From differential equations to trigonometric functions

Cole Zmurchok Math 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 in Math Annex 1118 and Thursday from 3-4 in LSK 300.

I WebWorK due Thursday

I OSH 6 due Friday Last time: Disease Dynamics

I Our model is a slight variant of the original model studied by Kermack and McKendrick in 1927

dS = βSI + µI dt − dI = βSI µI dt − N = I + S

I This type of model is called a system of differential equations Last time: Disease Dynamics

I We found that

dI µ = βI(K I),K = N dt − − β

I An epidemic occurs whenever µ µ K > 0 N > 0 N > ⇐⇒ − β ⇐⇒ β βN R = > 1 ⇐⇒ 0 µ

I Quick Euler’s Method Demo The Mathematics of Love

I From: Strogatz, S. H. (2014). Nonlinear dynamics and chaos: with applications to physics, 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 tumultuous relationship

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

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

I Romeo: My passion for Juliet increases in proportion 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

• I hate you I love you

-1 0 1 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 over time? Will they live happily ever after? Will their story end tragically? Juliet’s Love, x(t)

Q1. Juliet’s love for Romeo decreases at a rate proportional to his love (assume k1 > 0). dx A. dt = k1x B. dx = k x dt − 1 C. dy = k y dt − 1 D. dx = k y dt − 1 E. Send help Romeo’s Love, y(t)

Q2. Romeo’s love for Juliet increases at a rate proportional to her love (assume k2 > 0). dx A. dt = k2x dy B. dt = k2x dy − C. dt = k2y dy D. dt = k2y dy − E. dt = k2x Romeo and Juliet

dx Juliet: = k y dt − 1 dy Romeo: = k x dt 2

I This is a system of differential equations for two functions of time (x(t), y(t)). 13.2. The geometry of change 247

Example 13.6 Consider the differential equation dy =2y. (13.4) dt Compute some of the slopes for various y values and use this to sketch a slope field for this differential equation.

Solution: Equation (13.4) states that if a solution curve passes through a point (t, y),then its tangent line at that point has a slope 2y,regardlessofthevalueoft.Thisexampleis simple enough that we can state the following: for positive values of y,theslopeispositive, for negative values of y,theslopeisnegative,andfory =0the slope is zero. We provide some tabulated values of y indicating the values of the slope f(y),itssign,andwhatthis implies about the local behaviour of the solution and its direction. Then, in Figure 13.1 we combine this information to generate the direction field and the corresponding solution curves. Note that the direction of the arrows (rather than their absolute magnitude) provides the 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) in Slope FieldExample for 13.6. 1D y(t)

y y 1.5 1.5

1 1

0.5 0.5

0 0

-0.5 -0.5

-1 -1

-1.5 -1.5

-2 -2 0 1 2 3 4 t 0 1 2 3 4 t (a) (b)

How do you findFigure the slopes? 13.1. Direction field and solution curves for Example 13.6. In constructing the slope field and solution curves, the following basic rules should be followed: dx/dt = - k1 y dy/dt = k2 x Direction field Romeo’s love y(t)

Juliet’s love x(t) dx/dt = - k1 y (3) What do you think will dy/dt = k2 x happen? Romeo’s love y(t)

Juliet’s love x(t) dx/dt = - k1 y dy/dt = k2 x Solution curve Romeo’s love y(t)

Juliet’s love x(t) dx/dt = - k1 y dy/dt = k2 x Solution curve

Romeo and Juliet chase each other in an endless love-circle! Romeo’s love y(t)

Juliet’s love x(t) Shown in 3D with time axis: Juliet’s love x(t) Romeo and Juliet

Time t à 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 and Juliet

Time t à

These curves are x(t) = cos(t), y(t) = sin(t). 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 cos t = sin t dt − Sine: d sin t = cos t dt Derivative of sin x

d sin(x) sin(x + h) sin(x) = lim − dx h 0 h → sin(x) cos(h) + sin(h) cos(x) sin(x) = lim − h 0 h →  cos(h) 1 sin(h) = lim sin(x) − + cos(x) h 0 h h →  cos(h) 1  sin(h) = sin(x) lim − + cos(x) lim h 0 h h 0 h → → = cos(x) Two important limits

sin x lim = 1 x 0 x →

cos(x) 1 lim − = 0 x 0 x → Derivative of cosine and sine Cosine: d cos t = sin t dt − Sine: d sin t = cos t dt The special relationship between these functions means that they satisfy the Juliet-Romeo equations: dx = y dt − dy = x dt Second derivative of y(t) = sin(t)?

Q5. sin(t) is a solution to the following differential equation. 2 A. d y = y dt2 − d2y B. dt2 = y dy C. d2 = y 2 D. d y = x dt2 − E. Send help Second derivatives

d2 sin t dy2 = sin t = y dt2 − ⇒ dt2 − d2 cos t dx2 = cos t = x dt2 − ⇒ dt2 −

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

dy2 = y dt2 − Wait, what?

I I thought trig functions had to do with angles and 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 arc subtended by that angle: s r θ s = rθ Convention

I Angles increase counterclockwise

θ = 0 Convert from degrees to radians

Q6. In terms of radians, the angles 30, 45, 60, and 90◦ 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 angles 364 Appendix F. Trigonometry review

degrees radians sin(t) cos(t) tan(t) 0 0 0 1 0 30 π 1 √3 1 6 2 2 √3 π √2 √2 45 4 2 2 1 π √3 1 √ 60 3 2 2 3 90 π 1 0 2 ∞

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

b c θ

a

Figure F.2. Law of cosines states that c2 = a2 + b2 2ab cos(θ). − Here are other important relations between the trigonometric functions that should be remembered. These are called trigonometric identities:

Angle sum identities The trigonometric functions are nonlinear. This means that,forexample,thesineofthe sum of two angles is not just the sum of the two sines. One can use the law of cosines and other 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) − Connection with angle ( ) 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 = ω dt

I Motion can be described by either 1. 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 any value 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

sin t 1 tan t = , cot t = cos t tan t 1 1 sec t = , csc t = cos t sin t Derivative of tan x

Q7. The derivative of tan x is A. cot x B. cot x − 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

c

b θ a

c2 = a2 + b2 2ab cos θ − Law of cosines

c2 = a2 + b2 2ab cos θ − π Q8. In the special case, θ = 2 , the law of cosines reduces to which of these? A. c2 = a2 + b2 2ab − B. c2 = (a b)2 − C. c2 = a2 + b2 D. sin2(t) + cos2(t) = 1 Summary

I The mathematics of love leads to a system of differential equations

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

I Sine and cosine are related to a system of differential equations

I Sine and cosine are related to a second-order differential equation (y00 = y) − I Trigonometric functions are related to motion around 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 Related Exam Problems