Calculus (eBook)

Maths of the Gods

Bill Todorovich, Ed Walker (Autoren)

eBook Download: EPUB

2024 | 1. Auflage
200 Seiten
Vivid Publishing (Verlag)
978-1-923078-25-3 (ISBN)

Infos

Discover the intriguing world of calculus with this book, which aims to demystify and enrich your understanding of this fundamental branch of mathematics. While many find calculus both fascinating and intimidating, this guide strives to make it more accessible and engaging. Written in an informal tone, every effort has been made to maintain mathematical integrity, ensuring clarity without sacrificing depth. Through a range of tools and techniques, readers will uncover the reasons behind phenomena such as the elliptical orbit of Earth around the sun, the story of Newton's apple, and Einstein's insights into the atom bomb. Encouraging readers to unleash their imagination, this book invites contemplation of the infinitely small and large, fostering a deeper understanding of the universe. Ultimately, it underscores the vital role of imagination in scientific discovery, with calculus serving as the language that quantifies our boldest imaginings.

Ed Walker is the author of 'Calculus - Maths of the Gods'.

Textauszug

Chapter 1 Trigonometric Identities and Logarithms

It is impossible to study calculus without knowing the basics of trigonometry and logarithms.

sinesin θ=OppositeHypotenusecosinecos θ=AdjacentHypotenuse

tangent tan θ=OppositeAdjacent=sinθcosθ

From the graphs and diagram above, it is apparent that sin is positive in the first and second quadrants, and negative in the third and fourth.

Cos is positive in the first and fourth quadrants, and negative in the second and third.

Tan is positive in the first and third, while negative in the second and fourth quadrants.

secant sec θ=1cosθcosecant csc θ=1sinθcotangent cot θ=1tanθ

sinθ=cos90−θtanθ=cot90−θ

sin−θ=−sinθcos−θ=cosθ tan−θ=−tanθ

sin2θ+cos2θ=11+tan2θ=sec2θ

The inverse functions arcsine, arccosine, arctan, arcsec, arccosec and arccot may be written:

asin, atan or denoted by sin-1 tan-1.

Example: Express cos θ in terms of x, where arctan x = θ.

In the diagram above, the angle is represented as θ.

As θ= atan x⇒tanθ=x1 from which     cos θ=11+x2

The Unit Circle Showing Six Trigonometric Identities

(1) AB = sin θ

(2) OB = cos θ

(3) AD = tan θ

(4) OE = cosec θ

(5) OD = sec θ

(6) AE = cot θ

Note:i       OE2+OD2=ED2⇒cosec2θ+sec2θ=cotθ+tanθ2Also ii     BD2+AB2=AD2⇒secθ−cosθ2+sin2θ=tan2θ iii   OE−AB2+OB2=AE2⇒cosecθ−sinθ2+cos2θ=cot2θiv   Area ΔOAE+Area ΔOAD=Area OED

12 cot θ+12 tan θ=12sec θ cosec θ⇒cot θ+tan θ=sec θ cosec θ

Double Angle formulas

OQ=cos βPB=cos α sinβAQ=sin αcos βsinα+β=BC+PB=AQ+PBsinα+β=sin α cos β+sin βcos αsinα-β=sin α cos β-sin βcos α

cosα+β=OC=OA−CA=OA−BQcosα+β=cos α cos β-sin α sin βcosα-β=cos α cos β + sin α sin β

tanα+β=sin α cos β+sin β cos αcos α cos β−sin α sin β

Dividing both the numerator and the denominator by cos α cos β we get

tanα+β=tan α+tan β1-tan αtan βtanα-β=tan α-tan β1+tan αtan β

sin 2θ=2 sinθ cos θ

cos2θ=cos2θ−sin2θtan2θ=2tanθ1−tan2θ

Pythagoras’s Theorem

sinα=ACAB=ADAC⇒AC2=ADABsinβ=BCAB=BDBC⇒BC2=BDABAC2+BC2=ABAD+BDAC2+BC2=AB2

Sine Rule

d=b sin C=c sinB⇒bsinB=csinC Similarly asinA=bsinB⇒asin A=bsin B=csin C

Area of Triangle

Area=12hd+e=h2a=b sin C2a=12ab sin CSimilarly        Area=12bc sin A=12ac sin B

Cosine Rule

a2=c+AD2+CD2=c2+2cAD+AD2+CD2=b2+c2+2cAD=b2+c2+2cb cos180−A⇒a2=b2+c2-2bc cosA

Similarly b2=a2+c2−2ac cosB and c2=a2+b2−2ab cos C

The Radian

A radian is vital for calculus as many of the formulas for differentiation and integration only apply when angles are given in radians.

A radian is the angle which subtends an arc length equal to the radius.

The ratio CircumferenceArc Length subtended by one radian=2πrr=2π

Therefore there are 2π radians in a circle.

12π=One radian expressed in degrees360⇒Radian=3602π ≈ 57.29577951 degrees

Exercise: a=2 b=1.2 angle B=30o Find angle A

asinA=bsinB⇒sinA=asinBb=2121.2=56⇒A=asin 56=56.443 or 123.557 degrees

This example illustrates that when using the sine rule, if the side opposite the given angle is the lesser of the two then there may be no solution or two solutions.

Exercise (2) Cover over the right side and have a try before checking the answer. Do not use your calculator. Turn the radians into degrees, and then work out which quadrant we are in.

sinπ4=sin45o=22sin−π4=sin315o=−22sin3π2=sin270o=−1tan3π4=tan135o=−1cos−π4=cos315o=22cos8π=cos0o=1

sin7π=sin180o=0

sin5π2=sin90o=1cos2π3=cos120o=−12

Exercise (3) x = cos θ y = sin 2θ Express y in terms of x.

y=sin2θ=2sinθ  cosθ=2x1−x2

Exercise (4) y = atan x Express sec y in terms of x

secy=1+x2

Exercise (5) Find the three sides of the triangle if a=2Area=1B=π6

Area=12ac sinB⇒1=122c12⇒c=2b2=a2+c2−2ac cosB=4+4−832∴b=22−3≈1.035276

The three sides are a=2b=1.035276c=2

Exercise (6) In triangle ABC a = 4 b = 2 c = 3 Solve for angle A

a2=b2+c2−2bc cosA⇒16=4+9−12 cosA

∴cosA=−14⇒A≈104.477512  Degrees

Note: The cosine rule only gives one possible answer.

Logarithms and Exponentials

y = loga x This means y is the power by which we need to raise “a” to equal the value of x. “a” can be any number; however, there are three particularly useful bases.

1   logx=log10x   Common Logarithm2   lnx=logex   Natural Logarithm3   lgx=log2x  Binary Logarithm

Common Logarithms are the most useful for calculations.

Example: As 103 = 1,000 ⇒ log10 1000 = 3

Multiplying

100×1000=100,000102×103=105log100+log1000=log100,000The exponents are       2+3=5

Dividing

10,00001,000=100105103=102log100,000−log1,000=log100That is       5−3=2

The Power Function

1003=1023=102×3=1063log100=log1,000,000That is 23=6

Summary of the rules of logarithms

loga×b=loga+logblogab=loga−logblogab=b  log  a

The Natural or Napierian logarithm loge written ln is extensively used in calculus.

“e” is the transcendental number 2.71828…

If ln 7.38905 = 2 ⇒ e2 ≈ 7.38905 or (2.71828)2 ≈ 7.38905

The same rules apply

lna×b=ln a+ln blnab=lna−lnblnab=blna

Binary Logarithms have a base of 2 and are widely used in engineering.

Changing the base of Logarithms

y=lnx⇒ey=x∴     logey=logx⇒yloge=logx∴y=logxloge⇒lnx=logxlogeand similarly        logx=lnxln10

Summary: The logarithm in the new base equals the logarithm (in the old base) divided by the logarythm(in the old base) of the new base.

i.e..  lognx=logmxlogmn

Changing the log base can be useful for transforming a function into a convenient form for calculus as shown by the...

Produkt-Details

• Erscheint lt. Verlag: 1.6.2024
• Sprache: englisch
• Themenwelt: Mathematik / Informatik ► Mathematik
• ISBN-10: 1-923078-25-9 / 1923078259
• ISBN-13: 978-1-923078-25-3 / 9781923078253