Question

What are the critical points for the plane curve defined by the equations x(t)=t 33−t22−2t−4y(t)=t32−2t+2

Answer 1

Answer:

The  points are

        $t_1 =  0.7433$

        $t_2 = -0.299$

Step-by-step explanation:

From the question we are told that

   The  first equation is  $x(t) = 3t^3 -2t^2-2t -4$

     The second equation is  $y(t) =  3t^2 -2t -2$

Now differentiating the first and second equation

        $\frac{dx(t)}{dt} =  9t^2 -4t-2$

and

      $\frac{dy(t)}{dt} =  6t-2$

Now

     $\frac{dy(t)}{dx(t)}  =  \frac{\frac{dy(t)}{dt} }{\frac{dx(t)}{dt} } = \frac{9t^2 -4t-2}{6t-2}$

at critical point

     $\frac{dy(t)}{dx(t)}  = 0$

=>   \frac{9t^2 -4t-2}{6t-2}=0[/tex]

=>   $9t^2 - 4t-2 = 0$

solving using quadratic formula we have that

     $t_1 =  0.7433$

and $t_2 = -0.299$