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shutvik [7]
3 years ago
9

The equation r(t) = sin(6t) i + cos(6t) j, t > 0 describes the motion of a particle moving along the unit circle.

Mathematics
1 answer:
Alex17521 [72]3 years ago
7 0

Answer:

1. constant speed of 6 units per second

2. yes

3. clockwise

4. no it begins from the point ( 0 , 1 )

Step-by-step explanation:

Solution:-

- The position of a particle moving in a path of a unit circle is defined by the following vector equation:-

                    r ( t ) = sin ( 6t ) i + cos ( 6t )  j

- To determine the speed of the particle in the circular motion we will derivate the position vector ( r ) of the particle with respect to time ( t ) to get the velocity vector:

                   d r (t ) / dt = v ( t )

                   v ( t ) = 6*[ cos ( 6t ) i - sin ( 6t ) j ]

- We will determine the speed of the particle by determining the magnitude of the velocity vector v ( t ) as follows:

                 | v(t) | = \sqrt{6^2 * cos^2 ( 6t ) + 6^2 * sin^2 ( 6t )}

                 | v(t) | = \sqrt{36. [cos^2 ( 6t ) + sin^2 ( 6t ) ] }

                 | v(t) | = \sqrt{36.} \sqrt{1}  = 6

- The speed | v(t) | remians constant at 6 units per second.  

- To determine the acceleration vector a ( t ) we will derivate the velocity vector v ( t ) with respect to time t as follows

                d v(t) / dt = a ( t )

                a ( t ) = - 36 * [ sin ( 6t ) i + cos ( 6t ) j ]

                   

- To determine whether the two vectors v ( t ) and a ( t ) are orthogonal to each other we will apply the dot product test for orthogonal vectors to be equal to zero as follows:

              v(t) . a (t) = -6*36 [ cos ( 6t ) * sin ( 6t )  -  sin ( 6t ) * cos ( 6t ) ]

              v(t) . a (t) = -6*36 [ 0 ]  = 0  ... ( proven )

- The velocity and acceleration vectors are orthogonal at all times t.

- To determine the direction of particle motion we will plug in two consecutive values of t = 0 and t = π / 6 and determine the value of position vector r ( t ):

           r ( 0 ) = sin ( 0 ) i + cos ( 0 ) j

           r ( 0 ) = 0 i + 1 j

 

          r ( π / 12 ) = sin ( π/2 ) i + cos ( π /2 ) j

          r ( π / 12 ) = 1 i + 0 j

- Plot the two points r ( 0 ) and r ( π / 12 ) on a Cartesian coordinate system and join the two with a curve directed from [ 0 i + 1 j ] to [ 1 i + 0 j ]. We see the motion is clockwise and starts from point ( 0 , 1 ) not ( 1 , 0 )

         

                 

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