortest route in constructing the next best run that reaches a final state.^^9
The disadvantage of using best contexts is that it limits which semirings can be used. The technique is compatible with the tropical semiring and, as discussed in Section 2, with the equivalent
Viterbi semiring, but seemingly not with semirings that are not extremal. It is currently unclear to us whether an appropriate extension is possible, and we leave this question for future work.
However, such extensions seem only relevant for the N-best runs problem, because it appears highly unlikely that the N-best trees problem could ever be solved with reasonable efficiency in cases
where the weight semiring is not extremal. The reason is that, in that case, the best run on a tree does not determine the weight of that tree, making it unclear how the problem can be solved even if
disregarding efficiency aspects. However, because the tropical semiring and the Viterbi semiring are the most prominent ones used to rank hypotheses in NLP, the computational advantages seen in this
article seem to justify the restriction to these semirings, even when looking only at the N-best runs problem.
We are thankful to Andreas Maletti for providing the machine translation data set used in our experiments; to Jonathan May for his support in the application of Tiburon to the N-best problem; to
André Berg for sharing his expertise in mathematical statistics; to Aarne Ranta, Peter Ljunglöf, and Krasimir Angelov for introducing us to Grammatical Framework; and to the reviewers for suggesting
numerous improvements to the article.
This replacement had already been done in Tiburon, without an explicit remark.
We assume that enq(K[τ],U) enqueues all u ∈U in K[τ] except those already in it, thus skipping duplicates.
Recall that $(Tqj)ij$ denotes the i[j]-th element of the sequence $Tqj$.
In fact, the running time obtained by Huang and Chiang (2005) is $O(m+smaxNlogN)$, but this assumes (the graph representation of) M to be acyclic. When M is cyclic, line 2 must be implemented by
using Knuth’s algorithm, resulting in an additional factor $logn$ in the first term.
Allows for cyclic input wta; $smax$ is the size of the largest output.
Note that a factor m is removed, compared with when the same algorithm is used for finding the best trees. This is because all runs originating at the same rule queue are distinct (and naturally the
same also holds for different rule queues).
We have conducted a small experiment not mentioned in the previous section, by switching off that feature. It showed that the use of best contexts (in that particular, randomly chosen case) reduced
the size of rule queues by a factor of 10.
Mohan Sondhi
, and
, editors.
Springer Handbook of Speech Processing
, and
Between a rock and a hard place—parsing for hyperedge replacement DAG grammars
. In
10th International Conference on Language and Automata Theory and Applications
, pages
, and
A comparison of two n-best extraction methods for weighted tree automata
. In
23rd International Conference on the Implementation and Application of Automata (CIAA 2018)
Lecture Notes in Computer Science
, pages
, and
Efficient enumeration of weighted tree languages over the tropical semiring
Journal of Computer and System Sciences
, and
n-best parsing revisited
. In
Proceedings of the 2010 Workshop on Applications of Tree Automata in Natural Language Processing
, pages
Thomas H.
Charles E.
Ronald L.
, and
Introduction to Algorithms
The MIT Press
Edsger Wybe
A note on two problems in connexion with graphs
Numerische Mathematik
Finding the k shortest paths
SIAM Journal on Computing
Jenny Rose
Christopher D.
, and
Andrew Y.
Solving the problem of cascading errors: Approximate Bayesian inference for linguistic annotation pipelines
. In
Proceedings of the 2006 Conference on Empirical Methods in Natural Language Processing
, pages
Better k-best parsing
. In
Proceedings of the Conference on Parsing Technology 2005
, pages
Víctor M.
Computation of the N best parse trees for weighted and stochastic context-free grammars
. In
Advances in Pattern Recognition
, pages
Best Trees Extraction and Contextual Grammars for Language Processing
. Ph.D. thesis,
Umeå University
James H.
Speech and Language Processing: An Introduction to Natural Language Processing, Computational Linguistics, and Speech Recognition
Pearson Prentice Hall
An overview of probabilistic tree transducers for natural language processing
. In
International Conference on Intelligent Text Processing and Computational Linguistics
, pages
Donald E.
Computer Programming as an Art
Communications of the ACM
Donald E.
A generalization of Dijkstra’s algorithm
Information Processing Letters
Rune B.
Christian N. S.
The consensus string problem and the complexity of comparing hidden Markov models
Journal of Computer and System Sciences
Special Issue on Computational Biology 2002
Tiburon: A weighted tree automata toolkit
. In
International Conference on Implementation and Application of Automata
, pages
An efficient algorithm for the n-best-strings problem
. In
Proceedings of the Conference on Spoken Language Processing (ICSLP 02)
, pages
Bimorphism Machine Translation
. Ph.D. thesis,
Universität Leipzig
Grammatical Framework: Programming with Multilingual Grammars
CSLI Publications
Christopher D.
, and
Andrew Y.
Parsing with compositional vector grammars
. In
Proceedings of the 51st Annual Meeting of the Association for Computational Linguistics
, volume
, pages
Combinatory grammars and parasitic gaps
Natural Language and Linguistic Theory
, and
Gaussian mixture latent vector grammars
. In
Proceedings of the 56th Annual Meeting of the Association for Computational Linguistics, ACL 2018, Volume 1: Long Papers
, pages
© 2022 Association for Computational Linguistics. Published under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) license.
Association for Computational Linguistics. Published under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) license.
This is an open-access article distributed under the terms of the
Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License
, which permits you to copy and redistribute in any medium or format, for non-commercial use only, provided that the original work is not remixed, transformed, or built upon, and that appropriate
credit to the original source is given. For a full description of the license, please visit | {"url":"https://direct.mit.edu/coli/article/48/1/119/108848/Improved-N-Best-Extraction-with-an-Evaluation-on","timestamp":"2024-11-05T08:03:30Z","content_type":"text/html","content_length":"581409","record_id":"<urn:uuid:e641bc72-f345-406d-be34-b934f89e34d4>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00546.warc.gz"} |
The Home Math Environment and Children's Math Achievment: A Meta-Analysis
Mathematical thinking is in high demand in the global market, but approximately six percent of school-age children across the globe experience math difficulties (Shalev, et al., 2000). The home math
environment (HME), which includes all math-related activities, attitudes, beliefs, expectations, and utterances in the home, may be associated with children’s math development. In order to examine
the relation between the HME and children’s math abilities, a preregistered meta-analysis was conducted to estimate the average weighted correlation coefficient (r) between the HME and children’s
math achievement and how potential moderators (i.e., assessment, study, and sample features) might contribute to study heterogeneity. A multilevel correlated effects model using 631 effect sizes from
64 quantitative studies comprised of 68 independent samples found a positive, statistically significant average weighted correlation of r = .13 (SE = .02, p < .001). Our combined sensitivity analyses
showed that the present findings were robust, and that the sample of studies has evidential value. A number of assessment, study, and sample characteristics contributed to study heterogeneity,
showing that no single feature of HME research was driving the large between-study differences found for the association between the HME and children’s math achievement. These findings indicate that
children’s environments and interactions related to their learning are supported in the specific context of math learning. Our results also show that the HME represents a setting in which children
learn about math through social interactions with their caregivers (Vygotsky, 1978), and what they learn depends on the influence of many levels of environmental input (Bronfenbrenner, 1979) and the
specificity of input children receive (Bornstein, 2002).
Public Significance Statement: The findings of this meta-analysis suggest that children’s home math environments (e.g., parent-child math interactions) are positively associated with children’s math
achievement. To promote children’s math skills, it may be beneficial to support parents in providing positive home math experiences for their children.
Project Active From
October 2018 to April 2021
Funding Agency / Grant Number
Eunice Kennedy Shriver National Institute of Child Health and Human Development / R01HD052120 | {"url":"https://ldbase.org/projects/7f57a2bb-903c-4b58-a212-f36ab559ec69","timestamp":"2024-11-10T02:16:08Z","content_type":"text/html","content_length":"70562","record_id":"<urn:uuid:b91dce4b-9323-4333-9eff-e86155babb57>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00735.warc.gz"} |
Add and subtract two numbers that bridge through 10 | Oak National Academy
Hello everyone.
Welcome back to another maths lesson with me, Mrs. Pochciol.
As always, I can't wait to learn lots of new things and hopefully have lots of fun.
So let's get started.
This lesson is called add and subtract two numbers that bridge through 10 and it comes from the unit calculating within 20.
By the end of this lesson, you should be able to add and subtract using strategies to bridge through 10.
Let's have a look at this lesson's keywords.
Whole, part, partition, and bridging 10.
Let's practise.
My turn.
Your turn.
My turn.
Your turn.
My turn.
Your turn.
My turn.
Bridging 10.
Your turn.
Now that we've practised these words, let's use them.
Let's have a look at our lesson outline.
The first part of our learning, we are going to be adding two addends that bridge through 10.
And in the second part of our learning, we are going to be subtracting two numbers that bridge through 10.
Are we ready to get started? Let's start with the first part, adding two addends that bridge through 10.
In this lesson, we are going to meet Laura and Andeep.
They're going to help us with our learning today.
Are we ready guys? Let's go.
Laura and Andeep are exploring an equation.
7 plus 5 is equal to something.
Andeep knows that the sum will be more than 10.
Laura explains that when a sum is more than 10, we use a strategy called bridging 10.
Laura and Andeep now solve this problem using this known strategy.
First we partition 5 into 3 and 2.
Remember so we can make 10 first, and we know that 7 and 3 is equal to 10.
Then we make 10.
So 7 add 3 is equal to 10 and we record that equation underneath.
Then we bridge the 10 because now we are going to add to go more than 10.
We need to add the other parts when we partition 5.
So 10 plus 2 is equal to 12.
So we now know that 7 plus 5 is equal to 12.
Did you see where we bridged 10? That simply means going over 10.
They now represent the addition on a number line.
Can you see? We are going to start with 7.
We are gonna add 3 to 7 because we want to first make 10.
Then we bridge through 10, we go past 10, and we add 2 more because that's the other part that we still need to add when we partitioned our 5, remember? We now have a three addend edition.
Look at the number line.
We now have 7 plus 3 plus 2.
We are all superstars at three addend additions now, aren't we guys? 7 add 3 is equal to 10.
10 add 2 is equal to 12.
So 7 plus 5 is equal to 12.
How easy is it to bridge through 10 when you can see it on a number line like that.
We changed 7 plus 5 into 7 plus 3 plus 2 by partitioning the 5.
We have now recorded our strategy in one equation.
We partitioned 5 into 3 and 2, can you see that in our equation? And we added 7 and 3 to make 10.
Then we added 10 and 2 to find the sum, which was 12.
All of that information in our one equation.
Laura now shows a bridging 10 edition using this method.
She's going to do 9 plus 6.
Let's have a look at what she does.
Hmm, can you help her to create the equation to show her strategy? You can see that the equation's already been started for you.
Can you fill in the missing numbers to complete her equation? Pause this video, complete the equation, and come on back when you are ready to see how you got on.
Welcome back.
Let's have a look at how you got on.
Come on then, Laura, let's complete this equation.
First we had to partition 6 into 1 and 5.
So first we added 9 and 1 to make 10.
Can we see? We first added that 1, you can see that that adding one step is the first thing that we did.
Then 10 add 5 is equal to 15.
So the next part of our equation is plus 5 because we partitioned the 6 into 1 and 5.
We're now adding that second part and we can see that the sum will be 15.
So 9 plus 6 is equal to 15.
Let's have a look at Laura's equation to see if yours is the same.
9 plus 6 is equal to 9 plus 1 plus 5, which is equal to 15.
Well done if you manage to get that equation.
Laura and Andeep are now finding out how much fruit is left after their snack time.
There are 8 apples in the large box and 6 apples in the small box.
How many apples do they have all together? Hmm.
Laura notices that we need to add together 8 apples and 6 apples to find out how many they have altogether.
Andeep says that they're experts at bridging 10 now.
So let's do this.
They represent the addition as an equation and show their working out on the number line.
So we're going to start on 8.
What are we going to have to do first? We need to make 10 first, remember? So how are we going to partition our 6? Ooh, we know that adding 2 and 8 will make 10, so we need to partition the 6 into 2
and 4.
So 8 and 2 is equal to 10.
Then we need to add that other part which is 4.
So 10 add 4 is equal to 14.
Can you see that they've recorded that three addend addition now in their equation 8 plus 2 plus 4? And we now know that 8 plus 6 is equal to 14, so we can also add that to our equation.
Our completed equation is now 8 plus 6 is equal to 8 plus 2 plus 4 because we partitioned that 6, which is equal to 14.
So we can now confidently say that 8 plus 6 is equal to 14.
Well done guys.
A really good use of that bridging 10 strategy there.
So can we retell our story now then, guys? There were 8 apples in the large box and 6 apples in the small box.
There are now 14 apples all together.
Wow, well done guys.
You really are experts at this now.
Should we have another go? Andeep now finds out how many bananas they have left, but he needs a little bit of help because there's a part of his equation missing.
Laura notices that Andeep is adding 6 again just like they just did, but this time they're going to have to partition it differently.
We know that 7 and 3 pair to make 10, so he's going to need to partition his 6 into 3 and 3 this time, whereas remember before they did 2 and 4.
This time it's 3 and 3.
Now we need to add the other 3 because that's the other part, so that's the bit that you missed, Andeep.
We need to add 3.
We know that 10 add 3 is equal to 13, so Andeep must have had 13 bananas left.
Well done if you spotted that.
Over to you then.
They finally calculate how many oranges there are.
There are 5 oranges in the small box and 8 oranges in the large box.
How many oranges do they have altogether? Can you complete the equation to calculate how many oranges there are? Pause this video.
Have a think about how you might partition 8 to help you with this calculation and find the sum of how many oranges altogether.
Come on back once you've got an answer.
Welcome back.
I hope you enjoyed there calculating those oranges.
Should we have a look? There are 5 oranges in the small box and 8 oranges in the large box.
How many oranges do they have altogether? We know that 5 and 5 is equal to 10, so we need to partition 8 into 5 and 3.
Once we've added our 5 and 5 to make 10, we're going to add the 3, which we know 10 add 3 is equal to 13.
So we can now confidently say that there are 13 oranges altogether.
Well done to you if you manage to complete that equation correctly.
Over to you then for task A, let's keep practising this bridging 10 strategy.
Fill in the missing numbers to find the sum of the equation.
So you will see that A and B, you are finding that other part that you need to add to find the sum.
In C and D, you have to partition the second addend to complete the equation and find the sum.
And in E, you have to show all of that second expression to find the sum.
Pause this video, have a go at completing those equations and finding the sums and come on back to see how you get on.
Welcome back.
I hope like Laura and Andeep, you're now thinking that you are experts at this bridging 10 strategy.
Let's have a look at how we got on.
7 plus 6, we can see that we've already done the first part.
We know that 7 and 3 make 10, so what's the other part that we partition 6 into? Hmm, 3 and 3.
So now we know that 10 plus 3 is equal to 13.
Well done if you've got that one.
Let's have a look at 8 and 6.
You can already see that we've partitioned the 6 into 2 and something, what's the other part? We know that 2 and 4 make 6.
So 8 plus 2 is equal to 10 plus 4 more is 14.
So we can say that 8 plus 6 is equal to 14.
Well done if you've got that one right.
This next one then we need to partition the 6 by ourself.
So we know that we've got 5 as our first addend, so how do we need to partition 6? Well I know that 5 plus 5 is equal to 10 and 5 and 1 make 6.
So now we can see that it's 5 plus 5 plus 1.
5 and 5 make 10 plus 1 more, we know is 11.
Well done to you, if you got 11.
D then, 9 plus 6.
Oh we're adding 6 again, but this time I'm going to have to partition it differently because I've got 9 as my first addend.
I know that 9 and 1 make 10, so this time I'm going to have to partition 6 into 1 and 5.
9 and 1 are equal to 10, plus 5 is equal to 15.
And finally, 8 plus 8.
We need to do this whole second expression.
So we know we are going to start with 8 because that's our first addend, and 8 and 2 make 10.
So we've partitioned that second addend into 2 and 6.
So 8 plus 2 is equal to 10, plus 6 is equal to 16.
So 8 plus 8 must be equal to 16.
Welcome to you if you got those correct.
Oh, did you notice that 6 had to be partitioned in different ways each time to bridge through 10? Look, A, B, C, and D, all were adding 6 and each time we partitioned it differently to help us bridge
through 10.
A good spot there Laura.
Oh, and Andeep noticed that actually in E, he didn't need to bridge through 10 because he could have just used his doubling knowledge.
8 plus 8 is equal to 16.
Well done to you if you spotted that.
Let's move on then to the second part of our learning.
In the first part, we've been practising adding two addends that bridge through 10.
Now we're going to apply our learning into subtracting to bridge through 10.
Are we ready? Let's go.
There are 13 children on the bus.
5 children gets off the bus.
There are now 8 children on the bus.
Can we see? 13 subtract 5 is equal to 8.
Now let's have a look at this.
Andeep notices that we have just bridged through 10 just like we did when we were adding, but this time we're subtracting.
3 children have to go from the top deck, which leaves 10 children altogether.
Then 2 children leave from the bottom deck, which leaves us with 8.
Can you see how first the 3 children got off and then 2 children got off? Let's represent this on our 10 frame to have a look.
There were 13 children on the bus and 5 got off, 3 children got off the top deck, which leaves us with 10 children, and 2 children get off from the bottom deck, which leaves us with 8 children.
There are now 8 children on the bus.
Ooh, can we see our written strategy on the right-hand side there? It looks very similar to when we were adding weren't we? Should we have a look then? First we took away 3 counters to make 10.
Then we took away another 2 counters to make 8 because we partitioned 5 into 3 and 2.
We are still making 10 and then subtracting the other part.
Let's have a look on our number line.
We know that subtracting 3 from 13 is equal to 10.
We have partitioned the minuend 5 into 3 and 2.
So now we need to subtract 2, which is the other part.
If we subtract 2 from 10, that leaves us with 8.
So 13 subtract 5 is equal to 8.
Can you see how it's the same bridging strategy that we use when we were adding, but this time we are subtracting? This is a really great strategy to help us subtract through 10.
Let's have a practise then.
The next day on snack duty, Laura and Andeep handout the daily fruit.
There are 15 apples in the box.
They give out 8 apples.
How many apples are left in the box? We need to subtract 8 from 15.
Andeep thinks that we can do this.
Let's get our 10 frames out.
Let's represent 15 subtract 8 on our 10 frame and complete our written strategy at the same time.
We need to subtract 5 ones to equal to 10 because we had 15 remember? So we're gonna subtract to 10, which is 5.
So first we partition 8 into 5 and 3.
Can you see that there on our written strategy? First we're going to subtract the 5 from 15 to leave us with 10.
Can we see? Then we're going to subtract the other 3 because that's the other part from 10 which will leave us with 7.
So we know that 15 subtract 8 is equal to 7.
Can we see all of our steps there on our written method? First we partitioned the 8 into 5 and 3, we subtracted the 5 to make 10, and then we subtracted the other part which was 3, which left us with
So we now know that 15 subtract 8 is equal to 7.
Let's have a practise of this then over to you.
Can you use Andeep stem sentence to help you to complete this problem? 13 subtract 6.
Use your 10 frame to help you and complete the stem sentence and the written strategy to show what you've done.
Pause this video, have a go at finding an answer, and come on back once you're ready to see how you got on.
Welcome back.
I'm hoping you enjoyed practising that strategy there.
Shall we see how you get on? So 13 subtract 6.
Hmm, 13, if I want to make 10, that means I need to subtract 3.
So first I'm going to have to partition 6 into 3 and 3.
Then we're going to subtract 3 from 13, which will give us 10.
There we go.
And then we need to subtract the other part which is 3.
So 10 subtract 3 is equal to 7.
So we can now say that 13 subtract 6 is equal to 7.
Well done to you if you completed that strategy and found that the missing answer was 7.
Laura now shows the apple subtraction on a number line.
So we're gonna start with the number 15.
And what we're going to do first, Laura? We partitioned 8 into 5 and 3.
So first we had to subtract 5 to get to 10.
Then we subtracted 3 because that's the other part that we partitioned 8 into.
This has led us to 7.
So 15 subtract 8 is equal to 7.
Can you see those two steps there to bridge 10? Over to you then.
Can you now represent your strategy on a number line? Remember, what number do we start on? What do we subtract first to make 10? Then what do we subtract the other part? And what number do we
finally end up on? Pause this video, complete your number line, and come on back to see how you get on.
Welcome back.
Let's have a look at what our number line should have looked like then.
So we are going to start with 13 because that was our starting number.
Then we had to subtract 3 to get to 10.
There's the first step.
I now need to subtract the other part.
So 10 subtract 3 is equal to 7.
We can now show that 13 subtract 6 is equal to 7.
So well done to you if your number line looks like mine.
Laura and Andeep now both apply their learning to solve a missing number problem.
Laura's got a bar model and Andeep's got a part-part-whole model.
Laura notices that her missing number is a part, so she needs to subtract the other part from the whole to find the missing number.
We need to solve 11 subtract 5.
We can see that that's going to bridge 10, can't we? Because 1 is a smaller ones number than 5, so that's definitely going to bridge 10.
Andeep has both of his parts but not the whole.
So he knows that he needs to add the two parts together to find the whole.
He's going to solve 5 plus 6.
Hmm, is that going to bridge 10? I think it might do.
Let's have a look at then Laura.
Laura is going to solve 11 subtract 5.
She partitions her 5 into 1 and 4 because remember we need to subtract that 1 first to make 10.
11 subtract 1 will make 10.
And 10 subtract 4, the other part from where we partitioned 5 will leave us with 6.
So Laura's missing part is 6.
Andeep partitions 6 into 5 and 1.
5 plus 5 is equal to 10 and 10 plus 1 is equal to 11.
Andeep's missing whole is 11 because 5 plus 6 is equal to 11.
Hmm, do we notice something there? Laura and Andeep noticed something about their missing number problems. Laura notices that the parts and the wholes are the same, but Andeep did an addition and
Laura did a subtraction.
That's because subtraction and addition can undo each other, remember? They are inverse operations.
If we know that 5 plus 6 is equal to 11, then we know that 11 subtract 5 is equal to 6 and vice versa.
Oh, so remember, when we are thinking about missing number problems, we can use this knowledge of addition or subtraction to help us.
So let's have a practise of this then.
What is the missing number in this equation? 14 subtract something is equal to 8.
We're going to show this as a part-part-whole model because that's going to help us to visualise this problem.
We know that 14 is the whole and subtracting something is equal to 8.
So one part must be 8 and the other part is the unknown part.
What are we going to have to do to find the missing part? Pause this video, do some calculations to find my missing number, and come on back to see how you get on.
Welcome back.
I hope you managed to find my missing number.
Shall we see how Andeep solved it? We know that a whole subtracts the known part is equal to the unknown part.
So we can partition 8 into 4 and 4.
14 subtract 4 is equal to 10.
And 10 subtract 4 is equal to 6.
So we know that the missing part must be 6.
So 14 subtract 6 is equal to 8.
Well done to you if you found that missing number.
Andeep now checks his answer using this knowledge.
If we know that 14 subtract 6 is equal to 8, then 8 plus 6 should be equal to 14.
We can partition 6 into 2 and 4 because 8 and 2 make 10.
Then we add 10 plus 4, which is equal to 14.
So was Andeep correct? Was 6 the missing number? Yes, because look, 8 plus 6 is equal to 14, which is agreeing with our part-part-whole model.
So 14 subtract 6 must be equal to 8.
Wow, using an addition to check your subtraction Andeep, that's a really good way to check your working.
And well done to you if you also did this to check your working.
Over to you then for task B.
Part one, use your 10 frames to solve these problems and fill in the missing numbers.
So you'll see that slowly the models start losing some of the parts for you to complete more of them.
So have a go at A, B, and C, and then move on to question two.
Part two is to use bridging 10 strategies for addition and subtraction to find the missing numbers.
So you can see we've got some bar models and some part-part-wholes in A, B, and C.
And D, E, and F are all equations.
So you might want to draw your part-part-whole because remember that's going to help you to visualise the maths.
Pause this video, have a go at part one and part two, and come on back when you're ready to see how you got on.
Welcome back.
Should we see how you got on? Let's fill in the missing numbers then.
We can see that we've partitioned 8 into 5 and 3.
First step subtract the 5 to make 10, then subtract the 3, which leaves us with 7.
So 15 subtract 8 is equal to 7.
B, let's have a look then.
I know that I'm gonna have to subtract 1 to make 10 from 11.
So 1 must be one of my parts, so the other part is 2.
The first step is 11 subtract 1 will leave me with 10.
Then I need to subtract the other part from 10, which is 2.
10 subtract 2 is equal to 8.
So 11 subtract 3 is equal to 8.
And C, let's have a look.
So they haven't given me anything on this model here.
So 12 subtract 5.
I can see that I need to subtract 2 from 12 to make 10.
So that's going to be my part, 2 and 3.
12 subtract 2 is equal to 10.
And 10 subtract 3 is equal to 7.
So 12 subtract 5 is equal to 7.
Well done to you if you've got all of those correct.
Let's have a look at part two then, bridging 10 strategies for addition and subtraction to find the missing numbers.
I can see that I've got 14 as my whole and 5 as a part, so to find the other part, I would've had to complete 14 subtract 5.
And I know that 14 subtract 5 is equal to 9.
Well done if you've got that one.
B then, 13 is my hole and 6 is a part, so I'm going to have to do 13 subtract 6, which will leave me with 7.
C, again, we have our whole and a part, so we are going to subtract 9 from 6, which will leave us with 7.
Let's have a look then.
17 subtract something is equal to 9.
If I visualise my bar model, I know that I've got a whole and a part there.
So if I do 17 subtract 9, that will give me the missing part.
17 subtract 9 is equal to 8.
So I know that 17 subtract 8 will be equal to 9 because that's the other part.
Well done if you got that one.
Oh, now E, what have I got here then? I can see that something subtract 4 is equal to 9.
Hmm, I know that when we are subtracting, we start with the whole so the whole is missing.
So to find the whole, I'm gonna have to add 4 and 9 together to find the starting number of the subtraction.
9 plus 4 is equal to, so 1 more makes 10 plus 3 more, so that's 13.
13 was the missing number for E.
And let's have a look at F.
8 plus something is equal to 15.
So I can see here that I've got the whole and I've got a part.
So if I did 15 subtract 8, that would leave me the missing part.
15 subtract 8.
So I'm gonna subtract the 5, then subtract the 3, which would leave me with 7.
Welcome to you if you've got those correct.
Let's have a look at what we've learned today.
Partition the addend so that you can make a number pair to 10.
The remaining part can then be added onto 10.
Partition the subtrahend so that one of the parts is equal to the ones digit to make 10 when subtracted.
The remaining part can then be subtracted from 10.
Bridging through 10 can be represented on a number line.
So we can show this using addition or subtraction.
Well done for all of your hard work today.
Remember, keep practising and you'll become so much more confident at this bridging through 10 strategy.
I hope to see you all again soon for some more maths learning.
See you soon. | {"url":"https://www.thenational.academy/pupils/programmes/maths-primary-year-2/units/calculating-within-20/lessons/add-and-subtract-two-numbers-that-bridge-through-10/video","timestamp":"2024-11-06T05:24:45Z","content_type":"text/html","content_length":"144327","record_id":"<urn:uuid:9d9c59c2-ff47-4286-9c5f-c16d8389030f>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00242.warc.gz"} |
Do You Know About Wonderlic Test?
Questions and Answers
Do you want to assess your cognitive and problem-solving ability? Take this Wonderlic test to measure your abilities for the range of occupations. This test is generally conducted by the employees of
different organizations to see how good they are at problem-solving and how does their mind works in different scenarios. Here in this quiz, you will get a series of questions as a practice paper.
The final score will tell you how good you are at the mind-work. All the best!
• 1.
Round 907.457 to the nearest tens place
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. 910.00
To round 907.457 to the nearest tens place, we look at the digit in the ones place, which is 7. Since 7 is greater than or equal to 5, we round up the tens place. Therefore, the tens digit
becomes 1 and the ones digit becomes 0. The correct answer is 910.00.
• 2.
What is the average of 12, 15, 23, 10
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. 15
The average of a set of numbers is found by adding up all the numbers and then dividing the sum by the total count of numbers. In this case, the sum of 12, 15, 23, and 10 is 60. Since there are 4
numbers, dividing 60 by 4 gives an average of 15. Therefore, the correct answer is 15.
• 3.
How many of the six pairs of items listed below are exact duplicates? Nieman,K.M. Neiman,K.M. Hoff, J.P. Hoff, J.P. Thomas,G.K. Thomas, C.K. Pino, L.R. Pine, L.R. Hammar,C.G. Hamnar,C.G. Warner,
T.S. Wanner,T.S.
Correct Answer
A. 1
Only one pair of items listed is an exact duplicate, which is "Nieman,K.M." and "Neiman,K.M." The other pairs have slight differences in spelling or order of initials, making them not exact
• 4.
On Monday and Thursday, lunch cost $5.43 total. On Tuesday and Wednesday, lunch costs $3.54 each day. On Friday, lunch cost $7.89. What was the average daily cost?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
D. 4.08
The average daily cost can be calculated by adding up the total cost of lunch for each day and dividing it by the number of days. On Monday and Thursday, the total cost is $5.43. On Tuesday and
Wednesday, the total cost is $3.54 each day, so the total for both days is $7.08. On Friday, the cost is $7.89. Adding up all the costs, we get $5.43 + $7.08 + $7.89 = $20.40. Since there are 5
days in total, the average daily cost is $20.40 divided by 5, which equals $4.08.
• 5.
. What is 1230.932567 rounded to the nearest hundredth place?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
C. 1230.930
The number 1230.932567 rounded to the nearest hundredth place is 1230.930. This is because when rounding to the nearest hundredth, we look at the digit in the thousandth place (3 in this case).
Since 3 is less than 5, we do not round up. Therefore, the digit in the hundredth place remains the same (0), and all digits to the right of the hundredth place are truncated. Hence, the rounded
number is 1230.930.
• 6.
PRESENT RESENT
□ A.
□ B.
□ C.
Correct Answer
C. Not Related
The words "PRESENT" and "RESENT" may look similar, but they have different meanings. "PRESENT" refers to something that is happening now or being given, while "RESENT" means to feel bitterness or
anger towards someone or something. Therefore, the two words are not related in terms of meaning or usage.
• 7.
Subtract the following numbers rounded to the nearest tenth place. 134.679 45.548 67.8807
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
A. 21.30
To subtract the numbers rounded to the nearest tenth place, we need to round each number to the nearest tenth and then subtract them. Rounding 134.679 to the nearest tenth gives us 134.7,
rounding 45.548 gives us 45.5, and rounding 67.8807 gives us 67.9. Subtracting these rounded numbers, 134.7 - 45.5 - 67.9, we get 21.3. Therefore, the correct answer is 21.30.
• 8.
What is the mathematical average of the number of weeks in a year, seasons in a year, and the number of days in January?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
E. 29
The mathematical average of the number of weeks in a year (52), seasons in a year (4), and the number of days in January (31) can be calculated by adding these values together and dividing by the
total number of values (3). Therefore, the average is (52 + 4 + 31) / 3 = 87 / 3 = 29.
• 9.
TEAR TIER
□ A.
□ B.
□ C.
Correct Answer
C. Not related
The two words "tear" and "tier" are not related in terms of meaning or similarity. "Tear" refers to the act of ripping or pulling something apart, while "tier" refers to a level or rank in a
hierarchical structure. There is no connection or contradiction between these two words, making the answer "Not related" correct.
• 10.
OPTIONAL OPTICIAN
□ A.
□ B.
□ C.
Correct Answer
C. Not related
The word "optional" is not related to the word "optician" in any way. "Optional" means something that is not required or mandatory, while "optician" refers to a professional who specializes in
fitting and dispensing eyeglasses. There is no logical or semantic connection between the two words. Therefore, the correct answer is "Not related."
• 11.
Over the course of a week, Fred spent $28.49 on lunch. What was the average cost per day?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
A. 4.07
The average cost per day can be calculated by dividing the total amount spent on lunch over the course of a week by the number of days in that week. In this case, Fred spent $28.49 on lunch over
the course of a week, which is 7 days. Therefore, the average cost per day is $28.49 divided by 7, which equals approximately $4.07.
• 12.
A roast was cooked at 325° F in the oven for 4 hours. The internal temperature rose from 32° F to 145° F. What was the average rise in temperature per hour?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. 28.25
The average rise in temperature per hour can be calculated by dividing the total rise in temperature by the total time taken. In this case, the total rise in temperature is 145°F - 32°F =
113°F, and the total time taken is 4 hours. Therefore, the average rise in temperature per hour is 113°F / 4 hours = 28.25°F.
• 13.
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The correct answer is "Similar" because both words, "flammable" and "inflammable," have the same meaning. They both indicate that something is easily able to catch fire or burn. The prefix "in-"
in "inflammable" does not negate the meaning of "flammable" as one might expect. Instead, it serves as an intensifier, making the word even stronger. Therefore, "inflammable" and "flammable" are
synonyms and can be used interchangeably.
• 14.
PARTNER JOIN
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The words "PARTNER" and "JOIN" are related to each other as they both have a similar meaning in the context of a relationship or collaboration. They both imply a connection or association between
two entities. Therefore, the correct answer is "Similar".
• 15.
PRESENT ABSENT
□ A.
□ B.
□ C.
Correct Answer
B. Contradictory
The words "present" and "absent" are antonyms, meaning they have opposite meanings. Therefore, the relationship between the words "present" and "absent" is contradictory.
• 16.
In the number 743.25, which digit represents the tenth space?
Correct Answer
A. 2
In the number 743.25, the digit 2 represents the tenth space. The tenth space is the second digit after the decimal point. In this case, the digit 2 is the second digit after the decimal point,
so it represents the tenth space.
• 17.
Add 0.98 + 45.102 + 32.3333 + 31 + 0.00009
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
C. 109.41539
The given question asks for the sum of several numbers. By adding 0.98, 45.102, 32.3333, 31, and 0.00009 together, we get the answer of 109.41539.
• 18.
WISH ORDER
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The words "wish" and "order" are similar in the sense that they both represent a desire or a request for something. While "wish" implies a more hopeful or longing desire, "order" suggests a more
direct and authoritative request. However, both words convey the idea of wanting or requesting something, making them similar in meaning.
• 19.
EXIT ENTRANCE
□ A.
□ B.
□ C.
Correct Answer
B. Contradictory
The words "EXIT" and "ENTRANCE" are antonyms, meaning they have opposite meanings. "EXIT" refers to a way out or a point of departure, while "ENTRANCE" refers to a way in or a point of entry.
Therefore, the words are contradictory in meaning.
• 20.
Find 0.12 ÷ 1
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
C. .12
The given division problem is 0.12 ÷ 1. When dividing any number by 1, the result is always the same number. Therefore, the answer is .12.
• 21.
(9 ÷ 3) x (8 ÷ 4) =
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. 6
The given expression involves two divisions: (9 ÷ 3) and (8 ÷ 4). The first division, 9 ÷ 3, equals 3. The second division, 8 ÷ 4, also equals 2. Multiplying these results together, 3 x 2 =
6. Therefore, the correct answer is 6.
• 22.
6 x 0 x 5 =
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
D. 0
When multiplying any number by zero, the result is always zero. In this case, multiplying 6 by 0 gives 0, and then multiplying the result by 5 still gives 0. Therefore, the answer is 0.
• 23.
SHELVE TABLE
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The words "shelve" and "table" are similar in the sense that they both refer to a piece of furniture used for storage or display. Both words are nouns and can be used to describe objects that are
used for similar purposes. Therefore, the correct answer is "Similar."
• 24.
KINDLE ENKINDLE
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The correct answer is "Similar" because both "Kindle" and "Enkindle" are related to the act of igniting or sparking something. They have similar meanings and are derived from the same root word.
Therefore, they can be considered similar in terms of their meaning and usage.
• 25.
7.95 ÷ 1.5 =
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. 5.3
The correct answer is 5.3 because when you divide 7.95 by 1.5, you get the quotient of 5.3.
• 26.
-32 + 7 equals:
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
A. -25
When we add -32 and 7, we combine the two numbers. Since 7 is positive and -32 is negative, their sum will be negative. By subtracting 32 from 7, we get 25. However, since the negative sign is
attached to the -32, the final answer becomes -25.
• 27.
-37 + -47 equals:
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. -84
The given question asks to add -37 and -47. When adding two negative numbers, we add their absolute values and keep the negative sign. Therefore, the sum of -37 and -47 is -84.
• 28.
41% equals:
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. .41
The correct answer is .41 because when converting a percentage to a decimal, you move the decimal point two places to the left. So, 41% becomes .41.
• 29.
HIRE FIRE
□ A.
□ B.
□ C.
Correct Answer
B. Contradictory
The words "HIRE" and "FIRE" have opposite meanings and are used in contrasting situations. "HIRE" refers to the act of employing someone, while "FIRE" refers to the act of terminating someone's
employment. Therefore, the words are contradictory in meaning.
• 30.
SUITABLE WORTHY
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The words "suitable" and "worthy" both suggest that something is appropriate or deserving. Therefore, they can be considered similar in meaning.
• 31.
REST BREAK
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The words "REST" and "BREAK" are similar in the sense that they both refer to a pause or a period of relaxation. They are related in terms of their meaning and can be used interchangeably in
certain contexts. Therefore, the correct answer is "Similar".
• 32.
Which among the following is not a whole number followed by its square?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
E. 11, 144
The given options consist of pairs of whole numbers followed by their squares. However, the pair 11, 144 does not follow this pattern. The square of 11 is 121, not 144. Therefore, 11, 144 is not
a whole number followed by its square.
• 33.
Which number is the largest?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. 23000
Among the given numbers, 23000 is the largest because it has the highest value. The other numbers are smaller in comparison.
• 34.
TRAIN WEEP
□ A.
□ B.
□ C.
Correct Answer
C. Not related
The words "TRAIN" and "WEEP" are not related in terms of their meaning or any other similarity. "TRAIN" refers to a vehicle that runs on tracks, while "WEEP" means to cry or shed tears. There is
no logical or semantic connection between these two words, making them not related.
• 35.
ADD AND
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The correct answer is "Similar" because both "ADD" and "AND" are words used to join or combine elements together. They serve a similar purpose in connecting different parts of a sentence or
• 36.
LIFT WOODEN
□ A.
□ B.
□ C.
Correct Answer
C. Not related
The words "LIFT" and "WOODEN" do not have any obvious connection or similarity in terms of their meaning or relationship. They belong to different categories and do not share any common
characteristics. Therefore, the correct answer is "Not related."
• 37.
There are 12 more apples than oranges in a basket of 36 apples and oranges. How many apples are in the basket?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
C. 24
Let's assume the number of oranges in the basket is x. Since there are 12 more apples than oranges, the number of apples would be x + 12. The total number of fruits in the basket is given as 36.
So, we can write the equation as x + (x + 12) = 36. Simplifying this equation gives us 2x + 12 = 36. Subtracting 12 from both sides gives us 2x = 24. Dividing both sides by 2 gives us x = 12.
Therefore, the number of apples in the basket is x + 12, which is 12 + 12 = 24.
• 38.
Which among the following correctly identifies 4 consecutive odd integers where the sum of the middle two integers is equal to 24?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
C. 9, 11, 13, 15
The sum of the middle two integers (11 and 13) is equal to 24, which satisfies the condition given in the question. Additionally, all the integers in the sequence are consecutive odd integers.
Therefore, the correct answer is 9, 11, 13, 15.
• 39.
Which number is next in the sequence? 6, 12, 24, 48, ___
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. 96
The given sequence is a geometric progression with a common ratio of 2. Each term is obtained by multiplying the previous term by 2. Starting with 6, the next term is 12 * 2 = 24, then 24 * 2 =
48. Following the pattern, the next term should be 48 * 2 = 96.
• 40.
A train is traveling 20 feet in 1/5 seconds. At the same speed, how many feet will it travel in three seconds?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
D. 300
The train is traveling at a constant speed of 20 feet in 1/5 seconds. To find out how many feet it will travel in three seconds, we can multiply the speed by the time. Since 1/5 seconds is
equivalent to 0.2 seconds, we can calculate 20 feet * 0.2 seconds = 4 feet. Therefore, in three seconds, the train will travel 4 feet * 3 = 12 feet.
• 41.
When a rope is selling at $.15 afoot, how many feet can you buy for sixty cents?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. 4
If a rope is selling at $0.15 per foot, to find out how many feet can be bought for $0.60, we can divide $0.60 by $0.15. This calculation gives us 4, which means that 4 feet of rope can be bought
for $0.60.
• 42.
The eighth month of the year is:
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
E. AUGUST
August is the eighth month of the year. It comes after July and before September. Therefore, August is the correct answer for the eighth month of the year.
• 43.
Which of the following group of numbers represents the smallest amount?
□ A.
□ B.
□ C.
□ D.
□ E.
Correct Answer
B. .25
The correct answer is .25 because it is the smallest number among the given options. The other numbers, 6, .7, 1, and .33, are all larger than .25.
• 44.
BOTTLES SCISSORS
□ A.
□ B.
□ C.
Correct Answer
C. Not related
The given words "BOTTLES" and "SCISSORS" do not have any direct or obvious relationship or similarity. They are unrelated objects and do not share any common characteristics or functions.
Therefore, the correct answer is "Not related".
• 45.
TUNNEL ROUTE
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The words "TUNNEL" and "ROUTE" are similar in the sense that they both refer to a pathway or passage. They are related in terms of transportation and navigation, as a tunnel is a specific type of
route that is usually underground. Therefore, the correct answer is "Similar".
• 46.
SOCKET PICTURE
□ A.
□ B.
□ C.
Correct Answer
C. Not related
The words "SOCKET" and "PICTURE" are not related in any way. They do not share any similar characteristics or have any connection to each other. Therefore, the correct answer is "Not related".
• 47.
ELLIPSE OVAL
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The words "ellipse" and "oval" are both geometric shapes that have a similar elongated and curved appearance. They are often used interchangeably to describe a shape that is wider in the middle
and tapers at the ends. Therefore, the correct answer is "Similar."
• 48.
TOMB TOME
□ A.
□ B.
□ C.
Correct Answer
C. Not related
The two words "TOMB" and "TOME" are not related because they have different meanings and origins. A tomb refers to a burial place for a dead person, while a tome refers to a large, scholarly
book. They do not share any similarities in terms of their definitions or usage.
• 49.
WHEAT GRAIN
□ A.
□ B.
□ C.
Correct Answer
A. Similar
The relationship between "WHEAT" and "GRAIN" is that wheat is a type of grain. Therefore, the words are similar in meaning and can be considered synonyms.
• 50.
Last one, which of the following weighs the less?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Milligram
The milligram weighs the least out of the given options. It is a unit of measurement in the metric system, representing one thousandth of a gram. A kilogram is equal to 1000 grams, a gram is
equal to 1000 milligrams, and a ton is equal to 1000 kilograms. Therefore, the milligram is the smallest unit of weight among the options provided. | {"url":"https://www.proprofs.com/quiz-school/story.php?title=mtewnjcxhxfj","timestamp":"2024-11-08T04:34:58Z","content_type":"text/html","content_length":"619652","record_id":"<urn:uuid:67ba93e8-d6bb-4197-9761-596059f9c81e>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00542.warc.gz"} |
SECTION 1
Damping Metrics
The following metrics are demonstrated by example throughout this document. The viscous damping ratio is defined in terms of the damping coefficients as
The critical damping coefficient is
The amplification factor is
The amplification factor can be calculated from measured frequency response function data via the half-power method.
The logarithmic decrement is
SECTION 2
Four Damping Categories
The four damping types are summarized in the following table. The description is for the free vibration of a single-degree-of-freedom system due to initial displacement or velocity.
Table 2.1. Damping Types
Type Value Response Description
Undamped Simple harmonic motion, sinusoidal response
Underdamped Damped sine response with exponential decay
Critically damped Border between the overdamped and underdamped cases
Overdamped Sum of two decaying exponentials with no oscillation
The first three of these types are shown by examples in Figure 2.1. The underdamped type emphasized throughout this document.
Figure 2.1. SDOF Response to Initial Displacement for Three Damping Cases
SECTION 3
Damping Mechanisms
Damping occurs as vibration energy is convert to heat, sound or some other loss mechanism. Damping is needed to limit the structural resonant response. Common sources are:
• viscous effects
• Coulomb damping, dry friction
• aerodynamic drag
• acoustic radiation
• air pumping at joints
• boundary damping
The dominant source for assembled structures is usually joint friction. This damping mechanism may be nonlinear due to joint microslip effects. The damping value tends to increase at higher
excitation levels.
Damping may also decrease as the natural frequency increases, such that the amplification factor increases with natural frequency. The equation per Reference [10] for approximating Q for an
electronic system subjected to a sine base input is
Table 3.1.
= 1.0 for beam-type structures
A = 0.5 for plug-in PCBs or perimeter supported PCBs
= 0.25 for small electronic chassis or electronic boxes
Gin Sine Base Input (G)
Beam structures: several electronic components with some interconnecting wires or cables.
PCB: printed circuit board well-populated with an assortment of electronic components.
Small electronic chassis: 8-30 inches in its longest dimension, with a bolted cover to provide access to various types of electronic components such as PCBs, harnesses, cables, and connectors.
SECTION 4
Huntsville, Alabama Pedestrian Bridge Damping
Figure 4.1. University Drive Pedestrian Bridge
Figure 4.2. Slam Stick X, Triaxial Accelerometer & Data Logger, Shown on Book Shelf
The bridge in Figure 4.1 is near the University of Alabama in Huntsville. The author walked to the center of the bridge and mounted the triaxial accelerometer in Figure 4.2 on the deck floor using
double-sided tape. He was the only person on the bridge. The ambient vibration response of the bridge was negligibly low in each of the axes, while he remained standing. The wind was very light on
this day. The author then jumped up-and-down on the bridge to excite its vibration modes. The response in the vertical axis was significant. The response in each of the lateral axes remained
negligibly low.
Figure 4.3. Signal Identification via Damped Sine Curve-fit
The jumps were performed near the 80 second mark. The accelerometer data was band-passed filtered from 1 to 20 Hz. The response after 82 seconds represents the free vibration decay. A damped sine
curve-fit synthesis was performed on the vertical acceleration time history as shown in Figure 4.3, using trial-and-error with convergence. The bridge’s fundamental frequency is 2.2 Hz with 0.16%
damping. The damping ratio is very low. But note that the fundamental frequency and damping may be nonlinear. A pedestrian’s vertical forces correspond to each footfall, and typically occur at 2.0
Hz. This is very close to the bridge’s 2.2 Hz natural frequency.
Also note that there is a potential for pedestrians to synchronizing their steps with the bridge motion and with one another. This behavior is instinctive rather than deliberate. Pedestrians find
that walking in synchronization with the motion of a bridge is more comfortable, even if the oscillation amplitude is initially very small. This cadence makes their interaction with the movement of
the bridge more predictable and helps them maintain their balance. But the synchronization also causes the pedestrians’ gait to reinforce the bridge’s oscillation in a resonance-like manner. These
sorts of problems occurred after the opening of the London Millennium Bridge in 2000. Both passive and tuned mass dampers were added to the bridge for vibration control. Retrofitting the University
Drive Bridge with dampers is unnecessary due to its low pedestrian traffic volume. But this could be an interesting project for the nearby engineering students.
SECTION 5
Pegasus Launch Vehicle Damping
Drop Transient
Figure 5.1. Pegasus Launch Vehicle, Drop & Stage 1 Burn
Figure 5.2. Pegasus Fundamental Bending Mode Shape, Exaggerated
A modified L-1011 aircraft carries the Pegasus vehicle up to an altitude of nearly 40,000 ft and a speed of Mach 0.8, as shown in Figure 5.1. Pegasus is suspended underneath the aircraft by hooks,
where it develops an initial displacement due to gravity. The strain energy is suddenly released at the onset of the drop transient, causing Pegasus to oscillate nearly as a free-free beam. This is a
significant “coupled loads” event for the payload which is enclosed in the fairing at the front end of the vehicle. The Pegasus first stage then ignites. The payload is eventually delivered into a
low earth orbit. A certain Pegasus/payload configuration was analyzed via a finite element modal analysis. The resulting wire mesh model of the fundamental bending mode is shown in Figure 5.2. Note
that this was a different Pegasus configuration than that represented by the flight data in Figure 5.3.
Drop Transient Damped Sine Curve-fit
Figure 5.3. Pegasus Drop Transient, Flight Accelerometer Data, Free Vibration Response
The data was measured in the transverse axis at the payload interface. The response is nearly a textbook quality damped sinusoid, with an exponential decay. The natural frequency and damping ratio
are identified via a synthesized damped sine, curve-fit.
Drop Transient Logarithmic Decrement Method
Figure 5.4. Pegasus Drop Transient, Flight Accelerometer Data, Logarithmic Decrement
The logarithmic decrement for a starting peak and a peak n cycles later is
The logarithmic decrement for the two peaks shown in Figure 5.4 is
The logarithmic decrement value is equivalent to 1.2% damping in agreement with the previous damped sine curve-fit. The logarithmic decrement method has been included for historical reasons. The
damped sine curve-fit method is more robust.
SECTION 6
Transamerica Building Damping
Figure 6.1. Transamerica Pyramid
The Transamerica Pyramid is built from a steel frame, with a truss system at the base. The height is 850 ft (260 m). Reference [11] gives natural frequency and damping as obtained in the 1989 Loma
Prieta earthquake and from ambient vibration. The ambient vibration was presumably due to wind, low level micro-tremors, mechanical equipment, outside street traffic, etc.
Table 6.1. Transamerica Pyramid, Modal Parameters
Direction Loma Prieta Earthquake Ambient Vibration
fn (Hz) Damping fn (Hz) Damping
North-South 0.28 4.9% 0.34 0.8%
East-West 0.28 2.2% 0.32 1.4%
The results show non-linear behavior with an increase in damping during the severe earthquake relative to the benign ambient vibration. | {"url":"https://endaq.com/pages/damping","timestamp":"2024-11-14T12:03:47Z","content_type":"text/html","content_length":"143951","record_id":"<urn:uuid:62781062-75f3-4a92-9446-68a24dca1bab>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00402.warc.gz"} |
Blinding, Information Hiding and Epistemic Efficiency
This post is about the importance of information hiding in epistemic systems. It argues (though “argues” may be too strong a word) that hiding information from certain participants in an epistemic
system can increase the epistemic efficiency of the overall system. While this conclusion is not particularly earth-shattering, the method adopted to reach it is quite interesting. It uses some of
the formal apparatus from Roger Koppl’s work on
Epistemic Systems
, which combines game theory and information theory in an effort to better understand and intervene in certain social systems.
In what follows, I lay out some of the key elements from Koppl’s theory and then describe a simple model epistemic system (taken from
Koppl’s article
) that illustrates the importance of information hiding.
1. What is an Epistemic System?
An epistemic system is any social system that generates judgments of truth or falsity. The classic example might be the criminal trial which tries to work out whether or not a person committed a
crime. Evidence is fed into this system via witnesses and lawyers, it is then interpreted, weighed and evaluated by a judge and jury, who in turn issue a judgment of truth or falsity, either: “Yes,
the accused committed the crime” or “No, the accused did not commit the crime”. Although this may be the classic example, the definition adopted by Koppl is broad enough to cover many others. For
example, science is viewed as an epistemic system under Koppl’s definition.
The goal of epistemic systems
is to adopt some of the formal machinery from game theory and information theory in order to better understand and manipulate these epistemic system. In effect, the goal here is to develop simple
models of epistemic systems, and use these to design better ones. The first step in this process is to identify the three key elements of any epistemic system. These are:
Senders: A set of individual agents who choose the messages that are sent through the system.
Message Set: The set of possible messages that could be sent by the senders.
Receivers: A set of individual agents who receive the messages and determine whether they represent the truth or not.
In its more mathematical guise, an epistemic system can be defined as an ordered triple of senders, receivers and messages {S, R, M}
[S:[S:, with a formal symbology for representing the members of each set. I will eschew that formal symbology here in both the interests of simplicity and brevity. Full details can be found in
Koppl’s article. I will use some elementary mathematics and pictures, such as the following, which represents a simple epistemic system with one message, one sender and one receiver. :S]:S] [S:
The system issues a judgment, and this judgment will either be true of false. Whether it is in fact true or false is not determined by the beliefs of the senders or receivers.
Senders and Receivers are viewed as rational agents, sometimes locked in strategic battles, within these systems. As such they have utility functions which represent their preferences for particular
messages or conclusions and they act so as to maximise their utility. One of the key assumptions Koppl makes is that these utility functions will not usually include a preference for the truth. For
instance, he assumes that scientists will have a preference for their pet theory, rather than for the true theory; or that lawyers will have a preference for evidence that supports their client’s
case, not for the true evidence. In doing so, he adopts a Humean perspective on epistemic systems, believing we should presume the worst in order to design the best. He uses a nice quote from Hume to
set this out:
… every man ought to be supposed a knave, and to have no other end, in all his actions, than private interest. By this interest we must govern him, and, by means of it, make him, notwithstanding
his insatiable avarice and ambition, co-operate to public good.
This assumption of knavishness is fairly common in rational choice theory and I have no wish to question it here. What does need to be questioned, however, is what represents the “public good” when
it comes to the design and regulation of epistemic systems. One could argue about this, but the perspective adopted by Koppl (and many others) is that we want epistemic systems that reach true
judgments. To be more precise, we want
epistemically efficient
systems, where this is defined as:
Epistemic Efficiency: A measure of the likelihood of the system reaching a true judgment. Either: 1 minus the error rate of the system; or the ratio of true judgments to total judgments.
So the goal is to increase the epistemic efficiency of the system. The argument we will now look at claims that information hiding is one way of achieving this.
2. The Importance of Information Hiding
The argument, like all arguments, depends on certain assumptions. One of the advantages of the formal machinery adopted by Koppl is that these assumptions are rendered perspicuous. If you think these
assumptions are wrong, the strength of the argument is obviously diminished, but at least you’ll be able to clearly see where it’s going wrong as you read along.
So what are these assumptions? First, we are working with an extremely simple system. The system consists of one sender, one receiver, and
messages. It does not matter what these messages are, so we shall simply denote them as m1 and m2. We shall refer to the sender as S and the receiver as R. S must pick either m1 or m2 to send to R. R
does not question whether S is right or wrong. In other words, R always assumes that the message sent by S represents the truth. This is, roughly, illustrated in the diagram below. We assume that m1
has a 0.25 probability of being true, and m2 has a 0.75 probability of being true.
The second crucial assumption relates to the payoff functions of R and S. They are as follows:
Receiver’s Payoff Function = U(m) = 1 (if m=m1) or 0 (if m=m2)
Sender’s Payoff Function = V(m) = Pr(m is true) x E[U(m)]
In other words, we assume that R prefers to receive m1 over m2. And we assume that S’s payoff function is partly determined by what he thinks is the truth, and partly determined by what he expects
R’s payoff function to be (E(U) denotes
expected utility
. This looks like a fairly realistic assumption. Imagine, for instance, the expert witness recruited by a trial lawyer. He will no doubt wish to protect his professional reputation by picking the
“true” message from the message set, but he will also wish to please the lawyer who is paying for his services. So if he knows that the lawyer prefers one message over the other, he too may have a
bias toward that message. That such biases may exist has been confirmed experimentally, and they may be entirely subconscious.
This is where information hiding comes into play. Look first at the efficiency of the system when there is no information hiding,
when S knows exactly what R’s payoff function is. In other words, when E[U(m)] = U(m).
If S sends m1 then:
□ (1) U(m) = 1; P(m1) = 0.25
□ (2) V(m) = P(m1) x E[U(m)]
If S sends m2 then:
□ (5) U(m) = 0; P(m2) = 0.75
□ (6) V(m) = P(m2) x E[U(m)]
Since we assume S acts so as to maximise his payoff, it follows that S will always choose m1 in this system. And since m1 only has a one in four chance of being correct, it follows that the epistemic
efficiency of the system as whole is 0.25. Which is pretty low.
Can efficiency be improved by hiding information about R’s preferences from S? Well, let’s do the math and see. Assume now that S has no idea what R’s preferences are. Consequently, S’s adopts the
principle of indifference and assumes that R is equally likely to prefer m1 and m2. In other words, in this scenario E[U(m)] = (0.5)(1) = (0.5).
If S sends m1 then:
□ (1*) E[U(m)] = 0.5 ; P(m1) = 0.25
□ (2*) V(m) = P(m1) x E[U(m)]
□ (3*) V(m) = (0.25) x (0.5)
If S sends m2 then:
□ (5*) E[U(m)] = 0.5; P(m2) = 0.75
□ (6*) V(m) = P(m2) x E[U(m)]
□ (7*) V(m) = (0.75) x (0.5)
S’s preference now shifts from sending m1 to sending m2. And since m2 has a three in four chance of being correct, the epistemic efficiency of the system is increased from 0.25 to 0.75. This is a
significant improvement. And, if the assumptions are correct, illustrates one significant way in which to improve the overall efficiency of an epistemic system.
As I said at the outset, this is not a particularly earth-shattering conclusion. Indeed, it is what motivates blinding protocols in scientific experimentation. What’s nice about the result is the
formal apparatus underlying it. This formal apparatus is flexible, and can be used to model, evaluate and design other kinds of epistemic system. | {"url":"https://philosophicaldisquisitions.blogspot.com/2012/07/blinding-information-hiding-and.html?m=0","timestamp":"2024-11-07T13:56:29Z","content_type":"text/html","content_length":"129211","record_id":"<urn:uuid:0fcdcc45-ae0d-4d8e-b5c9-caa9e45920b2>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00687.warc.gz"} |
AIU Math Worksheet - Custom Scholars
AIU Math Worksheet
MATH133 – Unit 1 Point Values:Question
Point value
You earned
MATH133 – Unit 1 Individual Project
NAME (Required): Rachael Campbell
Assignment Instructions:
1. For each question, show all of your work for full credit.
2. Insert all labeled and titled graphs by using screenshots from Excel or desmos.com as
described in the Unit 1 Discussion Board (or other graph program).
3. Provide final answers to all questions in boxes provided.
4. Round all value answers to 3 decimal places, unless otherwise noted.
Formulas and Equations
You will use the following formulas and equations for this assignment. Linear equations are in
either point-slope form or slope-intercept form.
Point-slope form:
𝒚 − 𝒚𝟏 = 𝒎(𝒙 − 𝒙𝟏 )
Slope-intercept form: 𝒚 = 𝒎𝒙 + 𝒃
m is the slope of the line
𝑠𝑙𝑜𝑝𝑒 = 𝑚 = (𝑥2−𝑥1)
𝑏 is the y-intercept of the function’s graph
(𝑥1 , 𝑦1 ) represents some point on the line
(𝑦 −𝑦 )
Use the following scenario for Questions 1 through 4:
The information technology department found data irregularities in the transaction rate of a
server. Your supervisor asked you to analyze the data to find the linear equation that best
represents these rates, to compare the predicted transaction rate to the questionable data.
Observations suggest that this server can store 𝑦 megabits of data in 𝑥 microseconds (see Table
First Letter of Your
(microseconds) (megabits) (microseconds) (megabits)
Last Name
15 to 16.99
12.5 to 12.99
17 to 19.99
17.5 to 17.99
20 to 21.99
13.5 to 13.99
22 to 24.99
18.5 to 18.99
25 to 26.99
14.5 to 14.99
27 to 29.99
19.5 to 19.99
30 to 31.99
15.5 to 15.99
32 to 34.99
20.5 to 20.99
35 to 36.99
16.5 to 16.99
37 to 39.99
21.5 to 21.99
40 to 41.99
17.5 to 17.99
42 to 44.99
22.5 to 22.99
Table A: Observations of Bit-Storage Amounts and Times
1. Chose a number within each range provided for the x and y columns.
From Table A, use the row that matches the first letter of your last name to select one number
within each column’s specified range, creating two pairs of numbers (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ).
For example, if your last name starts with M, then you might choose the following:
(𝑥1 , 𝑦1 ) = (30.2, 15.6)
(𝑥2 , 𝑦2 ) = (32.5, 20.5)
In the table below, submit your two selected pairs of points based on your last name.
(10 points)
Your Chosen Points
(x1, y1)
(x1, y1) = (15, 12.5)
(x2, y2) = (18,17.75)
(x2, y2)
2. Use the two points from above to construct a linear function model. Create first the
point-slope form equation, and then convert the equation to slope-intercept form.
(20 points)
Hint: You will want to determine the slope first and then construct the point-slope form
of the equation before you convert it to the slope-intercept form.
Point-Slope Form
Round all value answers to 3 decimal
Y2 -Y1 = m(X2-X1)
Slope-Intercept Form
Round all value answers to 3 decimal
Show your work below for constructing both linear function forms:
3. The number of megabits (y) in terms of microseconds (x) is given as follows:
𝑦 = 32.75𝑥 − 57.67
How many microseconds will it take for your model to store 200 megabits? (20 points)
How many microseconds?
Round all value answers to 3 decimal
Show your work below using Equation Editor:
Linear forecasting is the process of extrapolating the given function values to an independent
variable value that is not observed in the data. In this question, you are predicting or
forecasting the number of megabits that can be stored in 400 microseconds without collecting
any data at that range.
4. The number of megabits (y) in terms of microseconds (x) is given by the following
𝑦 = 32.75𝑥 − 57.67
How many megabits can be stored in 400 microseconds? (20 points)
How many megabits?
Round all value answers to 3 decimal
Show your work below:
5. Laptop A and Laptop B have fully charged batteries and were used by company sales
staff while traveling until the laptops ran out of power (note that the laptops will not
operate when the battery voltage drops below 3.0). The two laptops have different
battery capacities and power consumption rates. Below are the graphs that show the
power consumption. The x-axis represents the time used (in hours), and the y-axis
represents the remaining voltage in the battery. Using these graphs, determine which
laptop is more efficient, and support your choice using appropriate math. (30 points)
Hint: Keep in mind that what you are looking to compare is in volts per hour.
Which laptop is more
Explain your answer below: | {"url":"https://customscholars.com/aiu-math-worksheet/","timestamp":"2024-11-14T07:38:52Z","content_type":"text/html","content_length":"55898","record_id":"<urn:uuid:5bc93605-3848-4c37-8d2d-36f6953017e9>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00156.warc.gz"} |
Non-Terminating Repeating Decimal to Fraction
Before going into the conversion of non-terminating and repeating decimal to fractions, let us understand what significance do these terms hold. What are terminating, non-terminating, repeating and
non-repeating decimals.
Terminating and Non-Terminating decimals-
A terminating decimal that has an ends. It is a decimal, which has a finite number of digits (or terms).
Eg. 0.15, 0.86 etc.
Whereas non-terminating decimals are the one that do not have an end term. It has infinite number of terms.
Eg. 0.5444444….., 0.1111111….., etc.
Repeating and Non-Repeating decimals-
Repeating decimals are the one, which have a set of terms in a decimal to be repeated in a uniform manner.
Eg. 0.666666…., 0.123123…., etc.
It is to be noted that the repeated term in a decimal are represented by bar on top of the repeated part. Such as \(0.333333….. = 0.\bar{3}\).
Whereas non-repeating decimals are the one that do have have repeated terms.
Non-Terminating and non-repeating decimals are said to be an Irrational numbers. Eg. \(\sqrt{2} = 1.4142135……\).
The square roots of all the terms (leaving perfect squares) are irrational numbers.
Non- Terminating and repeating decimals are Rational numbers and can be represented in the form of p/q, where q is not equal to 0.
Let us now learn to convert Non-Terminating and repeating decimals in rational form.
(i) Fraction of the type \(0.\overline{abcd}\)–
\(\overline{abcd} = \frac{Repeated \; term}{Number \; of \; 9’s \; for \; the ;\ repeated \; terms}\)
Example- Convert \(0.\overline{7}\) in Rational form.
Solution- Here the number of repeated term is only 7, thus number of times 9 to be repeated in the denominator is only one.
\(0.\overline{7} = \frac{7}{9}\)
Example- Convert 0.125125125….. in Rational form.
Solution- The decimal shown above can be written as \(0.\overline{125}\).
Here 125 consist three terms to be repeated in a continuous manner. Thus number of time 9 to be repeated in the denominator becomes three.
\(0.\overline{125} = \frac{125}{999}\)
(ii) Fraction of the type \(0.ab..\overline{cd} =\frac{(ab….cd…..) – ab……}{Number \; of \; time \; 9’s \; the \; repeating \; term \; followed \; by \; the \; number \; of \; times \; 0’s \; for \;
the \; non-repeated \; terms }\)
Example- Convert \(0.12\overline{34}\) in a Rational form.
Solution- In the given ratio we have 12 to be of the non-repeated form and 34 to be of the repeating form. Thus denominator becomes 9900.
\(0.12\overline{34} = \frac{1234 – 12 }{9900} = \frac{1222}{9900}\)
Example- Convert \(0.00\overline{69}\) in p/q form.
Solution- In the given ratio we have 00 to be of the non-repeated form and 69 to be of the repeating form. Thus denominator becomes 9900.
\(0.00\overline{69} = \frac{0069}{9900} = \frac{69}{9900}\) | {"url":"https://www.careerstoday.in/maths/non-terminating-repeating-decimal-to-fraction","timestamp":"2024-11-14T06:42:14Z","content_type":"text/html","content_length":"14633","record_id":"<urn:uuid:fab37703-16e3-428b-bf9e-042ca0e8ecef>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00548.warc.gz"} |
Sortino Ratio Calculator
Last updated:
Sortino Ratio Calculator
The smart Sortino ratio calculator is an efficient tool that indicates the return of an investment considering its drawdown risk.
This article will cover the definition of the Sortino ratio, see a formula for calculating it, and discuss what a good Sortino ratio is. As a bonus, we'll go through a real-life example of a Sortino
ratio calculation!
What is the Sortino ratio?
The Sortino ratio is a statistical tool that investment managers use in portfolio risk analysis. It considers the historic asset returns, the risk-free rate, and the negative asset volatility. It
indicates how much profit you could earn in exchange for the risk you are taking.
Wait, negative what? 😨
We understand how you feel. Do not worry. We have you covered. The Sortino ratio is an improvement of the sharpe ratio calculator because it only considers negative returns of the asset. In contrast,
the latter considers both (positive and negative returns). Negative volatility is another name investors use to refer to the swing in price that causes lower values.
What is a negative stock return?
It is when the stock reports a decrease in its price value from one date to the other. On the other hand, a positive stock return is when the stock price increases.
But what about the risk-free rate?
The risk-free rate refers to a profit you can make from an investment with literally no risk of losing money. Bonds from governments are usually this kind of investment.
In a nutshell, the Sortino rate subtracts the risk-free rate from the historical return and divides it by the negative return volatility. The result indicates how much extra return you are gaining,
given the risk you are taking.
How to calculate the Sortino ratio?
As we mentioned above, the Sortino ratio formula is:
$\footnotesize \rm {Sortino \ ratio = \frac {(Ra - Rf)}{STD}}$
• $\small \rm{Ra}$ — Average return on the asset;
• $\small \rm{Rf}$ — Risk-free rate; and
• $\small \rm{STD}$ — Standard deviation of the downside or any negative return the company experienced.
You can get $\small \rm{Ra}$ by using the average calculator with the stock's past returns. You can calculate past returns by checking the stock's historical prices and using the stock calculator.
The risk-free rate can be the interest rate that your government pays for its bonds. In our example, we will use the US 3-month treasury bill bond.
The standard deviation of the downside only considers the negative returns, replacing the positive values in the historical returns with 0.
Can you show a Sortino ratio calculation example?
Sure. Let's consider the following data:
Date Price Return
12-31-2001 0.34 12.88%
01-31-2002 0.38 -12.22%
02-28-2002 0.33 9.08%
03-28-2002 0.36 2.53%
04-30-2002 0.37 -4.00%
05-31-2002 0.36 -23.95%
06-28-2002 0.27 -13.88%
07-31-2002 0.23 -3.34%
08-30-2002 0.23 -1.69%
09-30-2002 0.22 10.83%
10-31-2002 0.25 -3.55%
11-29-2002 0.24 -7.55%
12-31-2002 0.22 0.21%
… … …
01-29-2021 131.15 -7.97%
02-26-2021 120.70 0.73%
03-31-2021 121.58 7.62%
04-30-2021 130.85 -5.05%
05-28-2021 124.24 9.91%
06-30-2021 136.56 6.50%
07-30-2021 145.43 4.25%
08-31-2021 151.61 -6.80%
09-30-2021 141.29 5.87%
10-29-2021 149.58 10.51%
11-30-2021 165.30 7.42%
12-31-2021 177.57 N/A
Over the last 20 years, the average return has been: $\footnotesize \rm{Ra = 3.11 \%}$.
The current : $\footnotesize \rm{Rf = 0.13 \% }$.
The next step is to find out the negative returns' standard deviation (STD). We only need to replace the positive returns with 0 and calculate the STD of all the remaining monthly returns. Thus, we
obtain the following:
$\footnotesize \rm{STD = 0.04891}$
Finally, by using the Sortino ratio formula, we get:
$\footnotesize \rm{Sortino \ ratio = (3.11 \% - 0.13 \%) / 0.04891 }$
$\footnotesize \rm{Sortino \ ratio = 0.60928}$
We can conduct the same Sortino ratio calculation example for Microsoft (). Actually, our Sortino ratio calculator can do the work for you. The MSFT data is:
$\footnotesize \rm{Ra = 1.35 \%}$
$\footnotesize \rm{STD = 0.03489}$
$\footnotesize \rm{Sortino ratio = 0.34967}$
In conclusion, we note that although MSFT's negative risk is lower, the average monthly return of APPL is so much larger that it makes its Sortino ratio better. Consequently, we can say that an
investment in Apple showed better returns for the risk taken compared to Microsoft over the last 20 years.
⚠️ Note: The risk-free rate we consider for our Sortino ratio calculator example is the three-month treasury bill rate which is constantly changing. Please use the updated risk-free rate valid when
you perform your calculations.
How do investors use Sortino ratio?
The question you might have had from the very beginning is: What is a good Sortino ratio? There is a simple answer to such a question: the higher, the better.
When you compare the Sortino ratio of different stocks, you look for the one with the larger value. It means it is the one that has provided the most significant return for the less drawdown risk.
You will probably find the most significant Sortino ratio in the stocks that report high revenue growth and high free cash flow margin. Both financial indicators show business strength and
profitability, which in the market means excellent returns.
Finally, always consider analyzing your assets as a whole. We recommend you check the portfolio beta calculator for such a job.
Which is better out of Sortino ratio vs Sharpe ratio?
Investors consider the Sortino ratio to be superior to the Sharpe ratio because it only considers the negative risk (drawdowns). In that sense, the former ratio shows the historic return compared to
the probability of getting losses, which is the only risk that matters.
How do I calculate the Sortino ratio?
Perform these steps to calculator the Sortino ratio:
1. Obtain the historical stock price considering at least the last five years.
2. Calculate the returns. You decide whether to use daily, monthly, or yearly, depending on the size of your dataset. Remember, the last price is divided by the previous price minus 1. Continue
multiplying the result by 100%.
3. Get the average of the returns (Ra). Also, obtain the risk-free rate (Rf).
4. Compute the standard deviation only of the negative returns (STD).
5. Subtract Rf from Ra and divide by STD.
What is a good Sortino ratio?
When comparing stocks, the one with a larger Sortino ratio is better. In any case, you want to keep in mind the list:
• Sortino ratio > 1: a good risk/return (R/R) profile.
• Sortino ratio > 2: a great R/R profile.
• Sortino ratio > 3: an excellent R/R profile.
What is a bad Sortino ratio?
A negative Sortino ratio is terrible because it indicates that the investor could have gotten a better return with no risk by investing in a risk-free option like government bonds (treasury bills).
In other words, the investor took more risks and still got poorer results. | {"url":"https://www.omnicalculator.com/finance/sortino-ratio","timestamp":"2024-11-04T05:43:48Z","content_type":"text/html","content_length":"541858","record_id":"<urn:uuid:3d3ea021-a276-4556-a346-5a6ed58a1afe>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00259.warc.gz"} |
Using the Equation of Motion in a Straight Line to Find the Magnitude of the Average Velocity
Question Video: Using the Equation of Motion in a Straight Line to Find the Magnitude of the Average Velocity
A person drove a car for 723 m on a straight road with a velocity of 9 km/h. He then continued for the same distance in the same direction, but with a velocity of 6 km/h. Find the magnitude of the
average velocity during the whole trip.
Video Transcript
A person drove a car for 723 metres on a straight road with a velocity of nine kilometres per hour. He then continued for the same distance in the same direction but with a velocity of six kilometres
per hour. Find the magnitude of the average velocity during the whole trip.
So, to solve this problem, what we can think about is the speed–distance–time triangle or, in this case, the velocity–displacement–time triangle. And that’s because what we need to do in this
question is find the average velocity during the whole trip. So therefore, if we look at velocity in our triangle, well we can see that velocity is gonna be equal to the displacement or distance
divided by the time.
So therefore, to work out the average velocity for the whole trip, what I’m gonna need to do is the total distance of the whole trip and the total time for the whole trip. So, the total distance is
gonna be equal to 723 multiplied by two. And that’s because we’re told that the person drove a car for 723 metres on a straight road with a velocity of nine kilometres per hour.
But then, we’re told that he continued for the same distance in the same direction but with a different velocity of six kilometres per hour. So, the total distance is gonna be 723 multiplied by two.
And when we do this, we get 1446 metres. So, that’s the total distance for the whole trip.
Now, to work at the time for the whole trip, it’s slightly more complicated cause we need to find out the value of this cause we’re not told in the question. Well, the time is equal to the distance
divided by the velocity. So therefore, we can work this out. We can work out the time for both sections of the journey.
So, when we do this, we’re gonna get time is equal to 0.723 divided by nine. We’ve got 0.723 because our distance is in metres. However, our velocity is in kilometres per hour. So therefore, we
convert metres to kilometres. So, 723 metres is 0.723 kilometres. And then, this is plus 0.723 divided by six, which gives us an answer of 0.200 continued hours. So now, we’ve got the total distance
for the whole journey. And we’ve got the total time for the whole journey.
So, now what we can do is calculate the average velocity. We can do this by dividing as we said in the beginning, the total distance by the total time, remembering that we need to turn the total
distance into kilometres not metres. So, it’s gonna be 1.446 kilometres. So, the average velocity is gonna be equal to 1.446 divided by 0.200 continued. Which will give us an answer of 7.2 kilometres
per hour, and I’ve rounded that to one decimal place.
So therefore, we can say that if is a person drove a car for 723 metres on a straight road with a velocity of nine kilometres per hour. Then, he continued for the same distance in the same direction
but with a velocity of six kilometres per hour. The magnitude of the average velocity during the whole entire trip would be 7.2 kilometres per hour. | {"url":"https://www.nagwa.com/en/videos/768126089159/","timestamp":"2024-11-04T14:33:05Z","content_type":"text/html","content_length":"245026","record_id":"<urn:uuid:c82861c9-be21-4c42-b99d-d63ab760bdc3>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00697.warc.gz"} |
HP-67: Gary Tenzer's Curve Fit / Best Function program
02-24-2014, 10:55 PM
Post: #1
Marcel Samek Posts: 54
Member Joined: Dec 2013
HP-67: Gary Tenzer's Curve Fit / Best Function program
Based on this thread:
, I found the code for the HP-67 version of Gary Tenzers curve fit / best function program here:
Could someone who has the information please send me the usage instructions. Its not working for me and I am not sure whether I entered it wrong, or whether I am misunderstanding how it should be
used. Thanks in advance.
02-24-2014, 11:14 PM
(This post was last modified: 02-24-2014 11:18 PM by Joe Horn.)
Post: #2
Joe Horn Posts: 2,018
Senior Member Joined: Dec 2013
RE: HP-67: Gary Tenzer's Curve Fit / Best Function program
Click on these thumbnails for the full-size original pages from the PPC Journal. The corrections handwritten here were specified in a subsequent issue. I'll bet that's what was causing your troubles.
. . .
02-24-2014, 11:21 PM
Post: #3
Marcel Samek Posts: 54
Member Joined: Dec 2013
RE: HP-67: Gary Tenzer's Curve Fit / Best Function program
(02-24-2014 11:14 PM)Joe Horn Wrote: Click on these thumbnails for the full-size original pages from the PPC Journal.
Thank you very much!
02-25-2014, 11:38 PM
Post: #4
Marcel Samek Posts: 54
Member Joined: Dec 2013
RE: HP-67: Gary Tenzer's Curve Fit / Best Function program
(02-24-2014 11:14 PM)Joe Horn Wrote: The corrections handwritten here were specified in a subsequent issue. I'll bet that's what was causing your troubles.
It turns out that I had entered the program wrong. I had misinterpreted the ISZI instruction. I only have an HP-67 and the listing has the HP-97 keycodes. It was only when I looked at a picture of
the 97 that I understood my error.
The link referenced in my original post does have all the handwritten changes incorporated.
Program works like a charm.
02-28-2014, 10:41 PM
Post: #5
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: HP-67: Gary Tenzer's Curve Fit / Best Function program
I wrote a series of "best regression models" for the HP 39gII for HP Solve. These programs also work for the HP Prime. They perform best models selection among hundreds and thousands of models. I
encourage you to search the HP Solve issues (look in late 2011 or early 2012 issues). YOU WILL NOT BE DISAPPOINTED!
03-01-2014, 02:19 AM
Post: #6
Marcel Samek Posts: 54
Member Joined: Dec 2013
RE: HP-67: Gary Tenzer's Curve Fit / Best Function program
(02-28-2014 10:41 PM)Namir Wrote: I wrote a series of "best regression models" for the HP 39gII for HP Solve. These programs also work for the HP Prime. They perform best models selection among
hundreds and thousands of models. I encourage you to search the HP Solve issues (look in late 2011 or early 2012 issues). YOU WILL NOT BE DISAPPOINTED!
Thanks for the reference. I took a quick look and was not disappointed! I really like the approach. Did you ever adapt any of those routines to other calculators besides the 39gII?
03-01-2014, 10:52 AM
Post: #7
Terje Vallestad Posts: 159
Member Joined: Dec 2013
RE: HP-67: Gary Tenzer's Curve Fit / Best Function program
(02-28-2014 10:41 PM)Namir Wrote: Marcel,
I wrote a series of "best regression models" for the HP 39gII for HP Solve. These programs also work for the HP Prime. They perform best models selection among hundreds and thousands of models. I
encourage you to search the HP Solve issues (look in late 2011 or early 2012 issues). YOU WILL NOT BE DISAPPOINTED!
I believe
is the link, very interesting
Cheers, Terje
03-01-2014, 03:03 PM
Post: #8
Marcel Samek Posts: 54
Member Joined: Dec 2013
RE: HP-67: Gary Tenzer's Curve Fit / Best Function program
(03-01-2014 10:52 AM)Terje Vallestad Wrote: I believe this is the link, very interesting
I was looking at
This article
, linked from
, originally found in this
03-02-2014, 12:48 PM
Post: #9
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: HP-67: Gary Tenzer's Curve Fit / Best Function program
Those are the ones! I am embarrassed to say that forgot about the copies of these articles being posted in my own web site!
User(s) browsing this thread: 1 Guest(s) | {"url":"https://www.hpmuseum.org/forum/showthread.php?mode=linear&tid=762&pid=5931","timestamp":"2024-11-03T20:17:46Z","content_type":"application/xhtml+xml","content_length":"40915","record_id":"<urn:uuid:0d24f1b9-dfe6-4892-8487-8522b7a6be1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00024.warc.gz"} |
Fibonacci: The Math of Agile
Agile planning poker^1 uses the Fibonacci sequence or in some cases a bastardized version of it. I do not know the exact reason that the series was chosen, I do know the intent is that there is a
standard set of numbers and using these deters the player from the temptation of using false precision, like a value of 10.5 for example. Also it captures growth so each next number is a "unit" of
growth larger than the last. I suspect that the sequence of powers of two (1,2,4,8,16,32,64...) would also work, and might be more apropos to the computer industry, but these two sequences are
similar in that they both increase by a constant factor, each subsequent power of two is two times larger and in the Fibonacci sequence each term is roughly 1.618... larger than the previous term,
also each has a closed form exponential formula for each term 2^n or for Fibonacci sequence it's Binet’s Formula, see below. I don’t know if the Fibonacci sequence is the right way to go, but it
seems to work and it is fairly prevalent in nature and art so it’s probably not a bad choice. I confess that while this in one of my math of programming series, I am using it primarily as an excuse
to write a post about some interesting results relating to the Fibonacci sequence and the Golden Mean, so if you are only interested in Agile you may want to treat this as a short post and stop here,
but I encourage you to read on, the math is mostly high school level and it really is interesting.
The Fibonacci sequence is generally attributed to a man, Leonardo Pisano Bigollo who went by Fibonacci and did not actually discover the sequence, it was known about five hundred years earlier in the
tenth century by Pingala and it was probably known much earlier. Ironically Fibonacci’s greatest mathematical contribution was the introduction of Hindu-Arabic Numeral System to Europe, just imagine
trying to do multiplication and fractions with Roman numerals.
The Fibonacci sequence is what is known as a recurrence relation or recursive sequence each term is defined as the sum of the two previous terms where the first two terms, which are the generator or
seed values F[0] = 0 and F[1] = 1, the formula is:
It can alternatively be written as:
This form will be more useful to us in a moment.
The sequence is:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393,...
The Fibonacci sequence has a very interesting relation with a very interesting number known as the Golden Mean which is sometimes referred to as phi and has the following value:
An irrational number that is approximately: 1.6180339887498948482045868343656...
The ratio of a Fibonacci number divided by the previous number in the sequence eventually converges to phi, the first few ratios are:
3/2 = 1.5
5/3 = 1.666666...
8/5 = 1.6
13/8 = 1.625
21/13 = 1.615384...
34/21 = 1.619047...
55/34 = 1.617647...
89/55 = 1.6181818...
As you can see it is already heading there. It is interesting that a ratio which is in the set of rational numbers converges to an algebraic irrational number. The following equation describes this
relationship, the Limit of consecutive quotients:
Additionally the following is true:
If we take the following formula:
And divide it by F[n] we get:
We then take the limit as n goes to infinity, replacing the terms using the two limit equations from above:
And if we multiply by phi:
Solving for zero gives us:
Using the quadratic equation we get the following two roots (conjugates):
This is phi and its conjugate, which we will call tau:
Now that we have defined phi and tau we can describe Binet’s formula, mentioned above, which is an exponential form closed formula:
Binet’s formula is an easy way to compute Fibonacci Sequence values without having to calculate all of the values prior to the value you wish to calculate.
Additionally there is an interesting matrix based calculation:
The above matrix raised to the nth power gives the above matrix of Fibonacci numbers, the first few are:
The determinant of this matrix is (-1)^n which is described by Cassini’s Identity:
The Fibonacci sequence and the Golden mean have many interesting identities and relationships with other mathematical disciplines and natural phenomena, for example there is the famous relation
rabbit breeding, it relates to bee ancestry, the packing of seeds in flowers, and other plant growth proportions. The Fibonacci sequence occurs in the shallow diagonals of Pascal’s Triangle, it
relates to Pythagorean Triples. It relates to the similar Lucas Sequence , the Fibonacci Spiral form above which is a Logarithmic Spiral and the list goes on, a lot of this can be found on the
previously linked Wikipedia Page and this BBC radio program.
^1 http://renaissancesoftware.net/papers/14-papers/44-planing-poker.html | {"url":"https://www.elegantcoding.com/2012/02/fibonacci-math-of-agile.html","timestamp":"2024-11-11T01:20:40Z","content_type":"application/xhtml+xml","content_length":"90752","record_id":"<urn:uuid:11919c5a-b32c-4626-9cac-17de768244eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00337.warc.gz"} |
Maharashtra Board 10th Class Maths Part 1 Practice Set 6.1 Solutions Chapter 6 Statistics
Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.
Practice Set 6.1 Algebra 10th Std Maths Part 1 Answers Chapter 6 Statistics
Question 1.
The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.
∴ The mean of the time spent by the students for their studies is 4.36 hours.
Question 2.
In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by ‘assumed mean’ method.
Let us take the assumed mean (A) = 550
∴ The mean of the toll paid by the drivers is ₹ 521.43.
Question 3.
A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.
∴ The mean of the milk sold is 2.82 litres.
Question 4.
A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by ‘assumed mean’ method.
Let us take the assumed mean (A) = 37.5
∴ The mean of the production of oranges is ₹ 35310.
Question 5.
A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of he funds by ‘step deviation’ method.
Here, we take A = 1250 and g = 500
∴ The mean of the funds collected is ₹ 987.5.
Question 6.
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by ‘step deviation’ method.
Here, we take A = 2500 and g = 1000.
∴ The mean of the weekly wages is ₹ 3070.
Question 1.
The daily sale of 100 vegetable vendors is given in the following table. Find the mean of the sale by direct method. (Textbook pg. no. 133 and 134)
The mean of the sale is 2150.
Question 2.
The amount invested in health insurance by 100 families is given in the following frequency table. Find the mean of investments using direct method and assumed mean method. Check whether the mean
found by the two methods is the same as calculated by step deviation method (Ans: ₹ 2140). (Textbook pg. no. 135 and 136)
∴ The mean of investments in health insurance is ₹ 2140.
Assumed mean method:
∴ The mean of investments in health insurance is ₹ 2140.
∴ Mean found by direct method and assumed mean method is the same as calculated by step deviation method.
Question 3.
The following table shows the funds collected by 50 students for flood affected people. Find the mean of the funds.
If the number of scores in two consecutive classes is very low, it is convenient to club them. So, in the above example, we club the classes 0 – 500, 500 – 1000 and 2000 – 2500, 2500 – 3000. Now the
new table is as follows
i. Solve by direct method.
ii. Verily that the mean calculated by assumed mean method is the same.
iii. Find the mean in the above example by taking A = 1750. (Textbook pg. no. 137)
i. Direct method:
∴ The mean of the funds is ₹ 1390.
ii. Assumed mean method:
Here, A = 1250
∴ The mean calculated by assumed mean method is the same.
iii. Step deviation method:
Here, we take A = 1750 and g = 250
∴ The mean of the funds is ₹ 1390. | {"url":"https://maharashtraboardsolutions.in/class-10-maths-solutions-part-1-chapter-6-practice-set-6-1/","timestamp":"2024-11-11T10:19:25Z","content_type":"text/html","content_length":"65112","record_id":"<urn:uuid:476117cb-1d5b-4645-ac94-2593d8db5c71>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00357.warc.gz"} |
04 Median and its measure of variability
Median as the center of distribution
In the last lessons, we learned about the mean as the center of distribution. We interpreted mean as representation of equal or fair reallocation of the total for all the data. We also saw the mean
corresponds to the balance point of the distribution, i.e. the total distances of the points on the left of the mean is the same as those on the right.
You might however have noticed that in some cases, the number of points on the left of the mean was not the same as those on the right of the mean. Like in the example below where mean is 21, and
there are 4 points to the left but only 2 to the right of the mean.
And you might have wondered “Shouldn’t the center of the distribution have an equal number of points on the left and right of it, since it’s the ‘center’?”
Well, good news! There is another way to think about the center of a distribution, whereby we identify a value with approximately half the data on each side. This quantity is called the median (also
called the 50th percentile), and it is the “middle value” when the data have been arranged in order. Half of the values in a data set are less than or equal to the median, and half of the values are
greater than or equal to the median.
Great! Seems easy so far. But let’s actually try it out to make more sense of it.
To find the median, we first need to order the data values from least to greatest. Then we find the number in the middle. Suppose we have 5 dogs whose weights, in pounds, are shown as below (already
in ascending order).
20, 25, 32, 40, 55
The median weight for this group of dogs is 32 pounds, since it is the middle value. There are three dogs that weigh less than or equal to 32 pounds and three dogs that weigh greater than or equal to
32 pounds.
Now suppose we have 6 cats whose weights, in pounds, are as shown below. Notice that there are two values in the middle this time: 7 and 8.
4, 6, 7, 8, 10, 10
So which one is the median?
In cases like these, when we have an even number of data points and there are two values in the middle, the median weight must be between 7 and 8 pounds. This is because half of the cats weigh less
or equal to 7 pounds and half of the cats weigh greater than or equal to 8 pounds. In general, when we have an even number of values, we take the number exactly in between the two middle values. In
this case, the median cat weight is 7.5 pounds because (7+8)/2 = 7.5. Basically, find the mean of the two to get the median!
Let’s look at an example with more data. Below is the data for the different times taken by Eli to travel to school in 15 different days.
6, 6, 6, 7, 7, 8, 8, 9, 10, 10, 10, 12, 14, 15, 15
They look like they are in ascending order, so let’s find the median.
There are 15 data points, so the middle value is at the 8th position, which is 9.
In the dot plot given, you can see that there are 7 data points both to the right and to the left of the median (shown in blue).
Interquartile range to measure variability
Earlier we learned that the mean is a measure of the center of a distribution and the MAD is a measure of the variability (or spread) that goes with the mean. There is also a measure of spread that
goes with the median. This is called the interquartile range (IQR).
Finding the IQR involves splitting a data set into fourths. How do we go about doing this? Let’s look at it one step at a time using the given data.
1. We first divide the whole data into half, which is done by the median.
12 18 20 22 22 | 33 (Median) | 35 37 40 40 49
2. We see that each side of the median now has 5 values each, these are the lower and upper halves. We now further divide the two halves with the middle value as well. 12 18 | 20 (First) | 22 22 |
33 (Median) | 35 37 | 40 (Third) | 40 49
We now have four different divisions, each with 2 values and the three numbers that divide the data into the four sections. Each of the three values that splits the data into fourths is called a
quartile. The first quartile (20) is Q1, the median (33) is the second quartile and also called Q2. The third quartile (40) is Q3.
So what do these numbers/quartiles tell us?
The median is 33 and it divides the data in two halves. The first quartile is 20, which is the median of the numbers that are less than 33. This means that a forth of the total data lies below 33 and
the remaining ¾ lie above this value. The third quartile 40, which is the median of the numbers greater than 33. This means that ¾ of the data lies below this value and ¼ lie above it.
In general, all data ranges from a minimum to a maximum value. The difference between the maximum and minimum values of a data set is the Range, which can also tell the spread of a dataset. Here, the
range is 49 - 12 = 37
Just like range, the interquartile range is a difference between two numbers in the dataset. Rather than the maximum and minimum, the difference between Q3 and Q1 is called Interquartile Range (IQR).
Here, IQR is calculated as Q3 - Q1 = 40 - 20 = 20.
The distance between Q1 and Q3 includes the middle two-fourths of the distribution (¼ + ¼), so the values between those two quartiles are sometimes called the middle half (or middle 50%) of the data.
The IQR of 20 indicates that any data in the middle 50% of the data will differ by at most 20. The IQR provides an additional measure of variability for a distribution and is used when median is
chosen as the measure of center. The bigger the IQR, the more spread out the middle half of the data values are. The smaller the IQR, the closer together the middle half of the data values are. This
is why we can use the IQR as a measure of spread.
There are several different methods for determining quartiles. For example, when n is odd, the ordered data cannot be evenly divided in half, since the single number in the middle is the median. For
this lesson, we exclude the median from the lower and upper “halves” when determining the quartiles Q1 and Q3. We excluded 33 when finding Q1 and Q3 in the previous example since we have 11 values.
So far, we have identified the minimum, first quartile, median, third quartile, and maximum of the data set. These measures together give us a “five-number summary”, which we can use to summarize a
distribution. The five-number summary is great because it includes both the measure of center (median) and the measure of variability (IQR) as well as the range of the values. For the previous
example, the five-number summary is 12, 20, 33, 40, and 49. These numbers are marked with diamonds on the dot plot. Median is marked as the bigger diamond.
Just like how we saw that a different dataset can have the same mean, different data sets can also have the same five-number summary. For instance, here is another data set with the same minimum,
maximum, and quartiles as the previous example, but we see that it looks different from the previous dataset.
Box plot to represent the five number summary
While it is perfectly fine to show the five number summary in the dot plot as we have seen earlier, there is another graphical representation for this. Let’s look at it first and then understand what
it represents. We will be using the same example from before.
Can you think where the five numbers are represented in the diagram given? Our five number summary was 12, 20, 33, 40, and 49. Can you see any pattern that could help us distinguish these 5 numbers
in the plot?
We see that the 5 vertical lines in the plot represent our five numbers.
This is how a box plot can be used to visually represent the five-number summary of a data set. It shows the first quartile (Q1) and the third quartile (Q3) as the left and right sides of a rectangle
or a box. The median (Q2) is shown as a vertical segment inside the box. On the left side, a horizontal line segment—a “whisker”—extends from Q1 to the minimum value. On the right, a whisker extends
from Q3 to the maximum value. The rectangle in the middle represents the middle half of the data. We know that Q3 - Q1 is the IQR of the data. This means that the length of this box or its width is
the IQR here. The whiskers represent the bottom quarter and the top quarter of the data set.
Let’s now look at a box plot for the weight of pugs.
The five number summary for this distribution is: Minimum -> 6 Q1 -> 6.6 Q2/Median -> 7 Q3 -> 7.4 Maximum -> 8
On the basis of the five-number summary and the box plot, what can we say about the distribution of weights of pugs?
Approximately 25% of pugs weigh between 6 and 6.6 kg. The middle 50% of weights are between 6.6 and 7.4 kg. The range of the middle 50% (the difference between the third and first quartiles) or the
IQR is 0.8 kg. The IQR indicates that any two weights from the middle 50% of the data will differ by at most 0.8 minutes.
We can also use the box plot to compare different groups.
Case I: Two groups with the different median but the same IQR
Here are two sets of plots that show the weights of some berries and some grapes.
The median berry weight is 3 grams and the median grape weight is 5 grams. But in both cases, the IQR is 1.5 grams. Because the grapes in this group have a higher median weight than the berries, we
can say a grape in the group is typically heavier than a berry. Since both groups have the same IQR, we can say that they have a similar variability in their weights.
In the case of berries, we can see that the middle rectangle is more to the left, meaning that the data is right skewed. In the case of the grapes, it seems to be slightly left skewed.
Case II: Two groups with the same median but different IQR
These box plots represent the length data for a collection of ladybugs and a collection of beetles. The medians of the two are the same, but the IQR of the ladybugs is much smaller (2 vs 11.5 for
beetles). This tells us that a typical ladybug length is similar to a typical beetle length, but the ladybugs are more alike in their length than the beetles are in their length. | {"url":"https://edukimath.com/grade-6/statistics-and-probability/median/","timestamp":"2024-11-03T07:03:26Z","content_type":"text/html","content_length":"49406","record_id":"<urn:uuid:0f5991c8-dbe4-4d8c-ab19-54da867b980e>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00530.warc.gz"} |
Mathematics of Sustainability
Carbon Dioxide from Deforestation
An article from the Union of Concerned Scientists explains how we measure the amount of CO2 from deforestation in our atmosphere. The study concludes that deforestation contributes to around 3
billion tons of carbon dioxide in our atmosphere per year; that is, about 10% of all CO2 emissions.
Simplifying these numbers and their units to a more easily understandable human term is extremely helpful in cases such as this. Explaining that 3 billion tons of carbon dioxide is different from 3
billion tons of bricks is essential in this understanding. So just how big (or small) is 3 billion tons of CO2, and how would we find out?
We can start by using the unit factor method to compute this number into human terms. For instance, we could express 3 billion tons of CO2 as the equivalent of 13 million railroad cars, stretching
around 125,000 miles or half way to the moon. By using a familiar term such as the size of a piece of land or animal, we get to see a big number from a different perspective. That being said, the
total amount of CO2 from deforestation is equivalent to the total emissions from all of Western Europe combined.
Finally, the article takes into consideration the approximation and estimations that we read about in the textbook this week. The author described two approaches that one might take to compute these
numbers. In one approach they explain using the most certain and comparable numbers, focusing on specific dates and measuring only one thing (CO2). The next approach brings in other aspects with
complete and up-to-date information, calculating all possible variables creating a more substantial result.
There are many different ways and units to measure numbers with. In this specific case, all calculations came out to around the same conclusion: 10% of all CO2 emissions are from deforestation
itself, or the equivalent of 600 million cars (twice as many than there are in the entire US).
4 thoughts on “Carbon Dioxide from Deforestation”
1. Thank you for putting the numbers into simpler terms, it perfectly embodies what we talked about in class today and helped me conceptualize what you were describing. My jaw dropped when you used
the railroad car example because that really helped me visualize the issue at hand. Great post!
2. This is an excellent job at attempting to paint a visual picture in the readers mind. While the process of deforestation is happening on a widespread level, the masses are generally unaware (or
choose to ignore) the significant extent to which this occurs. You do a great job at applying these seemingly in-quantifiable numbers into tangible comparisons. If this logic was applied when
gauging other factors of life, maybe problems would get solved faster!
3. This article is peaking a lot of questions in my mind like: How can we reduce the rate of deforestation? What objects in our consumeristic society contribute the most to the harmful effects of
deforestation? Is it too late to save these destroyed ecosystems by deforestation?
I think it was really interesting to ask that kind of question in regards to tons and I wonder if put into the terms you are explaining deforestation people would be more open to hearing it
because of how well communicated the message of this blog post is.
4. I thought you did an awesome job of putting CO2 emission into an easier-to-grasp perspective. The comparisons you made and the way you discussed the meaning of the numbers associated with CO2 is
extremely valuable. My favorite example you used was comparing CO2 emissions to the distance of halfway to the moon because it really showed how large of a number it really is and helped me to
conceptualize everything. Thank you!
You must be logged in to post a comment. | {"url":"https://muse.union.edu/mth-063-01-f18/2018/09/09/carbon-dioxide-from-deforestation/","timestamp":"2024-11-10T20:26:53Z","content_type":"text/html","content_length":"69202","record_id":"<urn:uuid:2416843f-35fa-4c94-bfa2-dd951585aed9>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00194.warc.gz"} |
Quantitative Spectroscopy Pitfalls and Practicalities, Part III: Calibration Validation and Calibration Checks
This column continues the discussion of the practicalities and pitfalls of quantitative spectroscopy started in the last two columns (1,2). The reason that a discussion of quantitative spectroscopy
matters for cannabis analysis is that many cannabis potency methods use high performance liquid chromatography (HPLC) along with ultraviolet-visible (UV-vis) spectroscopic detection (3). Also, there
exist cannabis potency analyzers based on mid-infrared spectroscopy (4).
The fundamental equation of quantitative spectroscopy is Beer’s Law (5), whose form is seen in Equation 1.
where A is the absorbance, the amount of light absorbed by a sample; ε is the absorptivity, a fundamental physical constant of a molecule; L is the pathlength or sample thickness; and C is the
I will dispense with any further discussion of Beer’s Law or the basics of quantitative spectroscopy and refer you to my previous columns and book on these topics (1,2,5-8). In this, the third in a
series of pitfalls to avoid and practicalities to know about, we will discuss the importance of validating and checking your calibration to ensure quality.
The Fundamental Assumption of Quantitative Spectroscopy
Our tendency once we have a calibration in our hands is to immediately begin using it because they are a lot of work, and we may be anxious to start getting results. This is, however, the wrong
approach. Recall (5-8) that when we generate a spectroscopic calibration, we have to make up samples of known concentration called standards, measure the absorbance of these standards, and plot their
absorbance versus concentration to obtain a calibration line. We can then use our calibration to predict concentrations in unknown samples using Equation 2:
where C[unk] is the concentration of analyte in the unknown sample, A[unk] is the absorbance of the analyte in the unknown sample, and εL is the product of the absorptivity and pathlength obtained
from the slope of the calibration line.
Note in Equation 2 that the concentration and absorbance refer to the unknown sample, but that εL is derived from a plot based on the standard samples. The assumption we are making here is that εL is
the same for the standards as it is for the unknown samples. This is what I call the fundamental assumption of quantitative spectroscopy (FAQS) because if it is violated it means εL is different for
the standards and unknowns. Applying an incorrect value of εL to unknown samples will give incorrect values of C[unk] because we are using a value of εL that applies to the standards but not to the
unknowns. Therefore, violating the fundamental assumption of quantitative spectroscopy can be very damaging. So then, how do we know if we are violating this assumption? This is where calibration
validation comes in.
What Is a Calibration Validation?
To check if we are violating the FAQS we must test whether εL is the same for the standard and unknown samples. Why would εL ever be different for the standards and unknowns? The pathlength, L, can
be different if the sampling cells or devices you are using for the standards and unknowns are not identical. The best way to ensure this is to use the same sample cell for all standards and
unknowns. Ensuring the absorptivity, ε, is the same for samples and unknowns is trickier. Recall that the absorptivity is matrix sensitive and changes with variables such as temperature, pressure,
concentration, and composition (5,6). This is why it is so important to ensure that all the experimental variables are the same when you measure the absorbance of the standard and unknown samples.
Given the vicissitudes of experiments and our inability to always perfectly control all variables, we must perform calibration validations before using them to analyze unknown samples. To test the
FAQS, we make up a standard sample of known concentration but do not use it in the calibration. We will call this sample a validation sample. We treat the validation sample as an unknown, measure its
absorbance, apply Equation 2, and predict its concentration. We then compare the predicted concentration of the validation sample to its known concentration to see how well they agree. If they agree
within the accuracy you are looking for this means you can go forward implementing your calibration. If not, back to the drawing board, and more on that in future columns.
Recall that accuracy is a measure of how far away a measurement is from its true value (9). The difference between the known and predicted concentration for the validation sample is the measure of
the accuracy of your calibration. It is important to use at least one validation sample to test that your calibration does not violate FAQS. It is even better to create and analyze several validation
samples. In this case for each sample, like above, measure its concentration, apply the calibration, and predict its concentration value. The beauty of analyzing multiple validation samples is that
it allows you to calculate what is, in my opinion, the best measure of calibration accuracy called the standard error of prediction (SEP), which is seen in Equation 3.
where SEP is the standard error of prediction, ∑[i] is the validation sample number index, C[p] is the predicted concentration, C[a] is the actual concentration, and n is the total number of
validation samples.
Let’s say there are three validation samples. To calculate the SEP the first thing to do is subtract the actual concentration from the predicted concentration for the three samples, square each of
the differences, and then add these together. This gives the numerator in Equation 3. You then divide this sum by the number of validation samples minus 1, this is the denominator in Equation 3.
Then, take the square root, and you have yourself a standard error of prediction. The SEP is the best measure of calibration quality because it shows you how well your calibration does on samples
that are not included in the calibration, which is exactly how a calibration is used in real life. The units of the SEP will be in the units of your concentration measurement. For example, if C[p]
and C[a] are in weight percent (Wt. %) then the SEP will be in Wt. %.
What value of the SEP tells me if my calibration does or doesn’t violate the FAQS? Strangely enough after all this math this ultimate answer is a judgement call. It depends on how accurate you need
the calibration to be, that is, the needed accuracy is application specific. For example, if you want to know the total tetrahydrocannabinol (THC) in a cannabis bud that is normally around 20 Wt. %,
an accuracy of ±1 Wt. % might be fine. However, if you are a hemp grower and have to ensure your crop is less than 0.3% weight total THC to comply with federal law, an accuracy of ±1 Wt. % total THC
is worthless because the error bar is bigger than 0.3 Wt. % and overlaps with zero. In this case, an acceptable total THC accuracy would need to be significantly less than 0.3% total THC. For
example, mid-infrared spectroscopy is capable of an accuracy of ±0.04 Wt. % for total THC in dried, ground hemp (10).
A mistake I find many people make is demanding more accuracy out of a calibration than is necessary. For example, for cannabis growers in states such as California where cannabis is legal, an
accuracy of ±0.04 Wt. % for total THC is not needed since the state allows a ±10% relative error between your label and the actual product (11). For example, a bud labeled at 20% total THC can
contain anywhere from 18% to 22% total THC and still be within the law. Because increasing accuracy is always time consuming and expensive, determine before you start calibrating what accuracy your
calibration requires, and then stop work once you achieve that accuracy level.
Calibration Checks
Now that we have a validated calibration, we can begin using it to legitimately predict concentrations in unknown samples. Does this mean our job is done? By no means. Remember that because of the
FAQS we must always ensure that εL is the same for the standards and the unknowns. Things change over time, and the only way to make sure you are not violating the FAQS over time is to run frequent
calibration checks. A calibration check is not a full calibration and is very much like a validation. To perform a calibration check, make up some validation samples, measure their absorbance,
predict their concentration, and compare the predicted and known values like we did above. Calculate an SEP if you used more than one calibration check sample. If things still agree within your
required accuracy you can continue to use your calibration. If not, you need to immediately stop using the calibration and investigate what went wrong and how to fix it (more on this in later
How often should you run a calibration check? The answer is as often as possible. Some laboratories will run a calibration check every time they use a quantitative spectroscopic calibration. This is
great, but it can be time consuming and expensive. If you are using your calibration every day certainly a calibration check every week at least should be performed. If you use your calibration less
often, you really should think about running a calibration check every time you use the calibration.
The fundamental assumption of quantitative spectroscopy (FAQS) was introduced, wherein we assume that the product of the pathlength and concentration, εL, is the same for standard and unknown
samples. We test this assumption by running a calibration validation where a standard sample of known concentration, a validation sample, is analyzed with the calibration and the known and predicted
values are compared. The difference between these two is the accuracy of your calibration. Ideally several validation samples should be run so that a standard error of prediction, the best measure of
calibration accuracy, can be calculated. Whether or not your calibration is accurate enough to be used depends upon your application. To insure the FAQS is not violated over time, calibration checks
should be run as often as is practicable.
1. B.C. Smith, Cannabis Science and Technology 5(5), 8-13 (2022).
2. B.C. Smith, Cannabis Science and Technology 5(6), 8-11 (2022).
3. M.W. Giese, M.A. Lewis, L. Giese, and K.M. Smith, Journal of AOAC International 98(6) (2015)1503.
4. B.C. Smith, Quantitative Spectroscopy: Theory and Practice (Elsevier, Boston, Massachusetts, 2002).
5. B.C. Smith, Cannabis Science and Technology 5(4), 8-15 (2022).
6. B.C. Smith, Cannabis Science and Technology 5(3), 10-14 (2022).
7. B.C. Smith, Cannabis Science and Technology 5(2), 10-13 (2022).
8. B.C. Smith, Cannabis Science and Technology 1(4), 12-16 (2018).
9. B.C. Smith, Cannabis Science and Technology 3(6), 10-13 (2020).
About the Columnist
Brian C. Smith, PhD, is Founder, CEO, and Chief Technical Officer of Big Sur Scientific. He is the inventor of the BSS series of patented mid-infrared based cannabis analyzers. Dr. Smith has done
pioneering research and published numerous peer-reviewed papers on the application of mid-infrared spectroscopy to cannabis analysis, and sits on the editorial board of Cannabis Science and
Technology®. He has worked as a laboratory director for a cannabis extractor, as an analytical chemist for Waters Associates and PerkinElmer, and as an analytical instrument salesperson. He has more
than 30 years of experience in chemical analysis and has written three books on the subject. Dr. Smith earned his PhD on physical chemistry from Dartmouth College. Direct correspondence to:
How to Cite this Article:
B. Smith, Cannabis Science and Technology® Vol. 5(7), 8-11 (2022). | {"url":"https://www.cannabissciencetech.com/view/quantitative-spectroscopy-pitfalls-and-practicalities-part-iii-calibration-validation-and-calibration-checks","timestamp":"2024-11-14T22:01:21Z","content_type":"text/html","content_length":"337489","record_id":"<urn:uuid:99cfde43-776e-409e-a401-74c106e94a8d>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00324.warc.gz"} |
Tensor Calculus
A tensor is a relation between one vector and another. If you start with one vector, such as a force, and mathematically apply it to a tensor, then you get another vector. That vector might be, for
example, the stress caused by the force. That’s the simplest thing they do. Tensors do other things too; for example, the metric tensor represent the geometry of space. A tensor can represent the
energy-momentum density. A tensor can represent a combination of electric and magnetic fields in a way that some of Maxwell’s equations greatly simplify. But the simplest and most basic connection is
the vector to vector one. A function relates one number (a “scalar”) to another one. A tensor does that for vectors.http://qr.ae/TUTNzc
A tensor is a relation between one vector and another. If you start with one vector, such as a force, and mathematically apply it to a tensor, then you get another vector. That vector might be, for
example, the stress caused by the force.
That’s the simplest thing they do. Tensors do other things too; for example, the metric tensor represent the geometry of space. A tensor can represent the energy-momentum density. A tensor can
represent a combination of electric and magnetic fields in a way that some of Maxwell’s equations greatly simplify. But the simplest and most basic connection is the vector to vector one.
A function relates one number (a “scalar”) to another one. A tensor does that for vectors.http://qr.ae/TUTNzc
A Brief on Tensor Analysis by James Simmonds - a concise but great introductory text.
Tensor calculus is a technique that can be regarded as a follow-up on linear algebra. It is a generalization of classical linear algebra. In classical linear algebra one deals with vectors and
matrices. Tensors are generalizations of vectors and matrices. Introduction to Tensor Calculus by Kees Dullemond & Kasper Peeters | {"url":"https://physicstravelguide.com/basic_tools/tensor_calculus","timestamp":"2024-11-11T23:39:56Z","content_type":"text/html","content_length":"75705","record_id":"<urn:uuid:dccc6a23-7501-40f7-b87d-11eaa71cfe95>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00886.warc.gz"} |
Chains (US survey) to Inches (US survey) Converter
Enter Chains (US survey)
Inches (US survey)
β Switch toInches (US survey) to Chains (US survey) Converter
How to use this Chains (US survey) to Inches (US survey) Converter π €
Follow these steps to convert given length from the units of Chains (US survey) to the units of Inches (US survey).
1. Enter the input Chains (US survey) value in the text field.
2. The calculator converts the given Chains (US survey) into Inches (US survey) in realtime β using the conversion formula, and displays under the Inches (US survey) label. You do not need to
click any button. If the input changes, Inches (US survey) value is re-calculated, just like that.
3. You may copy the resulting Inches (US survey) value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Chains (US survey) to Inches (US survey)?
The formula to convert given length from Chains (US survey) to Inches (US survey) is:
Length[(Inches (US survey))] = Length[(Chains (US survey))] / 0.0012626262636525275
Substitute the given value of length in chains (us survey), i.e., Length[(Chains (US survey))] in the above formula and simplify the right-hand side value. The resulting value is the length in inches
(us survey), i.e., Length[(Inches (US survey))].
Calculation will be done after you enter a valid input.
Consider that a land survey recorded a boundary line of 25 chains (US survey).
Convert this distance from chains (US survey) to Inches (US survey).
The length in chains (us survey) is:
Length[(Chains (US survey))] = 25
The formula to convert length from chains (us survey) to inches (us survey) is:
Length[(Inches (US survey))] = Length[(Chains (US survey))] / 0.0012626262636525275
Substitute given weight Length[(Chains (US survey))] = 25 in the above formula.
Length[(Inches (US survey))] = 25 / 0.0012626262636525275
Length[(Inches (US survey))] = 19800
Final Answer:
Therefore, 25 ch is equal to 19800 in.
The length is 19800 in, in inches (us survey).
Consider that a railway track section measures 30 chains (US survey).
Convert this distance from chains (US survey) to Inches (US survey).
The length in chains (us survey) is:
Length[(Chains (US survey))] = 30
The formula to convert length from chains (us survey) to inches (us survey) is:
Length[(Inches (US survey))] = Length[(Chains (US survey))] / 0.0012626262636525275
Substitute given weight Length[(Chains (US survey))] = 30 in the above formula.
Length[(Inches (US survey))] = 30 / 0.0012626262636525275
Length[(Inches (US survey))] = 23760
Final Answer:
Therefore, 30 ch is equal to 23760 in.
The length is 23760 in, in inches (us survey).
Chains (US survey) to Inches (US survey) Conversion Table
The following table gives some of the most used conversions from Chains (US survey) to Inches (US survey).
Chains (US survey) (ch) Inches (US survey) (in)
0 ch 0 in
1 ch 792 in
2 ch 1584 in
3 ch 2376 in
4 ch 3168 in
5 ch 3960 in
6 ch 4752 in
7 ch 5544 in
8 ch 6336 in
9 ch 7128 in
10 ch 7920 in
20 ch 15840 in
50 ch 39600 in
100 ch 79199.9999 in
1000 ch 791999.9994 in
10000 ch 7919999.9936 in
100000 ch 79199999.9356 in
Chains (US survey)
A chain (US survey) is a unit of length used primarily in land surveying in the United States. One US survey chain is equivalent to exactly 66 feet or approximately 20.1168 meters.
The US survey chain is defined as 66 feet, based on historical surveying practices and used for measuring and plotting land.
Chains (US survey) are used in land surveying for tasks such as property measurement, land division, and mapping in the United States. This unit ensures consistency and accuracy in surveying and land
measurement activities.
Inches (US survey)
An inch (US survey) is a unit of length used in various contexts in the United States. One US survey inch is equivalent to exactly 0.0254 meters.
The US survey inch is defined as 1/12 of a US survey foot, with the US survey foot being slightly different from the international foot. This unit provides precise measurement for a variety of
Inches (US survey) are commonly used in construction, manufacturing, and everyday measurements in the United States. The unit ensures consistency and accuracy in both practical and scientific
Frequently Asked Questions (FAQs)
1. What is the formula for converting Chains (US survey) to Inches (US survey) in Length?
The formula to convert Chains (US survey) to Inches (US survey) in Length is:
Chains (US survey) / 0.0012626262636525275
2. Is this tool free or paid?
This Length conversion tool, which converts Chains (US survey) to Inches (US survey), is completely free to use.
3. How do I convert Length from Chains (US survey) to Inches (US survey)?
To convert Length from Chains (US survey) to Inches (US survey), you can use the following formula:
Chains (US survey) / 0.0012626262636525275
For example, if you have a value in Chains (US survey), you substitute that value in place of Chains (US survey) in the above formula, and solve the mathematical expression to get the equivalent
value in Inches (US survey). | {"url":"https://convertonline.org/unit/?convert=chain_us_survey-inches_us_survey","timestamp":"2024-11-02T02:40:49Z","content_type":"text/html","content_length":"91782","record_id":"<urn:uuid:8fc6ccf9-5149-4e2f-910f-4d96f23097fd>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00893.warc.gz"} |
Big 10 Solving Quadratic Equations by Factoring Worksheet | Math = Love
Big 10 Solving Quadratic Equations by Factoring Worksheet
Your students will enjoy this self-checking solving quadratic equations by factoring worksheet which uses the Big 10 review activity structure.
Earlier this week, I shared a Big 10 Worksheet over Percents for middle school students (or a much-needed review of percents for my high school statistics students…)
This activity was originally published in the Fall 2023 Oklahoma Council of Teachers of Mathematics Newsletter.
About the Big 10 Worksheet Structure
A few years ago, I started teaching AP Calculus AB. It has been a wonderful learning opportunity for both my students and me. I am continually blown away by the number of free activities and
resources that are shared by members of the AP Calculus community.
I realized that several of the practice activities that were shared in the AP Calculus Teachers Facebook group could be modified for use in math courses of all levels. One of
these commonly used activities for teaching AP Calculus is called “Big 10.”
The majority of the Big 10 activities I have used in calculus have been written by Bryan Passwater. Vicki Carter has also been creating some Big 10 Activities for AP Precalculus. There is also a
collection of AP Calculus Big 10 Activities that can be downloaded on TPT from CR Calculus.
A Big 10 worksheet has 10 questions. At the beginning of the worksheet, there is a box containing either the numbers 0-9 or 1-10.
Students cross out a number in the box after solving each problem. If done correctly, students will have all 10 numbers crossed off without any repeats.
I like these activities because they are self-checking, and they aren’t too tricky to put together yourself if you are needing a review activity for a certain topic. You simply need to come up with
10 questions with answers of either 0-9 or 1-10.
Worksheet Instructions
Complete each problem. Cross off the answer in the box below.
If done correctly, you will have all 10 numbers crossed off with no repeats.
This worksheet features 10 different quadratic equations for students to solve. Each of these quadratics can be solved by factoring. The worksheet does not specify that students MUST solve the
quadratic equations by factoring.
Students could use the quadratic formula to solve these equations if they preferred to. For me, however, I would most likely include this worksheet in my solving quadratic equations by factoring
These problems are intended for students at an Algebra 2 level or above. I have not personally used this activity with students yet. I created it to be featured in the Oklahoma Council of Teachers of
Mathematics quarterly newsletter, and I decided I wanted to share it with you as well!
Puzzle Solutions
I intentionally do not make answers to the printable math puzzles I share on my blog available online because I strive to provide learning experiences for my students that are non-google-able. I
would like other teachers to be able to use these puzzles in their classrooms as well without the solutions being easily found on the Internet.
However, I do recognize that us teachers are busy people and sometimes need to quickly reference an answer key to see if a student has solved a puzzle correctly or to see if they have interpreted the
instructions properly.
If you are a teacher who is using these puzzles in your classroom, please send me an email at sarah@mathequalslove.net with information about what you teach and where you teach. I will be happy to
forward an answer key to you.
Not a teacher? Go ahead and send me an email as well. Just let me know what you are using the puzzles for. I am continually in awe of how many people are using these puzzles with scouting groups,
with senior adults battling dementia, as fun activities in their workplace, or as a birthday party escape room. | {"url":"https://mathequalslove.net/big-10-solving-quadratic-equations-worksheet/","timestamp":"2024-11-08T10:51:25Z","content_type":"text/html","content_length":"236206","record_id":"<urn:uuid:c319c1f4-b776-4652-88d5-19eede7b8e7b>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00315.warc.gz"} |
The Mathematics for Real World Systems (MathSys) Centre for Doctoral Training at the University of Warwick and the Data Science and Systems Complexity Centre at the University of Groningen are
co-organizing a PhD Summer School on Complex Networks: Dynamics and Control. The school will take place 3-5 July 2019 at the University of Warwick. I will give two lectures on the Dynamics and
Synchronization of Globally Coupled Oscillators. More information can be found on the school website.
I gave a presentation on Hamiltonian Monodromy at the Symplectic Dynamics workshop that took place 8-12 April 2019 at the Lorentz Center. The workshop was organized by Urs Frauenfelder, Holger
Waalkens, and Lei Zhao.
On the occasion of the PhD defense of Nikolay Martynchuk we are organizing a mini-workshop in Groningen on Thursday 20 September 2018 with talks by Alexey Bolsinov, Holger Dullin, Sonja Hohloch,
Andreas Knauf, San Vũ Ngọc, and Nikolay Martynchuk. More information is available on the workshop website. The PhD defense will take place on Friday 21 September at 11:00.
Together with Andrea Giacobbe and Tudor Ratiu I am organizing a special session on Geometry and Dynamics in the context of AIMS 2018 (the 12th AIMS Conference on Dynamical Systems, Differential
Equations, and Applications). AIMS 2018 takes place 5-9 July 2018 in Taipei, Taiwan.
Pulse coupled oscillators is one of the most widely used models for describing collective phenomena such as synchronization. The most well studied such system is the Kuramoto model and it consists of
first order oscillators. When there is no coupling, the phase of each oscillator increases at a constant speed, called the natural frequency. However, coupling the oscillators by letting them
influence each other’s speed leads (for sufficiently strong coupling) to synchronization.
I will be giving a talk at the conference Symmetry and Perturbation Theory which takes place 3-10 June 2018. The conference is organized by Mariano Cadoni, Giuseppe Gaeta, and Sebastian Walcher.
I will be giving a talk at the IBS-CGP workshop on Integrable Systems and Applications which takes place 2-4 May 2018 in Pohang, South Korea. The workshop is organized by Alexander Aleksandrov,
Yong-Geun Oh, and Christophe Wacheux at the IBS Center for Geometry and Physics.
On 25 April 2018 I will be giving a talk at the UvA-VU Dynamic Analysis Seminar .
On 23 April 2018 I will be giving a talk at the Analysis and Geometry Seminar at the University of Antwerp.
I will be giving a talk at the conference Geometric Aspects of Momentum Maps and Integrability which takes place 8-13 April 2018. The conference takes place at the CSF (Congressi Stefano Franscini)
in Ascona, Switzerland and is organized by Anton Alekseev, Sonja Hohloch, and Tudor Ratiu.
On the occasion of being elected Teacher of the Year for Chemical Engineering, Mónica Espinoza Cangahuala interviewed me for Het Chemisch Bindmiddel, the magazine of De Chemische Binding, the study
association for Chemistry and Chemical Engineering. The interview is available here.
I have written a short introduction to the global geometry of physical systems of physical systems and its quantum manifestations. The article appeared in issue 2017-2 of Periodiek, the magazine of
the FMF student association, and is also available on my website. | {"url":"https://www.efstathiou.gr/journal/page/3/","timestamp":"2024-11-11T00:17:40Z","content_type":"text/html","content_length":"12770","record_id":"<urn:uuid:6c9f6fda-0046-408f-8211-50a6921695ba>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00271.warc.gz"} |
Mechanical Engineering
1) Assumptions
a) Steady operation with no heat generation
b) Thermal conductivity k of the material to remain constant.
c) Convection heat transfer coefficient h to be constant and uniform over the entire surface of the fin
i) The value of h is usually much lower at the fin base than it is at the fin tip because the fluid is surrounded by solid surfaces near the base, which seriously disrupt its motion to the point of
“suffocating” it.
ii) Adding too many fins on a surface may actually decrease the overall heat transfer when the decrease in h offsets any gain resulting from the increase in the surface area.
θ is Temperature excess
The General Solution for the above differential equation is
3) Different Cases in Fins
Infinitely long Fin Insulated Fin tip Specified temperature at fin tip Convection at fin tip
1^st Boundary Condition At Fin base at x=0
2^nd Boundary Condition
Temperature Distribution
Heat transfer from entire fin
4) Fin efficiency
5) Fin Effectiveness
When determining the rate of heat transfer from a finned surface, we must consider the unfinned portion of the surface as well as the fins. Therefore the rate of heat transfer for a surface
containing n fins can be expressed as
a) The thermal conductivity k of the fin material should be as high as possible. Thus it is no coincidence that fins are made from metals, with copper, aluminum, and iron being the most common ones.
Perhaps the most widely used fins are made of aluminum because of its low cost and weight and its resistance to corrosion.
b) The ratio of the perimeter to the cross-sectional area of the fin p/A[c] should be as high as possible. This criterion is satisfied by thin plate fins and slender pin fins.
i) mL = 5 ® an infinitely long fin
ii) mL= 1 offer a compromise between heat transfer performance and the fin size | {"url":"https://mechanicalengineering.softecksblog.in/5780/","timestamp":"2024-11-05T12:44:08Z","content_type":"text/html","content_length":"136977","record_id":"<urn:uuid:2ebb7f46-9f35-42af-b0a8-df16dcf1e26a>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00821.warc.gz"} |
Primitive draw
Primitive draw --- Introduction ---
Primitive draw is a graphical exercise on the geometry of integrals of functions of one real variable. The server will give you the graph of a function, then you are asked to draw its anti-derivative
with the mouse. You will have a score according to the precision of your drawing. This page is not in its usual appearance because WIMS is unable to recognize your web browser.
Please take note that WIMS pages are interactively generated; they are not ordinary HTML files. They must be used interactively ONLINE. It is useless for you to gather them through a robot program.
• Description: given the graph of a function, draw that of an anti-derivative. This is the main site of WIMS (WWW Interactive Multipurpose Server): interactive exercises, online calculators and
plotters, mathematical recreation and games
• Keywords: wims, mathematics, mathematical, math, maths, interactive mathematics, interactive math, interactive maths, mathematic, online, calculator, graphing, exercise, exercice, puzzle,
calculus, K-12, algebra, mathématique, interactive, interactive mathematics, interactive mathematical, interactive math, interactive maths, mathematical education, enseignement mathématique,
mathematics teaching, teaching mathematics, algebra, geometry, calculus, function, curve, surface, graphing, virtual class, virtual classes, virtual classroom, virtual classrooms, interactive
documents, interactive document, analysis, graphing, functions, integral | {"url":"https://wims.univ-cotedazur.fr/wims/en_U1~analysis~primdraw.en.html","timestamp":"2024-11-03T09:58:44Z","content_type":"text/html","content_length":"8953","record_id":"<urn:uuid:73aa47ab-d8c7-4aa8-a7d4-01d90ae19db9>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00152.warc.gz"} |
Creating Northwoods League WAR
The Northwoods League^1 is one of the nation’s premier summer collegiate baseball leagues, providing hundreds of prospects from a range of collegiate levels to showcase their skills over a 72-game
summer season. I spent my summer as the Statistics & Analytics Intern for the Battle Creek Battle Jacks^2, operating as the official scorekeeper for each of the team’s 36 home games.
The Northwoods League’s Scorebook software is a phenomenal tool for tracking and displaying player statistics, but it doesn’t dig much deeper than the most elementary numbers. Ultimately, this led to
emphasis being placed on the wrong player evaluation metrics, with too much focus on Batting Average (AVG) and not enough on On-Base (OBP) and Slugging Percentage (SLG). Halfway through the season,
upon initial creation of Wins Above Replacement (WAR), this became evident when examining the league’s elected All-Stars^3. Only 25 of the 54 All-Stars (46%), were deserving of the spot based on WAR
alone^4. Naturally, there are other factors, but such a discrepancy makes it clear that coaches, owners, broadcasters, and staff around the league were looking in the wrong place when assessing
player performance.
My favorite example of this comes from two players in the Great Lakes Division:
2B Fisher Pyatt – Battle Creek Battle Jacks via San Diego State University
.241 AVG/.418 OBP/.353 SLG, 2 HR 23 RBI, 1.1 WAR (155 PA)
3B Grant Broussard – Wisconsin Rapids Rafters via Utah Valley University
.235 AVG/.321OBP/.347SLG, 3 HR 30 RBI, 0.1 WAR (194 PA) *ALL-STAR*
Wisconsin Rapids’ Broussard was elected to the All-Star Game, while Battle Creek’s Pyatt was not. Similar across most metrics, Pyatt’s elite 20% walk-rate led to an OBP of .418, giving him an entire
WAR above Broussard.
(For the full list of players and a more in-depth analysis of Pyatt vs. Broussard, click here)
Clearly, development of a WAR metric would be beneficial to the league and all of its players. The first step in developing such a metric is to create a Run Expectancy Matrix^5.
Run Expectancy Matrix
The Run Expectancy Matrix is the backbone of all of sabermetrics. The matrix contains 24 cells, one for each of the unique 24 base-out states in baseball. Inside each cell is the expected runs scored
after reaching this state in an inning. From this number, we can assign estimated run values to each event based on their change in run expectancy.
As I mentioned earlier, I served as the official scorekeeper for all 36 Battle Creek home games. While in this position, I logged the play-by-play of every home and away game for the team, leading to
72 full games and 6,632 total observations. From this play-by-play, I constructed the following matrix:
Runners 0 Outs 1 Out 2 Outs
000 0.724 0.338 0.132
100 1.246 0.673 0.252
020 1.364 0.991 0.269
003 1.610 1.281 0.504
120 1.926 1.251 0.601
103 2.203 1.458 0.484
023 2.800 1.916 0.786
123 3.079 1.854 1.025
Table 1: Northwoods League Run Expectancy Matrix, 2022
Offensive WAR
Fangraphs^6 uses the following formula to calculate WAR for position players:
WAR = (Batting Runs + Base Running Runs + Fielding Runs + Positional Adjustment + League Adjustment +Replacement Runs) / (Runs Per Win)
Due to lack of adequate fielding and positional data, I alter it slightly to make for an easier calculation.
WAR = (Batting Runs + Base Running Runs + League Adjustment + Replacement Runs) / (Runs Per Win)
The first component of WAR is Batting Runs, which goes by the formula:
Batting Runs = wRAA + (lgR/PA – (PF*lgR/PA))*PA + (lgR/PA – GL or GP wRC/PA))*PA
There are numerous factors in this formula, and in order to reach this step we must climb the ladder, beginning with a player’s Weighted On-Base Average, or wOBA.
wOBA is a linear combination of each batting outcome with its respective linear weight. This is where the Run Expectancy Matrix comes in. Each outcome’s linear weight is the value of its mean change
in run expectancy. Since wOBA is weighted, there are two adjustments to be made after this to arrive at the final value. For simplicity and to avoid negative numbers, each linear weight is not just a
change in overall run expectancy but is a change in run expectancy above an out. Finally, wOBA is put on the same scale as a more well-known statistic: OBP. In the MLB, this value is usually near
.310, but in the Northwoods League it is inflated to .372.
For example, using the matrix, a double may change the base out state from runner on first with one out (1_100) to second and third with one out (1_023), for a change in run expectancy of 1.916 –
0.673 = +1.243. When compared to the average out, this play has a linear weight of 1.243 – (-0.429) = +1.672. When weighted at the wOBA scale, the final linear weight is 1.672 * 0.966 = 1.615. This
is just an example for one specific plate appearance, the table below shows this process for the weight of each event.
Event Run Exp. Above Out Scaled
All Outs -0.429 0.000 0.000
Walk 0.438 0.867 0.837
Hit By Pitch 0.487 0.916 0.884
Single 0.587 1.016 0.981
Double 0.944 1.373 1.326
Triple 1.261 1.690 1.632
Home Run 1.428 1.856 1.792
Table 2: wOBA Linear Weights
Incorporating these values into player statistics, here are the top ten players in wOBA in the 2022 season, minimum 100 plate appearances.
# Player Team wOBA
1. Ross, B WIR .527
2. Sojka, A FDL .514
3. Hug, C WAU .504
4. Ross, S MAN .498
5. Jackson, O GB .494
6. Gomez, D ROC .486
7. Comia, B FDL .485
8. Lipsey, T KZO .484
9. Mitchell, N FDL .483
10. Nankil, N WIR .483
Table 3: Northwoods League wOBA Leaders (minimum 100 plate appearances, 2022)
wOBA is a rate statistic, describing how many runs a player will provide to his team per plate appearance. In order to use this in the calculation of Batting Runs, it needs to be translated into a
cumulative statistic, Weighted Runs Above Average, or wRAA, as is seen in the Batting Runs formula.
wRAA is the principal component in Batting Runs, to arrive at the final number only two adjustments need to be made. The first is for a player’s home ballpark. Different home ballparks have different
run environments and therefore different Park Factors^7, the primary example of this is Denver’s Coors Field favoring offense due to high altitude. A similar range of environments occur in the
Northwoods League, for a variety of reasons. Here are the Park Factors for each of the 21 unique ballparks, calculated on data from 2021-2022 using the method by Baseball Reference^8. 100 is league
average, higher numbers indicate a higher run environment. For example, a PF of 105 has 5% more runs than average.
Great Lakes Team Offense Defense Great Plains Team Offense Defense
Battle Creek BC 92.4 94.4 Bismarck BIS 93.6 95.8
Fond du Lac FDL 103.4 102.1 Duluth DUL 101.1 102.1
Green Bay GB 104.5 106.6 Eau Claire EC 95.6 95.7
Kenosha KEN 107.6 107.2 La Crosse LAC 107.2 108.5
Kokomo KMO 95.0 96.4 Mankato MAN 92.7 91.3
Kalamazoo KZO 99.1 99.4 Minnesota MIN 100.0 100.0
Lakeshore LAK 100.3 100.1 Rochester ROC 101.4 102.1
Madison MAD 93.8 94.9 St. Cloud STC 102.4 97.7
Rockford RFD 115.1 115.7 Waterloo WAT 105.6 106.8
Traverse City TVC 90.3 87.8 Willmar WIL 101.1 98.6
Wausau WAU 102.2 101.9
Wisconsin Rapids WIR 97.0 94.3
Table 4: Northwoods League Park Factors, 2021-2022
The final adjustment is league-based. The Northwoods League is split into two conferences: Great Lakes and Great Plains. There is no interleague play, so an adjustment must be made based on a
player’s league/conference. Here is the strength of each leagues run environment, in a format similar to park factors.
Great Lakes Great Plains
97.4 103.4
Table 5: Northwoods League Division Factors, 2022
Putting it all together, here are the leaders for Batting Runs:
# Player Team Batting Runs
1. Ross, S MAN 34.61
2. Ross, B WIR 34.44
3. Mitchell, N FDL 32.72
4. Nett, J STC 32.14
5. Dykstra, R KZO 25.97
6. Bobo, B WIR 23.82
7. Stephan, A KZO 23.76
8. Tuft, C WIR 22.84
9. Campbell, K DUL 22.48
10. Sojka, A FDL 22.37
Table 6: Northwoods League Batting Runs Leaders, 2022
The next component is Base Running Runs, which is found from Weighted Stolen Bases. Advancement on batted balls will be disregarded due to lack of data.
wSB = SB * runSB + CS * runCS – lgwSB * (1B + BB + HBP – IBB)
Naturally, a stolen base adds to run expectancy and a caught stealing takes runs away. wSB allows us to factor in how many runs a player subtracts from their team when they don’t attempt to steal a
base at all. For example, Eau Claire Express third-baseman Sam Kuchinski^9 accumulated 246 plate appearances in 53 games but did not make a single stolen base attempt. Because of this, his wSB was
-0.84. Although he was never caught stealing, the missed opportunity of advancing to the next base cost his team nearly an entire run.
Here are the leaders for wSB and, subsequently, Baserunning Runs:
# Player Team Baserunning Runs
1. Guardino, R WAT 5.27
2. Traficante, C TVC 4.52
3. Seegers, M LAK 4.47
4. Toole, M TVC 3.87
5. Atkinson, A TVC 3.75
6. Bateman, B WIL 3.75
7. Rogers, JD DUL 3.75
8. Kaiser, C LAK 3.29
9. Payne, K WIL 2.69
10. Donahue, J LAC 2.60
Table 7: Northwoods League Baserunning Runs Leaders, 2022
The next component is league adjustment, now factoring in baserunning runs as well. This figure was minor (< 0.1 for each league) so I will not explain any further.
The final component is replacement level. FanGraphs and Baseball Reference have agreed to the same replacement level, so I will follow it here: 1,000 total WAR in a full MLB season, with 57%
attributed to position players and 43% to pitchers. The 2022 Northwoods League season comprised of only 756 games, so there will only be around 311 total WAR, ~177 for position players and ~134 for
pitchers. Therefore, the following formula will output Replacement Level Runs:
Replacement Level Runs = (177.33 * (lgG/756) * (Runs Per Win/lgPA) * PA
lgG and lgPA describe the total games and plate appearances across the entire league to the current date of calculation. Runs per win (RPW) is a Pythagorean^10-based statistic that determines how
many extra runs a team needs to score to add an extra win. In the NWL, the value of RPW is 12.3. Ultimately, a replacement level position player is worth about 0.0349 runs per plate appearance.
After adding every component and dividing by Runs Per Win, the final calculation of Offensive WAR is complete. Here are the top ten players:
# Player Team NWL WAR 162G WAR
1. Nett, J STC 3.6 8.7
2. Mitchell, N FDL 3.6 8.2
3. Ross, S MAN 3.6 8.6
4. Ross, B WIR 3.5 7.9
5. Dykstra, R KZO 2.9 6.4
6. Stephan, A KZO 2.7 6.1
7. Campbell, Kr DUL 2.7 6.4
8. Bobo, B WIR 2.7 6.1
9. Schwabe, C EC 2.5 6.0
10. Tuft, C WIR 2.5 5.6
Table 8: Northwoods League Offensive WAR Leaders, 2022
The fifth column illustrates a player’s WAR expanded onto a full 162 game MLB season, rather than the shortened summer season. It does not decrease in perfect order since teams in the Great Lakes
played more games than those in the Great Plains (72 vs. 68) due to team count.
Wins Above Replacement and other advanced stats for every Northwoods League player can be found on my public database^11.
Pitcher WAR
WAR for pitchers is slightly less straightforward than it is for position players, and I use the methodology given by FanGraphs^12. Here is the formula:
WAR = [[([(League “FIP” – “FIP”) / Pitcher Specific Runs Per Win] + Replacement Level) * (IP/9)] * Leverage Multiplier for Relievers] + League Correction
The main change that I make is eliminating the leverage multiplier for relievers, mostly due to a lack of adequate data.
The first step is to calculate Fielding Independent Pitching (FIP). FIP is designed as a substitute for the more common Earned Run Average (ERA), leveling out batted ball luck. FanGraphs uses FIP
with infield-fly balls, but due to lack of data regular FIP will be used, with the following formula:
FIP = ((13*HR)+(3*(BB+HBP))-(2*K))/IP + constant
The constants 13, 3, -2 reflect run values and were calculated from MLB data. Out of curiosity, I fit a linear model on individual Northwoods League pitcher data (min 36 IP, 2017-2021) and received
the following coefficients:
Table 9: FIP Model Coefficients
Close enough to validate FIP as a Northwoods League statistic. The FIP constant exists to make FIP easily comparable to ERA, giving them the same league average. Here are the ten pitchers with the
best FIP, minimum 36 innings pitched.
# Player Team FIP
1. Wright, D MAD 2.42
2. Lanoux, C BC 2.83
3. Chalus, E LAK 2.84
4. Schultz, D MAN 2.98
5. Horvath, T KMO 3.05
6. Clark, D TVC 3.26
7. Novotny, T WIL 3.36
8. Battaglia, J STC 3.47
9. Candiotti, C KEN 3.50
10. Seebach, K RFD 3.58
Table 10: Northwoods League FIP Leaders, 2022
Just like ERA, FIP does not describe all of the runs scored, as some are unearned. A simple adjustment turns FIP into FIP Runs Per 9 innings (FIPR9) accounting for the random, unearned runs. FIPR9
then undergoes near identical park and league adjustments as wRAA did for position players, becoming Runs Above Average Per 9 Innings (RAAP9).
The next important component is each pitcher’s Dynamic Runs Per Win (dRPW). This is similar to RPW for position players, but the difference is pitchers have more control over their run environment.
If Pitcher A is a much better pitcher than Pitcher B, than Team A will need fewer runs to win than Team B does. Dividing the two metrics (RAAP9/dRPW) will output a player’s Wins Per Game Above
Average (WPGAA).
However, WAR is above replacement, not average, meaning the next step is to add in replacement level. Unlike position players, this is not the same for everyone, depending on whether a pitcher is a
starter or reliever, following this formula:
Replacement Level = 0.03*(1 – GS/G) + 0.12*(GS/G)
This method clearly favoritism to starting pitchers, giving them a much higher replacement leavel. This is usually accounted for by a reliever’s leverage index, which I am leaving out of the
calculation. To make up for this, I move each constant halfway to the mean replacement level of 0.75, giving the following formula:
Replacement Level = 0.0525*(1 – GS/G) + 0.0975*(GS/G)
Once the replacement level is found, WPGAA becomes WPGAR. This value is multiplied by total “games” (IP/9) pitched and given a minuscule adjustment to finally become WAR. Here are the top ten
pitchers in WAR:
# Player Team NWL WAR 162G WAR
1. Thompson, R RFD 1.7 3.9
2. Schultz, D MAN 1.7 4.1
3. Harrison, C LAC 1.5 3.6
4. Buhr, D MAN 1.5 3.5
5. Wright, D MAD 1.4 3.2
6. Lanoux, C BC 1.4 3.2
7. Habeck, J WIL 1.4 3.3
8. Gustafson, Ar TVC 1.4 3.1
9. Candiotti, C KEN 1.4 3.0
10. Chalus, E LAK 1.3 2.9
Table 11: Northwoods League Pitching WAR Leaders
Wins Above Replacement and other advanced stats for every Northwoods League pitcher can be found on my public database^11.
How well does WAR represent real team success? When compared with three different metrics (Win Percentage, Pythagorean Win Percentage, Simple Rating System^13), WAR has a correlation near 0.83-0.85
for each one, making for a strong relationship.
Figure 1: Northwoods League WAR Per Game and Team Win Percentages, 2022
In the Win Percentage plot, the variance remains relatively equal throughout, without any large residuals, further enhancing the relationship.
Figure 2: Northwoods League WAR Per Game and Team Pythagorean Win Percentages, 2022
The Pythagorean Win Percentage plot is similar, just on a smaller y-axis window, leading to a slightly higher correlation.
Figure 3: Northwoods League WAR Per Game and Simple Rating System, 2022
Baseball Reference’s Simple Rating System yields the lowest correlation and highest variance. This is likely due to the statistic’s importance of strength of schedule, As we know, WAR does not factor
in opponent strength, causing this factor to bump certain points further from the regression line.
The Wins Above Replacement statistic is an accurate and effective way to analyze both individual player and team success in the Northwoods League. This metric levels the playing field for hundreds of
amateur players, allowing for a more valid form of evaluation.
In the future, similar statistics can be developed for the dozens of collegiate summer baseball leagues from coast to coast. One other example was Humbert Kilanowski’s calculation of WAR for the Cape
Cod League in 2019^14. The ultimate goal is one, all-inclusive Amateur WAR value that adjusts for not only ballparks and divisions, but for the overall talent of each summer league. Such a metric
would be labor intensive and require a strong knowledge of amateur leagues, but it is one of my biggest aspirations in baseball research.
1. Northwoods League. (n.d.). Retrieved November 9, 2022, from https://northwoodsleague.com/
2. Fan’s best friend. Battle Creek Battle Jacks. (n.d.). Retrieved November 9, 2022, from https://northwoodsleague.com/battle-creek-battle-jacks/
3. Admin. (2022, July 14). Northwoods League announces 2022 all-star selections. Northwoods League. Retrieved November 9, 2022, from https://northwoodsleague.com/blog/2022/07/08/
4. Banks, J. (2022, July 20). Northwoods League All-War Team. Battle Creek Battle Jacks. Retrieved November 9, 2022, from https://northwoodsleague.com/battle-creek-battle-jacks/2022/07/20/
5. Tango, T. (n.d.). Run expectancy matrix, 1950-2015. Retrieved November 9, 2022, from http://www.tangotiger.net/re24.html
6. Slowinski, P. (n.d.). War for position players. Sabermetrics Library. Retrieved November 9, 2022, from https://library.fangraphs.com/war/war-position-players/
7. Statcast Park factors. baseballsavant.com. (n.d.). Retrieved November 9, 2022, from https://baseballsavant.mlb.com/leaderboard/statcast-park-factors/
8. Park adjustments. Baseball Reference. (n.d.). Retrieved November 9, 2022, from https://www.baseball-reference.com/about/parkadjust.shtml
9. Sam Kuchinski. Northwoods League. (n.d.). Retrieved November 9, 2022, from https://northwoodsleague.com/statistics/?param=%2Fstatistics%2Fplayer%2F4457%2F13%2F52
10. Pythagorean theorem of baseball. Baseball Reference. (n.d.). Retrieved November 9, 2022, from https://www.baseball-reference.com/bullpen/Pythagorean_Theorem_of_Baseball
11. Banks, J. (n.d.). Advanced database.xlsx. Google Sheets. Retrieved November 9, 2022, from https://docs.google.com/spreadsheets/d/1RKc4uYdQNFrKGZDfba0nH9CWbJhtNwvb/edit#gid=1149427237
12. Slowinski, P. (n.d.). War for pitchers. Sabermetrics Library. Retrieved November 9, 2022, from https://library.fangraphs.com/war/calculating-war-pitchers/
13. SRS calculation details. Sports Reference. (n.d.). Retrieved November 9, 2022, from https://www.sports-reference.com/blog/2015/03/srs-calculation-details/
14. Kilanowski, H. (2020, June 16). cWAR: Modifying Wins Above Replacement with the Cape Cod Baseball League. Society for American Baseball Research. Retrieved November 9, 2022, from https://sabr.org
15. Marchi, M., Albert, J., & Baumer, B. S. (2019). Analyzing baseball data with R (Second Edition). CRC Press. | {"url":"http://jackbanks.web.illinois.edu/2022/11/09/creating-northwoods-league-war/","timestamp":"2024-11-11T10:40:08Z","content_type":"text/html","content_length":"56487","record_id":"<urn:uuid:e8e3abc6-0480-4244-998b-1147e8e9d2e9>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00244.warc.gz"} |
Suchergebnis: Katalogdaten im Herbstsemester 2018
Elektrotechnik und Informationstechnologie Master
Master-Studium (Studienreglement 2018)
Systems and Control
The core courses and specialization courses below are a selection for students who wish to specialize in the area of "Systems and Control", see
The individual study plan is subject to the tutor's approval.
These core courses are particularly recommended for the field of "Systems and Control".
You may choose core courses form other fields in agreement with your tutor.
A minimum of 24 credits must be obtained from core courses during the MSc EEIT.
Advanced Core Courses
Nummer Titel Typ ECTS Umfang Dozierende
227-0225-00L Linear System Theory W 6 KP 5G M. Kamgarpour
Kurzbeschreibung The class is intended to provide a comprehensive overview of the theory of linear dynamical systems, stability analysis, and their use in control and estimation. The focus
is on the mathematics behind the physical properties of these systems and on understanding and constructing proofs of properties of linear control systems.
Lernziel Students should be able to apply the fundamental results in linear system theory to analyze and control linear dynamical systems.
- Proof techniques and practices.
- Linear spaces, normed linear spaces and Hilbert spaces.
Inhalt - Ordinary differential equations, existence and uniqueness of solutions.
- Continuous and discrete-time, time-varying linear systems. Time domain solutions. Time invariant systems treated as a special case.
- Controllability and observability, duality. Time invariant systems treated as a special case.
- Stability and stabilization, observers, state and output feedback, separation principle.
Skript Available on the course Moodle platform.
Voraussetzungen / Sufficient mathematical maturity with special focus on logic, linear algebra, analysis.
227-0697-00L Industrial Process Control W 4 KP 3G M. Mercangöz, A. Horch
Kurzbeschreibung Introduction to process automation and its application in process industry and power generation
Lernziel Knowledge of process automation and its application in industry and power generation
Introduction to process automation: system architecture, data handling, communication (fieldbusses), process visualization, engineering, etc.
Analysis and design of open loop control problems: discrete automata, decision tables, petri-nets, drive control and object oriented function group automation philosophy,
Inhalt Engineering: Application programming in IEC61131-3 (function blocks, sequence control, structured text); process visualization and operation; engineering integration from
sensor, cabling, topology design, function, visualization, diagnosis, to documentation; Industry standards (e.g. OPC, Profibus); Ergonomic design, safety (IEC61508) and
availability, supervision and diagnosis.
Practical examples from process industry, power generation and newspaper production.
Skript Slides will be available as .PDF documents, see "Learning materials" (for registered students only)
Voraussetzungen / Exercises: Tuesday 15-16
Practical exercises will illustrate some topics, e.g. some control software coding using industry standard programming tools based on IEC61131-3.
151-0563-01L Dynamic Programming and Optimal Control W 4 KP 2V + 1U R. D'Andrea
Kurzbeschreibung Introduction to Dynamic Programming and Optimal Control.
Lernziel Covers the fundamental concepts of Dynamic Programming & Optimal Control.
Inhalt Dynamic Programming Algorithm; Deterministic Systems and Shortest Path Problems; Infinite Horizon Problems, Bellman Equation; Deterministic Continuous-Time Optimal Control.
Literatur Dynamic Programming and Optimal Control by Dimitri P. Bertsekas, Vol. I, 3rd edition, 2005, 558 pages, hardcover.
Voraussetzungen / Requirements: Knowledge of advanced calculus, introductory probability theory, and matrix-vector algebra. | {"url":"https://www.vorlesungen.ethz.ch/Vorlesungsverzeichnis/sucheLehrangebot.view?abschnittId=78190&semkez=2018W&ansicht=2&lang=de&seite=1","timestamp":"2024-11-07T01:36:56Z","content_type":"text/html","content_length":"15103","record_id":"<urn:uuid:ba32c783-ad15-4e64-aa01-ac9c85ba57d3>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00380.warc.gz"} |
The air flow around a milling cutter investigated experimentally by particle image velocimetry
AIP Conference Proceedings 2323, 030006 (2021); https://doi.org/10.1063/5.0041860 2323, 030006
© 2021 Author(s).
The air flow around a milling cutter
investigated experimentally by particle image velocimetry
Cite as: AIP Conference Proceedings 2323, 030006 (2021); https://doi.org/10.1063/5.0041860 Published Online: 08 March 2021
Tereza Kubíková
Dimensional analysis parameters of turbulence in the wake of a square cylinder AIP Conference Proceedings 2323, 030003 (2021); https://doi.org/10.1063/5.0041434
Observation of flow structure past a full-stage axial air turbine at the nominal and off-design states
AIP Conference Proceedings 2323, 030004 (2021); https://doi.org/10.1063/5.0041491 Particle image velocimetry measurement inside axial air test turbine – Effect of window AIP Conference Proceedings
2323, 030005 (2021); https://doi.org/10.1063/5.0041492
The Air Flow around a Milling Cutter Investigated Experimentally by Particle Image Velocimetry
Tereza Kubíková
Department of Power System Engineering, Faculty of Mechanical Engineering, University of West Bohemia in Pilsen, Univerzitní 22, 306 14, Pilsen, Czech Republic.
a)Corresponding author: kubice-ter@seznam.cz
Abstract. The movement of air around a rotating milling cutter is investigated experimentally by using the optical non- intrusive method PIV (Particle Image Velocimetry). The milling cutter is
located next to a flat surface simulating the processed material. The Flow is studied in three planes in axial × radial direction with phases 90°, 180° and 270° from the point closest to the flat
desk. The PIV timing has been synchronized with the cutter rotation. We observed plumes governed by centrifugal force (as in Taylor-Couette flow) acting on the air teared with the rotating milling
cutter. In average, the air flows along the milling cutter from its front end forming a large-scale vortex mostly apparent at higher velocity and at the plane opposite to the processed surface. At
the 90°plane, the air is sucked towards the milling cutter, while at the 270°plane, the air is pushed away from it. Spatial correlation function shows a weak periodicity close to the cutter and
widening of its shape with increasing distance from the milling cutter.
The general problem of flow around a rotating cylinder is known under the term Taylor-Couette flow [1, 2], where the Taylor instability [3, 4] exerts. This instability relates with the more
generalized Rayleigh type of instabilities, where two volumes of fluid are forced in opposite direction. In the rotating case, such a role is played by the centrifugal force, which pushes the fluid,
which is closer to the rotating object, against that fluid, which is more far and thus less rotating. The Kelvin-Helmholtz instability [5] occurs at the shear layers of fluid volumes moving by
different velocities, and it can be observed e.g. in a jet [6, 7] or in a wake [8]. A similar mechanism causes the secondary flow of second kind in [9, 10], but not in the case of so called steady
streaming [11]. The most spectacular example of Rayleigh instability is the Rayleigh-Bénard instability leading to thermal convection [12, 13]. Similar to the rotating case, this type of convection
is characterized by the presence of plumes – a mushroom like vortex ring followed by a spike. This coherent structure is not universal for thermal convection as, in superfluids, the heat transfer is
realized via the thermal counterflow [14, 15] of normal and superfluid component flowing one through the other in opposite direction. In both systems – the classical one and the quantum one, the
developed turbulence displays universal features [16], although its microscopic nature is as different as possible [17]. The similarity of classical and quantum turbulence on large scales is
demonstrated by the existence of wakes [18], secondary flows [11] or cavitation bubbles [19] in both systems.
One of the most practical effect of turbulence is the enhancement of mixing, as the mixing occurs at all scales from the largest one in the system (called integral length-scale) down to the smallest
one, where the energy dissipates (called Kolmogorov scale). This has a large impact on the heat transfer from the hot object (which is in our case the milling cutter) into the surroundings.
Therefore, the investigation of the structures and scales of turbulence around this device during its job can help us to understand the heat transfer needed for effective cooling.
The milling cutter with 5 blades is situated on a flat plate simulating the processed object as sketched in Fig. 1. By using the ATOS 3D optical scanner, we measured the milling cutter major diameter
to be 25.3 mm, while the minor one to be 15.1 mm, the active length to be 65 mm and the pitch to be 22.6 mm (note there are 5 blades, thus one blade twists ones in 113 mm). The flow is investigated
in three planes with phase shifts from the point closest to the solid surface are 90°, 180° and 270°. The side size of the planes is 75 mm.
The milling cutter is powered by using a drill, which offers only two stable rotation speeds: the slower one is measured to be 7.72 Hz, while the faster one 19.7 Hz.
We use the Particle Image Velocimetry technique, which measures optically the motion of small particles (droplets of oil) carried by the fluid and illuminated by a laser sheet. We use a commercial
system from the company Dantec with Mk II Flow Sense camera and New Wave Solo solid state laser. The tracking particles are produced by a fog generator Safex. The maximum frequency of the PIV system
is 7.4 Hz, thus we are not able to detect the time evolution. Our data have character of an ensemble of statistically independent snapshots.
FIGURE 1. (a) 3D scan of the used milling cutter. Front view. It rotates in counterclockwise direction. (b) 3D scan situated together with the planes studied by using the PIV method (depicted as a
green transparent planes), the dark plane bellow represents the solid surface simulating the processed object, although there is not touch between them in the current setup.
The timing of the PIV system (the laser shots and the camera exposure) are synchronized with the rotation of the milling cutter by using a custom made electrical circuit based on the Arduino
microcontroller, which triggers the signal from a photodiode placed next to the rotating shaft with a small reflective strip.
We use PIV in configuration with a single camera, thus only the in-plane velocity components can be measured.
In the later we refer to the axial velocity component as u and the radial as v, the tangential one is not measured, although it can be energetically dominant. This has to be taken into account when
interpreting the results.
In order to prevent unwanted reflections of the laser beam, the milling cutter has been painted to black. Other optical disturbances have been suppressed by subtracting the spatial map of minimal
point intensities. The milling cutter was masked out in order to prevent spurious velocity vectors in the area, where cannot be any particles. Standard method “Adaptive PIV” in Dantec Dynamic Studio
software has been used to calculate the velocity vectors from the captured images. The ensemble of such velocity fields was filtered by the energy of small-scale fluctuations. The last method has
been developed by my great supervisor and published in article [7].
Plumes in Instantaneous Velocity Field
The instantaneous velocity fields often display one of the feature typical for all Rayleigh-type instabilities. This esthetic flow pattern is called plume and few examples of them is shown in Fig. 2.
We show the plumes in the 90°
plane (Fig. 2), because the plumes there are on the start of their development and thus are not yet turbulized and mixed, as it is apparent in the 180° plane (Fig. 3).
The vorticity displayed in Fig. 2 and 3 is calculated from the instantaneous velocity as
߱ ൌݑሬԦ (1) where instead of the differential operator nabla () we used the symmetric differentiation on the grid of spatial resolution ȟݔǡ ȟݕ, thus
߱ሺݔǡ ݕሻ ൌݑሺݔǡ ݕ ȟݕሻ െݑሺݔǡ ݕ െȟݕሻ
ʹȟݕ െݒሺݔ ȟݔǡ ݕሻ െݒሺݔ െȟݔǡ ݕሻ
ʹȟݔ (2)
FIGURE 2. Top row shows a few typical images taken during the experiment and the bottom row shows velocity vectors obtained from the images above and colored by the instantaneous in-plane vorticity
(color version online). All data are taken at
the slower rotational speed in the 90° plane (see Fig. 1 for reference).
FIGURE 3. Plumes in raw images (top row) and the vorticity (bottom row) observed in the 180° plane at the slower rotational speed.
Plume is a part of fluid, which is moved through its surrounding due to some force acting to it and not (or less) to the surroundings. They are typically observed in the system of thermal convection,
especially, when the heat source is localized, as it is e.g. in the case of an explosion (“nuclear mushroom”). In the rotating system, the leading force is the centrifugal force acting more strongly
to the fluid, which share rotational motion with the milling cutter.
Average Velocity Fields
The spatial distribution of ensemble-average velocity is shown in Fig. 4. We can see, that first the air flows towards the milling cutter, there are still observable some middle-scale vortices close
to the rotating cutter. Later, the flow pattern is dominated by a large-scale vortex at the middle of the milling cutter length and, at the phase before the processed surface, the air leaves the area
rotating with the body, but middle-scale average structures are apparent.
Close to the body, the flow is pushed along the axis (from right to left in the figures) due to the helicity of the milling cutter. This is the reason, why there is low turbulence intensity at the
milling cutter tip (see turbulent kinetic energy in Fig. 6 and 7) – there is sucked the air from quiet surroundings.
FIGURE 4. Average velocity displayed as vectors, color corresponds to its radial component (color version online). The panels (a), (b) and (c) show planes in different phase position around the
rotating thing, see Fig. 1 for reference. The mask is little bit larger than the milling cutter in order to cover the unwanted laser reflections, which causes troubles especially at the 270° plane.
FIGURE 5. Average velocity under the faster studied rotational speed, color corresponds to the radial component of average velocity (color version online).
The average behavior of the flow is quite different under the faster rotational speed: first note, that the large vortex (formed in the 180° plane at the slower rotation) is apparent since the first
90° plane. At the 180° plane, this vortex is
so large, it is barely visible, the outcoming region is shifted along the milling cutter axis behind the active length (left edge of the Fig. 5), while the sucking region covers more than half of the
active length. The scheme of sucking and pushing the fluid at different planes is now represented solely by shift in the position of the main vortex, therefore the overall flow in the 90° is towards
the milling cutter, while at 270° it goes away from it. The middle scale structures visible at slower velocity (Fig. 4) are covered by the dominant vortex, but their presence will be shown in next
Figures 6 and 7 shows the spatial distribution of the turbulent kinetic energy (TKE). TKE is calculated as the sum of variances of the velocity component ensembles in each point:
ʹሺۃݑ^ଶۄ െۃݑۄ^ଶሻ ͳ
ʹሺۃݒ^ଶۄ െۃݒۄ^ଶሻ (3)
where ۃڄۄ is the ensemble averaging. TKE is displayed preferentially than the intensity of turbulence, as the intensity of turbulence can diverge in the areas of low average velocity. Naturally, TKE
in our case is calculated only by using the measured velocity components, i.e. the axial one (u) and the radial one (v), although the third (tangential) component might be expected to dominate this
flow problem.
FIGURE 6. Turbulent kinetic energy calculated from the in-plane velocities and its fluctuations. Note that this TKE is computed by using only the in-plane velocity fluctuations. Color version online.
FIGURE 7. Turbulent kinetic energy at the faster rotational speed. Color version online.
Length-scale of the Fluctuations
The length-scale of fluctuations is characterized by the correlation function Rvv of two quantities, in this case we choose as the relevant quantity the radial component of velocity. Rvv in the
spatially resolved data is defined as
ܴ௩௩ሺȟݔǡ ȟݕሻ ൌۃ൫ݒሺݔǡ ݕሻ െܸሺݔǡ ݕሻ൯ ڄ ቀ൫ݒሺݔ ȟݔǡ ݕ ȟݕሻ െܸሺݔ ȟݔǡ ݕ ȟݕሻ൯ቁۄ[்]
ߪሾݒሺݔǡ ݕሻሿ ڄ ߪሾݒሺݔ ȟݔǡ ݕ ȟݕሻሿ (4)
where ߪ represents the standard deviation, ܸis the ensemble average at the current point, thus the difference ݒሺݔǡ ݕሻ െܸሺݔǡ ݕሻ is the Reynolds decomposition. Note that the correlation function makes
sense only when applied to the Reynolds decomposed velocity fields, otherwise it does not converge. The value of correlation coefficient reaches the values from -1 to 1, where 1 means, that the two
points are statistically same, while -1 means, they are opposite. 0 is usually interpreted as a statistical independency, but it can also signify a phase shift by ߨȀʹ (imagine the correlation
function of ݔ and ݔ – it is identically 0, although both are functions of the same variable and thus depends one on the other). Nothing to say, that the correlation function inherits some
properties of the original function, among others its periodicity.
(b)Rvv in 180° plane in different distances from the milling cutter
FIGURE 8. Correlation coefficient Rvv of the fluctuation of radial velocity component v in the 180° plane at the slower rotational speed. (a) the spatial distribution of Rvv with point at different
radial distance from the milling cutter. The milling cutter is located
under the bottom edge of that crops. (b) the profiles of Rvv along the axial direction; different lines represent different radial distance from the milling cutter starting from the 1.17 mm (darkest
color) to 18.7 mm (clearest color).
In turbulence research, the correlation function is often used to determine a size of the fluctuation, i.e. a distance, where fluctuations disappear. Figure 8 shows how the correlation function of
radial velocity fluctuations depends on the distance from the milling cutter. At closer distance, we see a periodicity (Rvv reaches negative values and then returns to around 0) of length about 6 – 8
mm, later this behavior disappears and the correlation function widens as the fluctuations are larger. The slope of the Rvv close to zero is theoretically zero as the fluid motion at smallest scales
(Kolmogorov scales) is smoothed due to finite viscosity. We do not observe this feature as our spatial resolution (1.17 mm per grid point) is insufficient and this remains as an open task for future
When comparing the two different rotational speeds (compare Fig. 9 with Fig. 10), we see the shape of the correlation function is sharper, while the distance of reaching zero remains similar (this is
probably due to the periodicity close to the milling cutter). The shape does not seem to change much among the different explored planes, it is widest in the 90° plane, where the flow is not yet
developed and thus it is characterized by sucking the surrounding non-turbulized fluid and by developing laminar plumes. Surprisingly, the 270° plane, where the fluid is pulled out due to approaching
the processed desk, displays little bit wider shape of the correlation function, than the 180° plane opposite to the processed surface.
-0,5 0 0,5 1
-20 -15 -10 -5 0 5 10 15 20
Axial profile of Rvv
Δx [mm]
(b)Rvv in different planes at slower rotational speed
FIGURE 9. Correlation coefficient Rvv of the fluctuation of radial velocity component v in different planes (see Fig. 1 for reference) at the slower rotational speed. (a) the spatial distribution of
Rvv The milling cutter is located under the bottom edge of
that crops. (b) the profiles of Rvv along the axial direction. (Color version online) (b)Rvv in different planes at faster rotational speed
FIGURE 10. Correlation coefficient Rvv of the fluctuation of radial velocity component v in different planes (see Fig. 1 for reference) at the faster rotational speed. (a) the spatial distribution of
Rvv The milling cutter is located under the bottom edge of
that crops. (b) the profiles of Rvv along the axial direction. (Color version online)
The flow around a rotating milling cutter has been investigated experimentally by using the Particle Image Velocimetry (PIV) technique in three planes and at two rotational speeds. The planes were
oriented in axial × radial direction covering the entire active length of the milling cutter (which is 65 mm). The planes have been situated in three different phases from the touching point with the
processed surface (the surface is not milled in this experiment).
The investigated phases were 90°, 180° and 270°. Only the axial and radial in-plane velocity components have been measured
-12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12
Axial profile of Rvv
Δx [mm]
90°s 180°s 270°s
-12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12
Axial profile of Rvv
Δx [mm]
90°f 180°f 270°f
The flow was teared together with the rotating body, therefore it felt the centrifugal force similarly as in the Taylor- Couette problem, which has been proven by the observation of plumes carrying
the faster rotating fluid outside into the slower fluid. In the average, the flow has formed a large-scale vortex sucking the air at the open end of the milling cutter, pulling it in axial direction
and pushing away at the shaft end. This vortex was stronger under the faster rotation.
In the 90° plane, the air is sucked towards the rotating milling cutter, while in the 270° we observed the air flowed away from it. This effect had a form homogeneous motion under the slower
rotation, while under the faster one, it just moved the center of the just mentioned large-scale vortex.
Spatial correlation function of radial velocity component has shown a weak periodicity close to the cutter, which disappeared in distances larger than 1 cm. The shape of correlation function peak
widened with increasing distance from the milling cutter. Larger rotational speed leaded to narrowing that peak, while the base width remains similar.
This work was financially supported by student project SGS-2019-021 (Improving the efficiency, reliability and service life of power machines and equipment 5). I thank to my supervisor RNDr. Daniel
Duda, Ph.D. for significant help with data analysis and article writing. I thank to doc. Vitalii Yanovych, DrSc. for technical help and to prof. Ing.
Václav Uruba, CSc. for discussions.
1. C. D. Andereck, S. S. Liu and H. L. Swinney, J. Fluid Mech. 164, 155-183 (1986) 2. W. W. Saric, Annu. Rev. Fluid Mech. 26, 379-409 (1994)
3. M. A. Fardin, C. Perge and N. Taberlet, Soft Matter 10, 3523-3535 (2014)
4. D. Duda, M. Klimko, R. Škach, J. Uher and V. Uruba, EPJ Web of Conf. 213, 02014 (2019) 5. G.E. Volovik, JETP Letters 75, 418-422 (2002)
6. D. Duda, AIP Conf. Proc. 2047, 020001 (2018)
7. D. Duda and V. Uruba, ASME J of Nuclear Rad Sci 5(3), 030912 (2019) 8. D. Duda and V. Uruba, AIP Conf. Proc. 2000, 020005 (2018)
9. M. Uhlmann, A- Pinelli, G. Kawahara and T. Sekimoto, J. Fluid Mech. 588, 153-162 (2007)
10. D. Duda, J. Bém, V. Yanovych, P. Pavlíček and V. Uruba, Eur. J. Mech. B Fluids 79, 444-453 (2020) 11. D. Duda, M. La Mantia and L. Skrbek, Phys. Rev. B 96, 2, 024519 (2017)
12. P. Urban, V. Musilová and L. Skrbek, Phys. Rev. Lett. 107, 014302 (2011)
13. P. Urban, P. Hanzelka, I. Vlček, D. Schmoranzer and L. Skrbek, Low Temp. Phys. 44, 1001-1004 (2018) 14. D. Duda, M. La Mantia, M. Rotter and L. Skrbek, J. Low Temp. Phys 175, 331-338 (2014)
15. M. La Mantia, D. Duda, M. Rotter and L. Skrbek, J. Fluid Mech 717, R9 (2013) 16. M. La Mantia, P. Švančara, D. Duda and L. Skrbek, Phys. Rev. B 94, 18, 184512 (2016) 17. C. F. Barenghi, V. S.
L'vov, and P. E. Roche, P. Natl. Acad. Sci. USA, 111:4683-4690 (2014) 18. D. Duda, P. Švančara, M. La Mantia, M. Rotter and L. Skrbek, Phys. Rev. B 92, 064519 (2015)
19. D. Duda, P. Švančara, M. La Mantia, M. Rotter, D. Schmoranzer, O. Kolosov and L. Skrbek, J Low Temp. Phys.
187, 5-6, 376-382 (2017) | {"url":"https://9pdf.info/document/qmj837o8-flow-milling-cutter-investigated-experimentally-particle-image-velocimetry.html","timestamp":"2024-11-03T06:03:02Z","content_type":"text/html","content_length":"160079","record_id":"<urn:uuid:33404d1b-c9e7-47b8-8846-8a15cd05e099>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00353.warc.gz"} |
Let f(x)=max(x,x2,x3) in −2≤x≤2. Then :... | Filo
Not the question you're searching for?
+ Ask your question
Was this solution helpful?
Video solutions (3)
Learn from their 1-to-1 discussion with Filo tutors.
7 mins
Uploaded on: 5/3/2023
Was this solution helpful?
4 mins
Uploaded on: 4/21/2023
Was this solution helpful?
Found 5 tutors discussing this question
Discuss this question LIVE for FREE
9 mins ago
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Practice questions from Advanced Problems in Mathematics for JEE (Main & Advanced) (Vikas Gupta)
View more
Practice more questions from Continuity and Differentiability
View more
Practice questions on similar concepts asked by Filo students
View more
Stuck on the question or explanation?
Connect with our Mathematics tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text Let in . Then :
Updated On May 3, 2023
Topic Continuity and Differentiability
Subject Mathematics
Class Class 12
Answer Type Text solution:1 Video solution: 3
Upvotes 325
Avg. Video Duration 5 min | {"url":"https://askfilo.com/math-question-answers/let-fxmax-leftx-x2-x3right-in-2-leq-x-leq-2-then-a-fx-is-continuous-in-2-leq-x","timestamp":"2024-11-10T20:50:18Z","content_type":"text/html","content_length":"617418","record_id":"<urn:uuid:aa0c9188-d589-435b-819d-ad57fdf01a4f>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00270.warc.gz"} |
Deriving the Maclaurin Series
Do you remember that formula from calculus that states
\[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots + \frac{f^{(n)}(0)}{n!}x^n + \ldots\]
Two questions: why is it true and why is it useful?
Why it’s true
What are we trying to accomplish with Maclaurin Series? We are trying to find a polynomial which equals the function \(f(x)\) (e.g. \(e^x\) or \(sin^2(x)\)).
When are two functions equal? Two functions are equal if they have the same value for all inputs (all values of \(x\), in this case). One way that could be true is that the functions have the same
value at \(x = 0\), as well as the same derivative, the same second derivative, the same third derivative, etc. This isn’t air tight (and indeed, it isn’t always true), but it holds for many
functions^1 and should seem somewhat intuitive. If a function is the same at a certain point, and the amount it changes around that point (its first derivative) is the same, and the amount that
changes around that point (its second derivative) is the same, all the way down, then how can these functions ever diverge? Well, often they don’t.
So, to recap, we’re going to look for a polynomial, \(p\), that has the same value as \(f\) at \(x=0\), as well as the same derivative, for all derivatives.
What’s the form of a polynomial? It looks something like this:
\[p(x) = a + bx + cx^2 + dx^3 + ex^4 + \ldots\]
0th derivative
We need \(p(0) = f(0)\). When \(x = 0\) all terms of the polynomial \(p\) go to zero, other than the first. In other words:
\[\begin{align*} f(0) &= p(0) \\ &= a + bx + cx^2 + dx^3 + ex^4 + \ldots |_{x=0} \\ &= a \end{align*}\]
We’ve discovered our first coefficient in \(p\): \(p(x) = f(0) + bx + cx^2 + dx^3 + ex^4 + \ldots\)
1st derivative
We need \(p'(0) = f'(0)\).
\[\begin{align*} f'(0) &= p'(0) \\ &= 0 + b + 2cx + 3dx^2 + 4ex^3 + \ldots |_{x=0} \\ &= b \end{align*}\]
We’ve discovered our second coefficient in \(p\): \(p(x) = f(0) + f'(0)x + cx^2 + dx^3 + ex^4 + \ldots\)
2nd derivative
We need \(p''(0) = f''(0)\).
\[\begin{align*} f''(0) &= p''(0) \\ &= 0 + 0 + 2c + 3 \cdot 2dx + 4 \cdot 3ex^2 + \ldots |_{x=0} \\ &= 2c \end{align*}\]
We’ve discovered our third coefficient in \(p\): \(p(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + dx^3 + ex^4 + \ldots\)
3rd derivative
We need \(p'''(0) = f'''(0)\).
\[\begin{align*} f'''(0) &= p'''(0) \\ &= 0 + 0 + 0 + 3 \cdot 2d + 4 \cdot 3 \cdot 2 ex + \ldots |_{x=0} \\ &= 3!d \end{align*}\]
We’ve discovered our fourth coefficient in \(p\): \(p(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{3!}x^3 + ex^4 + \ldots\)
And beyond
We could continue this pattern (seriously, try a few), but at this point you’re probably seeing a pattern emerge. The \(n\)th term of the polynomial seems to be
making the entire function
\[p(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\]
Why it’s useful
Take this section with a grain of salt. It’s very possible that I don’t know the most important or useful practical applications of the Maclaurin Series. But here is my answer: polynomials are easy!
They’re way nicer to deal with than arbitrary functions. In addition, derivative are (pretty) easy, and that’s all we need to turn an arbitrary function into polynomial.
For example, what’s the integral of \(ln(x + e^{a + x^2})\)? Uhh…. wolfram alpha, anyone?
How about the integral of:
\[p(x) = e^{-a} + (2 - e^{-2a})x + e^{-3a}(2 - 6e^{2a})x^2 + \ldots\]
Sure, it’s a big equation, but it’s completely trivial to take that integral (remember, \(a\) is just a constant).
\[\int p(x) = c + e^{-a}x + \frac{(2 - e^{-2a})}{2} x^2 + \frac{e^{-3a}(2 - 6e^{2a})}{3} x^3 + \ldots\]
And (big surprise) \(p(x)\) is the first few terms of the Maclaurin Series of \(ln(x + e^{a + x^2})\).
They’re also fast. Let’s say you have a function which is expensive to compute, and whose input is changing relatively quickly. Maybe instead of recomputing your expensive function every time your
input changes, you could approximate it with some large, but finite number of terms of its Maclaurin Series. Then, re-evaluating it will take almost no time at all!
Bonus section!
Hopefully you’ve followed the sections above, but if not, maybe a concrete example can help.
Let’s consider the function \(f(x) = sin(x) + 1\) in the domain of \([-10, 10]\). We can approximate this function using the first \(n\) terms of the Maclaurin Series. As \(n\) increases, our
approximation looks closer and closer to the original \(f(x)\).
Notice how the red line (the Maclaurin Series approximation of \(f(x)\)) only seems to change when going from an odd number of terms to an even number of terms. Can you figure out why that is?
1. Analytic functions is the technical term for the class of functions for which the Maclaurin Series (and the more general Taylor Series) holds. ↩ | {"url":"http://blog.russelldmatt.com/2019/06/26/Maclaurin-Series.html","timestamp":"2024-11-11T10:16:22Z","content_type":"text/html","content_length":"13707","record_id":"<urn:uuid:90f7495e-c403-4a52-adfd-b94234cd2620>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00135.warc.gz"} |
Thousand Locker High -
At Thousand Locker High, there are exactly 1000 students and 1000 lockers. It's the first day of school, and all 1000 lockers are initially open. The first student walks down the hall (beginning at
the first locker) and reverses the position (open/close) of each locker. Now all the lockers are closed. Next, the
student walks down the hall and reverses the position of every
locker. Now all the even numbered lockers are open. Next, the
student walks down the hall and reverses the position of every
locker. Next, the
student walks down the hall and reverses the position of every
locker and so forth until all 1000 students have walked down the hall with the 1000th student reversing only the position of the 1000th locker. How many lockers are now closed?
31 lockers. EXPLANATION
: The number of times a locker is opened or closed is equal to the number of factors the locker number has. For example, locker number 21 has four factors (1, 3, 7, and 21) and will be opened and
closed a total of four times. The first student will close it, the 3rd student will open it, the 7th student will close it, and finally, the 21st student will open it.
All lockers that are reversed an even number of times will end in the open position (the same position they started in). All lockers that are reversed an odd number of times will end in the closed
position. So the question is how many numbers between 1 and 1000 have an
number of factors?
If you take certain numbered lockers at random, you will notice that most have an even number of factors because the factors are "paired up". For example, the locker number 12 has six factors which
can be paired up as: 1 x 12, 2 x 6, and 3 x 4. It turns out that the only numbers that have an
number of factors are
square numbers (perfect squares)
such as 1, 4, 9, 16... Take locker number 16 for instance. The 1 pairs up with the 16, the 2 pairs up with the 8, but the 4 is unpaired (since it essentially pairs with itself). So locker number 16,
like all square numbers (pefect squares), has an odd number of factors.
The highest square number (perfect square) that is 1000 or less is 961 (31 squared) which is of course the 31st square number (perfect square). So 31 lockers will end up closed. With a calculator,
you can make a list of all the square numbers (perfect squares) between 1 and 1000. There are 31 in all: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441,
484, 529, 576, 625, 676, 729, 784, 841, 900, and 961. Each of those 31 lockers will end up closed. | {"url":"https://puzzlesandriddles.com/ForGeeksAndBrainiacs22.html","timestamp":"2024-11-05T19:01:42Z","content_type":"text/html","content_length":"15134","record_id":"<urn:uuid:9e389eb1-4816-4e8a-a009-cb6c7b34a60a>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00357.warc.gz"} |
Liquidity Ratios: Understanding Financial Health in Canadian Securities
B.4.1 Liquidity Ratios
Liquidity ratios are crucial financial metrics that measure a company’s ability to cover its short-term obligations with its short-term assets. These ratios provide insights into the financial health
of a company, indicating whether it can meet its immediate liabilities without raising additional capital. Understanding liquidity ratios is essential for investors, analysts, and stakeholders who
need to assess the company’s financial stability and operational efficiency.
Understanding Liquidity Ratios
Liquidity ratios are financial metrics used to determine a company’s ability to pay off its short-term debts as they come due. These ratios are vital for assessing a company’s short-term financial
health and operational efficiency. The two primary liquidity ratios are the Current Ratio and the Quick Ratio.
Current Ratio
The Current Ratio is a liquidity ratio that measures a company’s ability to cover its short-term obligations with its short-term assets. It is calculated using the following formula:
$$ \text{Current Ratio} = \frac{\text{Current Assets}}{\text{Current Liabilities}} $$
A higher current ratio indicates that the company has more current assets relative to its current liabilities, suggesting better liquidity and financial health.
Quick Ratio (Acid-Test Ratio)
The Quick Ratio, also known as the Acid-Test Ratio, is a more stringent measure of liquidity. It excludes inventory from current assets, as inventory is not always easily convertible to cash. The
formula for the Quick Ratio is:
$$ \text{Quick Ratio} = \frac{\text{Current Assets} - \text{Inventory}}{\text{Current Liabilities}} $$
The Quick Ratio provides a more conservative view of a company’s liquidity by focusing on assets that can be quickly converted into cash.
Numerical Example
Let’s consider a company with the following financial data:
• Current Assets: $150,000
• Inventory: $50,000
• Current Liabilities: $100,000
Using these figures, we can calculate the Current Ratio and Quick Ratio.
Current Ratio Calculation
$$ \text{Current Ratio} = \frac{\$150,000}{\$100,000} = 1.5 $$
This means the company has $1.50 in current assets for every $1.00 of current liabilities, indicating a healthy liquidity position.
Quick Ratio Calculation
$$ \text{Quick Ratio} = \frac{\$150,000 - \$50,000}{\$100,000} = \frac{\$100,000}{\$100,000} = 1.0 $$
This indicates that excluding inventory, the company has $1.00 in liquid assets for every $1.00 of current liabilities, suggesting it can cover its short-term obligations without relying on inventory
Interpreting Liquidity Ratios
Liquidity ratios provide valuable insights into a company’s short-term financial health:
• Current Ratio: A ratio above 1 indicates that a company has more current assets than current liabilities, suggesting it can cover its short-term obligations. However, a very high ratio might
indicate inefficient use of assets.
• Quick Ratio: A ratio of 1 or higher is generally considered satisfactory, as it shows the company can meet its short-term liabilities without relying on inventory sales.
Comparing Liquidity Ratios Across Companies and Industries
When analyzing liquidity ratios, it’s essential to compare them with industry norms and competitors. Different industries have varying liquidity requirements, and what is considered healthy in one
industry might be inadequate in another. For example, a retail company might have a lower quick ratio due to its reliance on inventory, while a tech company might have a higher ratio due to its
asset-light business model.
Limitations and Considerations
While liquidity ratios are useful, they have limitations:
• High Liquidity Ratios: May indicate excessive cash holdings or inefficient use of assets.
• Industry Variations: Liquidity needs vary across industries, so it’s crucial to compare ratios with industry peers.
• Short-term Focus: These ratios focus on short-term assets and liabilities, providing limited insight into long-term financial health.
Liquidity ratios are indispensable tools for assessing a company’s ability to meet its short-term obligations. By understanding and calculating the Current and Quick Ratios, stakeholders can gain
insights into a company’s financial health and operational efficiency. However, it’s essential to consider industry norms and other financial metrics for a comprehensive analysis.
Quiz Time!
📚✨ Quiz Time! ✨📚
### What do liquidity ratios measure? - [x] A company's ability to pay off short-term debts with short-term assets - [ ] A company's profitability over a fiscal year - [ ] A company's long-term
solvency - [ ] A company's market share > **Explanation:** Liquidity ratios specifically measure a company's capacity to meet its short-term obligations using its short-term assets. ### How is the
Current Ratio calculated? - [x] Current Assets divided by Current Liabilities - [ ] Current Liabilities divided by Current Assets - [ ] Total Assets divided by Total Liabilities - [ ] Net Income
divided by Total Revenue > **Explanation:** The Current Ratio is calculated by dividing Current Assets by Current Liabilities. ### What is another name for the Quick Ratio? - [x] Acid-Test Ratio - [
] Cash Ratio - [ ] Solvency Ratio - [ ] Profitability Ratio > **Explanation:** The Quick Ratio is also known as the Acid-Test Ratio because it measures a company's ability to meet its short-term
obligations without relying on inventory. ### In the example provided, what is the Quick Ratio? - [x] 1.0 - [ ] 1.5 - [ ] 0.5 - [ ] 2.0 > **Explanation:** The Quick Ratio is calculated as 1.0, using
the formula \\((\$150,000 - \$50,000) / \$100,000\\). ### Why might a very high Current Ratio be a concern? - [x] It may indicate excessive cash holdings or inefficient use of assets. - [ ] It shows
the company cannot cover its short-term obligations. - [ ] It suggests the company is highly profitable. - [ ] It indicates a high level of debt. > **Explanation:** A very high Current Ratio might
suggest that a company is holding too much cash or not using its assets efficiently. ### What does a Quick Ratio of 1.0 indicate? - [x] The company can meet its short-term liabilities without relying
on inventory. - [ ] The company is not able to meet its short-term obligations. - [ ] The company has excessive inventory. - [ ] The company is highly profitable. > **Explanation:** A Quick Ratio of
1.0 indicates that the company has enough liquid assets to meet its short-term liabilities without needing to sell inventory. ### Which of the following is excluded from the Quick Ratio calculation?
- [x] Inventory - [ ] Accounts Receivable - [ ] Cash - [ ] Current Liabilities > **Explanation:** Inventory is excluded from the Quick Ratio calculation to provide a more conservative measure of
liquidity. ### Why is it important to compare liquidity ratios with industry norms? - [x] Different industries have varying liquidity requirements. - [ ] To determine the company's market share. - [
] To assess the company's long-term growth potential. - [ ] To evaluate the company's profitability. > **Explanation:** Comparing liquidity ratios with industry norms is important because different
industries have different liquidity needs and standards. ### What might a low Quick Ratio indicate? - [x] The company may struggle to meet short-term obligations without selling inventory. - [ ] The
company is highly profitable. - [ ] The company has excessive cash reserves. - [ ] The company is expanding rapidly. > **Explanation:** A low Quick Ratio might indicate that the company could
struggle to meet its short-term obligations without relying on inventory sales. ### True or False: Liquidity ratios provide insight into a company's long-term financial health. - [ ] True - [x] False
> **Explanation:** Liquidity ratios focus on a company's short-term financial health, not its long-term financial stability. | {"url":"https://csccourse.ca/32/6/1/","timestamp":"2024-11-07T09:12:13Z","content_type":"text/html","content_length":"115289","record_id":"<urn:uuid:4fda8ddb-c791-4e1a-af5d-1718d5b012fe>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00442.warc.gz"} |
Statistics Lectures - 14: Linear Regression & Spearman Correlation
Related Pages
12: Poisson Distribution
13: Scatter Plots & Pearson’s r Correlation
15: Sampling Error & Central Limit Theorem
16: Sample Proportions & Confidence Intervals
Jump to Table of Contents
Statistics - Lecture 39: Linear Regression
If we know that two variables are strongly correlated, we can use one variable to predict the other.
Statistics - Lecture 40: Spearman Correlation
The Spearman correlation is used when:
1. Measuring the relationship between two ordinal variables.
2. Measuring the relationship between two variables that are related, but not linearly.
Statistics - Lecture 41: Correlation Vs. Causation
Causation means that one variable causes something to happen in another variable.
To say that two things are correlated is to say that they share some kind of relationship.
In order to imply causation, a true experiment must be performed where subjects are randomly assigned to different conditions.
Researches want to test a new ant-anxiety medication. They split participants into three conditions (0mg, 50mg, and 100mg), then ask them to rate their anxiety level on a scale of 1-10. Are there any
differences between the three conditions using alpha = 0.05?
Statistics Lecture Series - Table Of Contents
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. | {"url":"https://www.onlinemathlearning.com/statistics-lecture-14.html","timestamp":"2024-11-14T17:06:44Z","content_type":"text/html","content_length":"45870","record_id":"<urn:uuid:7cfd4389-928a-4d46-9535-199976cdb24b>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00357.warc.gz"} |
Kolmogorov-Smirnov test in Excel tutorial
The data correspond to scores (0 – 30) measuring the quality of two brands of shoes (brand A and brand B). Scores were computed based on a survey addressed to customers using either brand. 15
customers have answered for brand A and 8 different clients for brand B.
Goal of this tutorial
This tutorial is divided into two parts:
In the first part we compare the distributions of the two samples without making assumptions on underlying theoretical distributions (normal distribution for example). We use the non-parametric
Kolmogorov-Smirnov test, which is well suited in this case.
In the second part, we use the Kolmogorov-Smirnov test to compare the distribution of one sample to a theoretical distribution.
Part 1: Running a Kolmogorov-Smirnov test to compare two observed distributions in Excel
Here, we are interested in comparing the distributions of the two samples.
First of all, what do these distributions look like? Histograms are a good tool to visualize continuous distributions: XLSTAT / Visualizing data / Histograms.
In the General tab, select both samples inside the Data cell range. In the Options tab, activate the minimum option and enter 0 in the box. This will force histograms to have the same lower bound on
the x axis making their comparison easier. Click on the OK button.
The histograms appear in the results sheet:
Without making any theoretical assumption, we may say that the distribution of sample B is more skewed towards low values compared to the distribution of sample A. We will now use the
Kolmogorov-Smirnov non-parametric test to compare the two distributions. Go to XLSTAT / Nonparametric tests / Comparison of two distributions.
Select the Brand A column in Sample 1 and the Brand B column in sample 2. The Kolmogorov-Smirnov test allows samples to be unbalanced such as in our data: sample B contains fewer scores than sample
A. In the Options tab, notice it is possible to select a one-tailed alternative hypothesis and/or an exact computation of the p-value. In the Charts tab, activate the Cumulative histograms option.
Click on the OK button.
The results sheet contains the Kolmogorov-Smirnov statistic (0.475) that can be easily extracted (see further, the cumulative histograms chart). This statistic is associated to a p-value (0.190)
indicating that the two distributions are not significantly different at alpha = 0.05.
Confused with p-values and statistical significance? Do not hesitate to visit our tutorial.
The cumulative distributions chart presents the studied variable (survey scores) on the x axis. For a given point on the x axis, a brand’s cumulative relative frequency is the proportion of scores
smaller than this point among the scores of the brand. Thus, as previously suggested by the histograms, brand B seems to start cumulating scores earlier than brand A along the x axis. Let’s take a
look at the medians, which are the scores corresponding to a cumulative relative frequency of 0.5. The median score for brand B (~20) seems to be higher than the median score for brand A (~17).
Kolmogorov-Smirnov’s D test statistic is the highest deviation occurring between the two curves.
It is calculated by the following formula:
where F_n and F are the distribution functions.
In our example, this deviation value falls inside the median region, but this may not necessarily be the case when using other data. The higher the D statistic, the lower the p-value and the more
significant the difference between the two distributions.
Part 2: Running a Kolmogorov-Smirnov test to compare an observed distribution to a theoretical one in Excel
Suppose that the quality scores of brand A were obtained in France. For US customers, this score follows a normal distribution with a mean of 21.5 and a standard deviation of 2.3. We may ask
ourselves if the French scores distribution is significantly different from the theoretical distribution of the US scores.
Here again, we will use the Kolmogorov-Smirnov test. The only difference with the previous part is that we aim at comparing an observed distribution to a theoretical one instead of comparing two
different distributions.
To run the test, go to XLSTAT / Nonparametric tests / Distribution fitting.
In the General tab, select the brand A data, the normal distribution, activate the Enter option and enter the following parameters: µ = 21.5 and sigma = 2.3. In the Charts tab, activate the
Cumulative histograms option. Click on the OK button.
In the results sheet, the histogram (on the left below) shows that the observed distribution of our data lays on low score values compared to the theoretical curve reflecting the US scores
distribution (red line).
The Kolmogorov-Smirnov test is associated to a p-value of 0.000 suggesting that the null hypothesis should be rejected and that the observed distribution is significantly different from the
theoretical one at alpha = 0.05.
Not sure that you have chosen the right test? This will let you know. | {"url":"https://help.xlstat.com/6676-kolmogorov-smirnov-test-excel-tutorial","timestamp":"2024-11-07T20:23:38Z","content_type":"text/html","content_length":"34744","record_id":"<urn:uuid:9f8d312c-7aab-4b38-ab21-6a70ea0a7873>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00617.warc.gz"} |
Book Chapters:
1. HUA, Y., SARKAR, T.K., and HU, F., "The generalized pencil-of-function (GPOF) method for extracting poles from transient responses," Spectral Analysis in One or Two Dimensions, pp. 417-433,
edited by S. Prasad, and R. L. Kashyap, Oxford & IBM Publishing Co., 1990.
2. HUA, Y., "Blind identification and equalization of channels driven by colored signals," Chapter 4, Vol. 1, pp. 113-138, Signal Processing Advances in Wireless and Mobile Communications, edited by
G. Giannakis, Y. Hua, P. Stoica and L. Tong, Prentice-Hall, 2001.
3. MANTON, J., and HUA, Y., "Blind channel identifiability with an arbitrary linear precoder", Chapter 10, Vol. 1, pp. 339-366, Signal Processing Advances in Wireless and Mobile Communications,
edited by G. Giannakis, Y. Hua, P. Stoica and L. Tong, Prentice-Hall, 2001.
4. ABED-MERAIM, K., HUA, Y., and HAARDT, M., “Joint Schur decomposition: algorithms and applications,” pp. 1-14, Defence Applications of Signal Processing, edited by D. Cochran, B. Moran and L.
White, Elsevier, 2001.
5. ZHU, G., and HUA, Y., "Multidimensional NMR spectroscopic signal processing," pp. 509-543, Signal Processing for Magnetic Resonance Imaging and Spectroscopy, edited by H. Yan, Marcel Dekker,
Inc., 2002.
6. CHENG, Q., and HUA, Y., “A review of parametric high-resolution methods,” pp. 1-62, High-Resolution and Robust Signal Processing, edited by Y. Hua, A. Gershman, and Q. Cheng, Marcel Dekker, 2003. | {"url":"https://intra.engr.ucr.edu/~yhua/List%20of%20Book%20Chapters.html","timestamp":"2024-11-06T12:28:55Z","content_type":"text/html","content_length":"30069","record_id":"<urn:uuid:528542bb-b57d-47af-ad54-cee9baf4831c>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00775.warc.gz"} |
T15: Quantitative microbial risk assessment
The tool helps to quantify the pathogen occurrence in source water and their removal by various treatment steps at MAR facilities by using a probabilistic approach. The interactive web-based QMRA
tool supports the evidence-based risk assessment to minimize water-related infectious diseases.
This tool was developed within the frame of SMART-Control, a WaterJPI project. For further information please visit: www.smart-control.inowas.com
The documentation of the tool is based on the SMART-Control Deliverable D4.3.
Quantitative Microbial Risk Assessment (QMRA) is recognized as an evidence-based approach to minimize water-related infectious diseases. The risks caused by pathogenic microorganisms can be assessed
which supports decision-making related to the microbial safety of water systems. The web-based QMRA tool was developed to support the implementation of QMRA through an interactive, easy-to-use, and
guided web-browser based application. The QMRA tool allows the quantification of pathogen occurrence in source water and their removal by various treatment steps and is based on a probabilistic risk
Due to the many preset parameters, microbial risks at sites with little known information can also be calculated. Based on the findings from the model, problem and risk areas can be identified and
protective measures prepared.
Parameter input
Inflow concentration of pathogens
Inflow concentrations of the pathogen to the treatment scheme are entered as absolute minimum and maximum concentration per litre and are required for all pathogens that should be used for QMRA.
Currently, one of the following probability density functions (PDF) can be selected by the user:
1. Uniform: provides a constant probability density function, the provided absolute “min” and “max” values are in case minimum concentration is 0 replaced with 0.01; in case maximum concentration is
0 it is replaced with 0.1
2. Log[10] uniform: same as uniform but all values log[10] transformed
3. Normal: the required parameters mean and standard deviation are derived from the provided absolute min/max values as follows:
1. “min”: if the minimum concentration is 0 it is replaced with 0.01
2. “max”: if the maximum concentration is 0 it is replaced with 0.1
3. “mean”: (min+max) /2
4. “sdev”: abs(max – mean) / 1.644854, assuming that 90% of all random values lie between min and max
4. Log[10] normal: same as normal but all values log[10] transformed
5. Log normal: same as normal but all values log[2] transformed
Total variability in pathogen concentration for inflow concentration is difficult to assess, because it can be unclear whether measurements cover the full range of microbial concentration (possible
unobserved values). Low pathogen concentrations and the irregular occurrence of pathogens contributes to the difficulties in capturing this variability. Moreover, some sites may not have any pathogen
data and are dependent on assumptions. At data scarce sites, it is recommended to use “log[10 ]uniform”, because compared to the other it largely overestimate the provided concentration range. In
contrast, the “log[10] normal” distribution could be used for simulating “peak” system behavior as it overestimates the maximum concentration by approximately 250% compared to the maximum value
entered by the user. The impact of different PDFs on the inflow concentrations is shown in Figure 1.
Figure 1: Logarithmic concentration ranges vs count of Probability Density Functions (min log10= 0 and max log10 = 2).
For further details the reader is referred to the documentation website of the R function create_random_distribution[1] within the R package kwb.qmra[2].
Treatment steps
For each treatment process the log[10]-removals for at least one of the three different pathogen groups (i.e. bacteria, protozoa, viruses) need to be defined. The user can define a new treatment
process or select from pre-defined literature based removal performance associated to the different treatment steps. Pre-defined treatments are available by clicking on list icon next to “Add
process” button. There are over 25 treatment steps available covering the following groups:
• Coagulation, flocculation and sedimentation
• Filtration
• Natural Attenuation
• Pre-treatment
• Disinfection
• Primary and
• Secondary treatment
Natural attenuation e.g. includes bank filtration, soil-aquifer passage and wetlands. Each treatment step is briefly described (Figure 2) and substantiated with literature sources.
Figure 2. Selection of treatment steps
After selecting the required treatment steps it is necessary to enter Log Removal Values (LRV) for each pathogen and treatment step. The available PDF are limited uniform and normal distribution
function (Figure 3).
Figure 3. Selection of probability density function for treatment steps
Treatment train
Treatment steps can be combined to a treatment train or treatment scheme (Figure 4). The user may construct several treatment trains. All selected treatment trains will be considered in the QMRA
calculation. The order of the treatment steps does not play a role in the calculation.
Figure 4. Combination of treatment steps to a treatment scheme
Exposure Scenario
The number of exposures per year and the ingested volume per event can be defined as fixed value or by following a pre-defined distribution. There are eight pre-defined exposure scenarios from
drinking water to irrigation water and domestic end-use. The pre-defined exposure scenarios are accessible by clicking on the “Add Scenario” button. Once selected the exposure scenario can be
switched to active by clicking on the “Toogle” button. Only one exposure scenario per simulation can be selected at a time (Figure 5).
Figure 5. Exposure scenario
The dose-response models are based on experimental data. The dose-effect relationships can be approximated for each pathogen by exponential binomial formulae or beta Poisson distributions. Parameters
for each pathogen are taken from QMRAwiki (2016)[3] (Figure 6).
Dose-response models are defined for the pathogens that are toggled active in the inflow concentration section.
Figure 6. Dose-response relation
For all pathogens to be used for QMRA, the infection to illness factor and the disability-adjusted life years (DALY) per case need to be defined (Figure 7).
Figure 7. Definition of Health parameters
Stochastic runs
In Monte-Carlo simulation the number of random distribution generations can be specified here. At least 1000 runs are recommended.
Calculation, results and data export
The results are presented in tables and made available for download in json and csv format. A graphical display of results is currently not available. Four files (events, total, stats_total,
stats_logremoval) are available for download (Figure 8).
Figure 8. Export of results
The file “events” include the calculated logreduction for each event from each pathogen group. The output file “total” contains the values of the inflow median, logreduction median, volume sum,
exposure sum, dose sum, infection probability sum, illness probability sum, and dalys sum for each stochastic run from each pathogen group. The file “stats_total” include aggregated values (min, p05,
p25, mean, median, p75, p95, and max) of dalys sum, dose sum, events, exposure sum, infection probability sum, illness probability sum, and dalys sum for each treatment step and pathogen group. The
file “stats_logremoval” include aggregated values (min, p05, p25, mean, median, p75, p95, and max) of log removal for each treatment step and pathogen group.
[1]https://kwb-r.github.io/kwb.qmra/reference/create_random_distribution.html, 22.10.2021
[2]https://kwb-r.github.io/kwb.qmra/, 22.10.2021
[3]http://qmrawiki.canr.msu.edu/index.php/Dose_Response, 22.10.2021 | {"url":"http://inowas.webspace.tu-dresden.de/tools/t15-quantitative-microbial-risk-assessment/","timestamp":"2024-11-05T19:32:23Z","content_type":"text/html","content_length":"58583","record_id":"<urn:uuid:d4c8a4b7-3a8a-4f23-baf8-62c7febb2e14>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00504.warc.gz"} |
Binary Search
Binary Search
The last page covered some simple examples of recursion, which could have been coded with loops just as easily. In other cases, using recursion is considerably more 'natural' than using loops.
The Algorithm
You are given an array sorted in increasing order and would like to find the location of number n in it. While you could just search through the array in order, the goal is to find an optimal
algorithm that will find the number much faster. Think about what algorithm you would use, and then click on the button below.
This task is similar to the "guess-a-number" game where one person guesses a number and is told if the correct number is higher or lower. The goal is to repeatedly split the possible numbers in half
until you reach the correct number.
1. Check the middle of the given numbers, and see if it's the right number. Return that index if it's right, and otherwise continue.
2. If the number is higher than n, check the left side (of the number checked) in the same way. If the number is lower than n, check the right side.
The Code
Now that we've described an algorithm, let's convert it to code. We start with an empty method body (Hover over the parameter names for a description):
int binarySearch(int[] ar, int iMin, int iMax, int num){
//algorithm here
Next, let's fill in each step of the algorithm in code:
Step 1 - Check the middle Number. Let's declare a variable iMid to store the middle index^1:
int binarySearch(int[] ar, int iMin, int iMax, int num){
int iMid = (iMax+iMin) / 2;
if(ar[iMid] == num){
return iMid;
The binarySearch will now work when num happens to be right in the middle! When we're not that lucky, we'll need step 2:
Step 2. - Otherwise, check the 'correct' half next. If the number at the midpoint is greater than num, we'll want to check the left-hand side of the array. We can call binarySearch again with
adjusted indices:
//...previous code
else if(ar[iMid]>num){
return binarySearch(ar, iMin, iMid-1, num);
This will call binarySearch with the same minimum index, but a new maximum index which is right before the number we just checked. If that condition isn't true, that means num is to the right:
return binarySearch(ar, __, iMax, num);
We now have a working binary search! To get to work we just pass in a sorted array, along with the minimum and maximum starting indices and the number we're looking for. For example:
int[] ar = {1,3,4,7,9,11,13};
int index = binarySearch(ar, 0, ar.length-1, 4);
This code will print out 2, the index that 4 is located at.
Here's a diagram of a binary search in action:
What happens:
1. The initial call starts searching from cell 0 to the last cell in the array
2. binarySearch sees that 7 (located at index 3) is bigger than 4, so it calls binarySearch on 0 to 2.
3. 3 is less than 4, so binary search moves up and finally finds 4.
Finishing the Search
The search works, but there's one issue - what if the number isn't found in the array? There's nothing to stop the indices from 'passing' each other and the search will continue until there's a
StackOverflow error.
So we need to add another base case that checks for this condition:
if(iMax < iMin){
return -1;
this will return -1 when the iMax passed iMin, which means the number isn't in the array (assuming it's sorted).
It's important to cover all "exit possibilities" so that the recursive function doesn't try going on forever.
The complete binarySearch is below. To see it visualized, click here.
int binarySearch(int[] ar, int iMin, int iMax, int num){
if(iMax < iMin){
return -1;
int iMid = (iMax+iMin)/2;
if(ar[iMid] == num){
return iMid;
else if(ar[iMid]>num){
return binarySearch(ar, iMin, iMid-1, num);
return binarySearch(ar, iMid+1, iMax, num);
^1. This is the simplest way to find the middle number, though it won't work for very large integers in Java.
What value should be put in the blank above so the correct minimum index is passed to binarySearch?
Please sign in or sign up to submit answers.
Alternatively, you can try out Learneroo before signing up.
The method doStuff takes in one square number. Create a separate recursive method that returns the square root of a given number. Call that method from doStuff and return the answer to be printed
Do not use any built in methods for calculating the square-root and don't try searching through all the numbers. Instead, use a binary-style search to home in on the actual square root.
(To make it simpler, the input will just contain square numbers.)
Please sign in or sign up to submit answers.
Alternatively, you can try out Learneroo before signing up.
Dear admin,
When clicking on the Raw I/O option under Input I see "8":
8 <<<<No Square
Correct Output
Your Output (I still have not solved it, so currently empty)
But on the instructions it is mentioned "(To make it simpler, the input will just contain square numbers.)" Is this a typo on the Raw I/O or is it actually handling that? Because on one of my
tests i got Stack Overflow so currently not sure if expected...
Thanks in advance,
@David, the first number in raw input states how many cases there will be, it's not one of the actual cases.
how do i know what is desired number? Should I make anarray from 0 to a, than start at midle checking if that number to power 2is a?
@armands, you don't actually need an array, but that's the right idea (see the hint).
Hi guys, i have written this code https://ideone.com/eFWUiZ its kind of bad, i know it , i suck at recursion but it was a bit logical for me i really don't know where i'm missing up
See if you can keep it organized. If you don't have the right number, see if it's larger or smaller and call your recursive method accordingly. You don't want to have two methods calling each
other since that's just confusing. Look over the provided code for binary search and see how to adjust it for the "square root search" problem.
(PS - You can link to your learneroo code submission, no need to link elsewhere.)
WOOOHOOO! Man I'm pleased with my implementation of binary search.
However, you see how I got the range for iMid and iMax?
I made a crude estimation there... that the square root of any number could certainly be within the range = a/2
But thats not very efficient. How do I choose a smaller range and thus use less cycles on finding the square root?
After solving the question here, you can click on the "Featured Answers" link (next to the hint button) to see a possible solution. The idea is to use the same technique as binary search, but in
search of the square root (i.e. the number that multiplied by itself equals the original number).
Working answer. Refer after trying it out
public static int computeRoot(int mid,int a){
return 0;
if(mid*mid == a ){
return mid;
return computeRoot(mid-1,a);
public static int doStuff(int a){
int mid = a/2;
return computeRoot(mid,a);
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
for(int i=0; i<n; i++){
int a = in.nextInt();
int result = doStuff(a); | {"url":"https://www.learneroo.com/modules/71/nodes/399","timestamp":"2024-11-04T12:26:36Z","content_type":"text/html","content_length":"57523","record_id":"<urn:uuid:b98c42d3-5019-4c9c-a9b6-15c3221f7aef>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00738.warc.gz"} |
Vector Autoregression (VAR) Models
A vector autoregression (VAR) model is a multivariate time series model containing a system of n equations of n distinct, stationary response variables as linear functions of lagged responses and
other terms. VAR models are also characterized by their degree p; each equation in a VAR(p) model contains p lags of all variables in the system.
VAR models belong to a class of multivariate linear time series models called vector autoregression moving average (VARMA) models. Although Econometrics Toolbox™ provides functionality to conduct a
comprehensive analysis of a VAR(p) model (from model estimation to forecasting and simulation), the toolbox provides limited support for other models in the VARMA class.
In general, multivariate linear time series models are well suited for:
• Modeling the movements of several stationary time series simultaneously.
• Measuring the delayed effects among the response variables in the system.
• Measuring the effects of exogenous series on variables in the system. For example, determine whether the presence of a recently imposed tariff significantly affects several econometric series.
• Generating simultaneous forecasts of the response variables.
Types of Stationary Multivariate Time Series Models
This table contains forms of multivariate linear time series models and describes their supported functionality in Econometrics Toolbox.
Model Abbreviation Equation Supported Functionality
• Represent the model by using a varm object:
1. Create a template for estimation or a fully specified model by using varm.
2. Estimate any unknown parameters by using estimate.
Vector autoregression VAR(p) ${y}_{t}=c+\sum _{j=1}^{p}{\Phi }_{j}{y}_{t-j}+{\epsilon }_{t}$ 3. Work with a fully specified model by applying object functions.
• Obtain the coefficient matrices of a VAR model from the coefficient matrices of
its VARMA(p,q) equivalent by using arma2ar.
• Given coefficient matrices, perform dynamic multiplier analysis by using armairf
and armafevd.
Vector autoregression with ${y}_{t}=c+\delta t+\sum _{j=1}^{p}{\Phi }_{j}{y}_{t-j}+{\epsilon }_{t}$ Represent the model by using a varm object. estimate and all other object functions
a linear time trend VAR(p) support this model.
Vector autoregression with ${y}_{t}=c+\delta t+\beta {x}_{t}+\sum _{j=1}^{p}{\Phi }_{j}{y}_{t-j}+{\ Represent the model by using a varm object. estimate and all other object functions
exogenous series VARX(p) epsilon }_{t}$ support this model.
• Obtain coefficient matrices of a VMA model from the coefficient matrices of its
VARMA(p,q) equivalent by using arma2ma.
Vector moving average VMA(q) ${y}_{t}=c+\sum _{k=1}^{q}{\Theta }_{k}{\epsilon }_{t-k}+{\epsilon }_{t}$
• Given coefficient matrices, perform dynamic multiplier analysis by using armairf
and armafevd.
• Obtain coefficient matrices of a VAR or VMA model from the coefficient matrices
${y}_{t}=c+\sum _{j=1}^{p}{\Phi }_{j}{y}_{t-j}+\sum _{k=1}^{q}{\Theta }_ of its VARMA(p,q) equivalent by using arma2ar or arma2ma, respectively.
Vector autoregression VARMA(p, q) {k}{\epsilon }_{t-k}+{\epsilon }_{t}$
moving average • Given coefficient matrices, perform dynamic multiplier analysis by using armairf
and armafevd.
Structural vector ${\Phi }_{0}{y}_{t}=c+\sum _{j=1}^{p}{\Phi }_{j}{y}_{t-j}+\sum _{k=1}^{q}
autoregression moving SVARMA(p, q) {\Theta }_{k}{\epsilon }_{t-k}+{\Theta }_{0}{\epsilon }_{t}$ Same support as for VARMA models
The following variables appear in the equations:
• y[t] is the n-by-1 vector of distinct response time series variables at time t.
• c is an n-by-1 vector of constant offsets in each equation.
• Φ[j] is an n-by-n matrix of AR coefficients, where j = 1,...,p and Φ[p] is not a matrix containing only zeros.
• x[t] is an m-by-1 vector of values corresponding to m exogenous variables or predictors. In addition to the lagged responses, exogenous variables are unmodeled inputs to the system. Each
exogenous variable appears in all response equations by default.
• β is an n-by-m matrix of regression coefficients. Row j contains the coefficients in the equation of response variable j, and column k contains the coefficients of exogenous variable k among all
• δ is an n-by-1 vector of linear time-trend values.
• ε[t] is an n-by-1 vector of random Gaussian innovations, each with a mean of 0 and collectively an n-by-n covariance matrix Σ. For t ≠ s, ε[t] and ε[s] are independent.
• Θ[k] is an n-by-n matrix of MA coefficients, where k = 1,...,q and Θ[q] is not a matrix containing only zeros.
• Φ[0] and Θ[0] are the AR and MA structural coefficients, respectively.
Generally, the time series y[t] and x[t] are observable because you have data representing the series. The values of c, δ, β, and the autoregressive matrices Φ[j] are not always known. You typically
want to fit these parameters to your data. See estimate for ways to estimate unknown parameters or how to hold some of them fixed to values (set equality constraints) during estimation. The
innovations ε[t] are not observable in data, but they can be observable in simulations.
Lag Operator Representation
In the preceding table, the models are represented in difference-equation notation. Lag operator notation is an equivalent and more succinct representation of the multivariate linear time series
The lag operator L reduces the time index by one unit: Ly[t] = y[t–1]. The operator L^j reduces the time index by j units: L^jy[t] = y[t–j].
In lag operator form, the equation for a SVARMAX(p, q) model is:
$\left({\Phi }_{0}-\sum _{j=1}^{p}{\Phi }_{j}{L}^{j}\right){y}_{t}=c+\beta {x}_{t}+\left({\Theta }_{0}+\sum _{k=1}^{q}{\Theta }_{k}{L}^{k}\right){\epsilon }_{t}.$
The equation is expressed more succinctly in this form:
$\Phi \left(L\right){y}_{t}=c+\beta {x}_{t}+\Theta \left(L\right){\epsilon }_{t},$
$\Phi \left(L\right)={\Phi }_{0}-\sum _{j=1}^{p}{\Phi }_{j}{L}^{j}$
$\Theta \left(L\right)={\Theta }_{0}+\sum _{k=1}^{q}{\Theta }_{k}{L}^{k}.$
Stable and Invertible Models
A multivariate AR polynomial is stable if
$\mathrm{det}\left({I}_{n}-{\Phi }_{1}z-{\Phi }_{2}{z}^{2}-...-{\Phi }_{p}{z}^{p}\right)e 0\text{for}|z|\le 1.$
With all innovations equal to zero, this condition implies that the VAR process converges to c as t approaches infinity (for more details, see [1], Ch. 2).
A multivariate MA polynomial is invertible if
$\mathrm{det}\left({I}_{n}+{\Theta }_{1}z+{\Theta }_{2}{z}^{2}+...+{\Theta }_{q}{z}^{q}\right)e 0\text{for}|z|\le 1.$
This condition implies that the pure VAR representation of the VMA process is stable (for more details, see [1], Ch. 11).
A VARMA model is stable if its AR polynomial is stable. Similarly, a VARMA model is invertible if its MA polynomial is invertible.
Models with exogenous inputs (for example, VARMAX models) have no well-defined notion of stability or invertibility. An exogenous input can destabilize a model.
Models with Regression Component
Incorporate feedback from exogenous predictors, or study their linear associations with the response series, by including a regression component in a multivariate linear time series model. By order
of increasing complexity, examples of applications that use such models:
• Modeling the effects of an intervention, which implies that the exogenous series is an indicator variable.
• Modeling the contemporaneous linear associations between a subset of exogenous series to each response. Applications include CAPM analysis and studying the effects of prices of items on their
demand. These applications are examples of seemingly unrelated regression (SUR). For more details, see Implement Seemingly Unrelated Regression and Estimate Capital Asset Pricing Model Using SUR.
• Modeling the linear associations between contemporaneous and lagged exogenous series and the response as part of a distributed lag model. Applications include determining how a change in monetary
growth affects real gross domestic product (GDP) and gross national income (GNI).
• Any combination of SUR and the distributed lag model that includes the lagged effects of responses, also known as simultaneous equation models.
The general equation for a VARX(p) model is
${y}_{t}=c+\delta t+\beta {x}_{t}+\sum _{j=1}^{p}{\Phi }_{j}{y}_{t-j}+{\epsilon }_{t}$
• x[t] is an m-by-1 vector of observations from m exogenous variables at time t. The vector x[t] can contain lagged exogenous series.
• β is an n-by-m vector of regression coefficients. Row j of β contains the regression coefficients in the equation of response series j for all exogenous variables. Column k of β contains the
regression coefficients among the response series equations for exogenous variable k. This figure shows the system with an expanded regression component:
$\left[\begin{array}{c}{y}_{1,t}\\ {y}_{2,t}\\ ⋮\\ {y}_{n,t}\end{array}\right]=c+\delta t+\left[\begin{array}{c}{x}_{1,t}\beta \left(1,1\right)+\cdots +{x}_{m,t}\beta \left(1,m\right)\\ {x}_{1,t}
\beta \left(2,1\right)+\cdots +{x}_{m,t}\beta \left(2,m\right)\\ ⋮\\ {x}_{1,t}\beta \left(n,1\right)+\cdots +{x}_{m,t}\beta \left(n,m\right)\end{array}\right]+\sum _{j=1}^{p}{\Phi }_{j}{y}_{t-j}+
{\epsilon }_{t}.$
VAR Model Workflow
This workflow describes how to analyze multivariate time series by using Econometrics Toolbox VAR model functionality. If you believe the response series are cointegrated, use VEC model functionality
instead (see vecm).
1. Load, preprocess, and partition the data set. For more details, see Format Multivariate Time Series Data.
2. Create a varm model object that characterizes a VAR model. A varm model object is a MATLAB^® variable containing properties that describe the model, such as AR polynomial degree p, response
dimensionality n, and coefficient values. varm must be able to infer n and p from your specifications; n and p are not estimable. You can update the lag structure of the AR polynomial after
creating a VAR model, but you cannot change n.
varm enables you to create these types of models:
□ Fully specified model in which all parameters, including coefficients and the innovations covariance matrix, are numeric values. Create this type of model when economic theory specifies the
values of all parameters in the model, or you want to experiment with parameter settings. After creating a fully specified model, you can pass the model to all object functions except
□ Model template in which n and p are known values, but all coefficients and the innovations covariance matrix are unknown, estimable parameters. Properties corresponding to estimable
parameters are composed of NaN values. Pass a model template and data to estimate to obtain an estimated (fully specified) VAR model. Then, you can pass the estimated model to any other
object function.
□ Partially specified model template in which some parameters are known, and others are unknown and estimable. If you pass a partially specified model and data to estimate, MATLAB treats the
known parameter values as equality constraints during optimization, and estimates the unknown values. A partially specified model is well suited to these tasks:
☆ Remove lags from the model by setting the coefficient to zero.
☆ Associate a subset of predictors to a response variable by setting to zero the regression coefficients of predictors you do not want in the response equation.
For more details, see Create VAR Model.
3. For models with unknown, estimable parameters, fit the model to data. See VAR Model Estimation Overview and estimate.
4. Find an appropriate AR polynomial degree by iterating steps 2 and 3. See Select Appropriate Lag Order.
5. Analyze the fitted model. This step can involve:
Your application does not have to involve all the steps in this workflow, and you can iterate some of the steps. For example, you might not have any data, but want to simulate responses from a fully
specified model.
[1] Lütkepohl, H. New Introduction to Multiple Time Series Analysis. Berlin: Springer, 2005.
See Also
Related Topics | {"url":"https://se.mathworks.com/help/econ/introduction-to-vector-autoregressive-var-models.html","timestamp":"2024-11-11T09:45:35Z","content_type":"text/html","content_length":"103854","record_id":"<urn:uuid:ecd47db3-4a99-47f9-91b6-8bcca8c55b4b>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00657.warc.gz"} |
CHAPTER 5 DC TRANSIENT ANALYSIS 1 SUB TOPICS
Sign up to view full document!
CHAPTER 5 DC TRANSIENT ANALYSIS 1
SUB - TOPICS n n NATURAL RESPONSE OF RL CIRCUIT NATURAL RESPONSE OF RC CIRCUIT STEP RESPONSE OF RL CIRCUIT STEP RESPONSE OF RC CIRCUIT 2
OBJECTIVES n n To investigate the behavior of currents and voltages when energy is either released or acquired by inductors and capacitors when there is an abrupt change in dc current or voltage
source. To do an analysis of natural response and step response of RL and RC circuit. 3
FIRST – ORDER CIRCUITS n n n A circuit that contains only sources, resistor and inductor is called and RL circuit. A circuit that contains only sources, resistor and capacitor is called an RC
circuit. RL and RC circuits are called first – order circuits because their voltages and currents are describe by first order differential equations. 4
R R i – + Vs L An RL circuit vs – + i C An RC circuit 5
Review (conceptual) n Any first – order circuit can be reduced to a Thévenin (or Norton) equivalent connected to either a single equivalent inductor or capacitor. R Th - RN L VTh – + IN C In steady
state, an inductor behave like a short circuit. In steady state, a capacitor behaves like an open circuit. 6
n n The natural response of an RL and RC circuit is its behavior (i. e. , current and voltage ) when stored energy in the inductor or capacitor is released to the resistive part of the network
(containing no independent sources) The steps response of an RL and RC circuits is its behavior when a voltage or current source step is applied to the circuit, or immediately after a switch state is
changed. 7
NATURAL RESPONSE OF AN RL CIRCUIT n Consider the following circuit, for which the switch is closed for t<0, and then opened at t = 0: t=0 Is Ro i L + R V – n The dc voltage V, has been supplying the
RL circuit with constant current for a long time 8
Solving for the circuit n n For t ≤ 0, i(t) = Io For t ≥ 0, the circuit reduce to i Io Ro L + R v – Notation: Ø Ø is used to denote the time just prior to switching. is used to denote the time
immediately after switching. 9
Continue… n Applying KVL to the circuit: (1) (2) (3) (4) 10
Continue n From equation (4), let say; (5) n Integrate both sides of equation (5); (6) Where: v i(to) is the current corresponding to time to v i(t) ia the current corresponding to time t n 11
Continue n Therefore, (7) n hence, the current is 12
Continue n n From the Ohm’s law, the voltage across the resistor R is: And the power dissipated in the resistor is: 13
Continue n Energy absorb by the resistor is: 14
Time Constant, τ n n Time constant, τ determines the rate at which the current or voltage approaches zero. Time constant, (sec) 15
n The expressions for current, voltage, power and energy using time constant concept: 16
Switching time n ü ü ü For all transient cases, the following instants of switching times are considered. t = 0 - , this is the time of switching between -∞ to 0 or time before. t = 0+ , this is the
time of switching at the instant just after time t = 0 s (taken as initial value) t = ∞ , this is the time of switching between t = 0+ to ∞ (taken as final value for step response) 17
n The illustration of the different instance of switching times is: -∞ ∞ 18
Example n For the circuit below, find the expression of io(t) and Vo(t). The switch was closed for a long time, and at t = 0, the switch was opened. 2Ω i 0 t=0 20 A 0. 1Ω + i 2 H 10Ω L 40Ω V – 19
Solution : Step 1: Find τ for t > 0. Draw the equivalent circuit. The switch is opened. So; sec 20
Step 2: At t = 0 - , time from -∞ to 0 -, the switch was closed for a long time. 2Ω 20 A 0. 1Ω i (0 -) 10Ω 40Ω L The inductor behave like a short circuit as it being supplied for a long time by a dc
current source. Current 20 A thus flows through the short circuit until the switch is opened. Therefore; i. L(0 -) = 20 A 21
Step 3: At the instant when the switch is opened, the time t = 0+, 2Ω io(0+) + 20 A i. L(0+) 2 H 10Ω 40Ω vo(0+) – The current through the inductor remains the same (continuous). Thus, which is the
initial current. Only at this particular instant the value of the current through the inductor is the same. Since, there is no other supply in the circuit after the switch is opened, this is the
natural response case. 22
By using current division, the current in the 40Ω resistor is: So, Using Ohm’s Law, the Vo is: So, 23
NATURAL RESPONSE OF AN RC CIRCUIT n Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0: Vo + Ro C t=0 + v – R Notation: Ø 0 - is used to denote the
time just prior to switching Ø 0+ is used to denote the time immediately after switching. 24
Solving for the voltage (t ≥ 0) n n For t ≤ 0, v(t) = Vo For t > 0, the circuit reduces to i Vo + + Ro C v R – 25
Continue n Applying KCL to the RC circuit: (1) (2) (3) (4) (5) 26
Continue n From equation (5), let say: (6) n Integrate both sides of equation (6): (7) n Therefore: (8) 27
Continue n Hence, the voltage is: n Using Ohm’s law, the current is: 28
Continue n The power dissipated in the resistor is: n The energy absorb by the resistor is: 29
Continue n n The time constant for the RC circuit equal the product of the resistance and capacitance, Time constant, sec 30
n The expressions for voltage, current, power and energy using time constant concept: 31
n For the case of capacitor, two important observation can be made, 1) capacitor behaves like an open circuit when being supplied by dc source (From, ic = Cdv/dt, when v is constant, dv/dt = 0. When
current in circuit is zero, the circuit is open circuit. ) 2) in capacitor, the voltage is continuous / stays the same that is, Vc(0+) = Vc(0 -) 32
Example The switch has been in position a for a long time. At Time t = 0, the switch moves to b. Find the expressions for the vc(t), ic(t) and vo(t) and hence sketch them for t = 0 to t = 5τ. 5 kΩ a
b 18 kΩ t=0 90 V + 10 kΩ 0. 1μF 60 kΩ 12 kΩ + Vo – 33
Solution Step 1: Find t for t > 5τ that is when the switch was at a. Draw 18 kΩ the equivalent circuit. 0. 1μF 60 kΩ 12 kΩ 34
Step 2: At t = 0, the switch was at a. the capacitor behaves like An open circuit as it is being supplied by a constant source. 5 kΩ 90 V + 10 kΩ + Vc(0 -) – 35
Step 3: At t = 0+, the instant when the switch is at b. 18 kΩ + 60 V 0. 1μF – 60 kΩ 12 kΩ + Vo – The voltage across capacitor remains the same at this particular instant. vc(0+) = vc(0 -) = 60 V 36
Using voltage divider rule, Hence; 37
Summary No RL circuit 1 RC circuit 2 Capacitor behaves like an open circuit when being supplied by dc source for a long time Voltage across capacitor is continuous v. C(0+) = v. C(0 -) 3 Inductor
behaves like a short circuit when being supplied by dc source for a long time Inductor current is continuous i. L(0+) = i. L(0 -) 38
Step Response of RL Circuit n The switch is closed at time t = 0. i Vs + R + t=0 L v(t) – n After switch is closed, using KVL (1) 39
Continue n Rearrange the equation; (2) (3) (4) (5) 40
Continue n Therefore: (5) n Hence, the current is; n The voltage; 41
Example The switch is closed for a long time at t = 0, the switch opens. Find the expressions for i. L(t) and v. L(t). t=0 10 V + 2Ω 3Ω 1/4 H 42
Solution Step 1: Find τ for t > 0. The switch was opened. Draw the equivalent circuit. Short circuit the voltage source. 2Ω 3Ω 1/4 H 43
Continue Step 2: At t = 0 -, the switch was closed. Draw the equivalent circuit with 3Ω shorted and the inductor behaves like a short circuit. 10 V + 2Ω i. L(0 -) 44
Continue Step 3: At t = 0+, the instant switch was opened. The current in inductor is continuous. Step 4: At t =∞, that is after a long time the switch has been left opened. The inductor will once
again be behaving like a short circuit. 45
Continue 10 V + 2Ω 3Ω i. L(∞) Hence: 46
Continue n And the voltage is: 47
Step Response of RL Circuit n The switch is closed at time t = 0 + t=0 Is R C i n vc(t) – From the circuit; (1) 48
Continue n Division of Equation (1) by C gives; (2) n Same mathematical techniques with RL, the voltage is: n And the current is: 49
Example The switch has been in position a for a long time. At t = 0, the switch moves to b. Find Vc(t) for t > 0 and calculate its value at t = 1 s and t = 4 s. 3 kΩ 24 V + 5 kΩ a + Vc – 4 kΩ b t=0
0. 5 m. F + 30 V 50
Solution Step 1: To find τ for t > 0, the switch is at b and short circuit the voltage source. 4 kΩ 0. 5 m. F 51
Continue Step 2: The capacitor behaves like an open circuit as it is being supplied by a constant dc source. 3 kΩ 24 V + 5 kΩ + Vc (0 -) – From the circuit, 52
Continue Step 3: At t = 0+, the instant when the switch is just moves to b. Voltage across capacitor remains the same. Step 4: At t = ∞, the capacitor again behaves like an open circuit since it is
being supplied by a constant source. 4 kΩ + Vc(∞) – + 30 V 53
Continue Step 5: Hence, At t = 1 s, Vc(t) = 20. 9 V At t = 4 s, Vc(t) = 28 V 54 | {"url":"https://slidetodoc.com/chapter-5-dc-transient-analysis-1-sub-topics/","timestamp":"2024-11-04T17:33:19Z","content_type":"text/html","content_length":"202660","record_id":"<urn:uuid:bdefb81b-cd96-4f85-b636-93026712935e>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00690.warc.gz"} |
Underconstrained native arithmetic on scalars, instance 2
Audit of Silent Protocol Circuits
October 13th, 2023
In the two weeks from September 22nd to October 6th 2023, zkSecurity performed a security audit of Silent Protocol's Circom circuits.
A number of observations and findings have been reported to the Silent Protocol team. The findings are detailed in the latter section of this report.
Silent Protocol's circuits were found to be of high quality, accompanied with thorough documentation, specifications and tests. As of writing, all high and medium-severity issues we found were
patched by the Silent Protocol team, and zkSecurity confirms that these patches properly address our findings.
Note that security audits are a valuable tool for identifying and mitigating security risks, but they are not a guarantee of perfect security. Security is a continuous process, and organizations
should always be working to improve their security posture.
A consultant from zkSecurity spent two weeks auditing the Circom circuits for the Silent multi-asset shielded pool (SMASP) application. These circuits represent the privacy-preserving portion of the
overall SMASP application. They are used in a set of Solidity smart contracts that verify zero-knowledge proofs created from these circuits.
Smart contracts were reviewed by zkSecurity at an earlier date, and were only in scope insofar as they perform preparation and validation of public circuit inputs. The main focus of this audit were
the application's circuits written in Circom. These include circuits for the main SMASP protocol, circuits used by the compliance logic, and circuits representing future functionality not used in the
contracts at the time of review. An overview of audited circuits can be found below.
The audit also covers all auxiliary circuit templates and utilities maintained by Silent Protocol itself, including (but not limited to) templates for foreign-field arithmetic and ElGamal encryption.
Not covered by the audit are templates imported from circomlib, Circom's standard library; and functionality contained in snarkjs, such as the Groth16 prover.
An overview of the overall SMASP protocol can be found in zkSecurity's first report for Silent Protocol. Here, we focus on introducing recurring concepts that provide context for understanding our
findings below.
To understand the flow of assets through Silent's shielded pool, we focus on one example first: the deposit circuit.
Deposit. A public Ethereum account deposits assets into a shielded pool account.
• The deposited amount is added to a shielded balance, without revealing that balance.
• The sender's shielded account is anonymized by making indistinguishable dummy updates to 7 other accounts.
The privacy properties of this method are achieved by combining a zero-knowledge proof with ElGamal encryption of balances.
The ElGamal ciphertext representing a balance is defined as
$(C_1, C_2) = (rG, bG + rX)$
Here, $(C_1, C_2)$ is the ciphertext, $r$ is encryption randomness, $G$ is a public elliptic curve base point, $b$ is the balance, and $X$ is the public key of the account owner.
The deposit circuit performs two operations on ciphertexts:
• The sender account is updated by adding the deposited amount $a$ to the balance, exploiting additive homomorphism:
$(C_1, C_2) \leftarrow (C_1, C_2 + aG)$
• All 8 balance ciphertexts (on the sender's and 7 decoy accounts) are updated with new randomness $r'$:
$(C_{1i}, C_{2i}) \leftarrow (C_{1i} + r'G, C_{2i} + r'X_i)$
The re-randomization step is what makes the sender's balance update indistinguishable from updates to the 7 decoy accounts, which leave their balance in place. Note that we can perform both the
re-randomization and the balance update without knowing the accounts owner's private keys, and we also preserve their ability to decrypt their balances. Therefore, decoys can be real, active accounts
by other users, which is necessary for this scheme to provide actual anonymity for the sender. We have to ensure that the sender knows their own private key, by proving that we can rederive their
public key $xG = X$.
The ciphertexts before and after applying the deposit update are public inputs to the circuit, while the new randomness is a private input. The deposit amount is public as well, because the contract
needs to equate it to the amount received from the sender's Ethereum account. The circuit asserts correct execution of the ciphertext updates given above.
A second part of the circuit computes a senderHash, defined as
$senderHash = H(r''G + x X_\text{aml}).$
Here, $H()$ is the Poseidon hash function, $r''$ is randomness, $x$ is the sender's private key and $X_\text{aml}$ is the public key shared by the compliance committee. The senderHash and $r''G$ are
made available as public inputs. Note that the key $x X_\text{aml}$ is shared between the sender and compliance committee. Both the sender and the compliance committee can recompute the senderHash to
check whether this transaction belongs to the sender.
Other circuits to interact with the shielded pool make use of the same concepts outlined above for the deposit circuit, with slight variations:
DepositAndWithdraw. The withdraw logic shares its circuit with deposits, but uses a negative amount. The only difference, which is handled in-circuit, is that for withdrawals we also check that the
amount is smaller than the balance. In order to do this, the circuit takes the current balance as private input and verifies that it is encrypted correctly.
Transfer differs from deposits in that the sent amount is private, and that two accounts are anonymously updated: the sender and recipient. The circuit likewise encrypts transaction details in two
versions: One shared between sender and recipient, and one shared with the compliance committee. Both versions derive a shared secret using ECDH and use it as the key in MiMC-based encryption.
Register is the circuit that creates new shielded accounts. It generates new ElGamal ciphertexts of zero balances for four different assets.
FeeRegistration lets a user subscribe for zero-fee withdrawals. It uses the same technique as deposits to encrypt the end of the subscription period.
WithdrawFeeReduction is the method unlocked for a user after calling FeeRegistration. It is similar to withdraw, except that it also verifies the validity of the encrypted subscription period against
the current block number, which is a public input.
TransferToNonSilent, ClaimAndRegister and ClaimBatchPoints are not used by smart contracts in the audited version of the protocol. They use the same techniques as Transfer and Register to verify
encrypted balances and compute encrypted transaction details.
All of these circuits -- with the exception of Register, ClaimAndRegister and ClaimBatchPoints -- hide the sender in a size-8 anonymity set, using the same re-encryption technique as deposits.
Generally speaking, we found the implementation of encrypted and anonymized asset transfers to be solid, with consistent usage of the same patterns and well-documented core templates, like
BalanceVerify() and BalanceUpdate(). Only one major issue was found in this part of the code, which stems from a non-standard application of ElGamal encryption in the FeeRegistration logic, breaking
the sender's anonymity; see finding #00.
After joining the compliance committee, every member will create a secret that they share with all other members. Likewise, the other members send a secret share to them. The scheme uses a variant of
Shamir secret sharing that is suitable for threshold decryption, by avoiding the need for a single dealer; it also ensures that secret shares are verifiable against public commitments to the secret
generated by each member. See AHS20 for an overview of the scheme.
The SecretSharing template represents the scheme in circuit form. The template is used by a committee member when they share secrets with other members, by posting them in encrypted form to a
compliance smart contract. Along with encrypted shares, commitments to the underlying secret are also posted publicly; this enables members to verify their shares. The circuit's purpose is to prove
that encrypted shares and commitments are computed correctly and from the same polynomial.
In mathematical terms, the secret is an element of a finite field, $s \in \mathbb{F}_p$. The sharing entity constructs a polynomial $p(X)$ of degree $T-1$ which evaluates to the secret at 0:
$p(X) = \sum_{j=0}^{T-1} p_j X^j,\quad\text{where }s = p_0 = p(0).$
Polynomial coefficients are passed as private inputs, along with $S$ public evaluation points $x_i$, $i=0,\ldots,S-1$. The SecretSharing template evaluates the polynomial at each $x_i$ to obtain the
$i$th secret share, $ss_i = p(x_i)$. Note that $S \ge T$ (and uniqueness of the $x_i$) ensures that $T$ or more shares can be combined to reconstruct the secret. In practice, members will only
combine shares "in the exponent" so as to not reveal them, to collectively compute a curve point $s C$ for ElGamal decryption.
Besides evaluating the polynomial, the template also needs to compute commitments to the polynomial coefficients, which are defined as
$A_j = p_j G.$
The $A_j$ are broadcast by storing them on the smart contract. To validate the share they received, each member can check that
$ss_i G = p(x_i) G = \sum_{j=0}^{T-1} x_i^j A_j.$
To do scalar multiplications $p_j G$ efficiently in the circuit, the chosen curve is BabyJubJub, whose base field is the native circuit field. Note, however, that the polynomial lives in the scalar
field of that curve. This means we have to perform polynomial operations in non-native arithmetic modulo the curve order $p$; with coefficients, evaluation points and secret shares all represented as
Non-native arithmetic is a major source of complexity in the SecretSharing template. Indeed, out of the 6 high-to-medium findings reported, 5 are related to SecretSharing and non-native arithmetic
(see findings #01 through #05).
Below are listed the findings found during the engagement. High severity findings can be seen as so-called "priority 0" issues that need fixing (potentially urgently). Medium severity findings are
most often serious findings that have less impact (or are harder to exploit) than high-severity findings. Low severity findings are most often exploitable in contrived scenarios, if at all, but still
warrant reflection. Findings marked as informational are general comments that did not fit any of the other criteria.
feeRegistration.circom High
Description. The feeRegistration circuit is used when a user registers for zero-fee withdrawals. They pay a subscription fee and, in return, are able to use the complementary WithdrawFeeReduction
method for a period of time. The block number at which the subscription ends is a public input of the feeRegistration circuit, called validUntil.
Like other Silent Protocol methods, the interacting user (sender) is kept hidden in an anonymity set of 8 accounts. The sender's index within the array of size 8 is a private input. No-op account
updates are performed on all 8 accounts, such that the actual update to the sender's account -- setting a new validUntil value -- can't be distinguished from the outside.
However, contrary to the other Silent Protocol methods reviewed by zkSecurity, feeRegistration falls short of protecting the sender's anonymity.
For context, each user account stores ElGamal ciphertexts $(C_1, C_2)$ which represent $v :=$validUntil via
$(C_1, C_2) = (rG, vG + rX)$
where $r$ is encryption randomness and $X$ is the users public key. The circuit takes two versions of each of the 8 ciphertexts as public inputs: before and after applying the update.
The circuit takes new encryption randomness $r'$ as private input. The following operations are then performed to update ciphertexts:
• For the sender account, the new $v$ is entirely re-encrypted using $r'$: $(C_1, C_2) \leftarrow (r'G, vG + r'X)$
• All 8 accounts are rerandomized, using the same $r'$: $(C_1, C_2) \leftarrow (C_1 + r'G, C_2 + r'X)$
By contrast, in balance updates we don't encrypt the sender value from scratch, but just add an amount $aG$ to $C_2$ exploiting the additive homomorphism of ElGamal encryption.
The problem with the outlined method of resetting $v$ is that it's easy to distinguish the sender update from the others, just based on how the ciphertexts change. Note that for all other accounts,
$C_1$ changes by adding $r' G$. This is not the case for the sender account, which is reset to the new value $2r' G$. So, to find out the secret index in $0,\ldots,7$ which belongs to the sender,
just look at the difference $\Delta C_1$ for each account and note that one is different from all the others. Then, look up which public key belongs to that index, and you have successfully
deanonymized the sender.
Impact. Deanonymizing the sender breaks the core promise of the protocol, which is to hide users within an anonymity set.
Recommendation. We can mimic how balance updates work. The recommendation is to introduce the previous value $v_\text{old}$ as a private circuit input, so that the sender update can just add to $C_2$
• For the sender account, update: $C_2 \leftarrow C_2 + (v - v_\text{old})G$
• Rerandomize all 8 accounts: $(C_1, C_2) \leftarrow (C_1 + r'G, C_2 + r'X)$
For the $v$ update, the BalanceUpdate template can be reused:
component updateSender = BalanceUpdate();
updateSender.amount <== validUntil + SUBORDER - oldValidUntil;
Since validUntil is checked to be a smallish value by the smart contract logic that validates public inputs, we recommend to pass $v + p - v_\text{old}$ as input to BalanceUpdate, where $p$ is the
scalar field size. This expression will never overflow the native field since $p$ is much smaller than the native modulus. By adding $p$, we ensure that passing $v < v_\text{old}$ doesn't cause an
underflow, which would lead to a very long subscription period.
Client response. Silent Protocol fixed the issue, following our recommendation.
secretSharing.circom High
Description. To distribute secret shares publicly, they are encrypted using hashed ElGamal encryption:
$(C_1, c_2) = (rG, ss + \text{H}(rX))$
where $ss$ is the secret share, $X$ is the public key of its recipient and $r$ is encryption randomness.
In the protocol, $c_2$ and $ss$ are BabyJubJub scalars, not native field elements. The SecretSharing template represents them as bigints and computes them in non-native arithmetic.
To compute $\text{H}(rX)$, the protocol uses Poseidon over the native field. Since the native field is bigger than the scalar field, the hash output is reduced modulo the subgroup order to become a
valid scalar.
The template uses the following code:
secretHasher[i] = Poseidon(2);
secretHasher[i].inputs[0] <== rPubKeys[i].out[0];
secretHasher[i].inputs[1] <== rPubKeys[i].out[1];
var SUBORDER = 2736030358979909402780800718157159386076813972158567259200215660948447373041;
hashed[i] <-- secretHasher[i].out % SUBORDER;
sum === hashed[i];
Here, sum contains $c_2 - ss$, stored in a single field element. hashed[i] is supposed to represent $\text{H}(rX)$ reduced modulo SUBORDER. In the last line, the code asserts the equality $c_2 - ss =
\text{H}(rX)$ to show that $c_2$ is the correct encrypted secret share.
The problem is that the modulo reduction secretHasher[i].out % SUBORDER is computed outside the circuit, and stored in hashed[i] with the witness assignment operator <--. This means that hashed[i] is
not provably linked to the output of secretHasher in any way. From the circuit point of view, it is completely unconstrained and its value can be set arbitrarily by the prover.
At zkSecurity, we have called this common mistake a boomerang value: A value that leaves the circuit and gets reinserted later, without being linked to the original computation.
Impact. A malicious user can publish any values they want as encrypted secret shares and have them accepted by the smart contract. This severely breaks the compliance component of the protocol.
Client response. Silent Protocol patched the issue by redefining the protocol, such that hashed ElGamal encryption now happens in the native field. The secret share $ss$ is still a bigint scalar, but
it can naturally be embedded in the native field because the scalar field is smaller. The check $c_2 - ss = \text{H}(rX)$ is now a simple constraint between two native field elements and we no longer
need reduction modulo the subgroup order.
secretSharing.circom High
Description. Secret shares are computed by evaluating a polynomial $p(X)$ on a set of points $(x_i)$:
$ss_i = p(x_i) = \sum_{j=0}^{T-1} p_j x_i^j \qquad \text{for }i = 0,\ldots,S-1$
This evaluation happens in non-native arithmetic, so the polynomial coefficients $(p_j)$ and evaluation points $(x_i)$ are passed in as bigints: fixed-size arrays of signals which represent the limbs
of a bigint. Concretely, 4 limbs of 64 bits each are used, so a value like $x_i$ is represented as $(x_{i0}, x_{i1}, x_{i2}, x_{i3})$ where $x_{i} = \sum_{k=0}^3 x_{ik} 2^{64 k}$.
The polynomial evaluation circuit is built on top of non-native arithmetic gadgets like FpAdd and FpMultiply originally taken from circom-pairing, which themselves rely on bigint implementations
taken from circom-ecdsa.
These non-native gadgets constrain their outputs to be valid bigints with limbs in $[0, 2^{64})$. However, none of them constrain their inputs; so, input limbs to FpAdd and FpMultiply are assumed to
already be range-checked to 64 bits.
The choice to range-check outputs but not inputs is very natural. If each gadget would range-check its inputs, then range checks would be applied multiple times when you used different gadgets on the
same input values, wasting constraints. By following the convention of always constraining witnesses where they originate, we keep the code auditable and also ensure that the same constraints aren't
applied multiple times.
The issue is that secretShare.circom doesn't follow this convention: None of the input bigint limbs are range-checked. In the abbreviated code below, the signals c2, indexes, polynomial and
secretShare all represent bigints that are not range-checked.
template SecretSharing(THRESHOLD, SHARES, n, k, p) {
// Public Signals
// ...
signal input c2[SHARES][k];
signal input indexes[SHARES][k];
// Private Signals
signal input polynomial[THRESHOLD][k];
signal input secretShares[SHARES][k];
// ...
/// 4- evaluate the polynomial at index_i and
/// verify p(index_i) = secretShares_i
component evalPoly[SHARES];
// ...
for (var i = 0; i < SHARES; i++) {
evalPoly[i] = EvalPoly(THRESHOLD, n, k, p);
for (var j = 0; j < THRESHOLD; j++) {
for (var l= 0; l < k; l++) {
evalPoly[i].polynomial[j][l] <== polynomial[j][l];
for (var j=0; j < k; j++) {
evalPoly[i].eval[j] <== secretShares[i][j];
evalPoly[i].at[j] <== indexes[i][j];
// ...
Impact. To understand the impact at the hand of a simple example, let's look at the ModSum template which is used under the hood by FpAdd to add two limbs with a carry:
// addition mod 2**n with carry bit
template ModSum(n) {
assert(n <= 252);
signal input a;
signal input b;
signal output sum;
signal output carry;
component n2b = Num2Bits(n + 1);
n2b.in <== a + b;
carry <== n2b.out[n];
sum <== a + b - carry * (1 << n);
If both a and b are constrained to $n$ bits, this template indeed performs addition modulo $2^n$. However, if a and b can be any field element, we can make a + b overflow the native field. For
example, let $a = q - 1$ where $q$ is the native modulus, and $b = 1$. Then the sum is $0$, which is not the correct result mod $2^n$.
This demonstrates a general principle: Missing range checks often let us introduce overflow of $q$ in non-native arithmetic, where it's not supposed to happen. This can make room for the prover to
modify the output or input of non-native computations, by adding or subtracting multiples of $q$.
We explain a concrete attack which breaks the compliance protocol when exploiting this in combination with finding #03; see that finding for details. But the missing range checks alone can likely be
exploited to modify either commitments to polynomial coefficients or secret shares.
Recommendation. All limbs of the input signals c2, indexes and polynomial should be range-checked to 64 bits. An appropriate way to do this is to pass the checked limb as input to Num2Bits(64) from
circomlib/circuits/bitify.circom. Notably, secretShares does not need to be range-checked because it is forced to equal the output of the EvalPoly template, which is already range-checked.
Client response. Silent Protocol fixed the issue by adding range checks to indexes and polynomial as recommended. The signal c2 was refactored to be a native field element in response to finding #01,
so the need for a range check no longer applies to it.
secretSharing.circom Medium
Description. One of the objectives of the SecretSharing template is to prove the correctness of commitments to the polynomial which is used for secret sharing. Commitments are of the form $p_i G$
where $p_i$ is a polynomial coefficient and $G$ a constant base point on the BabyJubJub curve.
Polynomial coefficients $p_i$ are passed in as bigints (arrays of limbs). To perform scalar multiplication, the code collapses the limbs to obtain a single field element which is passed to the
MulBase template as the scalar.
/// 1 - compute polynomial coefficients p_i
/// p = p_0 + p_1 x + p_2 x^2 + ... + p_n-1 x^n-1
signal poly[THRESHOLD];
for (var i=0; i < THRESHOLD; i ++){
var sum = 0;
for (var j=0; j<k; j++){
sum += polynomial[i][j] * ((2 ** n) ** j);
poly[i] <== sum;
/// 2- check broadcast_i = p_i * G
component mulScalars[THRESHOLD];
for(var i=0; i<THRESHOLD; i++) {
mulScalars[i] = MulBase();
mulScalars[i].e <== poly[i];\
// ...
There is an interplay between arithmetic in two fields here: the native field and the scalar field. Let's define
• $q :=$ the native modulus
• $p :=$ the scalar modulus (i.e. $p$ is the order of the subgroup generated by $G$).
Note how poly[i] is computed: The limbs polynomial[i][j] are multiplied with their corresponding power of two and summed, in native field arithmetic. So we end up constraining poly[i] to be
$p'_i = \sum_{j=0}^k p_{ij} 2^{nj} \mod{q}$
But the value we actually would like to put in poly[i] is
$p_i = \sum_{j=0}^k p_{ij} 2^{nj} \mod{p}.$
This is the scalar that $p_{ij}$ are supposed to represent, and that is implicitly used elsewhere in the code where the polynomial is evaluated in scalar arithmetic. We also don't care about
multiples of $p$ added to $p_i$, since they are eliminated by scalar multiplication.
The problem is that the prover can easily make $p_i$ differ from $p_i'$, if the limbs $p_{ij}$ can represent values larger than $q$. Just write
$p_i = p'_i + c_i q$
and choose limbs $p_{ij}$ that represent $p_i$. $c_i$ is an arbitrary value with the requirement that $p'_i + c_i q < (c_i + 1)q$ is within the max value that the limb sum can represent. Note that
$p_{ij}$ are entirely controlled by the secret sharing prover and can be chosen to facilitate an attack.
In the audited version of the code, $p_{ij}$ are not range-checked and take values in $[0,q)$, see finding #02. Therefore, a limb sum can represent values up to $q 2^{(k-1)n}$ (plus a bit more),
which means that $c_i$ can be any number in $[0,2^{(k-1)n})$. Plugging in $k=4$ and $n = 64$, we see that $c_i$ can take one out of $2^{192}$ values.
Impact. When picking $c_i eq 0$, the scalar multiplication creates a modified commitment $A'_i = (p_i - c_i q)G = A_i - c_i qG$. The recipient of the share is supposed to use this commitment to
validate their share $p(x_i)$, by checking that
$p(x_j)G = \sum_{i=0}^{T-1} x_j^i A_i.$
This fails when one of the commitments is modified, breaking the protocol. Since there are $2^{192}$ ways for the prover to modify each commitment, there is no way for the share recipient to recover
the true commitments $A_i = p_i G$.
Assuming that finding #02 is fixed and each limb is constrained to the range $[0, 2^n)$, a limb sum can only represent values up to $2^{kn} = 2^{256}$. This means $c_i$ in our analysis can at most be
$\lfloor 2^{256} / q \rfloor = 5$. Since commitments can be modified independently and the polynomial degree is currently set to $5 - 1$, there are $6^5 = 7776$ ways for the prover to change the
broadcasted commitments. This should be recoverable by a custom trial and error algorithm -- assuming that the secret share $p(x_j)$ is received in untampered form; see finding #04 for why this is
not the case.
Recommendation. There is no need to collapse bigint limbs to a single field element before using them for scalar multiplication. The fact that this is done seems to be just an artifact of the MulBase
API which takes a field element. Under the hood, MulBase uses circomlib's BabyPk which unpacks the field element to an array of bits. The bits are passed to EscalarMulFix which contains the core
scalar multiplication circuit.
We recommend to unpack the bigint limbs polynomial[i][j] into bits directly. This can serve the double purpose of constraining those limbs to their correct bit size. The resulting bits can be passed
to EscalarMulFix to perform scalar multiplication.
Client response. The issue was fixed by Silent Protocol, following our recommendation of using EscalarMulFix directly.
secretSharing.circom Medium
Description. After the SecretSharing template evaluated a polynomial on several points to compute secret shares $ss_i$, the goal is to prove correct encryption of those shares by showing that
$c_{2i} - ss_i = \text{H}(rX).$
Here, $c_{2i}$ is the encrypted value which is a public input. The code represents $c_{2i}$ and $ss_i$ as bigints and computes the subtraction $c_{2i} - ss_i$ in foreign-field arithmetic, using the
BigSubModP template. Then, the bigint version of $c_{2i} - ss_i$ is collapsed into a single field element using native arithmetic, which finally is compared with the hash value on the right hand
/// 5- verify secret shares are encrypted correctly
subModP[i] = BigSubModP(n,k);
for (var j=0; j <k; j++){
subModP[i].a[j] <== c2[i][j];
subModP[i].b[j] <== secretShares[i][j];
subModP[i].p[j] <== p[j];
var sum = 0;
for (var j=0; j<k; j++){
sum += subModP[i].out[j] * ((2 ** n) ** j);
// ...
sum === hashed[i];
The way the sum is only constrained modulo the native modulus $q$ creates a problem analogous to finding #03. The bigint stored in subModP can hold values larger than $q$, which allows the prover to
provide multiple modified versions of the public output c2[i][]. Namely, we can use
$c_{2i}' = c_{2i} + d_i q$
for a few different values $d_i$, which in turn will cause the share recipient to obtain the modified decrypted secret share
$ss_i' = ss_i - d_i q \bmod{p}.$
The value $d_i$ has to be small enough such that subModP[i]$\in [0, 2^{256})$ can still hold $c_{2i} - ss_i + d_i q$. Typically, $d_i$ can take 6 different values $0,\ldots,5$. A crucial point here
is that while BigSubModP forces its outputs to be valid bigints with 64-bit limbs, it doesn't require the inputs or output to be smaller than $p$.
Impact. Allowing 6 different values for the main circuit output $c_{2i}$, exactly one of which lies in $[0, q)$, might seem non-ideal, but not catastrophic. Can't the recipient of the secret share
simply reduce their encrypted share modulo $q$ to obtain the correct value? The reason we count this as a medium-severity issue is due to its interplay with two other findings:
• Finding #05 shows that there is ambiguity in the value of the secret share $ss_i$: It can be modified by a multiple of $p$. The multiple of $p$ can also be added to $c_{2i}$ without affecting the
difference $c_{2i} - ss_i$. With this extra degree of freedom in $c_{2i}$, there is no longer a unique reduction that a share recipient can perform to obtain a decrypted share in the correct
range. A trial-and-error procedure using the broadcast polynomial commitments to validate secret shares might be possible.
• However, as finding #03 demonstrates, polynomial commitments can themselves be modified by the prover, which further complicates any recovery of the correct secret shares.
In combination, findings #03, #04 and #05 can likely be used to put the compliance protocol in a temporarily broken or hard-to-recover-from state.
Recommendation. There are several general recommendations to make. First, instead of collapsing the bigint $c_{2i} - ss_i$ to a native field element before comparing with another value, consider if
it is possible to just compare the bigint limbs directly. This requires both sides to be represented in bigint form.
Second, in the case of a public input, ambiguity can be mitigated by making the verifier (here: the smart contract) validate the input; this would mean checking that each $c_{2i}$ is in $[0, p)$.
Finally, a bigint comparison in the circuit can be used to ensure a value is in its canonical range. Constraining subModP[i] to $[0,p)$ would have prevented the issue.
Client response. In light of both this finding and finding #01, Silent Protocol redefined secret share encryption to happen in the native field. $c_{2i}$ is now a native field element and this
finding, as well as the recommendations, no longer apply.
polynomial.circom Medium
Description. The EvalPoly template computes the evaluation of a polynomial in non-native arithmetic:
$s = p(x) = \sum_{j=0}^{T-1} p_j x^j$
Multiplications use the FpMultiply gadget taken from circom-pairing. This gadget allows the reduction $b = a\bmod{p}$ by witnessing bigints $X$ and $b$, and proving that $a = pX + b$. There is no
check that $b < p$ which would fix the result $b$ to a unique value; let's call a $b$ canonical if $b \in [0, p)$. Since $X$ and $b$ are provided as a witness, the prover can easily add multiples of
$p$ to $b$ by subtracting from $X$.
There are good reasons to not do a canonical check after every multiplication. If you need many multiplications, nothing is gained by ensuring every intermediate result is canonical if you only care
that the final result is.
The EvalPoly template does not enforce that its final output is canonical. Therefore, the prover can modify the result to $s' = s + cp$, where $c$ can be as large as allowed by $s + cp < 2^{256}$. At
most, there are $\lfloor 2^{256} / p \rfloor + 1 = 43$ possible values for $c$.
The output of EvalPoly is used as the secret share by SecretSharing, so the prover can pick between 43 possible values per secret share.
Impact. There are two ways we found this ambiguity to impact the protocol at different iterations of the code base. In the audited version of the code, the secret share was encrypted in non-native
arithmetic. As explained in finding #04, the freedom to add multiples of $p$ helps the prover to modify the final encrypted secret share $c_2$ in ways that can't trivially be recovered.
A similar impact was found in a later version of the code after Silent Protocol fixed finding #01 and #04 by performing encryption of secret shares in the native field. To collapse the EvalPoly
output to a native field value, a sum in native arithmetic is performed:
// `result[N][i]` holds the polynomial evaluation
var sum = 0;
for (var j=0; j<k; j++){
sum += result[N][j] * ((2 ** n) ** j);
eval <== sum;
In this case, after summing in native arithmetic, the 43 possible values for eval$= s + cp \bmod{q}$ are no longer distinguishable by size or by reducing modulo $p$; some of them will end up being
smaller than the unmodified value $s$. So, yet again, the prover is able to publish incorrect secret shares.
Recommendation. We recommend to add a dedicated canonical check at the end of EvalPoly which constrains the eval output to a unique value by enforcing that result[N][j] is smaller than $p$.
Client response. Silent Protocol fixed the issue by following our recommendation.
depositAndWithdraw.circom Low
Description. The depositAndWithdraw circuit handles either deposits or withdraws. In the deposit case, it takes a positive amount which is added to the balance. In the withdraw case, amount contains
the negative amount, i.e. a value of the form $p - |a|$ where $p$ is the scalar field size, and the absolute value $|a|$ is stored in another variable called isWithdraw.
To check that the amount is within the range $[0, 2^{30})$, the circuit attempts to encode the following logic:
• In the deposit case, assert that amount is smaller than $2^{30}$.
• In the withdraw case, assert that isWithdraw is smaller than $2^{30}$.
The code looks like this:
/// 9- Checks that amount/withdraw amount is nonzero and less than 2^30
component isAmountZero = IsZero();
isAmountZero.in <== amount;
component isAmountLessThan = LessThan(252);
isAmountLessThan.in[0] <== amount;
isAmountLessThan.in[1] <== 1 << 30;
component isWithdrawAmountZero = IsZero();
isWithdrawAmountZero.in <== isWithdraw;
component isWithdrawAmountLessThan = LessThan(30);
isWithdrawAmountLessThan.in[0] <== isWithdraw;
isWithdrawAmountLessThan.in[1] <== 1 << 30;
signal depositCheck <-- !isWithdrawCase.out && !isAmountZero.out && isAmountLessThan.out;
signal withdrawCheck <-- isWithdrawCase.out && !isWithdrawAmountZero.out && isWithdrawAmountLessThan.out;
depositCheck + withdrawCheck === 1;
There are several issues, the most obvious of which is that depositCheck and withdrawCheck are completeley unconstrained: They only receive their values with the witness assignment <-- and the only
constraints put on them is that their sum is 1. So, nothing in the circuit relates those values to the deposit or withdraw cases, and the prover is therefore not forced to pass amount and isWithdraw
that are within the $2^{30}$ range.
A second issue is the circular logic used in the LessThan checks. The LessThan(30) gadget used for isWithdraw is only valid with inputs that are at most 30 bits. So, if the isWithdraw input is larger
than 30 bits, then the output is not necessarily correct, so the entire check is superfluous. The same is true for an amount that is larger than 252 bits.
Impact. The reason why this is finding was given a low severity is that range checks on amount and isWithdraw are present in the contract logic which validates the public inputs to this circuit;
therefore, the problem is not actually exploitable.
Recommendation. We don't recommend to leave broken circuit logic in place even if it is without effect. Deployed code is often used as example for other projects, copied from one place to another,
etc. Therefore, the recommendation is to either fix the range-checking logic or remove it.
Here is how the conditional check could be written instead:
// the number to be checked is either `amount` or `isWithdraw`, depending on the case
signal checkedAmount <== isWithdrawCase.out * (isWithdraw - amount) + amount;
// non-zero check
signal checkedAmountInv <-- checkedAmount != 0 ? 1/checkedAmount : 0;
checkedAmount * checkedAmountInv === 1;
// 30-bit range check
component rangeCheck = Num2Bits(30);
rangeCheck.in <== checkedAmount;
Client response. Silent Protocol has addressed the finding by removing the in-circuit range-check.
crypto/src/*.ts Low
Description. The insecure randomness source of Math.random() is used to select the index of the anonymity set we are using for all entrypoints to the protocol. Under certain circumstances, an
adversary might learn a bias for the random value and thus the anonymity set does not provide perfect indistiguishibility between the choices in the set.
Example of one such infraction:
const senderIndex = Math.floor(Math.random() * RING_SIZE);
Recommendation. We recommend using cryptographically secure randomness sources for the sender index selection. In NodeJS 19 or higher as well as in modern browsers, you can use crypto.getRandomValues
(new Uint8Array(1))[0], which returns a single random byte. Since we always want to pick between indices 0 through 7, we can look at the first 3 bits to get our securely random index value:
const senderIndex =
crypto.getRandomValues(new Uint8Array(1))[0] & (RING_SIZE - 1);
As long as RING_SIZE remains a power of 2, this code will still work when the anonymity set size is changed.
Client response. Silent Protocol pointed out that the finding is out of scope of the audit; it was going to be addressed anyway.
bigint_func.circom Informational
Description. The log_ceil() helper function is used by FpMultiply to determine the maximum number of bits that certain expressions can have. Expressions arise in the bigint multiplication code and
are of the form $S = \sum_{i=1}^k a_i$ where $a_i$ are known to be $n$-bit integers. To determine the max number of bits of $S$, we want to estimate
$S = \sum_{i=1}^k a_i \le k \cdot (2^n - 1) < 2^m$
Finding the smallest $m$ on the right hand side will give us a precise upper bound on the number of bits that $S$ needs. Assuming we knew that $k \le 2^{K}$, we could infer
$k \cdot (2^n - 1) < 2^{n + K}$
so we find $m = n + K$. Choosing the minimal $K$ will make $m$ minimal as well. Note here that the inequality on $k$ is not strict -- it's fine if $k = 2^K$, thanks to the $-1$ in the other factor.
Taking logs, we get the condition
$log_2(k) \le K$
and so the minimal $K$ can be written as $K = \lceil log_2(k) \rceil$. The maximum number of bits of $S$ is precisely $n + \lceil log_2(k) \rceil$. This motivates the use of a log_ceil() function for
this problem.
The log_ceil() function at the time of the audit was implementing something slightly different, namely, it was computing the number of bits of $k$.
Here are some examples:
Input $k$ Input (Binary) Expected output log_ceil(k) Actual output = number of bits
The difference is for $k$ a power of two, when the actual implementation returns something too large by 1.
Impact. In the current usages of the function, its incorrect result is not a security issue. Because of the specific use as an upper bound, returning something too large is just more conservative
than necessary. If used in other contexts in the future, it could lead to more serious issues.
Recommendations. We recommend fixing log_ceil(), first because off-by-one issues could become a problem when used in other circumstances, and second because using more bits than necessary implies
using more constraints than necessary.
Furthermore, at the time of audit log_ceil() defaults to 254 for any input larger than 254 bits. Again this is fine because inputs that large are not used in practice. We recommend adding an
assertion preventing this case from being exercised. | {"url":"https://zksecurity.xyz/reports/silent2/","timestamp":"2024-11-04T17:41:06Z","content_type":"text/html","content_length":"432082","record_id":"<urn:uuid:c15091bf-afef-4bbe-928c-005b559cd063>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00853.warc.gz"} |
Rcpp, Camarón de la Isla and the Beauty of Maths | R-bloggersRcpp, Camarón de la Isla and the Beauty of Maths
Rcpp, Camarón de la Isla and the Beauty of Maths
[This article was first published on
R – Fronkonstin
, and kindly contributed to
]. (You can report issue about the content on this page
) Want to share your content on R-bloggers?
if you have a blog, or
if you don't.
Desde que te estoy queriendo
yo no sé lo que me pasa
cualquier vereda que tomo
siempre me lleva a tu casa
(Y mira que mira y mira, Camarón de la Isla)
The verses that head this post are taken from a song of Camarón de la Isla and illustrate very well what is a strange attractor in the real life. For non-Spanish speakers a translation is since I’m
loving you, I don’t know what happens to me: any path I take, always ends at your house. If you don’t know who is Camarón de la Isla, hear his immense and immortal music.
I will not try to give here a formal definition of a strange attractor. Instead of doing it, I will try to describe them with my own words. A strange attractor can be defined with a system of
equations (I don’t know if all strage attractors can be defined like this). These equations determine the trajectory of some initial point along a number of steps. The location of the point at step
i, depends on the location of it at step i-1 so the trajectory is calculated sequentially. These are the equations that define the attractor of this experiment:
$x_{n+1}= a_{1}+a_{2}x_{n}+a_{3}y_{n}+a_{4} |x_{n}|^{a_5}+a_{6} |y_{n}|^{a_7}\\ y_{n+1}= a_{8}+a_{9}x_{n}+a_{10}y_{n}+a_{11} |x_{n}|^{a_{12}}+a_{13} |y_{n}|^{a_{14}}$
As you can see there are two equations, describing the location of each coordinate of the point (therefore it is located in a two dimensional space). These equations are impossible to resolve. In
other words, you cannot know where will be the point after some iterations directly from its initial location. The adjective attractor comes from the fact of the trajectory of the point tends to be
the same independently of its initial location.
Here you have more examples: folds, waterfalls, sand, smoke … images are really appealing:
The code of this experiment is here. You will find there a definition of parameters that produce a nice example image. Some comments:
• Each point depends on the previous one, so iteration is mandatory; since each plot involves 10 million points, a very good option to do it efficiently is to use Rcpp, which allows you to iterate
directly in C++.
• Some points are quite isolated and far from the crowd of points. This is why I locate some breakpoints with quantile to remove tails. If not, the plot may be reduced to a big point.
• The key to obtain a nice plot if to find out a good set of parameters (a[1] to a[14]). I have my own method, wich involves the following steps: generate a random value for each between -4 and 4,
simulate a mini attractor of only 2000 points and keep it if it doesn’t diverge (i.e. points don’t go to infinite), if x and y are not correlated at all and its kurtosis is bigger than a certain
thresold. If the mini attractor overcome these filters, I keep its parameters and generate the big version with 10 million points.
• I would have publish this method together with the code but I didn’t. Why? Because this may bring yourself to develop your own since mine one is not ideal. If you are interested in mine, let me
know and I will give you more details. If you develop a good method by yourself and don’t mind to share it with me, let me know as well, please.
This post is inspired in this beautiful book from Julien Clinton Sprott. I would love to see your images. | {"url":"https://www.r-bloggers.com/2019/01/rcpp-camaron-de-la-isla-and-the-beauty-of-maths/","timestamp":"2024-11-03T09:35:53Z","content_type":"text/html","content_length":"115352","record_id":"<urn:uuid:47b3da8b-bf2d-4fdf-a0d7-00042977992e>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00337.warc.gz"} |
inear regression
Linear regression in MATLAB [duplicate] Ask Question Asked 5 years, 2 months ago. Active 5 years, 2 months ago. Viewed 11k times 0. 1. This question already has an answer here: How do I determine the
coefficients for a linear regression line in MATLAB? [closed] (1 answer) Closed 5 years
Multiple Linear Regression | MATLAB Author Regression Code:- clc clear all close all x1=[3 5 6 8 12 14]; x2=[16 10 7 4 3 2]; y=[90 72 54 42 30 12]; n=length(x1); a=[ones(n,1) x1' x
Linear regression fits a linear data model in the model coefficients. The most frequent form of linear regression is a least square fit which can match polynomials and lines among other linear
models. L = loss (Mdl,X,Y) returns the mean squared error (MSE) for the linear regression model Mdl using predictor data in X and corresponding responses in Y. L contains an MSE for each
regularization strength in Mdl. L = loss (Mdl,Tbl,ResponseVarName) returns the MSE for the predictor data in Tbl and the true responses in Tbl.ResponseVarName. We develop the following Matlab code
(note that Matlab has its own built-in functions to make linear regression easier for all of us, but we'd like to show a step-by-step way to do it, to understand the inner concepts): Multivariate
Linear Regression Model The multivariate linear regression model expresses a d -dimensional continuous response vector as a linear combination of predictor terms plus a vector of error terms with a
multivariate normal distribution. Let denote the response vector for observation i, i = 1,, n. Learn how to take a model, linearize it and perform linear regression to fit "experimental data" in
MATLAB. In this example, we use the Antoine equation to m Examples of getting prediction interval and confidence interval for linear regression in matlab My Machine Learning playlist https://
www.youtube.com/playlist?list=PLkNswIK0bUDfw08PZohbaFvQeIQ1-QPdAThis video steps you through how to implement Linear reg You can check these 2 videos , if you want to understand the working of
pinv:Why pinv(a) ?https://youtu.be/DzAbRxZ_YOYMultiple Linear Regression from Scratch [r,m,b] = regression(t,y) calculates the linear regression between each element of the network response and the
corresponding target.
The variable we want to predict is called the dependent variable (or sometimes, the outcome, target or criterion variable). - yihanzhao/Multiple-Regression-matlab MATLAB에서는 mldivide 연산자를 사용
하여 를 구할 수 있습니다. 즉, B = X\Y 와 같은 형식을 사용합니다. 데이터셋 accidents 에서 교통사고 건수 데이터를 y 에 불러오고 주별 인구 데이터를 x 에 불러옵니다.
2.1 Enkel linjär regression . 2.2 Multipel linjär regression . Vi kan också skatta regressionsparametrarna med hjälp av MATLAB och MATLAB Central contributions by Bhartendu.
Jul 29, 2020 In this article, we will discuss a simple code to plot a Linear Regression (LR) curve. The code is written in MATLAB and can be downloaded
Active 9 years, 1 month ago. Viewed 6k times 4. 1.
Multiple Linear Regression Linear regression with multiple predictor variables; Les navigateurs web ne supportent pas les commandes MATLAB. Fermer.
Erik G. Larsson and Yngve Selén, "Linear Regression With a Sparse Parameter Vector BPM, BOSS : Here I have some MATLAB functions available for the two In this exercise, you are given a Matlab/Octave
script for doing a linear regression to a data set, using two methods: (1) ordinary least squares, Stockholm University MATLAB Student Ambassador at MathWorks Various techniques were applied, namely
linear regression, multi-regression, random av M Karlsson · 2015 — This thesis presents an attempt to use linear regression to predict the i Matlab. Detta verkade vid vissa tillfällen ändå ge bra
resultat, så det fick följa med till av R Hu · 2014 · Citerat av 3 — a linear regression model and a time series model, and then taking the mean of their such as SPSS, Excel and Matlab will be used.
set terminal svg fname "Helvetica" fsize 25 set output "LinearRegression.svg" set key left set xtics axis 4.0,0.5 set ytics (identical code also works in Matlab.) Predictive Modeling - Time-Series
Regression, Linear Regression Models. Learn how MATLAB can help to predict future outcomes by creating predictive av dokumentet Introduction to the Matlab language Examples and exercises
Matlab-script och Matlab-funktioner Diagram Introduktion till Linjär regression 4.
I have attached a picture of the graph and the linear fitting that I obtained. Any help is much appreciated! Thank you in advance!
Emma stenström vingåker
Adriaan. 15.9k 7 7 gold badges 35 35 silver badges 67 67 bronze badges. Here, we implement regularized linear regression to predict the amount of water flowing out of a dam using the change of water
level in a reservoir. In the next half, we go through some diagnostics of debugging learning algorithms and examine the effects of bias v.s.
1. I'm trying to work out the most efficient Also I've implemented gradient descent to solve a multivariate linear regression problem in Matlab too and the link is in the attachments, it's very
similar to univariate, so you can go through it if you want, this is actually my first article on this website, if I get good feedback, I may post articles about the multivariate code or other A.I.
stuff. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. linear regression.
Fiske washington island
2.8 Multilinear PLS Regression (N-PLS). Chapter 5. Multilinear PLS Analysis with Application to 3D QSAR 5.6 Matlab Code for Regression Coefficients .
Regression models describe the relationship between a response (output) variable, and one or more predictor (input) variables.
Create an -by- design matrix X. Add a column of ones to include a constant term in the regression. X = [ones (size (x)),x]; Fit the multivariate regression model. where and , with between-region
concurrent correlation. There are 18 regression coefficients to estimate: nine intercept terms, and nine slope terms.
Nonlinear Regression describes general nonlinear models. A special class of nonlinear models, called … In this short video I am showing you how to implement the Linear Regression (OLS) in MATLAB.If
you have any questions please feel free to comment below LinearModel is a fitted linear regression model object. A regression model describes the relationship between a response and predictors. The
linearity in a linear regression model refers to the linearity of the predictor coefficients. Use the properties of a LinearModel object to … Regression is the procedure for fitting models to data.
Linear regression assumes the relationship between the independent variable as well as the dependent variable. MATLAB is a robust computing environment and programming language widely used in finance
and statistics.
intervallet ”för hand” eller använd regress i MATLAB. 2.1 Enkel linjär regression . 2.2 Multipel linjär regression . | {"url":"https://investerarpengarzglcse.netlify.app/15473/45980","timestamp":"2024-11-06T10:49:06Z","content_type":"text/html","content_length":"12788","record_id":"<urn:uuid:c7a2a97e-fbc7-48c4-ae5d-903862a7a429>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00552.warc.gz"} |
Calculating matters less than setting up math problems
That’s what math feels like in school: rules and procedures that you follow to solve problems. It’s too bad, because that’s not what math means to adults in the real world.
Math problems start with questions
When carpenters, engineers, lawyers, architects, and accountants use math, they start with a question. For example:
• The carpenter might ask: how far from the left wall is the left edge of the new window?
• The engineer might ask: should I make a particular bike part out of steel or aluminum?
• The accountant might ask: what kind of crisis could cause this organization to go bankrupt?
Next, they turn their real-world problem into a math problem. For example the carpenter knows that the centerline of the window should be placed ⅔ of the way from the left wall toward the right wall.
The wall is nine feet wide. The new window is two feet wide.
Now he turns his real-world question into a math problem:
Distance from left wall = (⅔ X 9) – (2-1)
The answer is 5 feet.
The engineer thinks about her problem some more. Steel is stronger. Aluminum is lighter, but also more expensive. Her math equations are more complicated than the carpenter’s.
The accountant knows that the company has $1 million of cash on hand and is spending $500,000 per month. What kind of crisis could cause what kind of decrease in revenue? If revenue decreased by a
particular amount, how much would expenses decrease? To answer his question, he’ll develop some equations that are more complex than the carpenter’s, but simpler than the engineer’s.
With the problem set up, each of them does some calculations to get an answer.
Does the answer we calculated really answer our question?
Then, after they get their answer, they’ll think about it for a little while. Did the answer they calculated really answer their question? For the carpenter, it’s a quick check. The engineer and
accountant have to work a little harder. Does their equation capture what’s important? Is there a different and better way to think about the question, which would lead to a different calculation and
In all three cases, we can summarize the process in four steps:
1. Pose a real-world question
2. Translate the real-world question into a math problem
3. Calculate the answer
4. Consider: did we answer the question? Is there a better way to answer it?
In all cases, step three, calculating, is the easiest part. The carpenter can do the problem in his head. The engineer will probably use a computer. The accountant might use a spreadsheet.
Setting up math problems is the most interesting and challenging part
Steps two and four are the more challenging and interesting parts of the process for everyone. The second step involved translating a real-world question into a math problem. The fourth step involved
thinking about whether the answer we got is a good answer to our original question. If not, we try again.
If you want your child to grow up engaged with mathematics — and you do! — introduce them to parts one, two of this four-part process at an early age. Instead of giving them worksheets, draw them
into real-world questions and talk about them.
Ask your first grader questions like: “We need to bring enough dog food for our weekend away. Could you do that?” Ask your third grader question like: “Do you think we have enough eggs to last until
the weekend? “
Inspire your child to ask their own math-related questions
As your child gets older, look for ways to engage them with mathematical thinking on topics that interest them. Let’s say you have a soccer player. You could ask them: how far ahead of a teammate do
you kick the ball? How do they know where to aim?
Through conversation, see if you can inspire them to translate these kinds of real-world questions into simple math problems. Then, when they get an answer, you can ask: is this a good answer? Is
there a way to make the answer better? They may struggle a bit at first, and you may need to give them some help along the way.
With this kind of conversation, you’re laying the groundwork for your child to understand and engage with math like adults do in the real world.
Calculating matters less than setting up math problems | {"url":"https://raisereadykids.com/calculating-matters-less-than-setting-up-math-problems/","timestamp":"2024-11-04T07:24:13Z","content_type":"text/html","content_length":"51164","record_id":"<urn:uuid:e9ecc64d-02cd-41ba-86dd-a54b4496cd3c>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00838.warc.gz"} |
Celebrating the Wonders of Pi Day: A Journey into Mathematics and Beyond
Mathematics is a universal language, understood and appreciated in every corner of our ever-inquisitive world. March 14 is recognized as Pi (π) Day by mathematicians and enthusiasts worldwide to
acknowledge the significance of Pi, which has been an unwavering ally in humanity's scholarly and scientific endeavors.
The First Pi Day
Pi pops up in everything from rocket science-level math to the stuff in elementary school, so it’s gained a cult following. On March 14 (or 3/14) in 1988, a physicist at San Francisco's Exploratorium
held what is thought to be the first official Pi Day celebration, which smartly included the consumption of fruit pies. Math teachers quickly realized the potential benefits of teaching students
about pi while they ate pie, and it all caught on so much that in 2009, the U.S. Congress officially declared March 14 National Pi Day.
The significance of Pi goes beyond the classroom, where we are first introduced to its infinite, non-repeating decimal form. Pi, approximately 3.14159, is the ratio of a circle's circumference to its
diameter, a number that has teased and perplexed mathematicians throughout history with its non-repeating, non-terminating nature. It appears across various facets of science and engineering—from the
orbits of planets to the waves of the sea, the structure of DNA, and the design of powerful telescopes. Pi is the foundation of trigonometry, one of the fundamental mathematical concepts in Cadence's
computational software.
Eistein and Heisenberg
It is a component of foundational principles like Heisenberg's uncertainty principle, which articulates the limits of our knowledge about the state of the universe. Furthermore, as beautifully
explained by Albert Einstein through fluid dynamics and chaos theory, the meandering paths of rivers exhibit a meandering ratio that astonishingly aligns with Pi. This ratio reveals how the river's
actual length, compared to the straight-line distance from its source to the mouth, remarkably approaches Pi, underscoring the concept that even the seemingly random curvatures of rivers have a
mathematical explanation rooted in the ubiquitous Pi.
Unleashing Innovation with Pi
Pi Day is more than just a date (3/14) that coincides with the first three digits of this infinite number; it is a celebration of the limitless explorations and innovations that Pi represents in our
quest to understand our universe. On Pi Day, we commend the virtues of a number so fundamental to our understanding of the world that it transcends figures and equations—it becomes a metaphor for the
infinite pursuit of knowledge. Pi Day is also a testament to the human spirit and its relentless dedication to advancement.
Just as Pi continues infinitely without pattern or end, so does humanity's pursuit of innovation and growth. It's an invitation to young minds to look beyond the digits, to dream of the unknown, and
to step forward into a future of discovery bravely. On Pi Day, there is a palpable sense of excitement and wonder as people come together to celebrate this universal constant. Pi Day has evolved into
a cultural phenomenon showcasing mathematics’ versatility and beauty, from mathematical challenges and competitions to creative expressions of art and music.
But beyond the festivities lies a more profound message—that the pursuit of knowledge knows no boundaries. Pi Day is a reminder that the quest for understanding our universe and ourselves never ends;
it is an invitation to constantly push the boundaries of what we know and reach for new frontiers.
Celebrating Pi
So, this Pi Day, let us celebrate not just a number but the spirit of human curiosity and the boundless potential within each of us. Let us embrace the infinite possibilities that Pi represents and
continue to be inspired by its mystery and wonder. Happy Pi Day! So let us raise our slices of pie, whether sweet or savory, in honor of this remarkable number and its endless possibilities. Cheers
to a future filled with infinite discovery!
More Slices of Pi
• March 14 is also Albert Einstein’s birthday.
• Pi Day occurs on March 14 because the date is written as 3/14.
• Celebrate the day precisely at 1:59 a.m. or p.m. so you can reach the first six numbers of pi, 3.14159.
• Rajveer Meena holds the record for reciting the most significant number of decimal places of pi. In 2015, Meena recited 70,000 decimal places blindfolded. It took him almost 10 hours.
• Professor Kanada and a team of researchers set a new world record by calculating the value of pi to 1.24 trillion places.
• The mirror image of 3.14 written on paper is PIE! | {"url":"https://community.cadence.com/cadence_blogs_8/b/corporate-news/posts/piday","timestamp":"2024-11-05T01:12:33Z","content_type":"text/html","content_length":"74792","record_id":"<urn:uuid:35abf2f5-d7ba-4b93-b150-238de0be34f0>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00427.warc.gz"} |
Gaya APA
(2004). Introduction to probability and statistics : principles and applications for engineering and the computing sciences fourth edition (fourth edition). Boston: McGraw-Hill.
Gaya Chicago
Introduction to probability and statistics : principles and applications for engineering and the computing sciences fourth edition. fourth edition Boston: McGraw-Hill, 2004. MPK.
Gaya MLA
Introduction to probability and statistics : principles and applications for engineering and the computing sciences fourth edition. fourth edition Boston: McGraw-Hill, 2004. MPK.
Gaya Turabian
Introduction to probability and statistics : principles and applications for engineering and the computing sciences fourth edition. fourth edition Boston: McGraw-Hill, 2004. MPK. | {"url":"https://lib-pasca.unpak.ac.id/index.php?p=cite&id=2949&keywords=","timestamp":"2024-11-05T06:28:28Z","content_type":"text/html","content_length":"2537","record_id":"<urn:uuid:9df4b333-676b-4848-8e72-971823e13108>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00523.warc.gz"} |
Translation of Omega Automaton to LO2
+ General Questions (11)
Exam : 15.02.2017 : Q 8 d part 2 : 2017.02.15.vrs.solutions.pdf (uni-kl.de)
Problem : Translate below formula to LO2 :
I am first using boolean operations for disjunction of two automaton.
Then I am just translating it to LO2 presented in lecture (Module 8 - Slide 25)
Q. Could you please check if the steps I am following is correct ?
Q. Also could you please check if the solution is correct ?
Thanks in advance.
Update : 25.08.2022
Final resultant in LO2 of Product automaton
Well, there are many ways how you can solve this. As I can see, you used the sum automaton to compute the disjunction of the two automata, and then you translated the result to LO2. That is fine.
However, you do not need the sum automaton here. The disjunction of universal automata is dual to the conjunction of existential automata, and for the latter, the simple product automaton would be
sufficient. Hence, you don't need an additional state variable.
I would have solved it as follows without modifying the automata:
(∀t1. (t0≤t1 → (∃t0. t1<t0 ∧ (∀t2. (t1<t2 → t0≤t2)) ∧ p[t0]) ∧ p[t1] ∧ a[t1]))
(∀t1. (t0≤t1 → ¬p[t1]))
∨ ∀q.q[t0] | {"url":"https://q2a.cs.uni-kl.de/3120/translation-of-omega-automaton-to-lo2?show=3127","timestamp":"2024-11-07T18:59:54Z","content_type":"text/html","content_length":"59993","record_id":"<urn:uuid:1eaa6049-2d09-4a4d-8a20-bb3f5741fce5>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00475.warc.gz"} |
3.3 Filter Coefficients in the Graph
One of the main features of ShipWeight is the possibility to do logical selections of the coefficients to be plotted in the graph. This makes it easy to plot only the relevant coefficients in the
graph and thus obtain a good regression line and a good coefficient for estimating.
The first and natural way to filter the coefficients is to select which project types are to be plotted. Selecting Project Types… under Graph in the Estimation Window menu bar let you do this.
Alternatively, click the Project Type button in the Estimation Window.
As a default, all the project types under the main project type equal to the project’s main project type are selected. By navigating in the tree-structure, using the Include and Exclude buttons,
project types can be selected or deselected from the plot. When you are set, click Close to go back to the Graph.
The other way to put constraints on the coefficients to be plotted is to select a comparison parameter in the grid table fields in the Comparison area, and then select Filter limits… under Graph in
the Estimation window menu bar.
In this window you can set the upper and lower limit for the comparison parameter you have selected. Only projcects with parameter value within the given limits will be plotted.
Remember to set both upper and lower limits.
When you are satisfied with the selected coefficients and have estimated the weight and standard deviation you can go on with the estimation by one of the following options:
Estimate VCG and LCG for that weight group by selecting VCG or LCG in Estimate in the menu bar.
Do estimation on another weight group by navigating by selections in the Weight group menu in the Estimation Window.
Return to the main window by closing the window, navigate to another weight group in the main window, and then open the Estimation Window, now with the new weight group in focus.
To ensure an efficient use of time, selecting the Most uncertain function from Weight group in the menu bar in the main window will guide you to a new weight group. Further splitting and estimation
on lower levels of this weight group will give the greatest impact on the total lightweight. This works only if standard deviations have been calculated for each weight group estimated earlier. | {"url":"https://shipweight.com/wp-content/uploads/software/SWwebhelp/documents_filtercoefficientsinthegraph.html","timestamp":"2024-11-08T08:53:38Z","content_type":"text/html","content_length":"13361","record_id":"<urn:uuid:998b3d68-3445-440d-81f0-0b2670ea74a9>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00239.warc.gz"} |
next → ← prev
Spectral Bi-Clustering
A data analysis method called spectral biclustering seeks to concurrently cluster a matrix's rows and columns, which usually represent a dataset. In contrast to conventional clustering techniques,
biclustering-also referred to as co-clustering or two-mode clustering-identifies groupings of both rows and columns that have comparable patterns.
Using spectral techniques based on eigenvalues and eigenvectors of matrices, spectral biclustering reveals patterns buried in the data. The objective is to convert the data matrix into a spectral
domain such that submatrices with comparable spectral qualities can be recognized as biclusters.
The spectral biclustering procedure can be summarised as follows:
Spectrum Transformation: To convert the data matrix into the spectrum domain, use spectral techniques. Calculating the singular value decomposition (SVD) or other matrix factorization techniques is
frequently required for this.
Bicluster Identification: Using eigenvector analysis or other spectral features, locate biclusters in the spectral domain. Submatrices known as biclusters are collections of rows and columns having
comparable spectral properties.
Interpretation: Consider your particular application when interpreting the biclusters that have been detected. Within the domain of genomics, biclusters are associated with gene sets that co-localize
under specified conditions.
Among other domains, spectral biclustering has been applied in image analysis, text mining, and bioinformatics. When working with datasets where subsets of both rows and columns display coordinated
behavior that conventional clustering techniques would miss, it is especially helpful.
Remember that there are numerous methods for biclustering and that the efficacy of spectral biclustering relies on the particular objectives of the research as well as the characteristics of the
data. Spectral biclustering algorithms and methods might have different implementation details.
Certainly! Let's examine the ideas and factors surrounding spectral biclustering in more detail:
Singular Value Decomposition(SVD):
A crucial element of spectral biclustering is singular value decomposition (SVD). In order to uncover underlying patterns, it breaks down a matrix into three smaller matrices. Regarding a matrix
X, = where U provides the breakdown is the left singular vectors matrix, and T is the total.
A diagonal matrix with singular values is called ?Σ.
The correct singular vectors matrix is V.
Choosing a subset of the singular values and vectors that best represent the most significant patterns in both rows and columns is the main goal of biclustering.
Metrics for Bicluster Quality:
It is essential to assess bicluster quality. Among the metrics are:
Cohesiveness: gauges how related two things are inside a bicluster.
Distinctiveness: Indicates how different two biclusters are from one another.
Relevance: Uses statistical metrics to assess a bicluster's significance.
Algorithms for Spectral Clustering:
For biclustering, a number of spectrum clustering methods can be modified, including:
Normalized Cut: This technique was first created for picture segmentation, but it may also be used for biclustering jobs.
Multiplicative Update methods: By modifying rows and columns, these methods iteratively improve biclusters.
Sparse Singular Value Decomposition(SSVD): The SVD variation known as "sparse singular value decomposition" (SSVD) encourages sparsity in the factor matrices.
Spectral biclustering is used in a variety of domains, including:
Genomics and Transcriptomics: Discovering co-expressed genes in particular circumstances using transcriptomics and genomics.
Text mining: Identifying subjects that appear together in portions of text.
Image Analysis: Finding patterns in image data where pixels stand in for both rows and columns is known as image analysis.
Scalability: Because SVD is computationally expensive, spectral approaches may perform poorly on big datasets.
Parameter tuning: It can be challenging to choose the right parameters, like the number of biclusters.
Noise Sensitivity: The quality of biclusters may be impacted by spectral approaches' sensitivity to noise in the data.
Software and Libraries:
Numerical computing libraries like NumPy, SciPy, and scikit-learn in Python are frequently used in the implementation of spectral biclustering. Implementations may also be available from some
specialized biclustering libraries, such as R's BiBit and BicAT.
Studies on spectral biclustering are ongoing, and progress is being made. Scholars frequently suggest new methods and algorithms to solve certain problems related to biclustering various kinds of
In Conclusion, spectral biclustering is an effective method of data analysis that simultaneously groups a matrix's rows and columns, exposing hidden patterns in intricate datasets. Singular value
decomposition and other spectral techniques are used in spectral biclustering, which finds biclusters-subsets of rows and columns having comparable patterns. Applications for this method can be found
in image analysis, text mining, genomics, and other domains where coordinated behavior in both rows and columns has to be identified.
Although spectral biclustering provides insightful information, there are drawbacks. Academics and practitioners must address concerns about scalability, noise sensitivity, and parameter tweaking.
Meaningful interpretation requires assessing bicluster quality using parameters such as coherence, distinctiveness, and relevance.
The study of spectral biclustering is a dynamic topic where new methods and algorithms are always being developed through continuous research. The type of data and the analysis's objectives may
influence the approach that is selected. Numerical computing libraries and specialized biclustering tools found in major programming languages are frequently used in the implementation of spectrum
Spectral biclustering is an invaluable technique for delving into intricate datasets and identifying coordinated patterns in both rows and columns, giving analysts and researchers a better grasp of
the underlying structure in their data.
← prev next → | {"url":"https://www.javatpoint.com/spectral-bi-clustering","timestamp":"2024-11-10T05:52:22Z","content_type":"text/html","content_length":"83186","record_id":"<urn:uuid:2bbb4b5d-b74d-49f3-85c1-c12d222bcca8>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00344.warc.gz"} |
Understanding the Importance and Impact of the Variable ‘b’ – Unraveling the Value and Significance
Understanding the Value of ‘b’ in Mathematical Equations and Analysis
Have you ever come across the variable ‘b’ in mathematical equations and wondered what its significance is? In this blog post, we will unravel the importance and value of ‘b’ in various applications,
ranging from mathematical equations and scientific analysis to statistical inference and economic modeling.
Understanding the Importance of the Variable ‘b’
Before we delve into the specifics, let’s start by introducing the variable ‘b.’ In mathematical terms, ‘b’ commonly represents a coefficient or constant that plays a crucial role in equations and
formulas. The value of ‘b’ often determines the behavior, relationship, or outcome of the given equation.
The purpose of this blog post is to shed light on the significance of ‘b’ in different fields of study and applications, highlighting how different values of ‘b’ can impact the results and
Unraveling the Value of ‘b’ in Mathematical Equations
‘b’ holds great importance in mathematical equations and formulas, affecting the shape, slope, and direction of graphs as well as the values of dependent variables. In various algebraic equations,
‘b’ is used as a coefficient, controlling the magnitude and direction of the equation’s terms.
For instance, in a linear equation of the form ‘y = mx + b,’ ‘b’ represents the y-intercept, indicating the point where the line intersects the y-axis. Different values of ‘b’ can shift the line
vertically, altering its starting position and affecting the overall relationship between the independent and dependent variables.
In quadratic equations, ‘b’ determines the line of symmetry. It plays a significant role in parabolic curves, affecting the placement and orientation of the graph.
Even in more complex equations or models, ‘b’ can still exert its influence. Consider the equation for simple interest: ‘I = PRT,’ where ‘b’ represents the rate of interest. The value of ‘b’
determines the interest amount, illustrating how a higher or lower interest rate affects the final outcome.
The Role of ‘b’ in Statistical Analysis and Inference
Statistical analysis relies heavily on ‘b’ coefficients, particularly in regression analysis. In simple linear regression, ‘b’ represents the estimated slope of the line, which indicates the strength
and direction of the relationship between two variables.
By interpreting ‘b’ coefficients, analysts can determine the impact of changes in the independent variable on the dependent variable. For example, in a regression model exploring the relationship
between advertising expenditures and sales, ‘b’ represents the change in sales for each unit increase in advertising expenditure.
Furthermore, ‘b’ coefficients can be used to make predictions. By plugging in different values for independent variables, analysts can estimate the expected value of the dependent variable, allowing
for more informed decision-making.
The Impact of ‘b’ in Economic and Business Analysis
In economic and business analysis, ‘b’ plays a vital role in understanding the relationships between various factors and variables. Economic models, such as supply and demand equations, often
incorporate ‘b’ coefficients to signify the impact of different factors on the equilibrium quantity and price.
For instance, in the supply equation, ‘b’ represents the slope of the supply curve, indicating how changes in price lead to changes in quantity supplied. Understanding the implications of varying ‘b’
values can inform decisions regarding price levels and supply chain management.
In addition to economic models, ‘b’ is present in forecasting and planning. Business analysts utilize ‘b’ coefficients to predict future outcomes, identify trends, and assess the potential impact of
different scenarios.
In conclusion, the value of ‘b’ extends beyond its representation as a simple variable in mathematical equations. Its significance stretches into statistical analysis, economic modeling, and business
planning. Understanding the role and impact of ‘b’ is crucial for making informed decisions, predicting outcomes, and unraveling relationships between variables.
By exploring the examples and applications discussed in this blog post, you can gain a deeper appreciation for the value of ‘b’ in various fields and harness its power effectively. So, the next time
you come across ‘b’ in an equation, remember its importance and embrace its ability to provide insights, connections, and predictions. | {"url":"https://skillapp.co/blog/understanding-the-importance-and-impact-of-the-variable-b-unraveling-the-value-and-significance/","timestamp":"2024-11-06T01:22:05Z","content_type":"text/html","content_length":"110066","record_id":"<urn:uuid:bf8eddcb-9c79-4918-8b6d-8b1c180ad5f7>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00445.warc.gz"} |
Comp 251 Assignment 3 solved
Exercise 1. (25 points) Ford-Fulkerson
We will implement the Ford-Fulkerson algorithm to calculate the Maximum Flow of a directed weighted graph.
Here, you will use the files WGraph.java and FordFulkerson.java,
which are available on the course website. Your role will be to complete two methods in
the template FordFulkerson.java.
The file WGraph.java is similar to the file that you used in your previous assignment to
build graphs. The only differences are the addition of setter and getter methods for the
Edges and the addition of the parameters “source” and “destination”. There is also an
additional constructor that will allow the creation of a graph cloning a WGraph object.
Graphs are encoded using a similar format than the one used in the previous assignment.
The only difference is that now the first line corresponds to two integers, separated by
one space, that represent the “source” and the “destination” nodes. An example of such
file can be found on the course website in the file ff2.txt. These files will be used
as an input in the program FordFulkerson.java to initialize the graphs. This graph
corresponds to the same graph depicted in [CLRS2009] page 727.
Your task will be to complete the two static methods fordfulkerson(Integer source,
Integer destination, WGraph graph, String filePath) and pathDFS(Integer source,
Integer destination, WGraph graph). The second method pathDFS finds a path
via Depth First Search (DFS) between the nodes “source” and “destination” in the
“graph”. You must return an ArrayList of Integers with the list of unique nodes belonging to the path found by the DFS.
The first element in the list must correspond to
the “source” node, the second element in the list must be the second node in the path,
and so on until the last element (i.e., the “destination” node) is stored. The method
fordfulkerson must compute an integer corresponding to the max flow of the “graph”,
as well as the graph encoding the assignment associated with this max flow. The method
fordfulkerson has a variable called myMcGillID, which must be initialized with your
McGill ID number.
Once completed, compile all the java files and run the command line java FordFulkerson
ff2.txt. Your program must use the function writeAnswer to save your output in a file.
An example of the expected output file is available in the file ff226000000.txt. This
output keeps the same format than the file used to build the graph; the only difference
is that the first line now represents the maximum flow (instead of the “source” and “destination” nodes).
The other lines represent the same graph with the weights updated
to the values that allow the maximum flow. The file ff226000000.txt represents the
answer to the example showed in [CLRS2009] Page 727. You are invited to run other
examples of your own to verify that your program is correct.
Exercise 2. (25 points) Bellman-Ford
We want to implement the Bellman-Ford algorithm for finding the shortest path in
a graph where edges can have negative weights. Once again, you will use the object
WGraph. Your task is to complete the method BellmanFord(WGraph g, int source)
and shortestPath(int destination) in the file BellmanFord.java.
The method BellmanFord takes an object WGraph named g as an input and an integer
that indicates the source of the paths. If the input graph g contains a negative cycle,
then the method should throw an exception (see template). Otherwise, it will return
an object BellmanFord that contains the shortest path estimates (the private array of
integers distances), and for each node, its predecessor in the shortest path from the
source (the private array of integers predecessors).
The method shortestPath will return the list of nodes as an array of integers along
the shortest path from the source to the destination. If this path does not exists, the
method should throw an exception (see template).
Input graphs are available on the course webpage to test your program. Nonetheless, we
invite you to also make your own graphs to test your program. | {"url":"https://codeshive.com/questions-and-answers/comp-251-assignment-3-solved/","timestamp":"2024-11-14T07:19:50Z","content_type":"text/html","content_length":"101643","record_id":"<urn:uuid:19ca6a33-4704-4802-b623-bc8271baa8ef>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00023.warc.gz"} |
Chapter 1: Introduction
Chapter 1, Introduction to Probability and Statistics
(1) Information comes with uncertainty. For example
whether a friend will call today
the time that a train will arrive
the time that a machine will fail
In order to study these phenomena, we need probability and statistics
(2) Probability and statistics is a study of uncertain events. It gives us a quantitative way
to measure the uncertainty.
(3) An example: take a survey from the students in the front row:
record name
hours of study per week
hours of exercise per week
(4) What is the average?
the questions:
average hours of study per week
average hours of study per week of male students
average hours of study per week of female students
average hours of exercise per week
the measure of average:
the mean:
1 n
x xi
n i1
the medium:
sort the data in ascending order
the datum at the (n+1)/2 position is the medium
the mode
arrange the data by incremental intervals
the position at which most data is concentrated is the mode
in the above example
x 3 = 15.5 (the mean hours of study per week)
x 4 = 5.25 (the mean hours of exercise per week)
also, for male student:
x 3M = 15.5
x 4M = 5.75
for the female student
x 3F = 15.5
x 4F = 4.25
note different data sets may have a same mean, for example:
A = {1, 1, 1, 1}
B = {2, 2, 0, 0}
xA= xB=1
therefore, it is necessary to have another way to measure the uncertain events: variation.
(5) What is the variation?
the questions:
the variation of hours of study per week
the variation of hours of study per week of male students
the measure of variation
the variance
1 n
xi x 2
n 1 i1
the standard deviation: s
the range:
r = max{xi} - min{xi}
in the above example:
sx3 = 4.19
sx4 = 3.25
sx3M = 4.78
sx4M = 3.45
sx3F = 3.32
sx4F = 2.98
- each student is somewhat different
- the statistics would be different (e.g., the mean and the variance of male and
female students)
- female students are more consistent than the male students (because of the
variation is small).
- note the above statistics are true ONLY for the 12 students in the front row. What about
the students in the entire class? the students in the University? ....
(6) Unanswered questions:
the questions:
- what is the hours of study per week of a students in the university? (prediction)
- is there any correlation between the hours of study and the hours of exercise?
- ......
the answers can be found by the probability and statistical theory, which we will learn in
the first part of this course. | {"url":"https://studylib.net/doc/7863123/chapter-1--introduction","timestamp":"2024-11-05T16:45:27Z","content_type":"text/html","content_length":"60999","record_id":"<urn:uuid:6aa8d00e-56dd-4482-a588-9dc95dc75fc7>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00457.warc.gz"} |
How to solve the equation?
Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.
Community Tip - Want the oppurtunity to discuss enhancements to PTC products? Join a working group! X
How to solve the equation? Prime 7 can not solve the answers.
Mathcad never was the tool to solve diophantic equations (apart from an extension pack which worked only in a very early version of MC).You get a result in MC15 but its in no way helpful:
Maybe Luc can show us what MC11 with Maple can come up with
I got distracted. It is a Diophantine equation. See on Wikipedia.
Well, you can find the answer, but it's not really "solving" . . . | {"url":"https://community.ptc.com/t5/Mathcad/How-to-solve-the-equation/td-p/721335","timestamp":"2024-11-02T10:57:13Z","content_type":"text/html","content_length":"311398","record_id":"<urn:uuid:6ab2d596-eab6-4513-91a9-5aae7c7834ac>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00182.warc.gz"} |
Score Matched Neural Exponential Families for Likelihood-Free Inference
Lorenzo Pacchiardi, Ritabrata Dutta.
Year: 2022, Volume: 23, Issue: 38, Pages: 1−71
Bayesian Likelihood-Free Inference (LFI) approaches allow to obtain posterior distributions for stochastic models with intractable likelihood, by relying on model simulations. In Approximate Bayesian
Computation (ABC), a popular LFI method, summary statistics are used to reduce data dimensionality. ABC algorithms adaptively tailor simulations to the observation in order to sample from an
approximate posterior, whose form depends on the chosen statistics. In this work, we introduce a new way to learn ABC statistics: we first generate parameter-simulation pairs from the model
independently on the observation; then, we use Score Matching to train a neural conditional exponential family to approximate the likelihood. The exponential family is the largest class of
distributions with fixed-size sufficient statistics; thus, we use them in ABC, which is intuitively appealing and has state-of-the-art performance. In parallel, we insert our likelihood approximation
in an MCMC for doubly intractable distributions to draw posterior samples. We can repeat that for any number of observations with no additional model simulations, with performance comparable to
related approaches. We validate our methods on toy models with known likelihood and a large-dimensional time-series model.
PDF BibTeX code | {"url":"https://www.jmlr.org/beta/papers/v23/21-0061.html","timestamp":"2024-11-02T11:46:07Z","content_type":"text/html","content_length":"7894","record_id":"<urn:uuid:150147b3-0e03-4120-9ddf-a30db679b24a>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00274.warc.gz"} |