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posted by  el chanan on 11/4/2009 3:27:55 PM  |  status: Closed  |  Earned Karma: 25

vector definrd function

Course Textbook Chapter Problem Needs by
Calculus N/A N/A N/A 11/4/2009 at 8:00:00 PM
Question Details:
Find the velocity and acceleration vectors, the speed at the given value of   t  and the angle between the velocity and the acceleration vectors
 
r = [4 cos t,  sqrt2(sin t)], at the point where t = π/4
 
We are dealing with a vector defined function.
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posted by anonymousxyz on 11/4/2009 5:04:59 PM  |  status: Live
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Response Details:


The position function is:


r( t )  =  <  4 cos t,  √2 sin t  >


r( π / 4
              =  <  4 cos (
π / 4 ),  √2 sin ( π / 4 )  > 

              =  <  4 ( √2 / 2 ),  √2 ( √2 / 2 )  >


              =  <  2√2
,  1  >


The velocity function is:


v( t ) 
         =  r'( t ) 

         =  <  -4 sin t,  √2 cos t  >


v( π / 4
              =  <  -4 sin ( π / 4 ),  √2 cos ( π / 4 )  > 

              =  <  -4 ( √2 / 2 ),  √2 ( √2 / 2 )  >


              =  <  -2√2
,  1  >


The speed function is:


| v( t ) | 
            =  sqrt( 
( -4 sin t )2  +  ( √2 cos t )2  )

            =  sqrt(  16 sin2 t  +  2 cos2 t  )


| v( π / 4
) | 
                  =  sqrt(  16 sin2 ( π / 4 )
  +  2 cos2 ( π / 4 )  )

                  =  sqrt(  16 ( 1 / 2 )  +  2 ( 1 / 2 )  )

                  =  sqrt(  8  +  1  )

                  =  sqrt(  9  )

                  =  3


The acceleration function is:


a( t ) 
         =  v'( t ) 

         =  <  -4 cos t,  -√2 sin t  >


a( π / 4 ) 
              = 
<  -4 cos ( π / 4 ),  -√2 sin ( π / 4 ) >

              =  <  -4 ( √2 / 2 ),  -√2 ( √2 / 2 ) >

              =  <  -2√2,  -1  >


| a( t ) | 
            =  sqrt( 
( -4 cos t )2  +  ( -√2 sin t )2  )

            =  sqrt(  16 cos2 t  +  2 sin2 t  )



| a( π / 4
) | 
                  =  sqrt(  16 cos2 ( π / 4 )
  +  2 sin2 ( π / 4 )  )

                  =  sqrt(  16 ( 1 / 2 )  +  2 ( 1 / 2 )  )

                  =  sqrt(  8  +  1  )

                  =  sqrt(  9  )

                  =  3


Find the angle between the velocity and acceleration vectors when t =
π / 4 as follows:


v( π / 4 )  dot  a( π / 4 ) 
                                    =  <  -2√2,  1  >  dot  <  -2√2,  -1  > 

                                    =  (
-2√2 )( -2√2 )  +  ( 1 )( -1 )

                                    =  8
  -  1

                                    =  7


v( π / 4 )  dot  a( π / 4 )  =   | v( π / 4 ) | | a( π / 4 ) | cos θ  =>

7  =   ( 3 )( 3 ) cos θ  =>

cos θ  =  7 / 9  =>

θ  =  cos-1 ( 7 / 9 )  =>

θ  =  0.679673819 rad  =>

θ  =  38.9424413 deg


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