Math1023 - Lecture notes all PDF

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Week 1 and 2 ● Quick equations ○ F=ma○ a=dv/dt=d2 x d t 2 ○ v=dx/dt ● Free falling objects ○ dv/dt=g ○ dv/dt=g−k v 2○ vdv dt=−GMr 2+k v 2● Gravitational force○ F=−GMm r 2where Gm> 0○ v(r)=±√❑ and g=6 10 − 11 m/kgs 2 ● Constant g and air friction depending linearly to velocity ○ F=F 1 −F 2 or grav...

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Week 1 and 2 ● Quick equations F=ma ○ ○

●

v

dv −GM = 2 +k v 2 dt r

Gravitational force ○ ○

●

d2 x d t2

v =dx / dt ○ Free falling objects dv /dt = g ○ 2 ○ dv /dt =g−k v ○

●

a=dv /dt=

−GMm r2 v (r)=± √❑ F=

Gm>0

where and

g=6.67 x 10

−11

m /kg s

2

Constant g and air friction depending linearly to velocity F =F1− F 2 or gravitational force minus air friction ○ F=mg − kmv where m is mass ○ −kt

○ ●

v =g /k

Terminal velocity

The order of a DE is the order of the highest derivative in it e.g. ○

● ●

g− A e k

For very large t, the motion is approximately free fall with constant escape and terminal velocity ○ Escape velocity v o =√ ❑ ○

●

v (t )=

d 12 v would be the 12th order d t12

The degree of a DE is the highest power of highest derivative

The standard form of a first order DE is An nth order DE is linear with form

dy /dx=f (x , y )

❑

●

an (x )

d v dn v +...+a1 (x ) ❑ +a0 ( x ) y =b( x) n dx dx

○

All derivatives of y have power 0 or 1

○

Coefficient

○

Remaining term (RHS of DE) depends on x

k

a k of d vk dx

depends only on x for all k

Week 3 ● Radioactive decay dm / dt =−km , k >0 ○ Probability of decay is constant in time ○ The rate of change in mass is proportional to the current mass ○ K is the decay constant

○ ○

m (t)= A e−kt Half life, denoted by ■

● ●

t 1 /2 =

t1 /2 is the time such that

m (t1 /2 )=m 0 /2

ln 2 k

Population with x as population Linear x (t)=x o + kt

dx =k dt ○ If k >0 the population increases ○ If k =0 the population is constant ● Exponential x (t)= x o e kt dx =kx ○ dt ○ If k >0 the population increases ○

●

○ We assume birth rate minus death rate is constant Logistic ○

dx =g (t , x ) x dt

where g(t , x )

○ ○

If x is small, x(t) grows so g(t , x )> 0 If x is very large, x(t) levels so g(t , x )≈ 0

○

If x is too large, x(t) decreases so g(t , x )< 0

○

represents birth rate - death rate

dx =(k −a x o )x dt

dx =x (birth rate−death rate) dt ○ ○ ○

If (b.r. - d.r.) = f(t), then DE is separable If (b.r. - d.r.) = f(x), then DE is separable If (b.r. - d.r.) = f(x,t), then DE may or may not be is separable

dx =ax (b− x) dt ● Carrying capacity: b=k /a ; growth rate: k =ab b−x o b x (t)= where A= ○ −kt xo 1+ A e ●

Logistics equation in equivalent form is

○ ● ● Week 4 ●

If x = 0 and x=b are equilibrium solutions

Concentration = m/v (mass over volume) If the rate out = rate in, then the volume is constant

v (t )= v

where a , b>0

Week 5 Second order DE with constant coefficients ● y ' ' +ay '+by=0 has characteristic equation

2

m + am+ b=0 Case 1: if

●

a2−4 b >0 , then m 2+am + b=0 has distinct roots m 1 , m 2 −a ± √❑ ○ m 1,2= ❑ m x m x and y 2 =e are both solutions to y ' ' +ay ' +by =0 ○ y 1 =e 1

○

2

Then the general solution is

1

1

( A ,B∈R¿

a2−4 b distinct real roots from characteristic equation

●

Critical damping -> repeated real roots

●

Subcritical damping -> complex roots

●

Undamped, i.e. simple harmonic motion

●

Small angle approximation of pendulum motion

●

System of first order DE with constant coefficients (converting from system of first order DE into second order)

● ●

Converting second order DE into a system of first order DE

Week 6 ● Solving higher order inhomogeneous equations ○ ○

●

To find

yh

To find

yp

use the characteristic equation use educated guesses, then equate the

y

x to ¿¿ F¿

●

Amplitude of undamped oscillation with periodic forcing and resonance in undamped harmonic oscillator with periodic forcing

yp

●

Guessing

●

System of DE ○ Coupled systems have x, y in both equations

●

●

Types of systems ○ Uncoupled equations

○

Equations

■ Successively means to plug one into the other ○

System of first order DE

Week 7 Graphing in 3D j=(0,1,0 ) k =(0,0,1 ) Standard basis of 3D: i=( 1,0,0) - Use Right-hand rule where thumb is the z-axis and index is x-axis A curve in 2D can be represented by 1. An equation in x and y 2. The graph of a function f(x) 3. Parametric equations x=f(t) and y=g(t) Parametric equation examples

A curve in 3D can be parameterized by x=f (t ), y=g(t ), z=h( t ) ● To graph, make a table of values then join the points. ○ If t is linear, then only plot two points and join together by a straight line. Two ways of representing a surface in 3D 1. By equation x, y, and z 2. By graph of a function f(x,y) -

Rotation symmetry about the z-axis

2

2

2

z =x + y −1

or

2

2

Plane ax +by +cz = d where a, b, c, d are contained in real numbers ● We find three points on the plane, then connect them.

a y−¿ where ¿ 2 2 z −b ¿ =r ¿

Cylinder

x

is assumed to be contained in real

numbers ● The center is at (a , b) ● Infinitely long cylinder because x is not contained in interval 2

Cone

x y2 2 + 2 =z a2 b

2

z =x + y +1

z−c ¿2=r 2 2 Sphere y−b ¿ +¿ 2 x−a ¿ +¿ ¿

2

(Elliptic) Paraboloid

z=

x y2 + 2 +c a2 b

Hyperbolic paraboloid

z=

2 x2 − y +c a2 b2

red line sits inside the xz plane while the green sits in the y area

Hyperboloid

x2 + 2

y 2 z 22 =d , d ϵ {− 1,0,1 } −c b2

Let D be a subset of R2 . A real valued function of two variables is a exactly on real number, denoted by f (x , y ) to every (x , y )ϵ D .

If

● ●

The set D is the domain of the function f (x , y ) The natural domain is the largest set D for which f (x , y )

●

The range of the function is the set of all values of

function that assigns

is defined

f (x, y) where (x , y )ϵ D .

3 2 f : D ⊂ R → R , then the graph of f is the set of points ( x , y , f ( x , y ,))ϵ R where (x , y )ϵ D . - Equivalently, the graph of f (x, y) is the set of points in R3 such that z=f (x , y )

A level curve of a function f (x, y) is a curve in R2 satisfying the equation f (x , y )=c where c is a constant in the range of f. ● A level curve is the intersection of graph f (x , y ) with the plane z=c ● A level curve need not be a continuous curve. ○ It is just a set: called the level set ●

If c is not in the range of

f (x , y ) then {f (x , y )−c }=∅ , the empty set.

Week 8 Slope of tangent line to a surface intersected with a plane e.g. y=b ; thus this is with respect to x then is a curve ● If (a , b)ϵ D , Or we are intersecting the graph with the plane with the plane This curve γ 1 is the graph of g1 (x)=f (x , b) ○

●

y=b , aka the curve is a cross-section of the surface

● ● ●

Assuming

g1 (x) is differentiable, then the slope of the tangent line to

γ 1 at (x , y )=(a , b ) is

Given a function

f : D ⊆ R 2 → R , (x , y )→ f (x , y) , the partial derivative of f with respect to x at (x , y )=(a , b)

is

provided the limit exists. Consider

f : D ⊆ R 2 → R , (x , y )→ f (x , y)

. If

f x (a , b) exists for all (a , b)ϵ D , then the function is said to

be differentiable with respect to x on D, and the partial derivative of the function with respect to x is the function of two variables given by

Slope of tangent line to a surface intersected with a plane e.g. x=a ; thus this is with respect to y ● If (a , b)ϵ D , then is also a curve

γ 2 is the graph of g2 ( y )= f (a , y ) ● Assuming g2 ( y ) is differentiable, then the slope of the tangent line to γ 2 at ●

This curve

Differentiation Rules

(x , y )=( a , b ) is

f (x, y) is sufficiently smooth, then ● To find f x (x , y ) , then treat y as a constant and apply the differentiation rules for the x variable ● To find f y (x, y ) , then treat x as a constant and apply the differentiation rules for the y variable

Suppose

Multiple variables

Vector-valued function

Consider

and assume it is differentiable

x=a is y −f (a)=f '(a)( x −a) ● Geometrically, the tangent line just touches the curve y=f (x) at the point ( a , f ( a)) . ○ Note: the tangent line is never vertical since ¿ f ' (a)∨¿ ∞ ●

Consider exist. ●

at

f : D ⊆ R 2 → R , (x , y )→ f (x , y) , and assume the function is sufficiently smooth so that

Geometrically, the tangent plane to the surface surface at that point.

Equation of tangent plane ● The equation of a plane in

●

y =f ( x )

The tangent line to the curve

R

3

z=f (x , y ) at the point

containing the point

f x and f y

(a , b , f (a , b)) just touches the

(a , b , c) is

given by ○ Where a , b , c , α , β , γ ϵ R ( α , β , γ not all zero) The tangent plane to the surface z=f (x , y ) at (x , y )=( a , b ) contains the point the equation of the tangent plane is ○

(a , b , f (a , b)) . So

Calculating the unknown values α , β , γ ● Intersecting the tangent plane with the plane ●

Setting y=b

○

Where

l1 in the equation of the tangent plane, gives the equation of l 1 :

−α γ

is the slope of

y=b , then we get a line

l1 and we can assume that

γ ≠ 0 , otherwise

l1 is tangent to the curve z=f (x , b) at x=a , so l1 has slope −α −β =f x (a , b) and similarly, =f y (a ,b) ○ So γ γ ○

●

Therefore the tangent plane is given by the equation −f x (a , b)( x−a)− f y (a , b)( y−b )+ z−f (a ,b )= 0 ○

The Equation of the tangent plane to the surface

z=f (x , y ) at the point

(a , b , f (a , b))

l 1 would be vertical

f x (a , b)

Week 9 Linear Approximation ● One variable ● Let the tangent line at point

y=g(x ) be

x=a of curve

x 0 of the curve is near a , then g(x 0 ) is approximately

●

If

● ●

Two variables The equation of the tangent plane to the surface

z=f (x , y ) at

z=f (a , b)+ f x (a , b )( x−a)+ f y (a ,b)( y−b ) ● If (x 0 , y 0 ) is near (a , b) then f (x 0 , y 0 ) is approximately

(x , y )=(a , b ) is

Differentiation ●

Let z=f (x , y ) defined by

where

f is differentiable ( f x , f y

exist). The differential

○ ●

Differentiation and linear approximation ○ If dx= Δ x and dy= Δ y , then dz ≈ Δ z since dz=f x (x , y) Δ x +f y (x , y )Δ y≈ Δ z =f ( x+ Δ x , y + Δ y )−f (x , y) ○ ○

●

Δ z is the increment in z

dz of the function is

Chain Rule ● Let z=f (x , y ) with x=g(t ) and y=h (t) , where derivative of z=f (g(t), h(t )) is given by

f , g , h are differentiable. Then the total

dz ∂ z dx ∂ z dy + = dt ∂ x dt ∂ y dt

● ●

●

Chain rule for partial derivatives Suppose that z=f (x , y ) with are differentiable.

∂z ∂ z ∂ x ∂ z ∂ y = + ∂s ∂ x ∂s ∂ y ∂ s

x=g(s , t)

and y=h (s .t) , where

∂z ∂ z ∂ x ∂ z ∂ y = + ∂t ∂ x ∂ t ∂ y ∂ t

f , g,h

Representing curve in

● ●

R

2

as a graph

Implicit function theorem

C ⊂ R 2 be a curve defined by

Let

constant. ○ Suppose that

f (x , y )=k

(a , b) is a point on the curve

for

f (x , y ) is differentiable and k ∈ R is a

C ( so f (a , b)=k

) with the partial derivative

f y (a , b)≠ 0 . ○

● ● ●

Then there exists a region D around (a , b) such that the part of the curve is the graph of a differentiable function y=g(x ) , or the local result.

C in region D

Implicit differentiation (application of the implicit function theorem and chain rule)

y=g(x ) is defined implicitly by f (x , y )=k where k is constant, and if (without finding g(x) in explicit form) dy −f x ( x , y) = dx f y(x, y)

If

f y (x , y )≠ 0 , then

Week 10

●

●

Directional derivatives - motivation ○

Let the positive x-axis point east

○

Let

h(x , y) be the elevation above the ground level. Then h x ( x, y) is the rate of change in if you travel east, and h y (x , y) is the rate of change in h if you travel north

f (x , y ) is differentiable, then: ○ We compute f x (x , y) by fixing the y

h

Examples: given

○

positive x-axis; We compute f y (x , y ) by fixing the

variable, so

x variable, so

f x (x , y) is the derivative in the direction of f y (x, y) is the derivative in the direction

of positive y-axis;

●

A vector is a quantity having both

direction and magnitude ○ The standard basis of ○

Any vector

u∈R

2

R

2

is the set {i,j}

can be written as

u=u 1 i+ u2 j for some

u1 ,u 2 ∈ R

■ i is the vector pointing one unit in the positive x-axis, j y-axis ○

u=u 1 i+ u2 j is ¿ u∨¿ √❑ ¿ v ∨¿ 1 ■ A unit vector v has u , is the unit vector in the direction of u ● Any non-zero vector

The magnitude of a vector

( magnitude 1) ○

○ ○

The dot product of u and v is defined by u ⋅v =u 1 v 1 +u 2 v 2 (scalar) where theta is the angle between the vectors and ■ u ⋅v=¿u∨¿ v∨cos θ ■ The vectors must go away, not towards one another The vectors u and v are orthogonal / perpendicular if u ⋅v=0 Orthonormal vectors are unit vectors that are orthogonal to each other

0 ≤θ ≤ π

Directional derivatives

●

^u=u 1 i+ u2 j be the unit vector in the direction of a non-zero vector u. The directional derivative of a

Let

function

○

● ●

f (x , y ) in the direction of u at the point (a , b) is defined by (provided the limit exists:)

We use the unit vector to define/compute

D u f =(f x (a ,b)i+f y (a , b) j)⋅ u^

Theorem 1

f (x , y ) is differentiable, then the directional derivative of f in the direction of u (not necessarily a unit vector) at (a , b) is ● Where ^u=u 1 i+ u2 j is the unit vector in the

If

direction of u

●

●

If we don’t use

^u but use u=10u^ =10 u1 i+10 u2 j , then 10( f x ( a ,b)u1 +f y (a , b)u 2)

● ● ● ● ●

Geometric meaning of Suppose

D u f (a ,b)

f (x, y) is differentiable and there exists vector u with

g (t)=f ( a +u1 t , b+ u 2 t)

where t ∈(δ , δ ) for small

^u representing unit vector of u

δ> 0

x=a +u 1 t , y =b +u2 t , z=g(t ) define a curve γ ○ This curve is the intersection of the surface z=f (x , y ) and the plane containing vector u and z-axis D u f (a ,b) is the slope of the tangent line to γ at (a , b , f (a , b))

Then

Gradient and directional derivatives ● If f (x , y ) is differentiable, then the gradient of f 2

●

The gradient is a vector-valued function, also called a vector field in

●

The directional derivative of a differentiable function of f (x , y ) in the direction of u is ∇f (a , b) contains the ‘full’ info of 1st order derivative of f (x , y ) at (a , b) ; ○ gives

●

Du f

Properties of the nabla / del operator ∇(f + g)=∇f +∇g ○

∇

R

∇f (a , b)⋅ u^

○ ○

∇(α f )=α ∇f where ∇(fg)=f ∇g +g ∇f

α is a real constant

D u f (a ,b)=∇f (a , b)⋅ u^ −¿ ∇f (a , b)∨cos θ where 0 ≤θ ≤ π is the angle between ∇f (a , b) and ^u D u f (a ,b)>0 1. iff 0 ≤θ ≤ π /2 Steepest ascent/increase in f : D u f (a , b) achieves its maximum value ¿ D u f (a ,b)∨¿ when θ=0 , i.e. in the direction of ∇f (a , b) ●

2.

D u f (a ,b) 0 there exists δ > 0 such that ¿ f (x)−l∨¿ ε whenever

0 0 there exists δ> 0 such that ¿ f (x , y )−l∨¿ ε whenever 0...