2
$\begingroup$

At which points on the curve $\alpha(t):=(3t-t^3,3t^2,3t+t^3)$ the corresponding tangent lines are parallel to the plane $3x+y+z+2=0$?

2 Answers 2

3

Hints:

  1. The tangent line at $t$ has direction $\alpha'(t)$
  2. The normal of a plane given by $ax + by + cz + d = 0$ is $(a,b,c)$.
  3. In order for a line to be parallel to a plane, its direction vector must be orthogonal to the normal to the plane.

This will give you a polynomial equation in $t$.


The tangent line at $t$ has slope $\alpha'(t) = (3-3t^2,6t, 3 + 3t^2)$, and for it to be parallel to the plane given by the equation $3x+y+z +2= 0$, with normal $(3,1,1)$, we must have $(3-3t^2,6t, 3 + 3t^2) \cdot (3,1,1) = 0$.

Then

$\begin{align*}&(3-3t^2,6t, 3 + 3t^2)\cdot (3,1,1)=9-9t^2+6t+3+3t^2=-6t^2+6t+ 12=0\\ &\Leftrightarrow t^2-t-2=0 \Leftrightarrow(t+1)(t-2)=0\end{align*}$

And we get

$t = -1$ or $t=2$.

  • 0
    But ... it was homework.2012-06-23
  • 0
    @Gigili I am getting a bit carried away, aren't I? I've fixed it somewhat. I can't make the spoiler thing work.2012-06-23
  • 0
    Forget about the spoiler. It's the same as in Mohamed's answer anyway.2012-06-23
  • 0
    Weird, I've spent about five minutes and it doesn't work! Silly thing. +1 anyway, and thanks for editing your answer.2012-06-23
  • 0
    @Gigili It seems to work for me now. I think there has to be a non-empty line between two spoiler blocks (although I've seen other posts without this problem). Thanks for editing my answer, and for pointing out that I was giving away too much.2012-06-23
  • 1
    Thank you for your answer. I initially thought about approach similar to this but at the end I wasn't sure.2012-06-23
  • 0
    @Gigili Sorry for this comment on an old answer but does this give the times $t$ at which the space curve is tangential to the plane? I'm doing a similar problem and I'm not sure if it's related to this.2016-11-02
0

$n=(3,1,2)$ is a normal vecttor to the plane . We have : $\alpha'(t)=(3-3t^2,6t,3+3t^2)$ is a tangent line director vector is parallel to the plane if it's normal to $n$, thus : $n.\alpha'(t)=0$, thus : $$3(1-t^2) + 2t + 1+t^2=0$$ $$-2t^2 + 2t +4 =0 $$ $$t^2-t-2 =0 $$ gives : $t=-1$ or $t=2$