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    #1
    Hello!!!

    I want to show that the parabola $y=ax^2, a \neq 0$, has its largest curvature at its vertex and has no minimum curvature.


    I have thought the following:


    The parametrization of the parabola is $(a t^2, t), t \in \mathbb{R}$.

    $v(t)=(2at, 1), ||v(t)||=\sqrt{4a^2t^2+1}\\ T(t)=\frac{1}{\sqrt{4a^2t^2+1}} (2at,1) \\ \frac{dT}{ds}=\frac{1}{4a^2 t^2+1} (2a,0)$

    $\kappa=\frac{2a}{4a^2 s^2+1}$

    $\kappa'(s)=\frac{-16 \alpha^3 s}{(4 a^2 s^2+1)^2}=0\Rightarrow s=0$

    For $s>0$: $\kappa'(s)<0 \Rightarrow \kappa$ decreasing.

    For $s<0$: $\kappa'(s)>0 \Rightarrow \kappa$ increasing.

    So we get that $\kappa(s) \leq 2a$ for any $s$.

    $\kappa(s)=2a \Rightarrow s=0$.

    Is it right so far?

    How can we find the vertex of the parabola? Also, why does the parabola have no minimum curvature? Because there are no other roots rother than $0$ ?

  2. MHB Master
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    Euge's Avatar
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    #2
    Hi evinda,

    The parametrization $(at^2, t)$, $t\in \Bbb R$, is not a parametrization of the parabola $y = ax^2$, but rather the parabola $x = ay^2$. An appropriate parametrization would be $(t,at^2)$, $t\in \Bbb R$. Then

    $$\kappa(t) = \frac{\lvert \dot{x}\ddot{y}-\dot{y}\ddot{x}\rvert}{(\dot{x}^2+\dot{y}^2)^{3/2}} = \frac{|(1)(2a) - (2at)(0)\rvert}{(1 + 4a^2t^2)^{3/2}} = \frac{2\lvert a\rvert}{(1 + 4a^2t^2)^{3/2}}$$

    Therefore

    $$\dot{\kappa}(t) = 2|a|\cdot \left(-\frac{3}{2}\right)(1 + 4a^2t^2)^{-5/2}\cdot (8a^2 t) = -24|a|a^2t(1 + 4a^2t^2)^{-5/2}$$

    There is only one zero of $\frac{d\kappa}{dt}$, namely, $\kappa = 0$. It follows that $\kappa$ has a unique critical point at $t = 0$. Since $\dot{\kappa}(t)$ is positive for negative $t$ and negative for positive $t$, it follows from the first derivative test ensures that $\kappa$ has a maximum at $t = 0$ and no minimum (after all, $\kappa$ has no other extrema besides the one at $t = 0$). The maximum value of $\kappa$ is $\kappa_{\operatorname{max}} = \kappa(0) = 2\lvert a\rvert$. Keep in mind that the vertex of the parabola $y = ax^2$ is $(0,0)$, so the maximum indeed occurs at the vertex where $t = 0$.

    Edit: Here's an important side note. If $f(x)$ is a twice-differentiable function, the curvature of the graph of $y = f(x)$ is given by

    $$\kappa(x) = \frac{\lvert f''(x)\rvert}{(1 + [f'(x)]^2)^{3/2}}$$
    Last edited by Euge; April 14th, 2016 at 14:00. Reason: Give further information

  3. MHB Master
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    #3 Thread Author
    Quote Originally Posted by Euge View Post
    Hi evinda,

    The parametrization $(at^2, t)$, $t\in \Bbb R$, is not a parametrization of the parabola $y = ax^2$, but rather the parabola $x = ay^2$. An appropriate parametrization would be $(t,at^2)$, $t\in \Bbb R$. Then

    $$\kappa(t) = \frac{\lvert \dot{x}\ddot{y}-\dot{y}\ddot{x}\rvert}{(\dot{x}^2+\dot{y}^2)^{3/2}} = \frac{|(1)(2a) - (2at)(0)\rvert}{(1 + 4a^2t^2)^{3/2}} = \frac{2\lvert a\rvert}{(1 + 4a^2t^2)^{3/2}}$$

    Therefore

    $$\dot{\kappa}(t) = 2|a|\cdot \left(-\frac{3}{2}\right)(1 + 4a^2t^2)^{-5/2}\cdot (8a^2 t) = -24|a|a^2t(1 + 4a^2t^2)^{-5/2}$$

    There is only one zero of $\frac{d\kappa}{dt}$, namely, $\kappa = 0$. It follows that $\kappa$ has a unique critical point at $t = 0$. Since $\dot{\kappa}(t)$ is positive for negative $t$ and negative for positive $t$, it follows from the first derivative test ensures that $\kappa$ has a maximum at $t = 0$ and no minimum (after all, $\kappa$ has no other extrema besides the one at $t = 0$). The maximum value of $\kappa$ is $\kappa_{\operatorname{max}} = \kappa(0) = 2\lvert a\rvert$. Keep in mind that the vertex of the parabola $y = ax^2$ is $(0,0)$, so the maximum indeed occurs at the vertex where $t = 0$.

    Edit: Here's an important side note. If $f(x)$ is a twice-differentiable function, the curvature of the graph of $y = f(x)$ is given by

    $$\kappa(x) = \frac{\lvert f''(x)\rvert}{(1 + [f'(x)]^2)^{3/2}}$$
    Ok... Thanks a lot!!!

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