Math Final Flashcards by Melissa Wei

You have 10 meters of fence and want to build an enclosure that is as small as possible, but it can’t be narrower than one goat (1/2 meter). How long and wide should you make the enclosure?

1) Draw & Label a Picture

2) Write down what you wish to optimize:

3) Write an equation that expresses what you wish to optimize & in terms of variables

4) Reduce equation down to 1 variable using constraint in problem

5) Differentiate & solve for Global Extrema wanted: find f(x) at critical, singular & endpoints

6) Check if realistic within domain:

w=1/2m
l=9/2

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6 steps to performing Optimization

1) Draw & Label a Picture

2) Write down what you wish to optimize:

3) Write an equation that expresses what you wish to optimize & in terms of variables

4) Reduce equation down to 1 variable using constraint in problem

5) Differentiate & solve for Global Extrema wanted: find f(x) at critical, singular & endpoints

6) Check if realistic within domain

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Optimization on open vs closed interval

If f (x) is continuous on a closed interval, then f(x) has a global maximum and a global minimum on that interval

A continuous function on an open interval does not necessarily have a global maximum and a global minimum on that interval. However, you can still have a local and global extrema by determining where the function is increasing and where it is decreasing.

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100m of fencing to enclose a rectangular area against a straight wall. What is the largest area you can enclose?

1) Draw & Label a Picture

2) Write down what you wish to optimize:

3) Write an equation that expresses what you wish to optimize & in terms of variables

4) Reduce equation down to 1 variable using constraint in problem

5) Differentiate & solve for Global Extrema wanted: find f(x) at critical, singular & endpoints

6) Check if realistic within domain

A=1250
w=25

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Extremas & How to Evaluate for Them

Found evaluating the function at critical f’(x)=0, singular f’(x)=DNE & endpoints f(a)=#, f(z)=#, interval of [a,b]

Local Max x=c if f(c)f(x) for all x near c

Local Min x=c if f(c)f(x) for all x near c

2nd Derivative Test:
- If f’‘(c)<0 then c is a local max
- If f’‘(c)>0 then c is a local min

Global Max x=c if f(c)f(x) for all x
Global Min at x=c if f(c)f(x) for all x

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Find Global extrema of -x^4+x^2+1 on (-5,0)

Global Max at x=-sqrt(1/2)
Global Min at x=0

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1m long string cut into 2. One piece forms a square the other into a circle. Where to cut the string to maximize the total enclosed area of both shapes?

1) Draw & Label a Picture

2) Write down what you wish to optimize:

3) Write an equation that expresses what you wish to optimize & in terms of variables

4) Reduce equation down to 1 variable using constraint in problem

5) Differentiate & solve for Global Extrema wanted: find f(x) at critical, singular & endpoints

6) Check if realistic within domain

x=1/12

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Sketch x^2, x^3, logx, e^x, sinx, cosx, tanx, 1/x, sqrt(x), absolute x

See desmos

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Which term in a polynomial function will dominate for small x & large x

For small x, smaller degrees of power will dominate.

For large x, larger degrees of power will dominate

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Sketch & asymptotically show e^x dominates any power function

(e^x, x>k)>x^k

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Polynomial Power Function

f(x)=Kx^n, K ^ n are constants, can be negative or fraction

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Polynomial Functions

f(x)=anx^n+an-1x^n-1+…+a2x^2+anx+a0, a & k are constants

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Sketch f(x)=x^2-x^4, f(x)=3x+x^2, e^x-x^4 using asymptotic reasoning

See desmos + notes for reasoning

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Hill Function, for small & large x what does it look like

f(x)=(Ax^n)/(B+x^n)

Reason out denominator first, then simplify
For small x, f(x)=Ax^n/B
For large x, f(x)=A

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Asymptotic Reasoning Process

What is negligable at small x, simplify equation
What is negligable at large x, simply equation
Sketch out, find points where they intersect if required

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Explain with words & graphs what lim(x>a^+/-) f(x)=L

Limit of f(x) as x approaches a is equal to L means f(x) is arbitrarily close to L provided x is sufficiently close to a (but not equal to a)

Limit of f(x) as x approaches a from below is equal to L is arbitrarily close to L provided x is sufficiently close to a and x<a (left hand limit)

Limit of f(x) as x approaches a from above is equal to L is arbitrarily close to L provided x is sufficiently close to a and x>a (right hand limit)

If function does not approach a single number as x approaches a, the limit does not exist (DNE)

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Explain with words & graphs what lim(x>a^+/-) f(x)=+/- infinity

Vertical asymptotes

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Explain with words & graphs what lim(x>+/- infinity) f(x)=L

Horizontal asymptotes.

Limit of f(x) is arbitrarily large & +/- provided x is sufficiently close to a (but not equal to a)

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How to calculate limit

Easy, continuous function’s limits computed by substituting =limx->a f(x)=f(a)

Not nice (0/0, infinity) function’s limits computed by simplifying into nice function, finding holes, VA & HA asymptotes & finally substituting

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How to find features of quotient functions

(x+3)/(x+3)(x-2) Hole: x=-3, Vertical A: x=2 sub in x=1.9, 2.1 to find +/- limits, Horizontal A: n<d y=0, n=d y=coe/coe, n>d y=oblique, sub in x=+/- infinity

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Identify & explain what a function with a continuous domain is

f(x) is continuous at x=a if xaf(x)=f(x), unbroken, consistent graph, no abrupt changes in value (ex. Polynomials, trig, e^x, log(x))

Discontinuous (ex. piecewise, traffic light)

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Points of Discontinuity

Jump Discontinuity: piecewise

Removable Discontinuity: hole

Infinity Discontinuity: vertical asymptotes

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When given a function with paraments, how to select parameter values to make a function continuous

Set equations equal at x=a, if x<k=f(x), xk=g(x), f(k)=g(k), solve for variable

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Average Rate of Change

Slope of a Secant Line =[f(x1)-f(x0)]/x1-x0

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Solve for f’(1), if f(x)=x^2

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Linear Approximation + Use

Linear Approximation/Tangent Line to f(x) at x=a is L(x)=f(a)+f’(a)(x-a), used to approximate hard values of functions

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Linear approximation to f(x)=e^x at x=a=0, f’(0)=1

Find e^1/10

L(x)=f’(0)+f’(0)(x-0)=e0+1(x-0) = 1+x

L(1/10)=1+1/10=1.1

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Euleur’s Number e

lim(h>0) eh-1/h=1

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Solve for derivative of e^x

e^x

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Differential Equation + 2 Examples

Differential Equation y’(t)=y(t): a unknown function & its derivatives with a solution that satisfies it

y=e^x, y’=e^x, velocity vs acceleration, compound interest, bacteria growth, population growth
y=Ce^x, y’=Ce^x, h>0(Ce(x+h)-Ce^x)/h=h>0(Ce^xeh-Ce^x)/h=limh>0 Ce^x(eh-1)/h=Cex

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Instantaneous Rate of Change

Derivative of f(x) at x0 is f’(x0)=d/dxf(x)=lim(h>0)[f(x+h)-f(x)]/h

Doesn’t exist at corners, sudden changes in slope, jumps or where the limit is infinite number

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Linearity of Differentiation + Proof

Proof using the limit definition of derivative & use to break down derivatives of sums + constant multiples

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Power Rule + Proof

Proof using the limit definition of derivative & use for integer exponents

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Derivative of a function

f’(x) = df/dx =limh>0(f(x+h)-f(x))/h

Used to solve derivatives, show how derivatives can be equivalent, etc.

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Find derivative of f(x)=3x+x^2

f’(x)=3+2x

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Derivative of an exponent (a^x) & tan(x)

d/dx 4^x=4^xlog(4)
d/dx a^x=a^xlog(a)
d/dx a^f(x)=a^f(x)log(f(x))f’(x)
d/dx tan(x)=sec^2(x)

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Reciprocal Rule + Proof

See notes

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Product Rule + Proof

See notes

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Find derivative of f(x)=x^2+1/x^2

f’(x)=2x-2/x^3

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Find derivative of f(x)=x*e^x

f’(x)=e^x+x*e^x

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Quotient Rule + Proof

See notes

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Find derivative of f(x)=(x^3+1)(x^2+2)

f’(x)=3x(x^2+2)+(x^3+1)2x

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Graphically show trig f(x)s & their derivatives

See notes

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Addition Formulas

sin(a+b)=sin(a)cos(b)+sin(b)cos(a)
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
limh>0 sin(h)/h=1
limh>0 cos(h)-1/h=0

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Derivative sinx, cosx & tanx proof

See notes

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Chain Rule

d/dxf(g(x)) = f’(g(x))g’(x) = df/dgdg/dx

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Differentiate f(x)=sin(3x)

cos(3x)(3)

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Differentiate f(x)=(x^3-2x+1)^6

6(x^3-2x+1)^5(3x*2)

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Implicit Differentiation

Taking derivative of both sides using Chain rule
Solve for dy/dx

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Differentiate x^2+y^2=1, then find x where y’=1 using Algebra & Diagram

dy/dx=-2x/2y
x=+/- 1/sqrt(2)

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Logarithmic Differentiation

Applying the natural logarithm to both sides of equation to make it easier to differentiate, especially in f(x)^g(x), combining Chain Rule+Implicit Differentiation+Logarithmic Rules

Take log of both sides, simplifying using log rules
Take d/dx of both sides (implicit differentiation+chain rule)
Isolate for y’(x)

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Differentiate x^a using logarithmic differentiation

f(x)= x^a
logf(x) = logx^a
1/f(x) * f’(x) = alogx
1/(x^a) * f’(x)=a/x
f’(x) = a/x(x^a)
f’(x)= ax^(a-1)

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How to solve a related rate question?

Draw & label a picture + Write down what is known
Write down what you wish to find
Write down an equation that relates the thing whose rate y (reduce to 1 variable)
Differentiate & Solve
Check if answer makes sense

Rate of Change of Area = L’W + LW’
Rate of Change of Focal Length of Lense
Rate of Change of radius of expanding circle
Rate of change of water level entering inverted cone
Rate of Change of angle of a moving person across a room

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Wall = 5.9m
Rate of Change of Walking = 1/4m/s
x = opposite/walking distance
tangent = angle

At tangent=pi/4, what is the rate of change of tangent?

Draw label diagram
Find dtangent/dt or rate of change of tangent or f’(pi/4) f(x)=tangent
t * tan(tangent)=tx/L –> differentiate ^ solve
(1215.6)1.4

An inverted cylindrical cone, 28 m deep and 14 m in diameter across the top, is being filled with water at a constant rate of 13 m3/min. At what rate is the water rising in the tank when the depth of the water is 1m, 10m, 27m

208/pi(h)^2

Draw sin(x)=arcsin(y) & sin(y)=arcsin(x) at range[-pi/2,pi/2]<->domain[-1,1]

See Diagram

Draw cos(x)=arccos(y) & sin(y)=arccos(x) at range[-0,pi]<->domain[-1,1]

See diagram

Differentiate arcsin(x) in terms of x

d/dxsin(y)=dx/dx
cos(y)dy/dx=1

Triangle + SOH CAH TOA

dydx=11-x2

Differentiate arccos(x) in terms of x

d/dxcos(y)dy/dx=dx/dx
dy/dx=-1/sin(y)

Triangle + SOH CAH TOA

dy/dx=-1/sqrt(1-x2)

Differentiate arccos(e^2x) in terms of x

2e^(2x)/sqrt(1-e^4x)

Draw tan(x)=arctan(y) & tan(y)=arctan(x) at range[-pi/2,pi/2]<->domain[-infinity,infinity]

See diagram

Differentiate arctan(x) in terms of x

d/dxtan(y)=dx/dx
sec^2(y)dy/dx=1
dy/dx=1/sec2(y)
dy/dx=cos^2(y)

Triangle + SOH CAH TOA

dy/dx=1/1+x2

How to determine concavity?

If f(x)’‘>0 = slope is increasingly increasing = concave up = f(x) lies above its tangent lines on that interval
Diagram

If f(x)’‘<0 = slope is increasingly decreasing = concave down = f(x) lies below its tangent lines on that interval
Diagram

Degree 0 & 1(linear) approximation around x=0? Diagram?

T0(x) be the degree 0 approximation of f(x) around x=0: T0(x)=f(0)=1
Let T1(x)be the linear approximation to f(x) about x=0: T1(x)=f(0)+f’(0)(x-a)=1+1(x-0) = 1+x

The higher the degree of approximation, the better the approximation

Degree n Approximation to f(x) at x=a is ?

Tn(x)=f(a)+f’(a)(x-a)+f’‘(a)/2(x-a)^2+f’’‘(a)/3!(x-a)^3+…+f(n)(a)/n!*(x-a)^n

Taylor vs Maclaurin Polynomial

Taylor x=a
Maclaurin x=a=0

Proof of Degree nth Approximation

Find c0,c1,c2,c3,…,cn, such that the kth derivatives match: T^(k)n(x)=f(^k)(a) for k=0,1,2,3,…,n

Proof of Degree 3 e^x Approximation

Let T2(x)=c0+c1(x-0)+c2(x-0)2=c0+c1x+c2x2 be the degree 2 approximation to f(x)=ex about x=0
Find c0, c1 and c2 such that
T2(0)=f(0)=1 → c0+c1(0)+c2(0)2=c0=1
T’2(0)=f’(0)=1 → c1+2c2(0)=c1=1
T’‘2(0)=f’‘(0)=1 → 2c2=1, c2=1/2
Therefore: T2(x)=c0+c1x+c2x2 = 1+1x+1/2x

cos(x) nth Degree Maclaurin Polynomial + Show Work

Tn(x)=1-x2/2!+x4/4!+…

sin(x) nth Degree Maclaurin Polynomial:

Tn(x)=x-x3/3!+x5/5!+…

Find degree 3 MP of f(x)=4xe^2x:

expand: T3(x)=4x[1+(2x)+(2x)2/2!+(2x)3/3!]

Derive f(x)=a/1-x general formula of nth degree Maclaurin Polynomial

Tn(x)=a+ax+ax^2+ax^3+…+ax^n

Find limit of the partial sums Sn as n–>infinity

Sn(x)=a+ax+ax2+…+axn-1
xSn(x)=ax+ax2+ax3+…+axn-1+axn
Sn(x)-xSn(x)=a-axn
Sn=a(1-x^n)/(1-x)
lim n–>infinity a +ax+ax2+…+axn-1 = a/1-x if absolute x < 1 so x^n is negligeble

Make statement about 5-10/3+20/9-40/27+….

lim n–>infinity (5+5(-2/3)+5(-2/3)^2+5(-2/3)^3… )
=5/(1+2/3)=3

Find the sixth degree Taylor Polynomial T6(x) about x=0 of (e^3x^2)/(1+5^x^3)

T6(x)= 1 + 3x^2-5x^3+ 0x^3 + 9/2x^4 - 15x^5 +59/2x^6

water tank is in the shape of an inverted right circular cone with top diameter 6 metres and depth 8 metres. Suppose that the tank is leaking water. Note the volume of a right circular cone is V = 1/3πr2h, where r is the radius and h is the height of the cone.

a) Suppose that when the water is 4 metres deep, the water leaks out of the tank at a rate of 3 m^3//hr. What is the rate of change of the depth of the water at this point in time?

b) Suppose now that the water from the conical tank is leaking directly into another tank, which is in the shape of a cube, and has base area A m2. How fast is the depth of the water in the cubic tank increasing atthe same moment in time described in part (a)?

a) h’ = 4/3pi m/hr
b) h’c = 3A m/hr

Consider a container filled with some fixed amount of ideal
gas. Suppose that the container is perfectly isolating, the thickness of its wall is negligible and it is capable of expanding or shrinking while staying perfectly spherical (you may think of an ideal balloon). The pressure p, the volume V and the temperature T of the gas satisfy the ideal gas law pV = nRT, where n is the amount of the gas enclosed in moles and R is a universal constant.

You may use that the volume of a solid ball of radius r is V = 4/3πr3 and the surface area of a sphere of radius r is A = 4πr2

(a) Suppose that we heat up the gas at a constant rate of 3 K/s while keeping the pressure at a constant level 3/4 Pa. What is the rate of change of the radius of the container, when the surface area of the container is 0.4 m2? Your answer should be in terms of n and R.

b) Now suppose that we keep the temperature constant while we are expanding the container in such a way that its radius is increasing at a constant rate of 0.1 m/s . What is d/dt log(p) when the radius of the container is 0.6 m?

dr/dt = 10nR.

d/dt log(p) = − 0.3
0.6 = −1/2

Consider an isosceles triangle of base 10cm and
sides x centimetres. Suppose x is increasing at a rate of 4cm per second.
a) What is the rate of change of the height of the triangle when x = 13cm?
b) What is the rate of change of the area of the triangle when x = 13cm?

a) h’ = 13/3

b) A’= 65/3

Curve Sketching Process

1) f(x)
- Domain Restrictions: Vertical asymptotes, Horizontal asymptotes, Holes
- Intercepts
- Positive+Negative Intervals

2) f’(x)
- Critical Points, Singular Points
- Increase & Decrease Intervals
- Local Extremas (Min/Max)

3) f’‘(x)
- Concavity

4) Global Extrema
- End, critical & singular points

What info can be pulled from just f(x)

Domain, +ve,-ve, intercepts, DNE, even, odd, periodic, asymptotes

Domain Restrictions: roots x>0 unless x^even, log(x), 1/x

Vertical Asymptotes: restrictions in denominator/when is denominator equal to 0 or limx>? f(x)=+/-infinity

Horizontal asymptotes: limx>infinity f(x)=? and limx>-infinity f(x)=?
1. x-f(x)=, no horizontal asymptote
2. N=D, y=Coefficients of N/D
3. N>D, y=N/D
4. N<D, y=0

Holes: factor, simplify, mark down what terms cancel out, set them to 0, isolate for x, sub into f(x)

Intercepts:
- x-intercept: set y=0, isolate for x
- y-intercept, set x=0, isolate for y

Postive +Negative Intervals:
- Test signs of intervals b/w roots & domain restrictions using f(x), -ve=-ve interval,+ve=+ve interval
- If f(x) is continuous, then f(x) will only switch sign around where f(x)=0 or DNE

What info can be pulled from just f’(x)

positive (f increasing), negative (f decreasing), f ′ = 0 or f ′ DNE

Critical Points, Singular Points:
- Critical Point (c): f’(c)=0, round
- Singular Point (s): f’(s)=DNE, sharp pointy

Increase & Decrease Intervals:
- Test signs of intervals b/w critical & singular points using f(x),
- -ve=interval of decrease,
- +ve=interval of increase

Local Extremas (Min/Max): First Derivative Test: if a function is continuous and switches from decreasing to increasing at x=c, there is a local minimum at x=c. If f(x) switches from increasing to decreasing at x=c, there is a local maximum at x=c

What info can be pulled from just f’‘(x)

positive (f concave up), negative (f concave down), f ′′ = 0 or f ′′ DNE

Concavity:
- Find points where f’‘(x)=0 or DNE , then test signs of intervals b/w them
- +ve=Concave Up
- -ve=Concave Down
- Inflection Point: an existing point where concavity switches

How to find global extrema in open vs closed interval?

Evaluate f(x) at end, critical & singular points
Smallest = global minimum
Largest = global maximum
Check concavity to double check
If f(x) is continuous on a closed interval, then f(x) has a global max and min on that interval
If f(x) is continuous on an open interval, there doesn’t have to be a global max and min on that interval, can still be determined

Sketch graph x^2/(x^2+1)

See desmos
1) f(x)
- Domain Restrictions: Vertical asymptotes, Horizontal asymptotes, Holes
- Intercepts
- Positive+Negative Intervals

2) f’(x)
- Critical Points, Singular Points
- Increase & Decrease Intervals
- Local Extremas (Min/Max)

3) f’‘(x)
- Concavity

4) Global Extrema
- End, critical & singular points

Sketch graph
x^2+8/x

See desmos
1) f(x)
- Domain Restrictions: Vertical asymptotes, Horizontal asymptotes, Holes
- Intercepts
- Positive+Negative Intervals

2) f’(x)
- Critical Points, Singular Points
- Increase & Decrease Intervals
- Local Extremas (Min/Max)

3) f’‘(x)
- Concavity

4) Global Extrema
- End, critical & singular points

f(x)=1/(1+e^(-kx)), k>0

What would happen to the graph if we changed the 1 to a 2?

If numerator changes 1-> 2 horizontal asymptote goes from y=1>y=2, more and more far apart

If denominator changes 1->2 vertical asymptote goes from 1=e^-kx > 2=e^-kx

See Desmos

Extrema

Extrema: found evaluating the function at critical f’(x)=0, singular f’(x)=DNE & endpoints f(a)=#, f(z)=#, interval of [a,b]

Local Max x=c if f(c)f(x) for all x near c
Local Min x=c if f(c)f(x) for all x near c

2nd Derivative Test:
If f’‘(c)<0 then c is a local max
If f’‘(c)>0 then c is a local min

Global Max x=c if f(c)f(x) for all x
Global Min at x=c if f(c)f(x) for all x

Find extrema of f(x)=-x^4+x^2+1 on (-5,0)

x=-1/(sqrt(2))

Geometric Optimization: 1m long string cut into 2. One piece forms a square the other into a circle. Where to cut the string to maximize the total enclosed area of both shapes?

String should be cut at x=112 for largest enclosed area.

Optimization Process

1) Draw & Label Picture
2) Write down what you wish to optimize: Area of enclosed space
3) Write an equation that expresses what you wish to optimize & in terms of variables
4) Reduce equation down to 1 variable using constraint in problem
5) Differentiate & solve for Global Extrema wanted: find f(x) at critical, singular & endpoints
6) Check if realistic within domain:

Algebraic Equation vs Differential Equation (Examples)

Algebraic Equation: contains an unknown number that is solved for
- Ex. 4=y2, y=4=2, Solution:2, Check:4=(2)2,4=4, LS=RS

Differential Equation: contains an unknown function & its derivatives that is solved for
- Solution=Function that satisfies the differential equation dydx=y
ddxe^x=e^x, Check:d/dx e^x=ex,ex=ex, LS=RS
d/dxCe^x=Ce^x, Check: d/dxCe^x=Ce^x, Ce^x=Ce^x, LS=RS

How to check if a function satisfies a differential equation

There could be many solutions & that don’t necessarily look similar.
1) Take is derivative dydx
2) Sub in to see if LS=RS

Check which function satisfies differential equation: dy/dx+x2-1=y

A)y=x2+1 B)y=x2+2x+1

A)y=x2+1, y’=2x,
LS:dydx+x2-1-y , sub in y’=2x, 0=2x+x2-1
RS: y=x2+1
LSRS, A) is not a solution to the differential equation

B)y=x2+2x+1, y’=2x+2
LS: dydx+x2-1, sub in y’, =2x+2+x2-1=x2+2x+1
RS: y, sub in y=x2+2x+1
LS=RS, B) Is a solution to the differential equation

Check which function satisfies differential equation: dydx=xy,
A)-x,
B)x+5
C)sqrt(x^2+5)

A)-x, y’=-1
B)x+5,y’=1
C)x2+5,y’=1/2(x^2+5)^–1/2*(2x)

Ansatz + Process

Ansatz: guess of form of the solution involving parameters/can find parameter values that give solution

1) Take necessary derivatives
2) Set LS=RS
3) Isolate & Solve for unknown parameter

Ansatz: Given differential equation y’‘-9y=0, y=e^ax find values of a such that y is a solution

^Therefore, y=e^-3x & y=e3x are both solutions to the differential equation, all in the form of y=Ce^x

Ansatz: Given differential equation y’‘+4y=0 & ansatz y=sin(ax) find values of a such that y is a solution

Therefore, y=sin(2x),y=sin(0x) or sin(-2x) are solutions to the differential equation

Initial Value Problem + 3 Solutions

Differential equation with an initial condition
1) Soln to dy/dt=ay with initial condition y(0) is y(t)=y(0)e^at
2) Soln to dy/dt=a(y-b) with initial condition y(0) is y(t)=(y(0)-b)e^at+b
3) Soln to differential equation dydt=y(1-y) is the logistic model y=1/(1+e^-t)

Proof to Soln to dy/dt=ay with initial condition y(0) is y(t)=y(0)e^at

Sketch the solutions to the differential equation dydx=2y, with initial condition y(0)=100

For any t value, slope is 2y. y(0)=Ce^2(0)=C=100, so y=100e^2t satisfies the initial value problem.

See notes

Proof to Soln to dy/dt=a(y-b) with initial condition y(0) is y(t)=(y(0)-b)e^at+b

Transform the differential equation dydt=a(y-b) by setting u=y-b & solving the differential equation with initial condition y(0)

Let u(t)=y(t)-b -> du/dt=dy/dt=a(y-b)=au,
Since du/dt=au is familiar to dy/dt=ay with soln y=C=y(0)e^at, so u(t)=u(0)e^at
In terms of y(t): y(t)-b=(y(0)-b)e^at y(t)=(y(0)-b)e^at+b

Newton’s Law of Cooling: suppose y(t) is the temperature of your coffee, and b is the temperature of the room, y temp is hotter than >b room temperature

1) Sketch 3 solutions to differential equation dydt=a(y-b) given a<0, b>0, y(0)>b, for y(0)=b and for y(0)<b in terms of y(t)

2) Sketch phase line, assuming a=-3, b=2 so dydt=-3(y-2) in terms ofy as function of t & dydt=-3(y-2) as a function of y

Everything approaches b, see notes

If dydt>0 coffee temp is increasing as y<b +ve=a(y-b)=a(-ve), a=-ve, a<0
If dydt=0, coffee temp is staying same y=b 0=a(y-b)=a(0), a=anything
If dydt<0, coffee temp is decreasing y>b, -ve=a(y-b)=a(+ve), a=-ve, a<0

Stable steady state y=2

Temperature at the time of death was 37. Temperature 32.5 at 1:30. Temperature 30.3 at 2:30. Room temperature is 20. Time of death if bodies follow Newton’s Law of Cooling=dy/dt=a(y-b), y(t)=(y(0)-b)e^at?

Let y(t)=temp of body, t=time after death at 1:20
1) y(0)=37, y(t)=32.5, y(t+1)=30.3, b=20, unknown=a & t
2) y(t)=(y(0)-b)eat+b 32.5=(37-20)eat+20eat=12.517

3) y(t+1)=(y(0)-b)eat+b 30.3=(30.3-20)eat+20eat=10.317

4) ea(t+1)eat=10.3/1712.5/17a=In(10.312.5)

5) With a, Isolate for t*

So at 1:30 it’s been 95mins since time of death, so murder occurred at 11:55pm

Phase Line

Phase line records dy/dt values on a horizontal y-axis

Slope Fields: y vs t axis
Phase line: dy/dx vs y axis

Identify critical points & record movement in b/w
f(y)>0, the solution y(t) is increasing=right pointing arrows

f(y)<0, the solution y(t) is increasing=left pointing arrows

f(y)=equilibrium

Slope Field

Slope Field: values of dy/dt on y vs t axes

Steady State

Steady State: state in which a system is not changing, such as if y remains constant for all t (dydx=0) for a solution y(t) to a differential equation
- Stable Steady State: if all nearby solutions tend to it
- Unstable Steady State: if all nearby solutions tend away from it
- Semistable: otherwise

Soln to differential equation dy/dt=y(1-y) is the logistic model y=1/(1+e-t)

dydt=0 when y=0,1

Sketch slope field & phase line of

1) Soln to dy/dt=ay with initial condition y(0) is
2) Soln to dy/dt=a(y-b) with initial condition y(0) is
3) Soln to differential equation dydt=y(1-y) is

y(t)=y(0)e^at
y(t)=(y(0)-b)e^at+b
the logistic model y=1/(1+e^-t)

Find the distance fallen x(t) as a function of time analytically, experimentally, numerically?

Analytically - Newton’s Second Law of Motion: x(t), x’‘(t)=g=acceleration due to gravity
- if object starts from rest initial velocity is 0 =x(0)=x’(x)=0
- x(t)=12gt2 satisfies both differential equations
x’(t)=1/2g2t=gt,x’(0)=0, x’‘(t)=g

Experimentally: used to evaluate accuracy of model, estimate parameter values in a specific model or compare b/w different models
- Drop object at different heights, record time it takes, plot a graph, compare with analytical model graph

Numerically: allows us calculate numerical solutions to differential equations when we cannot solve them analytically, to solve for x(t) only knowing x’‘(t)=g
v(t)=x’(t), v’(t)=x’‘(t)=g
v(ti+1)=v(ti)+v’(ti)t=v(ti)+gt

Euler’s Method deriving equation + visual + computer

Euler’s Method y(ti+1)=y(ti)+y’(ti)t, t=ti+t-ti, linear approximation to the function y at the point t=ti+1
- To satisfy differential equation dydt=f(y), y(t0)=y0, let y’(t)=f(y) so - y(t1)=y(t0)+f(y(t0))t
Initial Condition y0=__
- Linear approximation to point t0 evaluated at t1: y1=y(t0)+f(y(t0))t
- Linear approximation to point t1 evaluated at t2: y2=y(t1)+f(y(t1))t
- Linear approximation to point t2 evaluated at t3: y3=y(t2)+f(y(t2))t

See notes

Sketch solution to dy/dt=y, y(0)=1 using Euler’s Method on paper + on computer

See notes

Sketch solution to dy/dx=y(1-y), y(0)=18 using Euler’s Method on computer

See notes

Issues with Euler’s Method

Underestimation or Overestimation: Numerical solution will keep underestimating as it uses tangent line for a concave up function or overestimating for a concave down function. Meaning tangent lines/linear approximations will always be below/below the function.

Second Order Taylor Approximation of x(ti+1) about ti: x(ti+1)=x(ti)+x’‘(ti)/2*(delta t)2
- Error depends on t and is added to each approximation,
D- ecreasing the time step t reduces the error in the numerical approximation

Newton’s Method: deriving equation + visual + computer

Newton’s Method: xn+1=xni-f(xn)/f’(xn) to estimate root (x-intercept) with only the expressions for f(x) & f’(x)
Derivation: L(x)=f(x0)+f’(x0)(x1-x0)=0x1=x0-f(x0)f’(x0)

See notes

Issues with Newton’s Method

Things could go wrong when using Newton’s Method as at times numbers don’t always converge
Beware of starting points near min or max where division could be by 0
Approximation is too far away from the roots to give an accurate estimate

Differences b/w Euler & Newton’s Method

Euler calculates linear approximation at all y values of f(x) using differential equations (& its solutions)

Newton calculates roots using linear approximations closer & closer.

Apply Newton’s method to f(x)=x^5-2x^4+3x^2+1

See desmos + computer

Plot points (x,y,x)=(2,1,3),(1,3,2)

Plot planes x, y, z

Evaluate z values for given pairs of f(1,2)=e^(x^2+y^2)

See notes

Determine domain & Range of f(x,y) process.

Determine domain of f, what combinations of x and y will f accept by thinking in terms of one variable function

Determine range of f, what values of f can be output by thinking in terms of one variable function and checking endpoints/boundaries of x & y

Determine domain & Range of
f(x,y)=e^(x^2+y^2)
f(x,y)=sqrt(9-x^2-y^2)
f(x,y)=log(arctan(x+y))

Domain:
ex2+y2ex2, no restrictions so domain is all x and all y

9-x2-y29-x2, 9-x20, 9x2, 3x . Domain x2+y29 or x+y3

No restrictions on arctan(x+y), log(x) only accepts x>0 so arctan(x+y)>0 when x+y>0 or y>-x

Range:
Range is [1,infinity) as all x2+y20 and e0 will only output range of 1 or higher

Range is [0,3] as domain is 0<x^2+y^2<9, so plug in x^2=0,f(x)=sqrt(9-0^2)=3 and x^2=9, f(x)=sqrt(9-3^2)=0

Range is (log(0),log(pi/2))=(-infinity,log(pi/2)) as x+y>0 so plug in x+y=0,f(x,y)=log(arctan(0))=log(0) and x+y=, f(x,y)=log(arctan(infinity))=log(pi/2), so0<arctan(x+y)<pi/2

Partial Derivatives + f(x,y)=3y^2sin(x+3)

A partial Derivative in x, at a point (x,y), written dfdxfx or fx(x,y) , describes the slope of the line tangent to f(x,y) along the x direction while keeping y fixed (x,z graph).
f(x,y)=3y2sin(x+3)fx=3y2cos(x+3)

A partial Derivative in y, at a point (x,y), written dfdyfy or fy(x,y) , describes the slope of the line tangent to f(x,y) along the y direction while keeping x fixed (y,z graph).
f(x,y)=3y2sin(x+3)fy=6ysin(x+3)

Second Order Partial Derivative with respect to x & y +
f(x,y)=3y^2sin(x+3)

See notes
fxx=-3y^2sin(x+3)
fxy or yx=6ycos(x+3)
fyy=6sin(x+3)

Level Curves + Given z=x2+y2, sketch level curves corresponding to z=4,1,0,-1

Level Curves (z constant): horizontal slices of a 3 dimensional surface on z-plane in terms of x & y

4=x2+y2
1=x2+y2
0=x2+y2
-1=x2+y2

Traces + Given z=x^2+y^2, sketch traces corresponding to y=0,x=0,x=1,y=2

Trace (x or y constant): vertical slices of a 3 dimensional surface on x or y-plane in terms of z and leftover dimension

z=x^2
z=y^2
z=1+y^2
z=4+x^2

Sketch surface of z=x^2+y^2

1) Domain + Range Restrictions
2) Level Curves
3) Traces
4) Combine

See notes

Critical, Singular & Saddle Points on f(x,y)

A function f(x,y) has a critical point at (a,b) if fx(a,b)=0 and fy=0. fx=0 and fy=0, slope in any x or y direction is 0.

A function f(x,y) has a singular point at (a,b) if fx(a,b)=DNE or fy=DNE
- f’(x)=#0,-ve, log(-ve),arcsincos(-1>#>1), etc.

A Saddle Point is a critical point at neither a max or min, is there anywhere that you can sit on.
- Inflection points (x3)
- Max in one direction, min in the other (x2-y2), think of a saddle (horse hump max, sitting area min)

Critical Points of functions

f(x,y)=x^2+y^2
f(x,y)=x^2-x^4+y^2
f(x,y)=x^2

f(x,y)=x^2+y^2: one critical point at (0,0)

f(x,y)=x^2-x^4+y^2 have: 3 critical points at (0,0),(-1/sqrt2,0),(1/sqrt2,0)

f(x,y)=x^2 have: critical points at (0,yR)

Singular Points of function
f(x,y)=/x/+(x-1)^2+y^2

f(x,y)=sqrt(x^2+y^2)

-x+(x-1)2, x>0
x+(x-1)2, x>0

1 critical point at (12,0), singular points at (0,yR)

1 singular point at (0,0)

Process to find absolute extrema in f(x,y) with a boundary

1) Find critical points
2) Find singular points
3) Boundary to 1 Variable
4) Find Glboal extrema of boundary
5) Sub in global extrema of boundary into function
6) Sub critical, singular & boundary points into original
7) Compare all z-values to find global extrema in form (x,y)

Find absolute extrema f(x,y)=(x-1)^2+y^2, to the region x^2+y^2<=4

Global Max: f(-2,0)
Global Min: f(1,0)

Find absolute extrema f(x,y)=x^2y+xy^2-3xy without boundary

Min: f(1,1)=-1
Max: f(4,4)=80

Process to find absolute extrema in f(x,y) without a boundary

1) Find max & min of boundary of shape, setting to 1 variable, first derivative to find extrema
2) Find max & min within shape by taking partial derivatives, combining into points where slope in all direction fy=fx=0
3) Subbing in endpoint, singular + critical points into original function

Calculate all four second-order partial derivatives of
f(x,y)=(3x+2y)e^y

fxx=0
fyx=3e^y
fyy=3xe^y+2e^y+2e^y+2ye^y
fxy=3e^y

Find critical points of f(x,y)=6x-x^2-3xy^2

(3,0) (0,-sqrt(2)) (0,+sqrt(2))

What f(x,y) have no global max or min?

A) f=xy,
B) f=x^3+y^3,
C) f=x^2+y^2,
D) f=sin(xy),
E) f=x^2-e^y

Has no limited range. Goes on to infinities. Can make function infinitely negative or positive.

A, B, E

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