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Key concepts

  • To calculate the derivative of a vector-valued function, calculate the derivatives of the component functions, then put them back into a new vector-valued function.
  • Many of the properties of differentiation from the Introduction to Derivatives also apply to vector-valued functions.
  • The derivative of a vector-valued function r ( t ) is also a tangent vector to the curve. The unit tangent vector T ( t ) is calculated by dividing the derivative of a vector-valued function by its magnitude.
  • The antiderivative of a vector-valued function is found by finding the antiderivatives of the component functions, then putting them back together in a vector-valued function.
  • The definite integral of a vector-valued function is found by finding the definite integrals of the component functions, then putting them back together in a vector-valued function.

Key equations

  • Derivative of a vector-valued function
    r ( t ) = lim Δ t 0 r ( t + Δ t ) r ( t ) Δ t
  • Principal unit tangent vector
    T ( t ) = r ( t ) r ( t )
  • Indefinite integral of a vector-valued function
    [ 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
  • Definite integral of a vector-valued function
    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

Compute the derivatives of the vector-valued functions.

r ( t ) = t 3 i + 3 t 2 j + t 3 6 k

3 t 2 , 6 t , 1 2 t 2

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r ( t ) = sin ( t ) i + cos ( t ) j + e t k

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r ( t ) = e t i + sin ( 3 t ) j + 10 t k . A sketch of the graph is shown here. Notice the varying periodic nature of the graph.

This figure is a 3 dimensional graph. It is a curve inside of a box. The curve starts at the bottom of the box and spirals around the middle, with upward orientation.

e t , 3 cos ( 3 t ) , 5 t

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r ( t ) = e t i + 2 e t j + k

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r ( t ) = i + j + k

0 , 0 , 0

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r ( t ) = t e t i + t ln ( t ) j + sin ( 3 t ) k

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r ( t ) = 1 t + 1 i + arctan ( t ) j + ln t 3 k

−1 ( t + 1 ) 2 , 1 1 + t 2 , 3 t

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r ( t ) = tan ( 2 t ) i + sec ( 2 t ) j + sin 2 ( t ) k

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r ( t ) = 3 i + 4 sin ( 3 t ) j + t cos ( t ) k

0 , 12 cos ( 3 t ) , cos t t sin t

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r ( t ) = t 2 i + t e −2 t j 5 e −4 t k

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For the following problems, find a tangent vector at the indicated value of t .

r ( t ) = t i + sin ( 2 t ) j + cos ( 3 t ) k ; t = π 3

1 2 1 , −1 , 0

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

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r ( t ) = 3 e t i + 2 e −3 t j + 4 e 2 t k ; t = ln ( 2 )

1 1060.5625 6 , 3 4 , 32

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r ( t ) = cos ( 2 t ) i + 2 sin t j + t 2 k ; t = π 2

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Find the unit tangent vector for the following parameterized curves.

r ( t ) = 6 i + cos ( 3 t ) j + 3 sin ( 4 t ) k , 0 t < 2 π

1 9 sin 2 ( 3 t ) + 144 cos 2 ( 4 t ) 0 , −3 sin ( 3 t ) , 12 cos ( 4 t )

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r ( t ) = cos t i + sin t j + sin t k , 0 t < 2 π . Two views of this curve are presented here:

This figure has two graphs. The first graph is inside a 3 dimensional box. It has a lattice-look to the graph in the middle of the box, crossing over itself. The second graph is the same as the first, with a different position of the box for a different perspective of the lattice-looking curve.
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r ( t ) = 3 cos ( 4 t ) i + 3 sin ( 4 t ) j + 5 t k , 1 t 2

T ( t ) = −12 13 sin ( 4 t ) i + 12 13 cos ( 4 t ) j + 5 13 k

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

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Let r ( t ) = t i + t 2 j t 4 k and s ( t ) = sin ( t ) i + e t j + cos ( t ) k . Here is the graph of the function:

This figure is a 3 dimensional graph. It is inside of a box. The box represents an octant. The curve in the graph starts at the lower left corner of the box and bends upward and out towards the other end of the box.

Find the following.

d d t [ r ( t 2 ) ]

2 t , 4 t 3 , −8 t 7

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d d t [ t 2 · s ( t ) ]

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d d t [ r ( t ) · s ( t ) ]

sin ( t ) + 2 t e t 4 t 3 cos ( t ) + t cos ( t ) + t 2 e t + t 4 sin ( t )

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Compute the first, second, and third derivatives of r ( t ) = 3 t i + 6 ln ( t ) j + 5 e −3 t k .

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Find r ( t ) · r ( t ) for r ( t ) = −3 t 5 i + 5 t j + 2 t 2 k .

900 t 7 + 16 t

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The acceleration function, initial velocity, and initial position of a particle are
a ( t ) = −5 cos t i 5 sin t j , v ( 0 ) = 9 i + 2 j , and r ( 0 ) = 5 i .
Find v ( t ) and r ( t ) .

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The position vector of a particle is r ( t ) = 5 sec ( 2 t ) i 4 tan ( t ) j + 7 t 2 k .

  1. Graph the position function and display a view of the graph that illustrates the asymptotic behavior of the function.
  2. Find the velocity as t approaches but is not equal to π / 4 (if it exists).

  1. This figure is a graph of a curve in 3 dimensions. The curve has asymptotes and from the above view, the curve resembles the secant function.

  2. Undefined or infinite
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Find the velocity and the speed of a particle with the position function r ( t ) = ( 2 t 1 2 t + 1 ) i + ln ( 1 4 t 2 ) j . The speed of a particle is the magnitude of the velocity and is represented by r ' ( t ) .

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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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