The value of t for which the particle is at rest is : t = 2
Given data:
X = t³ - 3t²
Y = 2t³ - 3t² - 12t
First step : Determine the first derivative of eac equations
dx = 3t² - 6t
dy = 6t² - 6t - 12
Next step : determine the value of slope ( dy/dx ) = o
dy / dx = ( 6 (t² - t - 2) ) / ( 3t ( t - 2) )
Therefore
dy / dx = ( 2 ( t + 1 ) (t - 2) ) / ( t ( t -2) )
= 0
Therefore at ; t = -1 and 2 the particle is at rest
Hence we can conclude that Neglecting the negative value the particle will be at rest when t = 2
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Attached below is the complete question
The position of a particle moving in the xy-plane is given by the parametric equations x = t3 - 3t2 and y = 2t3 - 3t2 - 12t. For what values of t is the particle at rest?
