calculus question please help

Question

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

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

Answer 1

For a curve defined parametrically in terms of t, dy/dx = (dy/dt) / (dx/dt). The curve will have horizontal tangent lines when dy/dt = 0, provided dx/dt ≠ 0.

dy/dt = 6t² +18t - 24

dx/dt = 3t² +12t + 9

So, dy/dx = [(6t² + 18t -24) / (3t² + 12t + 9)]

dy/dx = 0 when the numerator of that fraction equals zero:

6t² + 18t - 24 = 0

t² + 3t - 4 = 0

(t + 4)(t - 1) = 0

t = -4 or t = 1.

dx/dt does not equal zero at those values of t, so the graph has horizontal tangent lines at the points where t = -4 or t = 1.

If you wanted to know where the graph has vertical tangent lines (if any), that will be where the denominator (dx/dt) is equal to zero (and dy/dt ≠ 0).

desmos.com/calculator/q8msqih5sa

Use the slider on "v" to see the graph being traced out.