1. May 6, 2012

StudentofSci

1. The problem statement, all variables and given/known data
A curve C is defined by the parametric equation x=t^2, y=t^3-3t.
a) show that C has two tangents at the point (3,0) and find their equations.

b) find the points on C where the tangent is horizontal

2. Relevant equations
y-y1=m(x1-x), (dy/dt)/(dx/dt)=m, when dy/dt=0 and dx/dt ≠ 0 there is a horizontal tangent at f(t),g(t), and vice versa for vertical tangents

3. The attempt at a solution

a)
@ (3,0)
3=t^2, t= ±√3, 0=t^3-3t @ ±√3 and @ 0 but since 0 doesnt agree with x=t^2 we just stick with ± √3
(dy/dt)/(dx/dt)= 3t^2-3/2t
@√3 3(√3)^2-3/2√3= 6/2√3= √3=m
y-0=√3(x-3), y=√3x-3√3= tangent @ √3
3(-√3)^2-3/2(-√3)= -12/-2√3= 2√3=m
y-0=2√3(x-3), y=2√3x-6√3=tangent @-√3

End part A

B)
dx/dt=2t=0 when t=0, dy/dt=3t^2-3 ≠0
thus vertical tangent @ f(0), g(0)= (0)^2, (0)^3-3(0)= 0,0
dy/dt=3t^2-3 =0 when t= 1 but dx/dt=2t ≠ 0 false so no horizontal tangent.

End part B.

That is my attempt at solving the problem. Any comments on what I did wrong or just telling me if I'm right is helpful. Thank you for your time.

2. May 6, 2012

LCKurtz

I think you will find the two slopes are $\pm\sqrt 3$.

Remember that a tangent line to a curve $\vec R=\vec R(t)$ when $t=t_0$is given by $\vec T(t) = \vec R(t_0) + t\vec R'(t_0)$. Much easier and quicker to give parametric equations of tangent lines
For a horizontal tangent line, what you need is $\frac{dy}{dt}=0$.