231sec3_1 (1).pdf - Section 3.1 Calculus of Vector-Functions De…nition A vector-valued function is a rule that assigns a vector to each member in a

231sec3_1 (1).pdf - Section 3.1 Calculus of...

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Section 3.1 Calculus of Vector-Functions De nition. A vector-valued function is a rule that assigns a vector to each member in a subset of R 1 : In other words, a vector-valued function is an ordered triple of functions, say f ( t ) ; g ( t ) ; h ( t ) ; and can be expressed as ~ r ( t )= h f ( t ) ; g ( t ) ; h ( t ) i : For instance, ~ r ( t )= h 1+ t; 2 t; 2 ¡ t i ~ q ( t )= ¿ 1 t ¡ 1 ; ln( t ) ; p 2 ¡ t À are vector-valued functions. The domain of a vector-valued function is a subset of all real number at which the function is well-de ned, i.e., Domain of ~ r ( t )= f t j ~ r ( t )= h f ( t ) ; g ( t ) ; h ( t ) i is de ned g = f t j each of f ( t ) ; g ( t ) ; h ( t ) is de ned g = f t j f ( t ) is de ned g \ f t j g ( t ) is de ned g \ f t j g ( t ) is de ned g : So D ( ~ r )= D ( f ) \ D ( g ) \ D ( h ) : Any vector-valued function ~ r ( t )= h x; y; z i may be written in terms of its components as x = f ( t ) y = g ( t ) z = h ( t ) : Thus, the graph of a vector-valued function is a parametric curve in space. For instance, the function ~ r ( t )= h 1+ t; 2 t; 2 ¡ t i is de ned for all t: Its component form is x =1+ t y =2 t z =2 ¡ t: 1
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The graphis a straight line with a direction h 1 ; 2 ; ¡ 1 i passing through (1 ; 0 ; 2) : Example 1.1. Find the domain of ~ r ( t )= ¿ 1 t ¡ 1 ; ln( t ) ; p 2 ¡ t À : Sol: We know that D μ 1 t ¡ 1 = f t 6 =1 g =( ¡1 ; 1) [ (1 ; 1 ) D (ln( t ))= f t > 0 g =(0 ; 1 ) D ¡ p 2 ¡ t ¢ = f t · 2 g =( ¡1 ; 2] : So D ( ~ r )= D μ 1 t ¡ 1 \ D (ln( t )) \ D ¡ p 2 ¡ t ¢ =(( ¡1 ; 1) [ (1 ; 1 )) \ (0 ; 1 ) \ ( ¡1 ; 2] =(( ¡1 ; 1) [ (1 ; 1 )) \ (0 ; 2] =(0 ; 1) [ (1 ; 2] : ° O 1 2 ( ) Limits of vector-valued functions are de ned through compo- nents: For any vector-valued function ~ r ( t )= h f ( t ) ; g ( t ) ; h ( t ) i ; the limit lim t ! a ~ r ( t )= D lim t ! a f ( t ) ; lim t ! a g ( t ) ; lim t ! a h ( t ) E exists if and only if the limits of all three components exist. Example 1.2. Consider ~ r ( t )= h 2cos t; sin t; t i : 2
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(a) Find lim t ! 0
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