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

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

dy/dx = (dy/dt)*(dt/dx)

dx/dt = (3t^2 + 12*t + 9) --> dt/dx = 1/ (3t^2 + 12*t + 9), where you have to exclude the values of t that make the denominator equal to zero.

dy/dt = 6*t^2 + 18t - 24 = 6(t^2 +3t -4) = 6(t+4)(t-1), which is equal to zero when t = -4 or when t = +1.

since dy/dx = (dy/dt)*(dt/dx), where dt/dx cannot be equal to zero, and dy/dt equals zero for t = -4 and t = 1; then dy/dx = 0 when dy/dt = 0.