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Now for some examples using these properties.

Using the properties of derivatives of vector-valued functions

Given the vector-valued functions

r ( t ) = ( 6 t + 8 ) i + ( 4 t 2 + 2 t 3 ) j + 5 t k

and

u ( t ) = ( t 2 3 ) i + ( 2 t + 4 ) j + ( t 3 3 t ) k ,

calculate each of the following derivatives using the properties of the derivative of vector-valued functions.

  1. d d t [ r ( t ) · u ( t ) ]
  2. d d t [ u ( t ) × u ( t ) ]
  1. We have r ( t ) = 6 i + ( 8 t + 2 ) j + 5 k and u ( t ) = 2 t i + 2 j + ( 3 t 2 3 ) k . Therefore, according to property iv.:
    d d t [ r ( t ) · u ( t ) ] = r ( t ) · u ( t ) + r ( t ) · u ( t ) = ( 6 i + ( 8 t + 2 ) j + 5 k ) · ( ( t 2 3 ) i + ( 2 t + 4 ) j + ( t 3 3 t ) k ) + ( ( 6 t + 8 ) i + ( 4 t 2 + 2 t 3 ) j + 5 t k ) · ( 2 t i + 2 j + ( 3 t 2 3 ) k ) = 6 ( t 2 3 ) + ( 8 t + 2 ) ( 2 t + 4 ) + 5 ( t 3 3 t ) + 2 t ( 6 t + 8 ) + 2 ( 4 t 2 + 2 t 3 ) + 5 t ( 3 t 2 3 ) = 20 t 3 + 42 t 2 + 26 t 16 .
  2. First, we need to adapt property v. for this problem:
    d d t [ u ( t ) × u ( t ) ] = u ( t ) × u ( t ) + u ( t ) × u″ ( t ) .

    Recall that the cross product of any vector with itself is zero. Furthermore, u″ ( t ) represents the second derivative of u ( t ) :


    u″ ( t ) = d d t [ u ( t ) ] = d d t [ 2 t i + 2 j + ( 3 t 2 3 ) k ] = 2 i + 6 t k .

    Therefore,


    d d t [ u ( t ) × u ( t ) ] = 0 + ( ( t 2 3 ) i + ( 2 t + 4 ) j + ( t 3 3 t ) k ) × ( 2 i + 6 t k ) = | i j k t 2 3 2 t + 4 t 3 3 t 2 0 6 t | = 6 t ( 2 t + 4 ) i ( 6 t ( t 2 3 ) 2 ( t 3 3 t ) ) j 2 ( 2 t + 4 ) k = ( 12 t 2 + 24 t ) i + ( 12 t 4 t 3 ) j ( 4 t + 8 ) k .
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Given the vector-valued functions r ( t ) = cos t i + sin t j e 2 t k and u ( t ) = t i + sin t j + cos t k , calculate d d t [ r ( t ) · r ( t ) ] and d d t [ u ( t ) × r ( t ) ] .

d d t [ r ( t ) · r ( t ) ] = 8 e 4 t

d d t [ u ( t ) × r ( t ) ] = ( e 2 t ( cos t + 2 sin t ) + cos 2 t ) i + ( e 2 t ( 2 t + 1 ) sin 2 t ) j + ( t cos t + sin t cos 2 t ) k

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Tangent vectors and unit tangent vectors

Recall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function r ( t ) = cos t i + sin t j . The derivative of this function is r ( t ) = sin t i + cos t j . If we substitute the value t = π / 6 into both functions we get

r ( π 6 ) = 3 2 i + 1 2 j and r ( π 6 ) = 1 2 i + 3 2 j .

The graph of this function appears in [link] , along with the vectors r ( π 6 ) and r ( π 6 ) .

This figure is the graph of a circle represented by the vector-valued function r(t) = cost i + sint j. It is a circle centered at the origin with radius of 1, and counter-clockwise orientation. It has a vector from the origin pointing to the curve and labeled r(pi/6). At the same point on the circle there is a tangent vector labeled “r’(pi/6)”.
The tangent line at a point is calculated from the derivative of the vector-valued function r ( t ) .

Notice that the vector r ( π 6 ) is tangent to the circle at the point corresponding to t = π / 6 . This is an example of a tangent vector    to the plane curve defined by r ( t ) = cos t i + sin t j .

Definition

Let C be a curve defined by a vector-valued function r, and assume that r ( t ) exists when t = t 0 . A tangent vector v at t = t 0 is any vector such that, when the tail of the vector is placed at point r ( t 0 ) on the graph, vector v is tangent to curve C. Vector r ( t 0 ) is an example of a tangent vector at point t = t 0 . Furthermore, assume that r ( t ) 0 . The principal unit tangent vector    at t is defined to be

T ( t ) = r ( t ) r ( t ) ,

provided r ( t ) 0 .

The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative r ( t ) . Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.

Finding a unit tangent vector

Find the unit tangent vector for each of the following vector-valued functions:

  1. r ( t ) = cos t i + sin t j
  2. u ( t ) = ( 3 t 2 + 2 t ) i + ( 2 4 t 3 ) j + ( 6 t + 5 ) k

  1. First step: r ( t ) = −sin t i + cos t j Second step: r ( t ) = ( sin t ) 2 + ( cos t ) 2 = 1 Third step: T ( t ) = r ( t ) r ( t ) = −sin t i + cos t j 1 = −sin t i + cos t j

  2. First step: u ( t ) = ( 6 t + 2 ) i 12 t 2 j + 6 k Second step: u ( t ) = ( 6 t + 2 ) 2 + ( −12 t 2 ) 2 + 6 2 = 144 t 4 + 36 t 2 + 24 t + 40 = 2 36 t 4 + 9 t 2 + 6 t + 10 Third step: T ( t ) = u ( t ) u ( t ) = ( 6 t + 2 ) i 12 t 2 j + 6 k 2 36 t 4 + 9 t 2 + 6 t + 10 = 3 t + 1 36 t 4 + 9 t 2 + 6 t + 10 i 6 t 2 36 t 4 + 9 t 2 + 6 t + 10 j + 3 36 t 4 + 9 t 2 + 6 t + 10 k
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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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