A point in the xy-plane moves in such a way that Vx = 4t ^ 3 - 4t.? and Vy = 4t If the position of the point (

the position of the point (1,2) when t = 0 find the equation of the trajectory

Answer: and y^ 2 = 4 x

Vx = dx/dt so dx/dt = 4t³ - 4t.
Integrate:
x = t⁴ - 2t² + C
x = t²(t²-2) + C
When t=0, x= 1 so
1 = 0²(0²-2) + C
C = 1
x = t²(t²-2) + 1 (equation 1)

Vy = dy/dt so dy/dt = 4t.
Integrate:
y =2t² + D
When t=0, y=2 so
2= 2*0² + D
D = 2
y =2t² + 2
t² = (y-2)/2

Substitute for t² in equation 1:
x = ((y-2)/2)*((y-2)/2 - 2) + 1
4x = (y-2))*((y-2) - 4) + 4
= (y-2)*y+2) + 4
= y² -4 + 4
= y²