please help with a differentiation question 10 pts?
The curve C has parametric equations x=t^2+4t and y=t^3+t^2.
The tangent to the curve at the point A where t=2 is denoted by k.
(I) find the cartesian equation of k
(II) the tangent k meets the curve C again at point B. Use a non calculator method to find the coordinates of B.
I just need help with part II of the question.
- Ian HLv 78 years agoFavorite Answer
The curve C has parametric equations x = t^2 + 4t and y = t^3 + t^2.
dy/dt = 3t^2 +2t, dx/dt = 2t + 4
dy/dx = (3t^2 +2t)/(2t + 4)
(I) Let cartesian equation of tangent be y = mx + c
when t = 2, y = 12, slope m = dy/dx = 16/8 = 2, x = 12
12 = 2*12 + c, so, c = -12
y = 2x - 12
(II) Substituting parametric y and x values into the line equation,
all points where tangent meets the curve are solutions of
t^3 + t^2 = 2(t^2 + 4t ) - 12
t^3 - t^2 - 8t + 12 = 0
But we already expect the coincident solution t = 2 which helps
(t + 3)(t - 2)^2 = 0
Point B has t = -3, x = -3, y = -18
An alternative but more cumbersome approach is
t^2 + 4t - x = 0 select one of the roots, (both work)
t = √(x + 4) - 2
y = t^2*(t + 1) = [-4√(x + 4)+ x + 8 ][√(x + 4) - 1]
y = (x + 12)√(x + 4) - 5x - 24
The tangent k meets the curve C again when
2x - 12 = (x + 12)√(x + 4) - 5x - 24
(x + 12)√(x + 4) = 7x + 12
(x^2 + 24x + 144)(x + 4) = 49x^2 + 168x + 144
x^3 - 21x^2 + 72x + 432 = 0
We already expect the coincident solution x = 12
(x + 3)(x - 12)^2 = 0
Point B has x = -3, t = -3, y = -18
Regards - Ian
- knakeLv 44 years ago
u can take the divide via t^3 up and the capacity will become t^-3 so the answer is-: (6t - 12t^2) (-3t^-4) =-3(6t - 12t^2)/ t^4 [now take t^-4 down and the capacity is t^4] = (-18t + 36t^2) / t^4