From differential equations to trigonometric functions · From di erential equations to...

From differential equations totrigonometric functions

Cole ZmurchokMath 102 Section 106

November 23, 2016

Today...

1. The Mathematics of Love

2. Introducing Trigonometric Functions

3. Friday: Many Trig Examples!

I Course Evaluations

I MLC Hours Changed

I Reminder: Office Hours today from 3-4 pm inMath Annex 1118 and Thursday from 3-4 inLSK 300.

I WebWorK due Thursday

I OSH 6 due Friday

Last time: Disease Dynamics

I Our model is a slight variant of the originalmodel studied by Kermack and McKendrick in1927

dS

dt= −βSI + µI

dI

dt= βSI − µI

N = I + S

I This type of model is called a system ofdifferential equations

Last time: Disease Dynamics

I We found that

dI

dt= βI(K − I), K = N − µ

β

I An epidemic occurs whenever

K > 0 ⇐⇒ N − µ

β> 0 ⇐⇒ N >

µ

β

⇐⇒ R0 =βN

µ> 1

I Quick Euler’s Method Demo

The Mathematics of Love

I From: Strogatz, S. H. (2014). Nonlineardynamics and chaos: with applications tophysics, biology, chemistry, and engineering.Westview press.

I Steven Strogatz: Famous!! https://www.

abebooks.com/servlet/SearchResults?

an=steven+strogatz&sts=t

Romeo and Juliet have a tumultuousrelationship

I Juliet: My passion for Romeo decreases inproportion to his love.

I The more Romeo loves me, the more I run awayfrom him. When he dislikes me, I start to love him.

I Romeo: My passion for Juliet increases inproportion to her love.

I The more Juliet loves me, the more I love her.When she dislikes me, my love for her decreases.

Love quantified

• Ihateyou Iloveyou

-101

Romeo and Juliet

I Let x(t) = Juliet’s love for Romeo

I Let y(t) = Romeo’s love for Juliet

How will Romeo and Juliet’s relationship evolve overtime? Will they live happily ever after? Will theirstory end tragically?

Juliet’s Love, x(t)

Q1. Juliet’s love for Romeo decreases at a rateproportional to his love (assume k1 > 0).

A. dxdt = k1x

B. dxdt = −k1x

C. dydt = −k1y

D. dxdt = −k1y

E. Send help

Romeo’s Love, y(t)

Q2. Romeo’s love for Juliet increases at a rateproportional to her love (assume k2 > 0).

A. dxdt = k2x

B. dydt = −k2x

C. dydt = k2y

D. dydt = −k2y

E. dydt = k2x

Romeo and Juliet

Juliet:dx

dt= −k1y

Romeo:dy

dt= k2x

I This is a system of differential equations fortwo functions of time (x(t), y(t)).

Slope Field for 1D y(t)

13.2. The geometry of change 247

Example 13.6 Consider the differential equationdy

dt= 2y. (13.4)

Compute some of the slopes for various y values and use this to sketch a slope field for thisdifferential equation.

Solution: Equation (13.4) states that if a solution curve passes through a point (t, y), thenits tangent line at that point has a slope 2y, regardless of the value of t. This example issimple enough that we can state the following: for positive values of y, the slope is positive,for negative values of y, the slope is negative, and for y = 0 the slope is zero. We providesome tabulated values of y indicating the values of the slope f(y), its sign, and what thisimplies about the local behaviour of the solution and its direction. Then, in Figure 13.1we combine this information to generate the direction field and the corresponding solutioncurves. Note that the direction of the arrows (rather than their absolute magnitude) providesthe most important qualitative tendency for the slope field sketch.

y f(y) = 2y slope of tangent line behaviour of y direction of arrow-2 -4 -ve decreasing ↘-1 -2 -ve decreasing ↘0 0 0 no change in y →1 2 +ve increasing ↗2 4 +ve increasing ↗

Table 13.1. Table of derivatives and slopes for the differential equation (13.4) inExample 13.6.

-2

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4

y

t-2

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4

y

t(a) (b)

Figure 13.1. Direction field and solution curves for Example 13.6.

In constructing the slope field and solution curves, the following basic rules shouldbe followed:

How do you find the slopes?

Directionfielddx/dt =- k1 ydy/dt =k2 x

Juliet’slovex(t)

Romeo

’slovey(t)

(3)Whatdoyouthinkwillhappen?

dx/dt =- k1 ydy/dt =k2 x

Juliet’slovex(t)

Romeo

’slovey(t)

Solutioncurvedx/dt =- k1 ydy/dt =k2 x

Juliet’slovex(t)

Romeo

’slovey(t)

Solutioncurvedx/dt =- k1 ydy/dt =k2 x

Juliet’slovex(t)

Romeo

’slovey(t) Romeoand

Julietchaseeachotherinanendlesslove-circle!

Shownin3Dwithtimeaxis:Juliet’slovex(t)

Romeo andJuliet

Timetà

Romeo and Juliet

Q4. Do you recognize these functions?

Yes! These are

A. Polynomials

B. Exponentials

C. Power functions

D. Sine and cosine

E. Not sure

Romeo andJuliet

Thesecurvesarex(t)=cos(t),y(t)=sin(t).

Timetà

Introducing the trigonometric functions

I cos t

I sin t

What is special about these functions?

I They are periodic

I They describe oscillating systems

I They have “nice” derivatives

Derivative of cosine and sine

Cosine:d

dtcos t = − sin t

Sine:d

dtsin t = cos t

Derivative of sinx

d sin(x)

dx= lim

h→0

sin(x+ h)− sin(x)

h

= limh→0

sin(x) cos(h) + sin(h) cos(x)− sin(x)

h

= limh→0

(sin(x)

cos(h)− 1

h+ cos(x)

sin(h)

h

)= sin(x)

(limh→0

cos(h)− 1

h

)+ cos(x)

(limh→0

sin(h)

h

)= cos(x)

Two important limits

limx→0

sinx

x= 1

limx→0

cos(x)− 1

x= 0

Derivative of cosine and sineCosine:

d

dtcos t = −sin t

Sine:d

dtsin t = cos t

The special relationship between these functionsmeans that they satisfy the Juliet-Romeo equations:

dx

dt= −y

dy

dt= x

Second derivative of y(t) = sin(t)?

Q5. sin(t) is a solution to the following differentialequation.

A. d2ydt2 = −y

B. d2ydt2 = y

C. dyd2 = y

D. d2ydt2 = −x

E. Send help

Second derivatives

d2 sin t

dt2= − sin t⇒ dy2

dt2= −y

d2 cos t

dt2= − cos t⇒ dx2

dt2= −x

I The trig functions sin t and cos t are solutionsto the differential equation

dy2

dt2= −y

Wait, what?

I I thought trig functions had to do with anglesand triangles.

I They do!

Angles in radians

I Define a new measure for angles:

1 revolution around a cirle = 2π radians

I Angles are associated with the length of an arcsubtended by that angle:

r

s

θs = rθ

Convention

I Angles increase counterclockwise

θ = 0

Convert from degrees to radians

Q6. In terms of radians, the angles 30, 45, 60, and90◦ are

A. π/6, π/4, π/3, π/2

B. π/3, π/2, π/6, π

C. π/30, π/45, π/60, π/90

Special angles364 Appendix F. Trigonometry review

degrees radians sin(t) cos(t) tan(t)0 0 0 1 030 π

612

√3

21√3

45 π4

√2

2

√2

2 160 π

3

√3

212

√3

90 π2 1 0 ∞

Table F.1. Values of the sines, cosines, and tangent for the standard angles.

whereas division by sin2(t) gives us

1 + cot2(t) = csc2(t).

These will be important for simplifying expressions involving the trigonometric functions,as we shall see.

Law of cosines

This law relates the cosine of an angle to the lengths of sides formed in a triangle. (Seefigure F.2.)

c2 = a2 + b2 − 2ab cos(θ) (F.2)where the side of length c is opposite the angle θ.

a

bc

θ

Figure F.2. Law of cosines states that c2 = a2 + b2 − 2ab cos(θ).

Here are other important relations between the trigonometric functions that shouldbe remembered. These are called trigonometric identities:

Angle sum identities

The trigonometric functions are nonlinear. This means that, for example, the sine of thesum of two angles is not just the sum of the two sines. One can use the law of cosines andother geometric ideas to establish the following two relationships:

sin(A + B) = sin(A) cos(B) + sin(B) cos(A) (F.3)

cos(A + B) = cos(A) cos(B)− sin(A) sin(B) (F.4)

Connectionwithangle()

Trig identity

I Equation of circle of radius 1:

x2 + y2 = 1

I Point on that circle

(cos(t), sin(t))

I Thus,sin2(t) + cos2(t) = 1

Motion around a circle

I Angle increases

dθ

dt= ω

I Motion can be described by either1. Polar coordinates:

r = 1, θ(t) = ωt

2. Cartesian coordindates:

(x(t), y(t)) = (cos(ωt), sin(ωt))

Periodic functions

I sin t is 2π-periodic since for any value of t

sin(t+ 2π) = sin t

I A function is periodic with period T if for anyvalue of t

f(t) = f(t+ T )

Frequency and period

I Frequency: https://www.desmos.com/

calculator/n8irldojfy

I Amplitude: https://www.desmos.com/

calculator/ubpyqcjy5b

Other trig functions

tan t =sin t

cos t, cot t =

1

tan t

sec t =1

cos t, csc t =

1

sin t

Derivative of tanx

Q7. The derivative of tanx is

A. cotx

B. − cotx

C. cos2(x)

D. sec2(x)

E. Send help

Important Trigonometric Identities

I Sum of two angles

sin(α + β) = sin(α) cos(β) + sin(β) cos(α)

cos(α + β) = cos(α) cos(β)− sin(α) sin(β)

I Pythagorean identity

sin2(t) + cos2(t) = 1

tan2(t) + 1 = sec2(t)

Law of cosines

θ

c

a

b

c2 = a2 + b2 − 2ab cos θ

Law of cosines

c2 = a2 + b2 − 2ab cos θ

Q8. In the special case, θ = π2 , the law of cosines

reduces to which of these?

A. c2 = a2 + b2 − 2ab

B. c2 = (a− b)2C. c2 = a2 + b2

D. sin2(t) + cos2(t) = 1

Summary

I The mathematics of love leads to a system ofdifferential equations

I For a 2D system, a direction field describeshow solutions change NOT ON EXAM

I Sine and cosine are related to a system ofdifferential equations

I Sine and cosine are related to a second-orderdifferential equation (y′′ = −y)

I Trigonometric functions are related to motionaround a circle

I Frequency, Period, Amplitude, Identities, etc.(See Course Notes Appendix)

Answers

1. D

2. E

3. C

4. D

5. A

6. A

7. D

8. C

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