Find the value of t for which = 0 for the parametric equations

Question

 

Find the value of t for which  = 0 for the parametric equations x = t³ + 6t² + 9t + 1 and y = 2t³ + 9t² - 24t + 6. (5 points)

answer choices

	t = 1, t = 3 and t = 4
	t = -1
	t = -3
	t = 1

Answer 1

dx/dt = 3t²+12t+9 and dy/dt = 6t²+18t-24

dy/dx = (dy/dt)/(dx/dt) = (6t²+18t-24)/(3t²+12t+9) = 0

6t²+18t-24 = 0 --> t²+3t-4 = (t+4)(t-1) = 0 --> t = 1, -4

Select the last choice, t = 1