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Find the unit tangent vector for the vector-valued function

r ( t ) = ( t 2 3 ) i + ( 2 t + 1 ) j + ( t 2 ) k .

T ( t ) = 2 t 4 t 2 + 5 i + 2 4 t 2 + 5 j + 1 4 t 2 + 5 k

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Integrals of vector-valued functions

We introduced antiderivatives of real-valued functions in Antiderivatives and definite integrals of real-valued functions in The Definite Integral . Each of these concepts can be extended to vector-valued functions. Also, just as we can calculate the derivative of a vector-valued function by differentiating the component functions separately, we can calculate the antiderivative in the same manner. Furthermore, the Fundamental Theorem of Calculus applies to vector-valued functions as well.

The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function represents the velocity of an object at time t , then its antiderivative represents position. Or, if the function represents the acceleration of the object at a given time, then the antiderivative represents its velocity.

Definition

Let f, g, and h be integrable real-valued functions over the closed interval [ a , b ] .

  1. The indefinite integral of a vector-valued function     r ( t ) = f ( t ) i + g ( t ) j is
    [ f ( t ) i + g ( t ) j ] d t = [ f ( t ) d t ] i + [ g ( t ) d t ] j .

    The definite integral of a vector-valued function    is
    a b [ f ( t ) i + g ( t ) j ] d t = [ a b f ( t ) d t ] i + [ a b g ( t ) d t ] j .
  2. The indefinite integral of a vector-valued function r ( t ) = f ( t ) i + g ( t ) j + h ( t ) k is
    [ f ( t ) i + g ( t ) j + h ( t ) k ] d t = [ f ( t ) d t ] i + [ g ( t ) d t ] j + [ h ( t ) d t ] k .

    The definite integral of the vector-valued function is
    a b [ f ( t ) i + g ( t ) j + h ( t ) k ] d t = [ a b f ( t ) d t ] i + [ a b g ( t ) d t ] j + [ a b h ( t ) d t ] k .

Since the indefinite integral of a vector-valued function involves indefinite integrals of the component functions, each of these component integrals contains an integration constant. They can all be different. For example, in the two-dimensional case, we can have

f ( t ) d t = F ( t ) + C 1 and g ( t ) d t = G ( t ) + C 2 ,

where F and G are antiderivatives of f and g, respectively. Then

[ f ( t ) i + g ( t ) j ] d t = [ f ( t ) d t ] i + [ g ( t ) d t ] j = ( F ( t ) + C 1 ) i + ( G ( t ) + C 2 ) j = F ( t ) i + G ( t ) j + C 1 i + C 2 j = F ( t ) i + G ( t ) j + C ,

where C = C 1 i + C 2 j . Therefore, the integration constant becomes a constant vector.

Integrating vector-valued functions

Calculate each of the following integrals:

  1. [ ( 3 t 2 + 2 t ) i + ( 3 t 6 ) j + ( 6 t 3 + 5 t 2 4 ) k ] d t
  2. [ t , t 2 , t 3 × t 3 , t 2 , t ] d t
  3. 0 π / 3 [ sin 2 t i + tan t j + e −2 t k ] d t
  1. We use the first part of the definition of the integral of a space curve:
    [ ( 3 t 2 + 2 t ) i + ( 3 t 6 ) j + ( 6 t 3 + 5 t 2 4 ) k ] d t = [ 3 t 2 + 2 t d t ] i + [ 3 t 6 d t ] j + [ 6 t 3 + 5 t 2 4 d t ] k = ( t 3 + t 2 ) i + ( 3 2 t 2 6 t ) j + ( 3 2 t 4 + 5 3 t 3 4 t ) k + C .
  2. First calculate t , t 2 , t 3 × t 3 , t 2 , t :
    t , t 2 , t 3 × t 3 , t 2 , t = | i j k t t 2 t 3 t 3 t 2 t | = ( t 2 ( t ) t 3 ( t 2 ) ) i ( t 2 t 3 ( t 3 ) ) j + ( t ( t 2 ) t 2 ( t 3 ) ) k = ( t 3 t 5 ) i + ( t 6 t 2 ) j + ( t 3 t 5 ) k .

    Next, substitute this back into the integral and integrate:
    [ t , t 2 , t 3 × t 3 , t 2 , t ] d t = ( t 3 t 5 ) i + ( t 6 t 2 ) j + ( t 3 t 5 ) k d t = ( t 4 4 t 6 6 ) i + ( t 7 7 t 3 3 ) j + ( t 4 4 t 6 6 ) k + C .
  3. Use the second part of the definition of the integral of a space curve:
    0 π / 3 [ sin 2 t i + tan t j + e −2 t k ] d t = [ 0 π / 3 sin 2 t d t ] i + [ 0 π / 3 tan t d t ] j + [ 0 π / 3 e −2 t d t ] k = ( 1 2 cos 2 t ) | 0 π / 3 i ( ln ( cos t ) ) | 0 π / 3 j ( 1 2 e −2 t ) | 0 π / 3 k = ( 1 2 cos 2 π 3 + 1 2 cos 0 ) i ( ln ( cos π 3 ) ln ( cos 0 ) ) j ( 1 2 e −2 π / 3 1 2 e −2 ( 0 ) ) k = ( 1 4 + 1 2 ) i ( ln 2 ) j ( 1 2 e −2 π / 3 1 2 ) k = 3 4 i + ( ln 2 ) j + ( 1 2 1 2 e −2 π / 3 ) k .
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Calculate the following integral:

1 3 [ ( 2 t + 4 ) i + ( 3 t 2 4 t ) j ] d t .

1 3 [ ( 2 t + 4 ) i + ( 3 t 2 4 t ) j ] d t = 16 i + 10 j

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Practice Key Terms 5

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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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