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

Culminating Review for Vectors An Introduction to Vectors Applications of Vectors 4 2 1 Equations of Lines and Planes 0011 0010 1010 1101 0001 0100 1011 Relationships between Points, Lines and Planes 5

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. B Referred to as or 5 km Magnitude is or 30 o A

Directions may be represented as a bearing; which is a compass measurement where the angle is measured from the north in a clockwise direction. quadrant bearing; which is a measurement between 0 o and 90 o east or west from the north-south line. N N 30 o 50 o N60 o E or 060 o S40 o W or 220 o

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 If Then Then

Vector Addition Geometric Vectors Algebraic Vectors If If and and

Vector Subtraction Geometric Vectors Algebraic Vectors If If and 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 Algebraic Vectors and If Geometric Vectors OR 𝑏 2 𝑏 3 𝑏 1 𝑏 2 𝑐 3 𝑐 1 𝑐 2 𝑐 2 𝑦 𝑧 𝑨 down product minus the up product Note: Direction is determined using the right-hand rule 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 : If you equate the respective x , y and z components we obtain the parametric equations : 𝑦 = 𝑦 𝑝 + 𝑢𝑏 𝑧 = 𝑧 𝑝 + 𝑢𝑐 𝑨 = 𝑨 𝑝 + 𝑢𝑑

𝑦 = 𝑦 𝑝 + 𝑢𝑏 𝑧 = 𝑧 𝑝 + 𝑢𝑐 𝑨 = 𝑨 𝑝 + 𝑢𝑑 If you isolate each parametric equation for the parameter, 𝑢 , we have Combining these statements gives the symmetric equation of a line:

Equations of Planes Vector Equation: [ x , y , z ] = [ x o , y o , z o ] + s [ a 1 , a 2 , a 3 ] + t [ b 1 , b 2 , b 3 ] If you equate the respective 𝑦 , 𝑧 and 𝑨 components we obtain the parametric equations: x = x o + sa 1 + tb 1 y = y o + sa 2 + tb 2 z = z o + sa 3 + tb 3

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 l 1 : [ x , y , z ] = [ – 3, 1, 4] + s [ – 1, 1, 4] l 2 : [ x , y , z ] = [1, 4, 6] + t [ – 6, – 1, 6] Convert to parametric form l 1 l 2 x = – 3 – s x = 1 – 6 t y = 4 – t y = 1 + s z = 4 + 4 s z = 6 + 6 t Select any two of the three equations and equate them – 3 – s = 1 – 6 t 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]  : 4 x + 2 y – z – 8 = 0 Convert to line to parametric form x = 3 + s y = 1 – 4 s z = 2 – 8 s Equate 4(3 + s ) + 2(1 – 4 s ) – (2 – 8 s ) – 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 Analyze the normals; what do they tell you?  2 : 2 x + y – 2 z = 3 Eliminate one of the variables, for example, z , 2  1 +  2 gives 4𝑦 − 𝑧 = 9 𝑧 = 4𝑦 − 9 Isolate for a variable Introduce a parameter Let 𝑦 = 𝑢 𝑧 = 4𝑢 − 9 Substitute x = t and y = 4 t – 9 into one of the planes t – (4 t – 9) + z = 3 z = 3t – 6 The line of intersection is x = t y = 4 t – 9 z = 3t – 6

The Intersection of Three Planes  1 : 2𝑦 + 𝑧 − 𝑨 = −3  2 : x – y + 2 z = 0  3 : 3 x + 2 y – z = – 5 Create two new equations 2  1 : 4 x + 2 y – 2 z = – 6  1 : 2 x + y – z = – 3  2 : x – y + 2 z = 0  3 : 3 x + 2 y – z = – 5 2  1 +  3 : 5 x + y  1 –  3 : – x – y = – 6 = 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.

Culminating Review for Calculus Limits Derivatives 4 2 Velocity and Acceleration 1 0011 0010 1010 1101 0001 0100 1011 Extreme Values Optimization 5 Curve Sketching

Limits 𝑦→𝑏 𝑔 𝑦 = 𝑀 lim Tells us that the value of 𝑔 𝑦 approaches 𝑀 as 𝑦 approaches the value 𝑏 . 𝑦→1 − 𝑔 𝑦 = lim (a) 𝑦→1 + 𝑔 𝑦 = lim (b) 𝑦→1 𝑔 𝑦 = lim (c) 𝑔 1 = (d)

• Algebraic methods 𝑦→0 3𝑦 2 − 2𝑦 + 7 lim direct substitution 𝑦 2 + 7𝑦 + 12 lim factor and simplify 𝑦 + 4 𝑦→−4 𝑦 + 9 − 3 multiply by the conjugate 𝑦 + 9 + 3 lim 𝑦 𝑦→0 or use substitution, let 𝑣 = 𝑦 + 9 3 𝑦 + 27 − 3 3 𝑦 + 27 lim use substitution, let 𝑣 = 𝑦 𝑦→0

Continuity • A function 𝑔 is continuous at 𝑦 = 𝑏 if 𝑦→𝑏 𝑔(𝑦) lim 1. exists 𝑦→𝑏 𝑔 𝑦 = 𝑔 𝑏 lim 2.

Derivatives Derivative of a function using first principles : 𝑔 𝑦 + ℎ − 𝑔 𝑦 𝑔 ′ 𝑦 = lim ℎ ℎ→0 Power Rule 𝑔 𝑦 = 𝑦 𝑜 𝑔′ 𝑦 = 𝑜𝑦 𝑜−1 Power of a Function Rule 𝑜−1 𝑕 ′ 𝑦 𝑜 𝑔′ 𝑦 = 𝑜 𝑕 𝑦 𝑔 𝑦 = 𝑕 𝑦

Product Rule 𝑔 ′ 𝑦 = 𝑞 ′ 𝑦 𝑟 𝑦 + 𝑞 𝑦 𝑟 ′ 𝑦 𝑔 𝑦 = 𝑞 𝑦 𝑟 𝑦 Quotient Rule 𝑔′ 𝑦 = 𝑞 ′ 𝑦 𝑟 𝑦 − 𝑞 𝑦 𝑟 ′ 𝑦 𝑔 𝑦 = 𝑞 𝑦 2 𝑟 𝑦 𝑟 𝑦 Chain Rule 𝑔 ′ 𝑦 = 𝑕 ′ ℎ 𝑦 ℎ ′ 𝑦 𝑔 𝑦 = 𝑕 ℎ 𝑦

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

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

Velocity and Acceleration Position, velocity and acceleration • Given position as 𝑡 𝑢 then – Velocity is 𝑤 𝑢 = 𝑡 ′ 𝑢 – Acceleration is 𝑏 𝑢 = 𝑤 ′ 𝑢 = 𝑡 ′′ 𝑢

Extreme Values 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 of these values, and the minimum value of 𝑔(𝑦) on the interval is the smallest of these values

Optimization An Algorithm for Solving Optimization Problems • Identify the variables. • Determine a function that represents the quantity to be optimized. • 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.

Curve Sketching 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.

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.