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Vector-Valued Functions

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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.

The parameter t can lie between two real numbers: atb.
Another possibility is that the value 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. 


Checkpoint 

For the vector-valued function   r t = t 2 t i + 2 t + 2 j  , evaluate  r 0   r 1   and  r 4  

 

r0=

  i +   j 

r1=

  i +     j  

r4=

  i +     j  

 

Does this function have any domain restrictions?

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Graphing Vector-Valued Functions

A vector is considered to be in standard position if the initial point is located at the origin. When graphing a vector-valued function, we typically graph the vectors in the domain of the function in standard position, because doing so guarantees the uniqueness of the graph. This convention applies to the graphs of three-dimensional vector-valued functions as well. The graph of a vector-valued function of the form r t = f t i + g t j  consists of the set of all t , r t  and the path it traces is called a plane curve.

The graph of a vector-valued function of the form r t = f t i + g t j + h t k consists of the set of all   t , r t  , and the path it traces is called a space curve. Any representation of a plane curve or space curve using a vector-valued function is called a vector parameterization of the curve.


Checkpoint

Select the graph of the vector-valued function  r t = t 2 6 i + 2 t 7 j ,    0t3.

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Derivatives of Vector-Valued Functions

The derivative of a vector-valued function  r(t) is

 

  r ' t = lim Δt 0 r t + Δt r t Δt  


provided the limit exists. If   r ' t   exists, then  r  is differentiable at t. If   r ' t   exists for all t in an open interval a,b then   r   is differentiable over the interval a,b

 

 

Example: Calculate the derivative of the function

r t = t ln t i + 5 e t j + cos t sin t k .

Solution:

The first component of  r t = t ln t i + 5 e t j + cos t sin t k  is  f t = t ln t , the second component is  g t = 5 e t , and the third component is  h t = cos t sin t . We have   f t = ln t + 1  g t = 5 e t   h t = sin t cos t , so  r t = ln t + 1 i + 5 e t j + sin t cos t k .


Checkpoint

Use the definition to calculate the derivative of the function  r t = 5 t 2 + 3 i + 7 t 1 j .

 

Enter the exact answer in terms of i and  j .

  r t = 

Preview Change entry mode 

 

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Checkpoint

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 the following.

 

a.  Select for   d dt u t × r t  

 

b. Evaluate   d dt r t r t  .

 

Enter the exact answer. Enclose arguments of functions in parentheses. For example,  sin 2 x  .

   d dt r t r t = 

Preview Change entry mode 

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