9_VectorFunc - f ( t ), g ( t ), h ( t ) ( ) , it is...

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Vector Functions Let x = f ( t ), y = g ( t ), z = h ( t ) be parametric equations that determine the path of a particle. Then r r ( t ) = φ(τ 29 ι ρ + γ(τ29ϕ ρ + η(τ29κ ρ is a position vector. Also, we can write r r ( t ) = ξι ρ + ψϕ ρ + ζκ ρ . r r t ( 29 is a vector valued function. Domain: Range: The derivative of r r t ( 29 = r r (τ29 = φ (τ29 ι ρ + ρ + ρ . The derivative of position is So r ( t ) = The derivative is also called the tangent vector. Speed is the magnitude of velocity so speed = v .
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The second derivative of r r t ( 29 = r r (τ29 = φ (τ29 ι ρ + γ(τ29ϕ ρ + η(τ29κ ρ . The derivative of velocity is acceleration so r r (τ29 = α ρ τ ( 29 . ex. r ( t ) = t 2 , t 3
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Graphing r r ( t ), v r t ( 29 , α ρ τ ( 29 1. r ( t ) is a position function with its tail at the origin and its head at the point f ( t ), g ( t ), h ( t ) ( ) . 2. If you graph v t ( ) with its tail at the point
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Unformatted text preview: f ( t ), g ( t ), h ( t ) ( ) , it is tangent to the curve and points in the direction of motion. 3. If you graph a r t ( 29 with its tail at the point f ( t ), g ( t ), h ( t ) ( ) , it will point to the concave side of the curve. ex. r ( t ) = sin t , t In this example is there any point at which a particle on this path is at rest? Do: Let r ( t ) = ln t , 1 + t , 1 , t > 0 Find v t ( ) , a r t ( 29 , and speed at t = 3. Integration: r ( t ) = i- sin tj + e t k . Find r ( t ) . Then suppose r ( ) = 3, 1, - 1 . ex. t i + e-t j + 1 t 2 k 1 4 dt Do: Let a ( t ) = 1 t i-e-t j-2 t 3 k . Find r ( t ) if v 1 ( ) = 0, 1 e , 2 and r 1 ( ) = 1, 1, 1 ....
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This note was uploaded on 04/12/2008 for the course MATH 1224 taught by Professor Dontremember during the Fall '08 term at Virginia Tech.

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9_VectorFunc - f ( t ), g ( t ), h ( t ) ( ) , it is...

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