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I've currently a parametric equation which is the following $x=t^3-3t$, $y=-3t^2$, $z=t^3+3t$ in the orthonormal set $(O,i,j,k)$ The first and second derivative vector are the followings

  • $v'(3t^2-3,-6t,3t^2+3)$ and $||v'||=\sqrt{18}*(t^2+1)$

  • $v''(6t,-6,6t)$

    I already proved that the first derivative has a constant angle with the $Oz$ axis. The value is $π/4$.

Now, I've some difficulties to prove that it does exist a straight line D, go throught the origin $O(0,0,0)$, on which the the projected $m$ of the point $M$ has a second derivative null. I need to find something like $y=ax$ which represents the equation of the D straight line at the end.

I should propably use the dot product $Mm.Vector-director-of-D$=0

If sombody knows, it would be great.

Thanks!

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    you mean tangent vector makes constant angle to Oz axis? – 2017-02-28
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    Yes, it is. You're right it's the tangant vector that make the angle (π/4) with Oz axis – 2017-02-28
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    Hello anybody knows about my issue? – 2017-03-02

1 Answers 1

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Not clear what exactly you want.It is a regular space curve, like a non-circular helix. Where is the second derivative null ? In the (x-z) view what looks like an intersection or double point at center is not really so, it is so only in a projection.

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