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# Vectors

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#### Transcript of Vectors

Vectors A presentation by:

Danielle Ambrosio & Kevin Bui BEFORE WE GET STARTED THOUGH... Story Time! YAY!~ What exactly is a vector? Quantities that have a magnitude and direction, represented by arrows on a coordiante plane or in a component form as < x₂-x₁, y₂-y₁ > where (x₁,y₁) is the location where the vector begins and (x₂,y₂) is the location in which the vector terminates. PQ < 2, 4 > Finding Magnitude (|m|) |m| = √(x² + y²) PQ < 2, 4 > | PQ | = √(2² + 4²) = √(4 + 16) = √20 Vector Operations Addition: u+ v = <u₁ + v₁, u₂ + v₂>

Subtraction: u - v = <u₁ - v₁, u₂ - v₂>

Scalar Multiplication: u = < u₁, u₂> (x₁,y₁) = (0,0)

(x₂,y₂) = (2,4)

Component Form :

< x₂-x₁, y₂-y₁ >

< 2-0, 4-0> = <2,4> Where:

u = <u₁, u₂>

v = <v₁, v₂>

= Scalar k k k k Example

u = <-5, 3>

v = <3,6>

= 2

Addition: u+v = <-5+3, 3+6> = <-2,9>

Subtraction: u-v = <-5-3, 3-6> = <-8,-3>

Scalar Multiplication: u = < (-5), (3)> = <-10,6> k k 2 2 Calculating Angle Between Vectors = cos ¹ - x₁ x₂+ y₁y₂ |u||v| Dot Product If u = <x₁,y₁> and v = <x₂,y₂>

then the dot product is

u = (x₁)(x₂) + (y₁)(y₂) v Velocity -Just the derivative of position

v(t)= r'(t)= x'(t)i + y'(t)j +z'(t)k EX.

Find velocity if the position is r(t) = 3ti +2(t^2)j - sin (t) k Just derive!

v(t) = 3i +4tj + cos (t) k II v II Speed r(t) = 3i + 2tj + cos (t) k

Find the speed after pi/4

seconds So v(t) = 2j - sin (t) k

v(pi/4) = rad (4.5) Acceleration a(t) = x"(t)i +y"(t)j + z"(t)k u EX. Find acceleration at t=-1 if

r(t) = (2t - 2) i + (t2 + t + 1)j v = cos ¹ - (-5)(3) + (3)(6) (√34)(√45) v(t) = r'(t) = 2i + (2t + 1) j

v(-1) = 2i - j

a(t) = v'(t) = 2j = 85.6 o we is done now~

Full transcriptDanielle Ambrosio & Kevin Bui BEFORE WE GET STARTED THOUGH... Story Time! YAY!~ What exactly is a vector? Quantities that have a magnitude and direction, represented by arrows on a coordiante plane or in a component form as < x₂-x₁, y₂-y₁ > where (x₁,y₁) is the location where the vector begins and (x₂,y₂) is the location in which the vector terminates. PQ < 2, 4 > Finding Magnitude (|m|) |m| = √(x² + y²) PQ < 2, 4 > | PQ | = √(2² + 4²) = √(4 + 16) = √20 Vector Operations Addition: u+ v = <u₁ + v₁, u₂ + v₂>

Subtraction: u - v = <u₁ - v₁, u₂ - v₂>

Scalar Multiplication: u = < u₁, u₂> (x₁,y₁) = (0,0)

(x₂,y₂) = (2,4)

Component Form :

< x₂-x₁, y₂-y₁ >

< 2-0, 4-0> = <2,4> Where:

u = <u₁, u₂>

v = <v₁, v₂>

= Scalar k k k k Example

u = <-5, 3>

v = <3,6>

= 2

Addition: u+v = <-5+3, 3+6> = <-2,9>

Subtraction: u-v = <-5-3, 3-6> = <-8,-3>

Scalar Multiplication: u = < (-5), (3)> = <-10,6> k k 2 2 Calculating Angle Between Vectors = cos ¹ - x₁ x₂+ y₁y₂ |u||v| Dot Product If u = <x₁,y₁> and v = <x₂,y₂>

then the dot product is

u = (x₁)(x₂) + (y₁)(y₂) v Velocity -Just the derivative of position

v(t)= r'(t)= x'(t)i + y'(t)j +z'(t)k EX.

Find velocity if the position is r(t) = 3ti +2(t^2)j - sin (t) k Just derive!

v(t) = 3i +4tj + cos (t) k II v II Speed r(t) = 3i + 2tj + cos (t) k

Find the speed after pi/4

seconds So v(t) = 2j - sin (t) k

v(pi/4) = rad (4.5) Acceleration a(t) = x"(t)i +y"(t)j + z"(t)k u EX. Find acceleration at t=-1 if

r(t) = (2t - 2) i + (t2 + t + 1)j v = cos ¹ - (-5)(3) + (3)(6) (√34)(√45) v(t) = r'(t) = 2i + (2t + 1) j

v(-1) = 2i - j

a(t) = v'(t) = 2j = 85.6 o we is done now~