2
$\begingroup$

Let $\alpha : I \to \mathbb{R^2}$ be a regular curve. Suppose that all the tangent lines intersect in a fixed point. Show that $\kappa = 0$.

My attempt:

Let $q$ be the fixed point.

So there exists a funtion $\lambda : I \to \mathbb{R}$ such that

$$q = \alpha(t) + \lambda(t)T(t) \quad(1)$$

where $T(t) = \alpha'(t)/\|\alpha'(t)\|$

Taking the derivative in $(1)$, we have:

$$\alpha'(t) + \lambda'(t)T(t) + \lambda(t)T'(t) = 0 $$ $$\Rightarrow \|\alpha'(t)\|T(t) + \lambda'(t)T(t) + \lambda(t)T'(t) = 0 $$ $$\Rightarrow \big(\|\alpha'(t)\|+ \lambda'(t)\big)T(t) + \lambda(t)T'(t) = 0 $$

From Frenet formula we get $T'(t) = \kappa(t)N(t)$, then:

$$\big(\|\alpha'(t)\|+ \lambda'(t)\big)T(t) + (\lambda(t)\kappa(t))N(t) = 0 $$

Since $\{T,N\}$ is a basis of $\mathbb{R^2}$, we must have $\lambda(t)\kappa(t)=0$

I couldn't see that $\kappa(t)= 0$, $\forall t \in I$. Why can't $\lambda$ be zero for some $t\in I$?

$\endgroup$
  • 1
    $\begingroup$ If $\lambda(t)=0$ then $q=\alpha(t)$. Since $\alpha$ is regular, this can only happen for distinct $t$'s, so by continuity of $\kappa$ it has to equal zero everywhere. $\endgroup$ – String Aug 28 '16 at 14:06
  • $\begingroup$ If you want to apply the Frenet formulas, you should start by assuming the curve is arclength parametrized (so $\|\alpha'(t)\|=1$). Otherwise, your formula for $T'(t)$ is incorrect. $\endgroup$ – Ted Shifrin Aug 28 '16 at 17:14
  • $\begingroup$ @String: Better to say "isolated," rather than "distinct." I can choose all $t$ in an interval to be distinct, but not isolated. $\endgroup$ – Ted Shifrin Aug 28 '16 at 17:15
  • $\begingroup$ @TedShifrin: Point taken! $\endgroup$ – String Aug 28 '16 at 17:48
0
$\begingroup$

Let me suggest another approach. Let $O$ be the point of common intersection of all tangents of the curve $\alpha(t)$. Now, define the vector field $V(x)$ on the real plane $\mathbb{R}^2$ as follows: at each point $x$ on the plane define the vector $V(x) = \overrightarrow{Ox} = x-O$. In Cartesian coordinates centered at $O$, this vector field is simply $V(x) = x$, i.e. $$V(x) = x^1 \frac{\partial}{\partial x^1} + x^2 \frac{\partial}{\partial x^2},$$ where $x = (x^1,x^2)$. Then the curves tangent to this vector field are obtained by solving the system of ODE $\,\,\frac{dx}{ds} = x$. Thus, the solutions are $x(s) = e^s\, x$ which are straight lines passing through the point $O$. Now, observe that by assumption $\alpha'(t)$ is a vector collinear with the vector $\alpha(t) = V(\alpha(t))$. In other words, $\alpha'(t)$ and $V(\alpha(t))$ are collinear which means that $\alpha(t)$ is a curve tangent to the vector field $V(x)$. However, all curves tangent to $V(x)$ are straight lines passing through $O$. Therefore, $\alpha(t)$ is also a (portion of a) straight line passing through $O$, which means its curvature is $0$.

A similar approach, basically the same philosophy, is to go as follows. Let $\beta_0$ be the common point of intersection of all tangents to $\alpha(t)$ (if you wish, for simplicity, you may think $\beta_0 = 0$ is the origin of your coordinate system). This means that $\alpha'(t)$ and $\alpha(t) - \beta_0$ are collinear. Hence, there exists a scalar function $\lambda(t)$ such that $\alpha'(t) = \lambda(t) \big(\alpha(t) - \beta_0\big)$. Write the equations as $$\alpha'(t) - \lambda(t) \big(\alpha(t) - \beta_0\big) = 0$$ and multiply both sides by $e^{-\int \lambda(t)}$ getting $$\Big(e^{-\int \lambda(t)}\Big) \, \alpha'(t) - \Big( e^{-\int \lambda(t)}\Big)\lambda(t) \, \big(\alpha(t) - \beta_0\big) = 0.$$ But this is the derivative $$\frac{d}{dt}\left( \Big(e^{-\int \lambda(t)}\Big) \, \big( \alpha(t)-\beta_0\big)\right) = - \Big( e^{-\int \lambda(t)}\Big)\lambda(t) \, \big(\alpha(t) - \beta_0\big) + \Big(e^{-\int \lambda(t)}\Big) \, \alpha'(t) = 0$$ which after integration yields the existence of a constant vector $\sigma_0 $ such that $$\Big(e^{-\int \lambda(t)}\Big) \, \big( \alpha(t)-\beta_0\big) = \sigma_0. $$ Therefore $$\alpha(t) = \beta_0 + \Big(e^{\int \lambda(t)}\Big) \, \sigma_0.$$ If you reparametrize the latter expression by $s = e^{\int \lambda(t)}$ you get $\alpha(s) = \beta_0 + s \sigma_0$ which is a (portion of a) straight line.

$\endgroup$

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.