Here is
the solution for this specific problem:
Based from the graph,
the curve will intersect itself at the y-axis, i.e. x = 0.
t^3 - 6t = 0
t(t^2 -
6) = 0
t = 0 or
t = ± √6
dx/dt =
3t^2 - 6
dy/dt =
2t
dy/dx =
2t/(3t^2 - 6)
@ t = 0,
dy/dx = 0.
x = 0, y
= 0
y = 0
@ t = √6,
dy/dx = 2√6/12 = √6/6
x = 0, y
= 6
y - 6 =
(√6/6) x
y =
(√6/6)x + 6
@ t =
-√6, dy/dx = -2√6/12 = -√6/6
x = 0, y
= 6
y - 6 =
(-√6/6) x
y = (-√6/6)x
+ 6
So the equations of the tangent
line at the point where the curve crosses itself are: y = (√6/6)x + 6 and
y = (-√6/6)x + 6. I am hoping that these answers have
satisfied your queries and it will be able to help you in your endeavors, and
if you would like, feel free to ask another question.
