Geometric Vectors A geometric vector is a representation of a vector - - PowerPoint PPT Presentation

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Culminating Review for Vectors

An Introduction to Vectors Applications of Vectors Equations of Lines and Planes Relationships between Points, Lines and Planes

An Introduction to Vectors Geometric Vectors

A geometric vector is a representation of a vector using an arrow diagram, or directed line segment, that shows both magnitude and direction.

Referred to as or Magnitude is or

30o A B 5 km

Directions may be represented as a bearing; which is a compass measurement where the angle is measured from the north in a clockwise direction.

30o N 50o N

quadrant bearing; which is a measurement between 0o and 90o east or west from the north-south line.

N60oE or 060o S40oW or 220o

Cartesian (Algebraic) Vectors We identify as a Cartesian vector because its endpoints can be defined using Cartesian coordinates.

position vector

In 3-space

Operations with Vectors

Scalar Multiplication Geometric Vectors Algebraic Vectors If Then If Then

Vector Addition Geometric Vectors Algebraic Vectors If and If and

Vector Subtraction Geometric Vectors Algebraic Vectors If and If and

Applications of Vectors

The Dot Product Geometric Vectors Algebraic Vectors If and We can use the dot product to solve for the angle between any two vectors:

The Cross Product Geometric Vectors Algebraic Vectors

Note: Direction is determined using the right-hand rule

If and OR 𝑏2 𝑏3 𝑏1 𝑏2 𝑐2 𝑐1 𝑐3 𝑐2 𝑦 𝑧 𝑨

down product minus the up product

The magnitude of the cross product is the area of a parallelogram defined by and

Scalar and Vector Projections The scalar projection of onto is The vector projection of onto is Recall:

Equations of Lines and Planes

Equations of Lines

Vector Equation:

𝑦 = 𝑦𝑝 + 𝑢𝑏

If you equate the respective x, y and z components we

• btain the parametric equations:

𝑧 = 𝑧𝑝 + 𝑢𝑐 𝑨 = 𝑨𝑝 + 𝑢𝑑

𝑦 = 𝑦𝑝 + 𝑢𝑏

If you isolate each parametric equation for the parameter, 𝑢, we have

𝑧 = 𝑧𝑝 + 𝑢𝑐 𝑨 = 𝑨𝑝 + 𝑢𝑑

Combining these statements gives the symmetric equation of a line:

Equations of Planes

Vector Equation:

x = xo + sa1 + tb1

If you equate the respective 𝑦, 𝑧 and 𝑨 components we

• btain the parametric equations:

y = yo + sa2 + tb2 z = zo + sa3 + tb3

[x, y, z] = [xo, yo, zo] + s[a1, a2, a3] + t[b1, b2, b3]

Cartesian Equation:

The Cartesian (or scalar) equation of a plane is of the form 𝐵𝑦 + 𝐶𝑧 + 𝐷𝑨 + 𝐸 = 0 with normal 𝑜 = 𝐵, 𝐶, 𝐷 . The normal 𝑜 is a vector perpendicular to all vectors in the plane. Given two directions vectors the normal to a plane is determined by calculating the cross product of the direction vectors.

Relationships between Points, Lines and Planes

The Intersection of Two Lines l1 : [x, y, z] = [–3, 1, 4] + s[–1, 1, 4] l2 : [x, y, z] = [1, 4, 6] + t[–6, –1, 6] x = –3 – s y = 1 + s z = 4 + 4s x = 1 – 6t y = 4 – t z = 6 + 6t Convert to parametric form Select any two of the three equations and equate them l1 l2 –3 – s = 1 – 6t 1 + s = 4 – t Solve for s and t and sub these results into both lines to check your solution

The Intersection of a Line and a Plane l : [x, y, z] = [3, 1, 2] + s[1, –4, –8]  : 4x + 2y – z – 8 = 0 x = 3 + s y = 1 – 4s z = 2 – 8s Convert to line to parametric form Equate 4(3 + s) + 2(1 – 4s) – (2 – 8s) – 8 = 0 Solve for s sub this result into the line to determine the point of intersection

The Intersection of Two Planes 1 : x – y + z = 3 2 : 2x + y – 2z = 3 Analyze the normals; what do they tell you? Eliminate one of the variables, for example, z, 21 + 2 gives 4𝑦 − 𝑧 = 9 Isolate for a variable 𝑧 = 4𝑦 − 9 Introduce a parameter Let 𝑦 = 𝑢 𝑧 = 4𝑢 − 9 Substitute x = t and y = 4t – 9 into one of the planes t – (4t – 9) + z = 3 z = 3t – 6 The line of intersection is x = t y = 4t – 9 z = 3t – 6

The Intersection of Three Planes 1: 2𝑦 + 𝑧 − 𝑨 = −3 2 : x – y + 2z = 0 3 : 3x + 2y – z = –5 Create two new equations 21 + 3 : 5x + y = –6 1 : 2x + y – z = –3 3 : 3x + 2y – z = –5 21 : 4x + 2y – 2z = –6 2 : x – y + 2z = 0 1 – 3 : –x – y = 2 Solve for x and y, and use these values and one of the planes to determine the value of z. Verify the results using the other planes.

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Culminating Review for Calculus

Limits Derivatives Velocity and Acceleration Extreme Values Optimization Curve Sketching

Limits

Tells us that the value of 𝑔 𝑦 approaches 𝑀 as 𝑦 approaches the value 𝑏. lim

𝑦→𝑏 𝑔 𝑦 = 𝑀

(a) lim

𝑦→1− 𝑔 𝑦 =

(b) lim

𝑦→1+ 𝑔 𝑦 =

(c) lim

𝑦→1 𝑔 𝑦 =

(d) 𝑔 1 =

• Algebraic methods

lim

𝑦→0 3𝑦2 − 2𝑦 + 7

lim

𝑦→−4

𝑦2 + 7𝑦 + 12 𝑦 + 4 lim

𝑦→0

𝑦 + 9 − 3 𝑦 lim

𝑦→0

3 𝑦 + 27 − 3

𝑦 direct substitution factor and simplify multiply by the conjugate 𝑦 + 9 + 3

• r use substitution, let 𝑣 =

𝑦 + 9 use substitution, let 𝑣 =

3 𝑦 + 27

Continuity

• A function 𝑔 is continuous at 𝑦 = 𝑏 if

1. lim

𝑦→𝑏 𝑔(𝑦)

exists 2. lim

𝑦→𝑏 𝑔 𝑦 = 𝑔 𝑏

Derivative of a function using first principles: 𝑔′ 𝑦 = lim

ℎ→0

𝑔 𝑦 + ℎ − 𝑔 𝑦 ℎ

Derivatives

Power Rule 𝑔 𝑦 = 𝑦𝑜 𝑔′ 𝑦 = 𝑜𝑦𝑜−1 Power of a Function Rule 𝑔 𝑦 = 𝑕 𝑦

𝑜

𝑔′ 𝑦 = 𝑜 𝑕 𝑦

𝑜−1𝑕′ 𝑦

Product Rule 𝑔 𝑦 = 𝑞 𝑦 𝑟 𝑦 𝑔′ 𝑦 = 𝑞′ 𝑦 𝑟 𝑦 + 𝑞 𝑦 𝑟′ 𝑦 Quotient Rule 𝑔 𝑦 = 𝑞 𝑦 𝑟 𝑦 𝑔′ 𝑦 = 𝑞′ 𝑦 𝑟 𝑦 − 𝑞 𝑦 𝑟′ 𝑦 𝑟 𝑦

2

Chain Rule 𝑔 𝑦 = 𝑕 ℎ 𝑦 𝑔′ 𝑦 = 𝑕′ ℎ 𝑦 ℎ′ 𝑦

Exponential Rules 𝑔 𝑦 = 𝑓𝑦 𝑔′ 𝑦 = 𝑓𝑦 𝑔 𝑦 = 𝑓𝑕 𝑦 𝑔′ 𝑦 = 𝑓𝑕 𝑦 𝑕′ 𝑦 𝑔 𝑦 = 𝑐𝑦 𝑔′ 𝑦 = 𝑐𝑦 ln 𝑐 𝑔 𝑦 = 𝑐𝑕 𝑦 𝑔′ 𝑦 = 𝑐𝑕 𝑦 (ln 𝑐) 𝑕′ 𝑦

Trigonometric Rules 𝑔 𝑦 = sin 𝑦 𝑔′ 𝑦 = cos 𝑦 𝑔 𝑦 = cos 𝑦 𝑔′ 𝑦 = − sin 𝑦 𝑔 𝑦 = sin 𝑕 𝑦 𝑔′ 𝑦 = cos 𝑕 𝑦 𝑕′ 𝑦 𝑔 𝑦 = cos 𝑕 𝑦 𝑔′ 𝑦 = − sin 𝑕 𝑦 𝑕′ 𝑦

Position, velocity and acceleration

• Given position as 𝑡 𝑢 then

– Velocity is 𝑤 𝑢 = 𝑡′ 𝑢 – Acceleration is 𝑏 𝑢 = 𝑤′ 𝑢 = 𝑡′′ 𝑢

Velocity and Acceleration

Algorithm for Finding Maximum and Minimum (Extreme) Values: If a function 𝑔(𝑦) has a derivative at every point in the interval 𝑏 ≤ 𝑦 ≤ 𝑐, calculate 𝑔(𝑦) at:

• all points in the interval 𝑏 ≤ 𝑦 ≤ 𝑐, where 𝑔′ 𝑦 = 0 (critical

values)

• the endpoint 𝑦 = 𝑏 and 𝑦 = 𝑐

The maximum value of 𝑔(𝑦) on the interval 𝑏 ≤ 𝑦 ≤ 𝑐, is the largest

• f these values, and the minimum value of 𝑔(𝑦) on the interval is the

smallest of these values

Extreme Values

An Algorithm for Solving Optimization Problems

Optimization

• Identify the variables.
• Determine a function that represents the quantity to be
• ptimized.
• Determine the domain of the function to be optimized.
• Use the algorithm for extreme values to find the absolute

maximum or minimum in the domain.

• Conclude.

An Algorithm for Curve Sketching Using 𝑔 𝑦

• Determine any intercepts.
• Determine any vertical asymptotes and determine

the behaviour on either side of the asymptote.

• Determine any horizontal or oblique asymptotes.

Identify if the function values approach the asymptote from above or below. Using 𝑔′ 𝑦

• Determine any critical numbers by finding where

𝑔′ 𝑦 = 0 or where 𝑔′ 𝑦 is undefined.

• Determine the intervals of increase/decrease, and

identify any local or absolute extrema.

Curve Sketching

Using 𝑔′′ 𝑦

• Determine possible points of inflection by finding

where 𝑔′′ 𝑦 = 0 or where 𝑔′′(𝑦) is undefined.

• Determine the intervals of concavity, and identify

any points of inflection.

• Sketch the curve.