Math 77 Notes
Sukhjeet Sekhon 2024-25
Instructions
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These notes are based off of De Leon’s lecture notes
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Try to avoid deleting slides
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Extra slides that have extra information are better than less information
Try to find good diagrams, models, drawings, or examples that have good explanations for
difficult concepts
Don’t cram notes
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Add notes from there or other reliable sources
Make new slides if you need to continue a topic into multiple slides
These notes are not made to teach someone about each topic, so don’t make full-blown
lessons with the slides (i might make another for future use that is styled that way)
These notes are made to be very shorthand and quick to review
Try to make review slides at the end of units or end of groups of slides that have “Review”
in the title for fasting studying
Tutorials
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Use mathispower4u - Calculus III for resources and tutorials & videos explaining topics
and examples
Calculus (OpenStax) LibreTexts Textbook
Notation
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Mrs. Leon is picky about notation
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Make sure to write vector arrows every time you write a vector variable like v, u, w, i, j ,k
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the zero vector also needs a vector arrow above the 0
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write the vector arrow above the letter of vector valued functions
Use angle brackets for vectors every time
use vector arrows over gradient symbols
Calculators
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Surface Integral over a surface (Flux over a surface/plane)
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Vector Cross product
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https://www.geogebra.org/m/EW3dYCeR
https://www.emathhelp.net/calculators/linear-algebra/cross-product-calculator/
Unit Vector (Normalization)
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https://www.emathhelp.net/calculators/linear-algebra/unit-vector-calculator/?u=x%2Cy%2Cz
Vector Notation
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Vector variables have a horizontal right arrow above the letter
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Components
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Angle Brackets contain the components
i,j,k equation form
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the x, y, z values in the vector
Angle Brackets in Component Form
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You could also use a vector hat, but it’s rarer
the vector components are coefficients of the unit vectors in the direction of each axis
Example
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This would also extend to 3D vectors with a “z” after “y”
What is a Vector?
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A vector is a series of values that corresponds to a coordinate point relative to the origin
This vector “points” to that point starting from the origin
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A 2D vector takes an x and a y value to “point” to the point (x,y) from the origin
Vectors have a direction and magnitude (length)
The magnitude of a vector is the hypotenuse of a triangle with legs of length x and y in 2D
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This simplifies to the magnitude equaling sqrt(x^2 + y^2 + z^2)
If the magnitude of the vector is 1, we call it a Unit Vector
There are 3 unit vectors that point in the direction of the x,y,z axes
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Magnitude of a vector is notated as double bars of the vector variable: ||v||
In 3D, the magnitude is the hypotenuse of a triangle with legs of length sqrt(x^2 + y^2) and z
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Vectors can also start from a point other than the origin only if specified
x, y, z -> i, j, k -> <1,0,0>, <0,1,0>, <0,0,1>
If the magnitude of the vector is 0, we call it a Zero Vector
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It also has no direction, but that’s obvious because it basically doesn’t point to anything
Magnitude of 3D Vector
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The magnitude for a vector in 2D plane is just the hypotenuse of a
triangle
The magnitude for a vector in 3D plane is also the hypotenuse of a
triangle
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One leg of the triangle is also a hypotenuse and the other leg is the z
value
You can think of finding the magnitude of a 3D vector as finding
the magnitude of the vector of the 2D plane using only the x and y
components and then finding the hypotenuse of the triangle with
a magnitude of the xy vector and the z vector
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Memorizing the formula is easy too, but understanding it is better
Vectors (copy)
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Have magnitude and direction
○ Magnitude is represented as the length from the initial point to the terminal point
Magnitude is notated as a value in between double bars ||v||
A vector with initial and terminal points at the same point is called a “Zero vector”
“Zero vector” is notated as the number zero with a horizontal right arrow above the
zero
Vectors are equivalent if they have the same magnitude and direction
Magnitude of vector v with initial point (x1,y1) and terminal point (x2,y2)
○ sqrt((x2-x1)^2+(y2-y1)^2) = ||v||
i. Same as using pythagorean theorem for the hypotenuse of a triangle
Vectors Review
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Magnitude ||v||
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Length of Vector or strength of force vector
Direction
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Where a vector is pointing
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Zero Vector
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○ vector with no direction or magnitude
Unit Vector
○ vector with a magnitude of 1
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Magnitude Equation
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square root of the sum of the squares of each vector component
sqrt ( x^2 + y^2 ) OR sqrt ( x^2 + y^2 + z^2 )
it’s pythagorean theorem
Combining Vectors
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Multiplying vector v by scalar k
○ Multiply ||v|| by k
■ If k is 0 then ||v|| = 0; if k > 0 then ||v|| is in the same direction; if k < 0 then ||v|| is
in the opposite direction
Vector addition of vector v + w
○ Triangle Method (tip-to-tail)
■ Set the initial point of w to the terminal point of v
■ Initial point of sum vector is initial point of v and terminal point of sum vector is
terminal point of w
○ Parallelogram Method
■ Draw first v and first w from the same initial point
■ Draw another second v with initial point of first w terminal point
■ Draw another second w with initial point of first v terminal point
■ Initial point of sum vector is initial point of first v and first w and terminal point of
sum vector is terminal point of second v and second w
Combining vectors cont.
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Subtracting vectors v - w
○ Subtracting vectors is the same as adding vector v to vector
w multiplied by -1
■ (v-w) = v + (-1)*w
Triangle inequality
○ Because the magnitude of v + w gets the sum vector, it’s like
a triangle
○ Any two sides of a triangle must be more than the remaining
side, so…
○ ||v + w|| <= ||v|| + ||w||
Parallelogram Method
Triangle Method
Combining Vectors Review
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Multiplying vector by scalar
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Adding two vectors
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add corresponding components and put sum into the sum-vector
components
Subtracting two vectors
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multiply each vector component by the scalar
multiply the subtraction vector by the scalar -1
add the two vectors together
Triangle Inequality
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The sum vector will have a magnitude either equal to or less than the sum the
magnitudes of both addend vectors
Combining Vectors Review
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Multiply by scalar
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a * <x, y> = <ax, ay>
Terms & Component Form
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The plane is called R2 because it’s made of 2 real dimensions
Standard position is the origin or the point (0,0)
Vector Component form is v = <x,y> (x and y being the
coordinates on the plane)
○ Let v’s initial point be (x1,y1) and terminal point be (x2,y2)
○ Component form is v = <(x2-x1),(y2-y1)>
○ Magnitude from component from is ||v|| = sqrt(x^2+y^2)
○ Addition: v = <x1,y1>,w = <x2,y2> -> v + w =
<(x1+x2),(y1+y2)>
○ Multiplication by a scalar: k*v = <kx,ky>
Unit Vectors have a magnitude of 1
Algebraic properties of vectors
This is the same as
scalar “normal”
numbers
Component Form from angle
and magnitude
● X-component
○ Let m be the magnitude and r be the angle
○ X = m * cos(r)
● Y-component
○ Let m be the magnitude and r be the angle
○ Y = m * sin(r)
● This is the same as converting from polar coordinates to cartesian
coordinates
○ m being the radius and r being the angle from the origin in polar
Unit Vectors
● Unit vectors have a magnitude of 1
● To find the unit vector from a nonzero vector v
○ Multiply v by the reciprocal of ||v|| to cancel
magnitude to 1
○ V is left with the original direction and
magnitude of 1
○ This is called normalization
Standard Unit Vectors
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There are two standard unit vectors
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i = <1,0>
j = <0,1>
Any vector v can be written as a linear combination of the unit
vectors
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V = xi + yj
3D Vectors
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Each vector now has 3 coordinates
The X-axis is the axis pointing to you
Y-axis is vertical and Z-axis is sideways
3D function planes
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The 3D graph is divided into 8 octants like the 4 quadrants in 2D
The distance between 2 3D points is the square root of the sum of their
differences squared
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Y = 0 is the xz plane because it has every point on the x and z axis
Z = 0 is the xy plane…every point on the x and y
X = 0 is the zy plane…every point on the z and y
d = sqrt( (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 )
In planes, we know at least 1 variable is constant
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The constant variable is the one that is not in the plane’s name
■ xz -> y is const, xy -> z is const, yz -> x is const
3D Planes & Graph
Sphere Equation
● The equation for a sphere
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(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2
(a,b,c) being the center point of the sphere
● Finding the midpoint of 3D coordinates
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Take the average of each point’s x,y,z coordinates
● The Cylinder equation omits the a coordinate variable from
the equation
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x^2 + y^2 = r^2
■ It’s like a circle in 2D that extends through the 3D plane
infinitely
3D vectors
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v = <x,y,z>
Unit vectors are i, j, k
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Magnitude is the square root of the sum of each component
squared
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i = <1,0,0>; j = <0,1,0>; k = <0,0,1>
Use unit vectors to create an equation with each component as a
coefficient of the corresponding unit vector
■ xi + yj + zk
m = sqrt (x^2 + y^2 + z^2)
All other 2D vector properties apply to 3D
Solution Planes of 2 3D
Vectors
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Any two non-collinear vectors create a plane
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If a solution does NOT exist between two vectors
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Imagine making two vectors in 3D space
■ You can make a slice in 3D space that goes through both
vectors completely
The calculated vector is on the plane
■ But the given solution vector is not on the plane
If a solution does exist between two vectors
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The given solution vector is on the solution plane and matches the
calculated vector
Solution vector on a plane
Red: vector u = <2,1,3>
Purple: vector v = <0,1,-2>
au + bv = <4,1,8>?
a=2
b = -1
Black: au + bv = <4,1,8>
This solution does exist because the given
vector is on the plane with the other
vectors
Non-Solution vector off a
plane
Red: vector u = <2,1,3>
Purple: vector v = <0,1,-2>
au + bv = <3,-6,4>?
a = 3/2
b = -15/2
Black: au + bv = <3,-6,19.5>
Green: <3,-6,4>
This solution does not exist because the
given vector is off of the plane
The Dot Product
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The Dot product is notated with a multiplication dot “⋅” (I will use a period in these notes)
The Dot product of two vectors is the sum of the products between each corresponding
component
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u = <3,1,4>
v = <1,5,9>
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u ⋅ v = (3*1) + (1*5) + (4*9)
Properties of the dot product follow normal algebra
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The property of magnitude is the only new one
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it’s just the inside of the square root when we find the magnitude of a vector, so the square
is just saying to undo the square root if you have the magnitude
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Magnitude = sqrt(v1^2 + v2^2 + v3^2)
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v . v = (v1*v1) + (v2*v2) + (v3*v3)
Dot Product & Angle
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u . v = ||u||*||v||*cos(θ)
The law of cosines can get us the angle between the two vectors
○ The work that shows why we use the law of cosines is on the
next slide
Orthogonal vectors
○ vectors are orthogonal if the dot product = 0 between them
■ Orthogonal means “perpendicular” or has a right angle
Angle between vector and
x-axis counterclockwise
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angle from x-axis
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arctan( j / i ) + pi
vector from angle and magnitude
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vector i = magnitude * cos ( angle )
vector j = magnitude * sin ( angle )
Law of Cosines & Vectors
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The law of cosines doesn’t need
multiple angles to solve unlike the law
of sines
we can extrapolate lengths to use the
law of cosines
We can find the length between the
two vectors at their terminal points
by taking their difference
Finding the 3rd length of a
triangle made of 2 vectors
This is the
easy-to-remember way,
but I will show why this
works.
The two vectors are v
and u, red and blue
respectively. Green is
the 3rd vector of the
triangle, which can
also form an open
parallelogram if you
do the tip-to-tail
triangle method of
subtraction. Duplicate
the difference and
make its initial point
the terminal point of
v, and you complete
the parallelogram
Expansion of law of
cosines with dot product
Angle & Vectors & Projections
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Direction Angles for the vector
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Direction Cosine
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the cosine for the direction angles for the vector
Vector Projections
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angles formed by a nonzero vector and the unit vector
A length along an axis or another vector if we imagine a vector’s shadow below it with a light
directly above it
We use a formula to get the magnitude of the projection and then we multiply the vector by a unit
vector in the direction of the projection to find the vector projection
Scalar Projection
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Length of the vector projected onto another vector
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i.e. scalar projection of v onto u
Projections and Orthogonal Vector Complement
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The vector projection of v onto w
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dot product divided by the magnitude of the “onto” vector squared
scalar multiplied to the “onto” vector
This returns the vector projection
Take the magnitude to find the length of the projection
The vector complement of v orthogonal w
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Subtract the vector projection from v
This returns an orthogonal vector to the projection
Cross Product
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Unlike the dot product, the cross product returns a vector
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That returned vector will be orthogonal to the 2 original vectors
uXv=w
because w is orthogonal to u and v
■ u . w = 0 and v . w = 0
■ you don’t need to memorize the cross product equation!
Any parallel vectors have a cross product of the 0 vector
Cross Product Properties
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Most of these resemble usual algebraic properties, but slight differences.
Easy: iv, v,
Hard: i, ii, iii, iv
Unit Vector Cross Products
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You can use the right-hand rule to
make these calculations
Imagine the 3D graph as your hand
and point in the direction of the unit
vector and the “crossing” vector
the cross product vector is the one
that is “orthogonal” to the other
vectors or “perpendicular”
Memorize the left 3 equations and
use the “anti communicative”
property to find the right 3
equations
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The left side order kinda has a component sequence
of going x then y then z then x
i is <1,0,0> then j is <0,1,0> then k is <0,0,1> then it
goes back to i.
Memorizing Unit Cross
Products
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If you can see the pattern
○ The left side goes up in dimensions in the unit vector i -> j -> k
○ the right side goes down in the dimensions of the unit vector k -> j -> i
Memorize the left side, then just flip each equation and the make the product
negative to get the right side
Magnitude of the Cross
Product
● The magnitude of the cross product
○ ||u X v|| = ||u|| * ||v|| * sin(θ)
Matrices
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Matrices can be m x n
○ m being the number of rows
○ n being the number of columns
Square matrix
○ the number of rows is the same as the number of columns
Determinant
○ for a 2x2 matrix the determinant is (a*d - b*c) in a criss cross pattern
○ for a 3x3 matrix you extend the matrix out 2 columns and draw those same
diagonal groupings
■ The downward lines are going to be positive
■ The upward lines are going to be negative
Matrix Determinant Formula
Expansion along the first
row Determinant Formula
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This way is easier to get the determinant in a 3x3 matrix
Multiply each index in the first row by the 2x2 matrix that is down one and to the right one
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If the 2x2 matrix goes out of the matrix, then go back to the far left column. it sorta wraps around the matrix
MAKE SURE THE MATRICES ROW NUMBERS GO FROM LOW TO HIGH
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Although a2 wraps around and makes the matrix out of order when you go left to right, change the order to b1, b3 and c1,c3
OR IMAGINE THE SECOND MATRIX AS LEFT TO RIGHT AND SPLIT BETWEEN COLUMN 1 AND 2
Alternate between positive and negative
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the first index coefficient is positive -> 2nd negative -> 3rd positive
The green 2x2 matrix starts at the 3rd column and wraps back to the 1st column
Cross Product using
determinant
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The 1st row are the 3 unit vectors
The 2nd row are the components of the 1st vector that crosses with the 2nd
The 3rd row are the components of the 2nd vector that is being crossed by the 1st
Apply the same expansion along the first row to get the cross product
Applications of Cross Product
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This would be a ton of example questions using vectors to calculate things like the area of
a parallelogram or triangle
These examples are in the lecture video, so I’m not gonna copy them here
Another example is calculating torque, which is in the textbook and not in the video
The application just requires understanding of what each operation does like dot and
cross products with magnitudes.
REMEMBER
○ the cross product returns a vector that is orthogonal to both of the input vectors
■ the magnitude of the cross product is the area of the shape formed by the
two input vectors
○ the dot product gives you the angle between two vectors
■ the dot product goes from -1 to 1
● -1 being collinear in the opposite direction
● 1 being collinear in the same direction
● 0 being orthogonal to the vector (90 degrees)
● The closer to 1 the closer the vectors with -1 being opposite directions
Examples of Applications
Dot Product | Desmos
Triple Scalar Product
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The triple scalar product of vectors u v and w
○ u . (v X w)
○ This can also be calculated as the determinant of a matrix of each vector’s
components
■ 1st row u, 2nd row v, 3rd row w
The scalar product has a geometric interpretation
○ it returns the volume of a parallelepiped
■ V = |u . (v X w)|
○ We can use this to show that 3 vectors are coplanar (on the same plane)
■ If we calculate the triple scalar product and return 0
● this means that the 3D parallelepiped has no volume, therefore it’s flat
Triple Scalar Product
Application
The shape is a parallelepiped
Equations of lines in space
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Vectors are parallel if they are in the
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Vector from two points
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same direction
opposite direction
initial points minus final points (x1-x0,y1-y0,z1-z0)
■ You can think of subtracting vectors
■ vector u = <x1,y1,z1> and vector v = <x0,y0,z0>
■ vector w = u - v
point P and point Q create a line
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we can get the vector from these points
vector PQ is a vector in a direction with an arbitrary length
■ Therefore, vector PQ = t*v. (scalar t)
■ <x,y,z> - <x0,y0,z0> = t<a,b,c>
■ <x,y,z> = t<a,b,c> + <x0,y0,z0>
● This is the Vector equation of a Line ( r = r0 + tv)
Vector Equation of a Line
Vector equation of a line
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<x,y,z> = t<a,b,c> + <x0,y0,z0>
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Assuming that a,b,c != 0
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vector r = (1-t) * vector p + t * vector q
For the line segment we find the limits of t
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We can solve for t
These equations are the symmetric equations of a line (next slide)
A line passing through two points
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this evaluates to <(x0 + ta),(y0 + tb),(z0 + tc)>
Each component expression is the parametric equation for a line
if t is 0, then vector r = vector p
if t is 1 then vector 4 = vector q
■ Therefore: 0 <= t <= 1
Components of vector are the coefficients of t
Components of 3D point are the # in x = # - ta, y = # - tb, z = # - tc
Parametric and symmetric equations are not unique
The direction vector is not unique
Vector equation of a line
notes
● The vector that is being multiplied by the parameter variable
is the direction vector of the line
● To get parametric equations from the equation
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set x(t) to the x component of the vector without the coefficient
and add the component of the coefficient vector multiplied by t
do the same for y(t) and z(t)
● a better equation is L(t) = p + dt
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p is the position vector of a point of the line
d is the direction vector of the line
Acute angle between 2
lines
1.
2.
3.
4.
Find the direction vectors of both lines
calculate the dot product of both vectors
calculate each d-vector’s magnitude
use the equation cos(theta) = dot product divided by the
product of the magnitudes
5. Take the arccosine of that answer
6. Subtract the angle from 180 degrees if the angle is greater
than 90 to get the acute angle
Vector from line
orthogonal to point
1. Find the direction vector of the line and point on the line
2. Solve for the vector PQ from the line’s point to given point
a.
given point - line’s point = PQ
3. Calculate the projection of PQ onto d
a.
( (PQ . d) / (d . d) ) * d = Proj_d(PQ)
4. Find the orthogonal complement
a.
Subtract the projection from PQ
5. That is the orthogonal vector from the line to the point
Distance from line to a point (shortest / not
orthogonal)
Distance between 2 skew
lines
https://www.toppr.com/guides/maths/three-dimensional-geometry/distance-between-skew-li
nes/
Solving for vector equation
of a line from 2 points
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The vector equation of a line is a vector adding another vector that is multiplied by a
scalar parameter t
Imagine that when t is 0
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Imagine that when t is 1
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the vector sum is pointing to the second point of the line
Therefore, the equation is r = r0 + tv
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1.
2.
3.
4.
the vector sum is pointing to the first point of the line
r0 is the vector that points to the first point with an initial point of the origin
v is the vector that must be added to r0 for it to point to the second point of the line
set vector r0 to the first point components
set vector r to the second point components
solve the for the direction vector of the line (P1 - P0) and set it to vector v
set up the equation r = r0 + tv
Equations of a line
r0 is the 3D
point on the
line
(x0,y0,z0) is any
point on the line
<a,b,c> are the
components of
the vector of
the line
Vector equation of a line
(visual diagram)
Think of the equation as a tip-to-tail addition of two vectors. The first vector is
to a point on the line, and the second vector is the direction vector with an
adjustable magnitude using the variable t of the function. By adjusting the
magnitude of that 2nd vector, you can point to every point of the line.
Distance between a point
and line
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distance equation
Relationships between lines
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Skew lines
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If direction vectors are equal
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lines that do not intersect
lines are parallel, but not equal
or they are the same line (equal)
When solving
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set Line equations to parametric equations
(x+a) = t, (y+b) = t, (z+c) = t
solve for x,y,z in terms of t
Find the direction vector: components are the coefficients of t
Check if direction vector of each line are equal (normalization of each vector)
Rewrite equations for each line x,y,z equations using r and s as parameters instead
of t
Isolate r and s onto one side of the equation and set equal to a number
If all equations are equal
■ infinitely many solutions and lines are equal
Equation of Line Segment
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P0 = (x0,y0,z0), P1 = (x1,y1,z1)
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Why limit t?
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let direction vector v = vector P0P1
direction vector v = <x1-x0,y1-y0,z1-z0>
vector equation for line segment (not unique)
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<x,y,z> = <x0,y0,z0> + t<x1-x0,y1-y0,z1-z0> : 0 <= t <= 1
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<x0,y0,z0> is the position vector with initial point (0,0,0) and terminal point P0
A line segment is finite and ends
if t = 0: <x,y,z> = <x0,y0,z0> + 0<x1-x0,y1-y0,z1-z0>
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<x,y,z> = <x0,y0,z0> (Position vector for P0)
if t = 1: <x,y,z> = <x0,y0,z0> + 1<x1-x0,y1-y0,z1-z0>
■
<x0+x1-x0,y0+y1-y0,z0+z1-z0>
■
<x1,y1,z1> (position vector for P1)
As t goes from 0 -> 1, the position vector goes along the line segment
parametric form (NO symmetric form because no scalar t)
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x = x0 +t(x1-x0)
y = y0 + t(y1-y0)
z = z0 + t(z1-z0)
Position Vector
● A position vector is a vector equivalent
to a line that has the initial point of the
origin
Equations of planes
● say we have a vector from two points called PQ =
<x1-x0,y1-y0,z1-z0>
● we have a vector <a,b,c> and we take the dot product
between PQ and <a,b,c> and set it to 0 because the
plane is orthogonal
○ a(x1-x0) + b(y1-y0) + c(z1-z0) = 0
● This is the scalar equation of a plane
Get equation of plane
● To get equation of a plane we need
○
○
normal vector to the plane vector with components =
<a,b,c>
a point on the plane P0 = (x0,y0,z0)
● if P(x,y,z) is any other point on the plane then
○
○
○
P0P = <x-x0,y-y0,z-z0>
is a vector in the plane and orthogonal to normal vector
vector P0P . normal vector = 0
● So we get a(x-x0) + b(y-y0) + c(z-z0)
Relationships of planes
● normal vectors are n2 or n1
● When are 2 planes parallel
○ Both their normal vectors are parallel
○ n2 = c*n1, c != 0 OR cross product between normal
vectors is 0 (order of cross doesnt matter)
● When 2 planes orthogonal
○ normal vectors are orthogonal
○ n2 . n1 = 0
●
Planes and Lines relationships
● line has direction vector, d, and plane has normal vector, n
● When is a line parallel to plane?
○
○
The direction vector is orthogonal to the normal vector
d.n=0
● When is a line orthogonal to a plane
○
○
○
the direction vector is parallel to the normal vector
cross product between direction and normal vector is 0
dXn=0
Parallel and Intersecting
Planes
●
2 planes are parallel
○
○
○
●
●
The normal vector components are the coefficients of x,y,z
Take dot product and cross product of both normal vectors to check if orthogonal or
parallel
○
●
●
if both products are 0, then the planes intersect
Solve for variables y in terms of z
solve for z in terms of parameter t
○
●
if both normal vectors are parallel
else if normal vectors are orthogonal
■
both planes are orthogonal
else both planes intersect in a line
substitute y, z into first or second equation to get z
Line of intersection is the parametric equations
Types of Quadric Surfaces
& Formulas
Ellipsoid: x²/a² + y²/b² + z²/c² = 1
Hyperboloid of one sheet: x²/a² + y²/b² - z²/c² = 1
Hyperboloid of two sheets: x²/a² - y²/b² - z²/c² = 1
Cone: x²/a² + y²/b² - z²/c² = 0
Elliptic Paraboloid: z/c = x²/a² + y²/b²
Hyperbolic Paraboloid: z/c = x²/a² - y²/b²
Quadric Surfaces
Vector-Valued Functions
(Vector VF)
●
●
Functions of each component of a vector
t is the parameter of each function
○
●
2 notations for functions
Functions can have domains and ranges
○
the intervals of t must apply to all functions
Graphing Vector VF
●
Plane curve
○
●
Space curve
○
●
the representation of a plane or space curve using vector-valued functions
You can rewrite vector-valued functions by representing the function as a line sometimes
○
●
●
the graph of a 3-dimensional vector valued function
vector parameterization of a curve
○
●
The graph of a 2-dimensional vector valued function
r(t) = <2,-1,3) + t<-4,5,1>
■
the first vector is made of the constants and the second vector is made of the coefficients
of t
Solve for the variable in terms of a function of t to solve for the graph
Vector-valued function limit
Vector VF Domain
● The domain of a vector-valued function is the
intersection of the domains of each component’s
function
○ Find the domain for each individual function
○ find the intersection of all of the domains
○ that is the domain of r(t)
●
Vector-Valued Functions limits
Limit Derivatives of Vector VF
●
Limit Definitions of vector valued function
○
Left and Right derivatives are applied the same
Derivatives of Vector VF
●
Derivatives of each component function is just the derivative of each function
Derivative of Vector VF Propty.
Factoring out
Distributes
product rule
Cross product is
like product rule
Vector VF vectors
● Tangent Vector
○ the derivative of r(t) at a given point t=t0
○ Then you add the r(t) at t=t0
○ T(t) = r’(t0) + t<r(t0)>
● Principal Unit Tangent Vector (unit tangent)
○ Just normalize the tangent vector
Integrals of Vector VFs
●
●
●
Integration constant is a vector
Indefinite Integrals are integrals of each component
<integral(x(t))dt + c1, integral(y(t))dt + c2, integral(z(t))dt + c3>
Definite Integrals of Vector VF
● Definite integrals are just definite integrals of each
component without the integration constant vector
Integrating a vector VF
1. Take the integral of each component
2. Plug in <+c1, +c2,+c3> into each component
3. Calculate t=0 (or whatever t value the given vector uses) for
each component
4. plug in c1,c2,c3 into each component
5. Set each “plugged-in” component to original given vector
6. solve for c values in each component
7. Plug c values into integrated components
Arc Length
●
This is integrating the magnitude of the tangent vector
●
●
Smooth function (uses another variable on the inside)
uses u because t is in the limit
Arc Length
Parameterization
1.
2.
3.
4.
5.
6.
Find the magnitude of the derivative of r(t)
Integrate the magnitude of the derivative of r(t) from 0 to t
Set that answer from the integral equal to s
Solve for t
Plug t back into r(t) and change it to r(s)
remember to define the variable limits
Distance on an arc length
1. Find r(s) from arc length
parameterization
2. evaluate r(s) at the given distance by
plugging the distance into s
3. that point is where you are on the arc
length at that distance
Tangent vector formula
1. Find the derivative of r(t)
2. normalize the derivative of r(t)
Curvature
●
●
The curvature of a circle is the reciprocal of its radius
If inscribed circle big
○
●
not sharp
if circle small
○
sharp
Kappa is the curvature
T is the unit tangent vector
Curvature Theorems
Curvature Theorems cont.
Solving for Curvature
1.
2.
3.
Solve for the derivative of r(t)
Find the magnitude of the derivative of r(t)
Solve for the unit tangent vector function
a.
4.
5.
6.
T(t) = the derivative of r(t) divided by the magnitude of the derivative of r(t)
Solve for the derivative of the unit tangent vector function
Solve for the magnitude of the derivative of the unit tangent vector function
Solve for kappa
a.
divide the magnitude of the derivative of the unit tangent vector function by the magnitude of the
derivative of r(t)
i.
Formula given
Solving for curvature of a curve
at a point
1.
2.
3.
4.
5.
Find the first and second derivatives of the function r(t)
Take the cross product between the first and second derivatives
Find the magnitude of the cross product
Solve for the magnitude of the first derivative
Solve for kappa
a.
The magnitude of the (cross product of the first derivative and the second derivative of r(t))
divided by (the magnitude of the derivative of r(t)) cubed
Solve for curvature over interval
1. Find the derivative of r(t)
2. Find the magnitude of the derivative
3. integrate that magnitude over the interval
4. This returns the arc length
Solving for curvature of a
function
1.
2.
3.
Find f’(x) and f’’(x)
| f’’(x) | /[1+f’(x)^2]^(3/2)
Plug in x-value to find curvature at a point
Normal & Binormal Vectors
●
The principal unit normal vector (unit normal vector)
○
●
is the normalized version of the derivative of the tangent vector function
The binormal vector
○
is the cross product of the tangent vector and unit normal vector
How to identify space curves
●
●
c is a real number
Tornadoes
○
○
●
Sinusoidal ring
○
○
●
●
c(t)trig(c(t)), c(t)trig(t(t)),c(t)
z goes from the bottom of the tornado to the top
ctrig(ct),ctrig(ct),ctrig(ct)
the two planes that make a circle when viewed perpendicular will have two different
trig functions. The last will be a duplicate
Helix / Spring
○ ctrig(ct),ctrig(ct),ct
○ z goes from the bottom of the helix to the top
Saddle
○ ctrig(t),ctrig(t),ctrig(ct)
Space Curves (my names not official)
Helix/spring , sinusoidal ring, saddle, tornado
Motion in space
●
The speed of an object is the magnitude of the vector value function
○
Technically, no position in time is required to find the speed if the magnitude is a constant
Tangential & Normal vector Components of
Acceleration
Normal Component of acceleration
Centripetal Component in physics
Applications
●
Effect of gravity as an object travels through the air
Solving for a Tangent line
with a vector VF and given
point
1.
Figure out what value of t was used to find the point
a.
2.
3.
4.
5.
6.
set a component equal to the same component of the point
Find the derivative of the vector VF
Plug the t-value of the point into the derivative of the vector VF
Each component of the r’(t0) is the <a,b,c> coefficients of t of the parametric equations
The given point is the (x1,y1,z1) of the line equation
the line equation will be <x1,y1,z1> + t<a,b,c> using the values from the given point and
derivative of the vector VF at the t value used to find the point
Solving for the acute angle of
the intersection of 2 vector VF
1.
2.
3.
4.
5.
Find the derivative of both vector VFs
Plug the t value that both vector VFs intersection into their derivatives
Find the dot product between the two vectors
Find the product of both of their magnitudes
Solve for the angle using the angle between two vectors formula
a.
b.
cos(theta) = dot product / product of magnitudes
take the arccos ( dot product / product of magnitudes) to get the angle
Solving for the acute angle
between two curves
1.
2.
3.
Take the derivative of both functions
Plug in the x value of the point that they intersect
Make a vector using the point of the first derivative of the function at that point
a.
b.
4.
5.
6.
7.
a = <x1,f’(x1)>
b = <x1,g’(x1)>
Find the dot product of them
Find the product of their magnitudes
Solve using the angle between two vectors formula
Take the arccos( dot product / product of magnitudes)
Find the value t when r(t)
is perpendicular to r’(t)
1. Find the derivative of r(t)
2. Find the dot product of r(t) and r’(t) and set
equal to 0
3. Solve for t using a graphing calculator or
something
4. Multiple t values are possible
Cylindrical Coordinates
Converting cylindrical coordinates
●
Cylindrical to Cartesian
○
○
○
●
x = rcos(theta)
y = rsin(theta)
z=z
Cartesian to Cylindrical
○
○
r^2 = x^2 + y^2
tan(theta) = y / x
Properties of cylindrical coordinates
Spherical Coordinates
Solving for spherical coordinates from 3D point
1. Solve for rho (rho is the magnitude of the vector
to that point)
a. rho = sqrt(x^2+y^2+z^2)
2. Solve for phi
a. rho*cos(phi) = z
b. phi = arccos(z/rho)
3. Solve for theta
a. theta = arctan(x/y) (angle from positive z axis)
b. if z = 0, theta = 0
Converting spherical coordinates
phi is angle
from
positive z
axis
theta
can be
any
angle
tan(theta) = y/x
phi is written as a circle with a diagonal!!
theta is written as a circle with a horizontal!
Properties of spherical coordinates
Functions of 2 variables
● z = f(x,y)
● the function is a
surface in 3D
space
Level Curves (contour curves)
●
●
●
The level curves are all curves in the x-y plane
You can think of it as a 3D function that intersects with the xy plane where z=c
The level curve are the intersections of the plane and function
Functions with more than 2 variables
● Finding the domain is the same as finding any
other
○ just find where there is a complex or undefined
output
● These are represented as curves in 4D space
○ we define it as a level surface where f(x,y,z) = c
where c is a real value
Finding the vertical traces of a 2 variable function
1. Set 1 variable to the variable c
2. set function to z
3. solve for c
4. set the other variable to c
5. set the function to z
6. solve for c again
Definition of a disk
Limits of functions of 2 variables
●
Properties of Limits of 2 variable-functions
Taking the limit of a multivariable function
1. Try plugging in values
2. Factoring & cancelling
3. Try taking limit along an axis
a.
b.
make the y value go to 0
make the x value go to 0
4. Try taking the limit along a line
a.
b.
make y=x
make x=y
Continuity of two variable
Functions
●
function f(x,y) is continuous at a point (a,b) in its domain if these conditions are all satisfied
○
○
○
f(a,b) exists
the limit as (x,y) -> (a,b) of f(x,y) exists
the limit as (x,y) -> (a,b) of f(x,y) = f(a,b)
Continuity of Multivariable function theorems
●
a function that is continuous and nested inside another function that is continuous is also
continuous
Functions of 3 or more
variables
●
●
delta ball can be centered at any point (a,b,c)
Take limits by trying to plug in then factor then other stuff
Partial Derivatives
●
●
these use a cursive “d” to show respect only to one variable
Basically pretend the other variable is a constant like a number just a plain number
○
DO NOT BE SCARED TO DELETE A VARIABLE THROUGH PARTIAL DERIVATION
■
IT IS OK!
Partial Derivatives of functions with 3 variables
●
Same thing as 2 variable-partial-derivatives but with 3 instead
Higher-Order Partial Derivatives
●
●
Basically taking the partial derivative multiple times sometimes mixing which one
Mixed partial derivatives are the f_xy or f_yx, etc.
Clairaut’s Theorem
●
●
Doing a mixed partial derivative doesn’t matter what order you do it in
doing x,y is the same as y,x
Partial Differential Equation (PDE)
Chain Rule for 1 independent variables
●
The sum of the derivative of z multiplied by both x and y in respect to t
Chain Rule for 2 independent variables
●
Do chain rule for 1 independent variable twice to z but with respect to 2 variables
Generalized Chain Rule
Implicit Differentiation
Implicit Differentiation
How to take the implicit differential equation
1.
2.
Diff both sides in terms of the variable
Tangent Planes Notes
Tangent Planes Notes cont.
Tangent Planes Example
Tangent Plane Instructions
1.
2.
3.
find dz/dx and dz/dy
plug x and y into both derivatives and f(x,y)
plug values into the tangent plane formula
Linear Approximations
Linear Approximations
Differentiability
Differential
●
total differential
○
●
this is the differential dz, an approximation of the difference
dz = f_x(x0,y0)*dx + f_y(x0,y0)*dy
○
dx and x-x0 and dy = y-y0
Differential Example
Differentiability of a function of 3 variables
Directional Derivatives
Directional Derivatives: Slope of the secant
Limit Definition of the directional derivative
Directional Derivatives Formula
Directional Derivatives Vector Formula
Directional Derivatives Instructions How to Solve
●
To solve for the directional derivative function
without solving for anything
do the dot product between the gradient function
and the unit vector to the point
To solve for the directional derivative with an angle
○
just do the dot product between the gradient and
the vector <cos(theta), sin(theta)>
To solve for directional derivative with a point
○
just plug the point into the gradient instantly
○
●
●
Directional Derivatives Example
The Gradient
●
Properties of the gradient and theorem
Gradient Example
Gradients and Level Curves
Property of Gradient and Level Curve Theorem
Gradient and Level Curves Example
3-Dimensional Gradients and Directional
Derivatives
3-Dimensional Gradients and Directional
Derivatives Example
Critical Points and Local Extrema
●
Critical Points and Local Extrema Instructions
1.
Find partial derivatives of f(x,y)
a.
2.
3.
4.
Set both partials equal to 0
a. (you don’t need to solve all equations)
Solve for x
Set x equal to one of the values and solve for y
a. plug this x value into one of the equations you used to find y
b. solve by factoring or quadratic equation etc.
c.
5.
6.
Find f_x(x,y) and f_y(x,y)
y might be in terms of x, therefore plug in a critical x value into that x value and
solve for y
Find the corresponding y value for each critical x-value
Format into points (x,y)
Critical Points and Local Extrema
Local Maxima and Minima
Fermat’s Theorem for Functions of Two Variables
Local Maxima and Minima
● Any local extremum at (x0,y0) for f(x,y) is also a critical point
Proof for Fermat’s theorem
Second Derivative Test
Second Derivative Test Conditions
Diagram of Local extrema and Saddle point
Critical Points and Local
Extrema instructions
1.
2.
Find the critical points for when x equals a value and when y equals a value
Find the double partial derivatives for x and y and the mixed partial
differential
a.
3.
4.
5.
f_xx, f_yy, and f_xy
i. f_xy will always be the same as f_yx
Plug derivatives into the 2nd derivative test formula
Evaluate what each point is by using the 2nd derivative test conditions
Label each point with their corresponding type of local extrema
Critical Points Example
Critical Points Example Cont.
Absolute Maxima and Minima
Extrema diagram
How to find the absolute extrema
Example of absolute extrema
Example of absolute extrema cont.
Example of absolute extrema cont.
Example of absolute extrema cont.
Example of absolute extrema cont.
Example of absolute extrema 2
Example of absolute extrema 2 cont.
Example of absolute extrema 2 cont.
Example of absolute extrema 2 cont.
Note for absolute extrema and applications
Example of extrema on a shape
●
●
●
●
●
●
●
●
●
●
●
●
Find the absolute max and min of f(x,y) = 192x^3 + y^2 -4xy^2 on the triangle with
vertices (0,0), (4,2), (-2,2).
f_x(x,y) = 192(3x^3)-4y^2 = 576x^2 - 4y^2
f_y(x,y) = 2y-8xy
576x^2 - 4y^2 = 0
2y-8xy = 0
2y(1-4x) = 0
x=¼&y=0
(0,0) is one critical point
576(¼)^2 - 4y^2 = 0
36 - 4y^2 = 0
y = +/- 3
(¼, 3), (¼, -3) are critical points
Example of extrema on a shape cont.
●
●
●
●
The only relevant critical point is at the boundary, (0,0).
f(0,0) = 0
Determine candidates for the absolute extrema on the boundary of D.
Top:
○
●
Right:
○
●
y = x/2 {0 <= x <= 4}
Left:
○
●
●
●
●
y = 2 {-2 <= x <= 4}
y = -x {-2 <= x <= 0}
Since y = 2, f(x,y) = f(x,2) = 192x^3 - 16x + 4
g(x) = 192x^3 -16x+4
g’(x) = 0 @ x = +/- ⅙
abs. max/min of g(x) occur at x = +/- ⅙ or on the boundary of -2<=y<=4
Example of extrema on a shape cont.
●
●
●
●
●
●
●
●
●
●
●
●
possible maxes or mins are at (-⅙,2),(⅙,2),(-2,2),(4,2)
f(-⅙,2) = 52/9
f(⅙.2) = 20/9
f(-2,2) = -1500
f(4,2) = 1228
since y = x/2
f(x,y) = f(x,x/2) = 192x^3 + ¼ * x^2
set that equal to g(x)
x=0
x=-1/1146 (not in 0<=x<=4, therefore not a critical value)
possible locations are (0,½(0)), (4,½(4)) because y= x/2
(0,0),(4,2) (same points as before)
Example of extrema on a shape cont.
●
●
●
●
●
●
●
●
●
●
since y = -x
f(x,y) = f(x,-x) = 192x^3 + (-x)^2 - 4x(-x)^2 = 188x^3 + x^2
g(x) = f(x,-x)
get g’(x) and solve for g’(x) = 0
x = 0, -1/282 (both work because they’re between -2<=x<=0)
possible locations: (0,0), (-1/282, - (-1/282)), (-2, -(-2)) because y = -x
f(0,0) = 0
f(-2,2) = -1500
f(-1/282,1/282) = 1/238572
looking at all of the candidates, we see that:
○
○
absolute max of 1228 at (4,2)
absolute min of -1500 at (-2,2)
Lagrange multipliers: one constraint
●
●
f(x,...) is the objective function
g(x,...) is the constraint function
Lagrange Multipliers: One constraint expln.
How to solve Lagrange Multiplier problem
Lagrange Multiplier example
Lagrange Multiplier example cont.
Lagrange Multiplier example cont.
Lagrange Multiplier Example 2
Lagrange Multiplier Example 2 cont.
Lagrange Multiplier Example 2 cont.
How to solve for lagrange multipliers
1.
2.
3.
4.
5.
6.
7.
8.
9.
Find the gradient of the objective function
Find the gradient of the constraint function
Multiply the constraint gradient by lambda
Set the objective gradient equal to the lambda constraint gradient
Set both series of equations equal to lambda if possible
After setting both equal to lambda, set both equations equal to each other
Set a variable equal to another variable if possible.
Plug a solved variable into the constraint function
Solve for other variables using the constraint function
Lagrange one constraint example
Lagrange one constraint example cont.
Lagrange one constraint example cont.
Lagrange one constraint example 2
Lagrange one constraint example 2 cont.
Lagrange one constraint example 2 cont.
Lagrange one constraint example 2 cont.
Lagrange Multiplier with 2 constraints
Lagrange with 2 constraints example
Lagrange with 2 constraints example cont.
Lagrange with 2 constraints example cont.
Double Integrals over Rectangles
Double Integrals over Rectangles
Double Integrals over Rectangles sum definition
Double Integrals over Rectangles example
Double Integrals over Rectangles example cont.
Properties of Double Integrals
Properties of Double Integrals cont.
Double Integrals Example
Iterated integrals
Fubini’s Theorem
Iterated Integrals Example
Iterated Integrals Example cont.
Applications of double integrals
Applications of Double Integrals example
Double Integrals over general regions
●
Type I
○
●
lies between 2 vertical lines
Type II
○
lies between 2 horizontal lines
How to integral double integrals over general
regions
How to solve for double integrals over general
regions
How to solve for double integrals over general
regions cont.
How to solve for double integrals over general
regions cont.
Double integrals over general regions Example
Double integrals over general regions Example
cont.
Double integrals over general regions Example
cont.
Double integrals over general regions Example
cont.
Double integrals over general regions Example
cont.
Double integrals over general regions Example
cont.
Changing the order of integration
●
Changing the order can make integration easier
Changing the order of integration example
Changing the order of integration example cont.
Changing the order of integration example cont.
Applications of double integrals
Improper double integrals
Improper Integrals example
Double Integrals in Polar Coordinates
Double Integrals in Polar Coordinates
Double Integrals in Polar Coordinates Example
Convert from Cartesian to polar example
Convert from Cartesian to polar example
Polar Areas and volume
Polar area example
Polar area example cont.
Polar area example cont.
Polar volume example
Polar volume example cont.
Polar volume example cont.
Triple Integrals
Riemann Sum definition of the triple integral
Fubini’s theorem
Fubini’s theorem triple integral example
Triple Integral over a general region
Triple Integral over general region example
Triple Integral over general region example cont.
Triple Integral over general region example cont.
Triple Integral over general region example cont.
Triple Integral over a general region cont.
Applications of triple integrals
Applications of triple integrals cont.
Applications of triple integrals cont.
Applications of triple integrals cont.
Changing the order of integration and coordinate
systems
Application of triple integrals: Average value of a
function
Application of triple integrals: Average value of a
function
Triple Integrals in cylindrical coordinates
Fubini’s Theorem in cylindrical coordinates
Triple Integral in cylindrical using projections
Triple Integral in Cylindrical coordinates example
Triple Integral in Cylindrical coordinates example
cont.
converting cartesian to cylindrical triple integral
example
Triple Integrals in spherical coordinates
Triple Integrals in spherical coordinates cont.
Fubini’s theorem in spherical coordinates
Triple integral in spherical coordinates example
Triple integral in spherical coordinates example
cont.
Triple Integrals in spherical coordinates part 2
example
Triple Integrals in spherical coordinates part 2
example
Converting to spherical coordinates example
Converting to spherical coordinates example
cont.
Converting to spherical coordinates example
cont.
Spherical Coordinate Notes
●
variables
○
○
○
○
○
p is rho
x = p*cos(theta)*sin(phi)
y = p*sin(theta)*sin(phi)
z = p*cos(phi)
r=p^2*cos(phi)
●
a sphere is x^2 + y^2 + z^2 = p^2
●
phi = 0
●
phi = pi/2
○
○
○
on the interval 0 <= phi <= pi
on xy-plane
on positive z-axis
●
phi = pi
●
●
Phi is the z-axis with the positive axis like the positive x-axis on a 2D circle
It only goes to pi because it wraps around the z-axis from the positive z-axis to phi(0 - pi) radians
○
on negative z-axis
Change of variables in double integrals
Plane Transformations
Plane Transformations cont.
Plane Transformations Example
The Jacobian
The Jacobian cont.
The Jacobian cont.
The Jacobian cont.
The Jacobian of the transformation
Change of Variables for Double Integrals
Change of variables for double integral
1.
Determine the change of variables x = g(u,v), y = h(u,v)
a.
2.
3.
4.
5.
6.
sometimes u = w(x,y), v = k(x,y)
Solve for x and y in terms of u and v
find partial derivatives of x and y
Plug them into a matrix
solve for the determinant
Take absolute value of determinant
Change of Variables in Double Integrals Example
Change of Variables in Double Integrals Example
cont.
Change of Variables in Double Integrals Example
cont.
Change of Variables in Double Integrals Example
cont.
Change of Variables in Double Integrals Example
cont.
Change of Variables in Double Integrals Example
cont.
Change of Variables for Triple INtegrals
Fubini’s theorem for any integral
●
●
over a rectangular region (or square)
You get the two intervals of the rectangle
○
○
○
●
●
●
x is the width of the rectangle
y is the height of the rectangle
The format will often be in the interval form
○
○
●
●
a <= x <= b
c <= y <= d
e <= z <= f
[a,b] X [c,d] X [e,f] (extended to z)
Each pair is the interval of x,y,z
Each pair of intervals go onto the limits of the integrals with their corresponding
variable of integration
You just put the interval of each variable into the limits of the integral
Vector Fields
●
Applications of representation
○
●
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Electric fields, fluid flow, gravitational fields
A vector field in R2 is a function F(x,y) that assigns a vector to each point (x,y) of a subset D
of R2. The subset D is the domain of the vector field.
A vector field in R3 is a function F(x,y,z) that assigns a vector to each point (x,y,z) of a
subset D of R3. Subset D is the domain of the field
F(x,y) = <P(x,y), Q(x,y)>
F(x,y) is a function that returns a vector
F is continuous if both P and Q are continuous
The gradient is a vector field
Vector fields are either radial (pointing toward or away from the origin) or rotational
(pointing tangent to a circle with all vectors either clockwise or counterclockwise)
Unit vector field
○
○
○
a vector field of unit vectors
Take the magnitude of F(x,y)
Divide F(x,y) by its magnitude to get the unit vector field function
Gradient Fields
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●
●
●
The Gradient Fields are sometimes called “conservative fields” because it can be used to model a
physical system in which energy is conserved
A vector field F of R2 or R3 is a gradient field if there exists a scalar function f such that grad f = F
in F(x,y) = <2x,2y> from f(x) = x^2 + y^2
○
When the gradient field solutions are closer to each other, the steeper the 3D object is
○ the vectors are orthogonal to the level curves
conservative vector field
○
a vector field F is conservative if there exists a function f such that grad f = F
●
potential function
●
●
Note: in some physical systems, the potential function f is defined as a function such that -grad f =
F
simply connected region
●
two regions
○
○
○
●
if F is a conservative vector field then a function f such that grad f = F is a potential function
no holes
not connected
not simply connected region
○
has holes in the shape
Uniqueness of Potential functions
●
Let F be a conservative vector field on an open and connected domain and let f and g be
functions such that grad f = F and grad g = F. Then there is a constant C such that f = G + C.
Cross-Partial Property of Conservative Vector
Fields
●
●
Let F be a conservative vector field whose components have continuous second-order
derivatives in the domain of F.
if F(x,y) = P(x,y)i + Q(x,y)j, then
○
●
If F(x,y,z) = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k, then
○
●
Py(x,y) = Qx(x,y)
Py = Qx, Qz = Ry, Rx = Pz
To prove this, we just need to recall that mixed partials are equal and in the case of F(x,y),
e.g., there is a function f(x,y) such that grad f = F; i.e., <fx,fy> = <P(x,y),Q(x,y)>. So,
○
○
○
(d/dy)fx = P, (fy = Qx)
fxy = Py, fyx = Qx
This is because mixed partials are the same from Clairaut’s theorem
Check if vector field is conservative example
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●
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Determine if the following vector fields are not conservative
(1) F(x,y) = -y + xj
(2) F(x,y,z) = x^2i + yzj + zyk
(1) if F is conservative then Py = Qx, but Py = -1 and Qx = 1
(1) F(x,y) is conservative
(2) check Py = Qx, Qz = Ry, Rx = Pz
(2) Py = 0, Qx = 0
(2) Qz = y, Ry = x THESE DO NOT MATCH
(2) F(x,y,z) is not conservative
Line Integrals
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●
●
●
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●
Gives us the ability to integrate multivariable functions and vector fields over arbitrary
curves in a plane or in 3D space
Scalar-line Integrals
Vector-line integrals
We parameterize the curve and partition the curve to get riemann sums
ds denotes that we are moving along the curve C, rather than along one of the axes
Sometimes called line integrals with respect to arc length
How to find line integral from line segment
parameterized vector function
1.
2.
3.
4.
5.
6.
Let f(x,y,z) = the integrand of the original line integral
Take the initial point of the line segment and call it vector “A”
Take the final point of the line segment and call it vector “B”
Calculate: (1-t)A + (t)B
This will be called r(t)
plug the components of r(t) into the integrand expression of the integral
a.
b.
7.
Calculate the gradient of r(t)
a.
8.
9.
11.
technically not the gradient but take the derivative in terms of t of all the components
Calculate the magnitude of the gradient of r(t)
Multiply the magnitude of the gradient of r(t) by dt
a.
10.
r(t) = <x(t), y(t), z(t)>
f(x,y,z) -> f( x(t), y(t), z(t) )
just stick the dt to the value. This will equal ds. (arc length)
Calculate the integral of f( x(t), y(t), z(t) ) on the interval [0,1] multiplied by the magnitude
of the gradient of r(t) multiplied by dt.
This returns the line integral
How to compute line integrals
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●
●
convert everything to parametric equations. a <= t <= b
delta s = int [t_(i-1)][t_i] ||r’(t)||dt
if delta t_i = t_i - t_(i-1)
○
●
●
●
●
●
●
the integral = ||r’(t)||delta t
find the derivative of the parameterized vector
find the magnitude of of the derivative of the param. vector
plug each parameterized component into f(x,y,z)
solve for limits of the integral
plug values into integrals
integrate
How to find the line integrals
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Find the vector versions of the equations of the functions
Find the derivative of the r(t) equation/vector
Find F(r(t) (plug in the components of r(t) into F(x,y,z)
Find the dot product between F(r(t) and r’(t)
take the integral of the dot product with the limits of the t value (probably [0,1])
calculate the integral
Vector line Integrals
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orientation of a curve
●
Oriented curve
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Oriented Closed Curve
○
○
○
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the direction of the curve
a single direction curve
direction loops
F is a constant force
d is the direction and length of travel
Work is (F)dot(d)
If F is not constant, then you can look at the tangent to the curve
W = line integral of (F dot T) ds
ds is the arc length of the parameterized function times dt
T(t) = normalized r’(t)
Work = integral [a,b] of (F(r(t) * (T(t))* ||r’(t)|| dt
Work = integral [a,b] of (F(r(t)*r’(t))dt
How to find the vectors and terminal points
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Make a table with columns: x, y, z, F(x,y,z), terminal point
Plug in an initial point (x,y,z)
Calculate the vector at (x,y,z) by plugging the point into F(x,y,z)
Find the terminal point by adding the vector to the initial point
Example:
(x,y,z)
F(x,y,z) = <-y,z,x>
terminal point
●
(3,1,5)
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<-1,5,3>
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(2,6,8)
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(1,2,1)
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<-2,1,1>
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(-1,3,2)
Applications of Line Integrals
●
Finding the mass of a wire on a curve C having a density p(x,y,z)
○
●
flux is the rate at which a fluid is going across
○
○
●
●
the line integral of (F dot N) ds
line integral of ( F dot n/magnitude(n) )
r(t) = < x(t), y(t) >
n(t) = < y’(t) , -x’(t) >
○
●
take the line integral of p for its mass
n is the normal vector
Calculating Flux
1.
2.
n(t) is the normal vector to r(t)
n(t) = <y’(t), -x’(t)>
a.
b.
c.
3.
line integral of (F dot N) ds
integral [a,b] (F dot n) * ||r’(t)|| dt
integral [a,b] [ F(r(t)) * {n(t)/||n(t)||} * ||r’(t)|| ] dt
integral [a,b] [F(r(t) * n(t)] dt
Conservative Vector Fields
●
Closed
○
●
no closed
○
●
starts and
ends at
different
points
simple
○
●
starts and
ends at the
same point
does not
cross itself
not simple
○
crosses
itself
Fundamental theorem of line integrals
●
Theorem with proof ( F is the parameterized version of f)
Line Integral notes
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●
●
We don’t need to find grad f to find the integral
We don’t need to know the path of C because it’s path independent where F is a
conservative function
if C is a closed curve then line integral of gradf * dx = 0
○
this is because the starting point is the same as the end point and there is NO area to solve
Cross-partial property of conservative fields
●
●
●
Let F = <P,Q,R> be a vector field on an open simply connected region D. Then Py = Qx, Px = Rx,
and Qz = Ry throughout D “iff” F is conservative
Find those partial derivatives
Check if they are equal to their counterparts
If they all match
●
If any of them do NOT match
●
●
A potential function of F is <fx,fy,fz> = <P,Q,R>
The constant of integration is a function of the two constant variables (the variables that are
treated like constants)
integral of P dx = integral of P + g(y,z)
int P dy = iP + g(x,z)
int P dz = iP = g(x,y) etc.
Find the partial derivatives of f(x,y,z)
Set those partial derivatives equal to the original <P,Q,R> and solve for the constant functions of
integration
●
○
○
●
●
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●
●
F is a conservative function
F is NOT a conservative function
Finding the potential function
Set partial derivatives of
potential function equal to
original function
Integrate each of the partials (add
the constant functions)
set the partial derivative of
the partial integral plus the
partial derivative of the
constant function equal to
the original function. Solve
for constant function
through integration (+ C)
Green’s Theorem, Circulation Form
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Let D be an open, simply, connected region with a boundary curve C that is a piecewise
smooth, simple closed curve oriented counterclockwise. Let F = <P,Q,> be a vector field
with component functions that have continuous partial derivatives on D.
The line integral of F dot (dr) = line integral of Pdx + Qdy = double integral (Qx-Py)dA
How to calculate using Green’s Theorem
Counterclockwise
●
C is clockwise, -C is counterclockwise
○
●
●
Identify each component function of the original F vector function
Find the corresponding partial derivative of each function (i.e. Qx and Py)2
○
○
●
●
●
●
apply Green’s Theorem using -C for clockwise (to make it act like it’s counterclockwise)
It’s gonna have the opposite letter in the original integral function
so the function with “dx” is gonna use the d/dy partial and “dy” is d/dx partial derivative
Find the limits of x,y in the shape using the lines/planes that make the object
take the integral using the limits of x and y
the integrand is (Qx - Py)
Integrate the answer
Flux Form of Green’s Theorem
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●
D is an open simply connected region with a boundary curve C that is a piecewise smooth,
simple, closed curve that is oriented COUNTERCLOCKWISE. F = <P,Q> let it be a vector
field with component functions that have continuous partial derivatives on an open
region containing D.
Flux form of Green’s Theorem proof
Stream function
●
F is source free if Px +Qy = 0
○
○
●
●
●
g(x,y) is a stream function for F.
gy = P, gx = -Q
g(x,y) = integral of xdy = xy + h(x)
○
●
●
Where P = x, Q = -y
Px + Qy = 1 + -1 = 0
h(x) is the integration constant
gx = -Q, gx = d/dx ( xy+h(x) ) = y + h’(x)
gx = -Q -> y + h’(x) = -(-y) -> h’(x) = 0 -> h(x) = integral of 0 dx = C
Green’s Theorem on General Regions
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●
●
This is for shapes that have holes, since you can’t use Green’s Theorem on regions with
holes
We split the shape into multiple shapes that do not have holes to use the theorem
The vector functions must stay counterclockwise
Green’s Theorem over general region proof
Extension of flux green’s theorem logic
Green’s theorem over general region diagram
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●
●
●
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Cut the shape to
eliminate the hole
It turns into two shapes
Both must have
counterclockwise
vectors around them
There will be vectors
that would touch if you
put the cut shape back
together
Those vectors travel in
opposite directions
Therefore, you must
cancel them out
Green’s Flux and Circulation
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A vector Field F = <P,Q> with a simply connected domain is source free if Px + Qy = 0
Circulation: line integral of Fdr
○
○
●
Flux: line integral of F dot N ds
○
●
= double integral of (Px + Qy) dA
F as the velocity field of a fluid. The divergence of F at a point P0 represents the net
change of mass of fluid flowing away from P0
○
○
●
= line integral of Pdx + Qdy
= double integral of (Qx - Py) dA
Something like a drain sucking in water has a high negative divergence
A fountain that pushes water away has a high positive divergence
if F = <P,Q,R> and Px, Qy, and Rz all exist
○
the divergence of F (div F) is
■
div F = Px + Qy + Rz
■
(if F = <P,Q>) div F = Px + Qy
Green’s circulation and flux cont.
●
●
●
grad f = <df/dx , df/dy, df/dz>
the nabla symbol for grad is <d/dx. d/dy, d/dz>
if nabla dot F = <d/dx, d/dy, d/dz> . <P,Q,R>
○
○
●
then it equals = dP/dx + dQ/dy + dR/dz = Px + Qy + Rz = div F
so div F = nabla dot F
Example given
Divergence Source-free theorem
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●
●
F = <P,Q> is source-free, the Px + Qy = 0
F is source-free if div F = 0.
Theorem: F = <P,Q> be a continuous vector field with differentiable components with a
simply-connected domain. Then div F = 0 iff F is source-free.
○
○
○
●
●
●
●
There is no point in the field where there is additional fluid coming into the field
Basically you have a container of fluid moving
A container with a pipe flowing fluid into the container is NOT source-free
Example given
flux form = double integral of (Px + Qy) dA
Px + Qy is the divergence
divergence is a pointwise flux
Divergence example
●
●
nabla is the gradient vector symbol (the upside down triangle)
If nabla is a velocity field of a fluid, then at a point P0
○
○
○
●
●
●
●
if div nabla (P0) > 0
■
more fluid is flowing away from P0 than towards it
■
Positive Net Flow
if div nabla (P0) < 0
■
more fluid is flowing into P0 than away from it
■
Negative Net Flow
if div nabla (P0) = 0
■
the fluid flow is balanced between flowing in and out
■
No Net Flow
Example given
so div nabla = 0 at the points: (0,y) for y is any real number
(x,2pi/3 + 2kpi) for any x is real and any integer k
(x, 4pi/3 + 2kpi) for any x is real and any integer k
Divergence of a function with 3 variables
1.
Take the partial derivative of each component with respect to the component’s variable
a.
b.
2.
3.
each component’s variable is <x,y,z>
So do <d/dx, d/dy, d/dz>
Find the sum of the partial derivatives
That is the divergence
Curl of a vector field
●
Meaning
○
○
●
Definition
○
○
○
●
If F = <P,Q,R> (a vector field in R3) (all partial derivatives of F exist)
the curl of F is = (given)
Remember Curl is a VECTOR because it’s the return of a cross product
Note/Tip
○
○
○
○
○
○
●
●
The curl of a vector field measures how much the field “spins” around and in which direction
How fast a hurricane’s winds are and what direction the wind is flowing
If F = <P,Q> pretend that F = <P,Q,0> (zero vector)
then nabla X F = (Qx - Py)k\
If F = <P,Q> then curl F = (Qx - Py)k
so Curl F dot k = (Qx-Py)k dot k = (Qx-Py)(k dot k)
(k dot k) = 1 so (Qx -Py) * 1
(Qx - Py)
circulation = double integral of (curl F dot k)dA
curl is “circulation at a point”
Calculating Flux of a function with 2 variables
1.
Take the integral of the first component with respect to x
a.
2.
3.
4.
5.
Take the derivative of that integral with respect to y
Set that derivative equal to the second component
Solve for g’(y)
integrate both sides with respect to y
a.
6.
7.
8.
9.
10.
remember to add +g(y) integration constant
remember to add +C integration constant
Plug g(y) with the +C back into the first integral of the first component
Plug in a given point
Solve for C
Plug the C value into equation with +C
Solve
Conservative Vector field
●
a conservative vector field has a curl of <0,0,0>
○
●
zero vector
LaPlace Operator & Harmonic Functions
●
LaPlace operator is nabla(gradient) squared
○
●
1.
2.
3.
it looks like a triangle
LaPlace of f = 0 if it’s a harmonic function
Find the gradient of f
Calculate Nabla dot gradient of f
If the dot product is 0, then f is harmonic
Key Equations for Green’s Theorems
Surface Integrals Pre-Note definitions
Surface Integrals parameterization
●
●
●
Parameterize shapes or equations into vectors
Usually use Cylindrical coordinates or Spherical
Usually set x = u and y = v
○
○
●
y = u, z = v etc.
the last function of the r(u,v) function is the variable that is not included in the equation or the
variable that a function returns the value for.
■
(in that case you still just set that variable to u or v)
Regular Parameterization / Smooth Surfaces
●
●
Shapes can be “piecewise smooth” if the shape has smooth faces and corners or edges
Shapes that have no “edges” or “corners” then it’s just “Smooth”
Tangent Plane Parameterization
1.
2.
3.
4.
5.
6.
7.
8.
You have a given r(u,v)
The given point is (a,b,c)
Find the partial derivatives of r(u,v)
Take the cross product of the partial derivatives
Solve for u and v with the given point
Plug the (u,v) point into the cross product = this is the vector p
p = <p1,p2,p3>
p1(x-a) + p2(y-b) + p3(z-c) = tangent plane parameterization
Smooth parametrization of Surface
●
There is technically infinite parameterizations.
○
●
it just has to be understandable to you and easy
Surface area of shape parameterization
1.
Figure out the parameter domain of r(u,v)
a.
2.
3.
Hold either u or v constant and imagine what the shape will be
Increase and decrease u or v to imagine what it does to the shape
a.
b.
4.
5.
6.
7.
This is the set of points in the uv-plane that work with the function
Plug in the constant K into either u or v and get the point from the function
From the points, find the 2D shapes and combine them to find the shape
(The parameter domain is a rectangular prism that contains the entire graph/shape)
Find the partial derivatives of r(u,v) and call it t_u(u,v) and t_v(u,v)
Find the cross product of the two partials
Take the double integral of f(r(u,v)) times the magnitude of the cross product
Parameterization of a surface Finding the
Parameter domain
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Let variables of x and y equal u and v
Find the equations for the sides of the shape in terms of x and y
Find them in terms of u and v
Find the equations with just u = ? on the left side
Find the equations with just v = ? on the left side
Find the parameter domains for integrating with respect to u first
Find the parameter domains for integrating with respect to v first
Chose which parameter domain to use
Write the parameterization of the function in terms of u and v
Compute the magnitude of the cross product of the two partials
Orientation of a surface
●
●
●
Parameterize the graph as r(u,v) = <u, v, f(u,v)>
N = cross product of the partial derivatives of u divided by the cross-product magnitude
The surface integral of a vector field is the double integral of F dot product with N
○
●
●
●
Capital N is the unit normal vector
lowercase n is the normal vector
The computation for the surface integral of a vector field
○
●
F dot N is a scalar function
is given in the double integral
It simplifies to
○
this:
Parameterization of the vector field
1.
2.
3.
Find parameterization of surface with u and v
Find the partials
Find the cross product of them
a.
4.
5.
6.
this is n
Find F(r(u,v))
Take the dot product of F(r(u,v)) and the cross product of the partials
Integrate over the shape
Stokes’ Theorem
●
Remember that the flux of a conservative vector field is 0
Stokes’ Theorem normal vectors Example
1.
S is the upper hemisphere of the unit sphere
a.
2.
The boundary of the circle in the xy-plane is x^2 + y^2 = 1
a.
3.
4.
5.
x^2 + y^2 + z^2 = 1 {z >= 0}
this is the curve C
The normal vectors on the surface point away from the origin
On the boundary of the circle at z = 0, The vectors go straight up in the positive z direction
These are the normal vectors
Stokes’ Theorem example
●
●
●
Evaluate the curl of F where F = <z^2, -3xy, x^3y^3> and S is the part of z = 5-x^2 - y^2
above the plane z=1 and S is oriented outward
The direction of travel is counter clockwise.
The boundary at z = 1 on the paraboloid is 1 = 5 - x^2 - y^2
○
●
r(t) = <2cos(t), 2sin(t), 1> {0 <= t <= 2pi}
○
●
●
●
●
x^2 + y^2 = 4
parameterization of C (the circle)
r’(t) = <-2sin(t), 2cos(t), 0>
F(r(t)) = <1, -12cos(t)sin(t),64cos^3(t)sin^4(t) >
F(r(t)) . r’(t) = -2sin(t) -24cos^2(t)sin(t)
Integral[0,2pi] F(r(t)) * r’(t) dt = 0
Consequence of Stokes’ Theorem
●
A full unit sphere with normal vectors has vectors pointing to the origin from the negative
z-hemisphere
Stokes’ Theorem example
●
●
●
●
●
●
●
Draw shape
Establish counter-clockwise direction
Find direction of normal vectors using right hand rule
Find equation of the plane
Find curl of F
integral [S] (curl of F) dot (N) dS
parameterize S (the plane)
○
○
●
●
●
●
●
●
x = u, v = y, z = 1-u-v
r(u,v) = <u,v,1-u-v>
Take partials of r(u,v)
Find the cross product of the partials
check if normal vectors = r_u X r_v
Find G(u,v) = (curl F)(u,v)
G(x,y) = <0,2z-1,0>
G(u,v) = <0,2(1-u-v)-1,0>
Stokes’ theorem Parameter Domain
●
●
●
xy-plane projection
convert xy units to uv
Find u and v limits for the integral
Stokes’ Theorem
●
●
●
Parameterize the boundary if F (vector field) is hard to parameterize
plug parameterizations of each variable into r(u,v)
calculate the cross product of the partial derivatives of r(u,v)
○
○
●
●
●
●
This is n (the normal vectors)
n = ru X rv
Calculate the curl of F
calculate (curl F)(r(u,v))
take the dot product of (curl F) and n
Interpretation of Curl
(curl F) . N ~~ (curl F)(P0) . N(P0)
●
direction of Curl
○
○
○
●
Magnitude of the Curl
○
●
angular velocity of the paddle wheel
= ½ ||(curl F)(P0) * N(P0)||
= ½||(curl F)(P0)|| * ||N(P0)||
twice the maximum angular
velocity
Remember that
○
○
u . v = ||u|| * ||v||cos(theta)
where theta is the angle between u
&v
Divergence Theorem
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Let S be a piecewise smooth closed surface that encloses a solid E in space. Assume that S
is oriented outward, and let F be a vector field with continuous partial derivatives on an
open region containing E. Then
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N are normal vectors
S is the boundary surfaces
The theorem is also called Gauss’s theorem
if there is net flow out of the surface then the integral is positive. If the integral is
negative, there is net flow into the surface. If the integral is 0, then the net flow into the
surface is equal to the net flow out of the surface.
Divergence Theorem example
On the top cylinder
on the sides
z = 0 on the bottom
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Solve for when z = 0
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net flow is out
of S, since the
integral is
greater than 0
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Note that F has continuous partial derivatives
Let E be the solid enclosed by the surface S
Take div of F
Find limits of integration
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Divergence Theorem
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Divergence
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Let S be a (piecewise) smooth closed surface that encloses a solid region E in space. ASsume that S
is oriented outward, and let F be a vector field with continuous partials for its components
Divergence Theorem Example
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Find the flux of the vector field F over the unit sphere centered at the origin
S (sphere):
E (ball):
We want to find
By the divergence theorem:
div F = nabla . F = 0 + 1 + 0 = 1
Find limits
evaluate:
= 4pi/3