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

The points are

`t_1 =  0.7433`

`t_2 = -0.299`

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

`(dx(t))/(dt) =  9t^2 -4t-2`

and

`(dy(t))/(dt) =  6t-2`

Now

`(dy(t))/(dx(t))  =  ((dy(t))/(dt) )/((dx(t))/(dt) ) = (9t^2 -4t-2)/(6t-2)`

at critical point

`(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`