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  • Write the general equation of a vector-valued function in component form and unit-vector form.
  • Recognize parametric equations for a space curve.
  • Describe the shape of a helix and write its equation.
  • Define the limit of a vector-valued function.

Our study of vector-valued functions combines ideas from our earlier examination of single-variable calculus with our description of vectors in three dimensions from the preceding chapter. In this section we extend concepts from earlier chapters and also examine new ideas concerning curves in three-dimensional space. These definitions and theorems support the presentation of material in the rest of this chapter and also in the remaining chapters of the text.

Definition of a vector-valued function

Our first step in studying the calculus of vector-valued functions is to define what exactly a vector-valued function is. We can then look at graphs of vector-valued functions and see how they define curves in both two and three dimensions.

Definition

A vector-valued function    is a function of the form

r ( t ) = f ( t ) i + g ( t ) j or r ( t ) = f ( t ) i + g ( t ) j + h ( t ) k ,

where the component functions     f, g, and h , are real-valued functions of the parameter t. Vector-valued functions are also written in the form

r ( t ) = f ( t ) , g ( t ) or r ( t ) = f ( t ) , g ( t ) , h ( t ) .

In both cases, the first form of the function defines a two-dimensional vector-valued function; the second form describes a three-dimensional vector-valued function.

The parameter t can lie between two real numbers: a t b . Another possibility is that the value of t might take on all real numbers. Last, the component functions themselves may have domain restrictions that enforce restrictions on the value of t. We often use t as a parameter because t can represent time.

Evaluating vector-valued functions and determining domains

For each of the following vector-valued functions, evaluate r ( 0 ) , r ( π 2 ) , and r ( 2 π 3 ) . Do any of these functions have domain restrictions?

  1. r ( t ) = 4 cos t i + 3 sin t j
  2. r ( t ) = 3 tan t i + 4 sec t j + 5 t k
  1. To calculate each of the function values, substitute the appropriate value of t into the function:
    r ( 0 ) = 4 cos ( 0 ) i + 3 sin ( 0 ) j = 4 i + 0 j = 4 i r ( π 2 ) = 4 cos ( π 2 ) i + 3 sin ( π 2 ) j = 0 i + 3 j = 3 j r ( 2 π 3 ) = 4 cos ( 2 π 3 ) i + 3 sin ( 2 π 3 ) j = 4 ( 1 2 ) i + 3 ( 3 2 ) j = −2 i + 3 3 2 j .

    To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is f ( t ) = 4 cos t and the second component function is g ( t ) = 3 sin t . Neither of these functions has a domain restriction, so the domain of r ( t ) = 4 cos t i + 3 sin t j is all real numbers.
  2. To calculate each of the function values, substitute the appropriate value of t into the function:
    r ( 0 ) = 3 tan ( 0 ) i + 4 sec ( 0 ) j + 5 ( 0 ) k = 0 i + 4 j + 0 k = 4 j r ( π 2 ) = 3 tan ( π 2 ) i + 4 sec ( π 2 ) j + 5 ( π 2 ) k , which does not exist r ( 2 π 3 ) = 3 tan ( 2 π 3 ) i + 4 sec ( 2 π 3 ) j + 5 ( 2 π 3 ) k = 3 ( 3 ) i + 4 ( −2 ) j + 10 π 3 k = −3 3 i 8 j + 10 π 3 k .

    To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is f ( t ) = 3 tan t , the second component function is g ( t ) = 4 sec t , and the third component function is h ( t ) = 5 t . The first two functions are not defined for odd multiples of π / 2 , so the function is not defined for odd multiples of π / 2 . Therefore, dom ( r ( t ) ) = { t | t ( 2 n + 1 ) π 2 } , where n is any integer.
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Practice Key Terms 8

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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