Here is
the solution for this specific problem:
<span>Based from the graph,
the curve will intersect itself at the y-axis, i.e. x = 0. </span><span>
</span><span>t^3 - 6t = 0 </span><span>
<span>t(t^2 -
6) = 0 </span>
<span>t = 0 or
t = ± √6 </span>
<span>dx/dt =
3t^2 - 6 </span>
<span>dy/dt =
2t </span>
<span>dy/dx =
2t/(3t^2 - 6) </span>
<span>@ t = 0,
dy/dx = 0. </span>
<span>x = 0, y
= 0 </span>
<span>y = 0 </span>
<span>@ t = √6,
dy/dx = 2√6/12 = √6/6 </span>
<span>x = 0, y
= 6 </span>
<span>y - 6 =
(√6/6) x </span>
<span>y =
(√6/6)x + 6 </span>
<span>@ t =
-√6, dy/dx = -2√6/12 = -√6/6 </span>
<span>x = 0, y
= 6 </span>
<span>y - 6 =
(-√6/6) x </span>
y = (-√6/6)x
+ 6</span>
So the equations of the tangent
line at the point where the curve crosses itself are: <span>y = (√6/6)x + 6 and
</span>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.
