Let $C$ the curve given by $\gamma:\mathbb R\to\mathbb R^n$, $\gamma(t)=(t,t^2,t^3,\ldots,t^n)$.

  1. Show that $C$ is contained in no hyperplane.

2.Find the tangent line and osculating plane at $(0,0,\ldots,0)$.

For (1) my idea is to find the curvature of $\gamma$. We have,

$$\gamma'(t)=(1,2t,3t^2,...,nt^{n-1})\quad \gamma''(t)=(0,2,6t,...,n(n-1)t^{n-2}).$$

Then $k(t)=||\gamma''(t)||=\sqrt[]{4+36t^2+\cdots+(n(n-1)t^{n-2})^2}>0$ so it is not contained in any hyperplane. Is it ok? If so, is there another way to solve this without using curvature?

To get the tangent line and osculating plane, I just take the binormal vector $B(t)=\gamma'(t)\times \gamma''(t)$ and them i get the plane.

Is it ok?

Thanks for your help.

  • 3
    $\begingroup$ The title isn't the same thing as what's in the body. $\endgroup$ – mathematician Oct 11 '16 at 6:40

Your solution involving the curvature doesn't work as positive curvature doesn't guarantee that the curve doesn't lie on a hyperplane if $n > 2$. For example, the curve $\gamma(t) = (\cos(t), \sin(t), 0)$ lies on the $xy$-plane but has curvature $k(t) = ||(-\cos(t), -\sin(t), 0)|| \equiv 1$. Also, your calculation of the curvature is not correct as the curve is not parametrized by arclength.

The correct generalization of the statement and a sufficiently nice curve lies on a line if and only if the curvature vanishes is that the curve lies on a hyperplane if and only if the last generalized curvature vanishes. For $n = 3$ this is the torsion of the curve, not the curvature.

Instead, you can argue directly. A hyperplane in $\mathbb{R}^n$ is described by a linear equation of the form

$$ a_1 x_1 + \dots + a_n x_n = c $$

for $ (0, \dots ,0) \neq (a_1, \dots, a_n) \in \mathbb{R}^n$ and $c \in \mathbb{R}$. If $\gamma(t)$ lies on such hyperplane, we must have

$$ a_1 t + \dots + a_n t^n = c $$

for all $t \in \mathbb{R}$. However, this is a polynomial equation in $t$ and so has at most $n$ solutions. Thus, we have shown that at most $n$ points on the image of $\gamma$ are contained in any given hyperplane.

For $(2)$, the tangent line at $t = 0$ will be $\{ \gamma(0) + t \dot{\gamma}(0) \, | \, t \in \mathbb{R} \}$. To find the osculating plane, again, you can't compute $\dot{\gamma}(0) \times \ddot{\gamma}(0)$ since the cross product of two vectors in $\mathbb{R}^n$ isn't even defined! Instead, the osculating plane is the plane spanned by the first two vectors in the generalized Frenet-Serret frame. Explicitly,

$$e_1(t) = \frac{\dot{\gamma}(t)}{||\dot{\gamma}(t)||}, e_2(t) = \frac{\ddot{\gamma}(t) - \left< \ddot{\gamma}(t), e_1(t) \right>}{||\ddot{\gamma}(t) - \left< \ddot{\gamma}(t), e_1(t) \right>||} $$

and the osculating plane at $\gamma(t)$ is $\gamma(t) + \operatorname{span} \{ e_1(t), e_2(t) \}$.

  • $\begingroup$ Thank you very much Levap, for (2) this question is from a multivariable calculus course, and i can not use Frenet frame, is there any other way to solve this? $\endgroup$ – John Oct 11 '16 at 14:16
  • $\begingroup$ Well, what is your definition of the osculating plane? $\endgroup$ – levap Oct 11 '16 at 18:15

Forget the curvature.

Let's take $n+1$ different points along the curve, the first of them being at $t=0$, just to make things look nice. Draw $n$ vectors from that point to all other points. Now, the non-(hyper)planarity of the curve is equivalent to rank of that system being $n$. In other words, these vectors must be a basis of all $\mathbb R^n$. The easiest way to verify that is to check the determinant of a matrix composed of these vectors - it must not be 0. But wait, that's the Vandermonde matrix, and as long as all points are different (which I said they are), $\det\ne0$.


In my opinion your idea does not work for a dimension greater than $2$.

Hint for (1). Try to show that the polynomials $t,t^2,\dots,t^n$ are linearly independent functions (see https://en.wikipedia.org/wiki/Wronskian).

Moreover, if there are constants $a_1,\dots, a_n$ such that $$a_1 t+a_2 t+\dots +a_nt^n=0$$ for ALL $t\in \mathbb{R}$, what may we conclude?

  • $\begingroup$ My answer has been downvoted. May I ask why? Anything wrong? $\endgroup$ – Robert Z Oct 11 '16 at 10:16
  • $\begingroup$ I can't see your answer is downvoted, but somehow, sometime, one will be without any apparent reaon. This site allows this behaviour and anyone can downvote anything just because. We all have to learn to take it easy...though it can be tough sometimes, I know. $\endgroup$ – DonAntonio Oct 11 '16 at 12:42
  • $\begingroup$ @DonAntonio Thank you for your answer. I will follow your advice. $\endgroup$ – Robert Z Oct 11 '16 at 19:49

If it is in some hyperplane then $\gamma(t)$ is between some two hyperplanes

That is $$ |v=(v_1,\cdots ,v_n)|=1,\ (\gamma (t)-p)\cdot v \geq 0 $$ and $$ (\gamma(t)-(-p))\cdot (-v) \geq 0 $$

Hence $$ C_1\leq \gamma(t)\cdot v \leq C_2 \ \ast$$

so that $$ \frac{C_1}{|t|^n }\leq \frac{\gamma(t)\cdot v}{|t|^n} \leq \frac{C_2}{|t|^n} $$

When $|t|$ goes to $\infty$, then $v_n=0$ And in $\ast$ we divide by $|t|^{n-1}$ so that we conclude $v_{n-1}=0$ Hence $v=0$ It is a contradiction


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.