My path is defined as follows:

$\gamma:[-1,1]\rightarrow \mathbb R, \space \gamma(t):= \begin{cases} \ (0,0) & \text{if $t$=0} \\[2ex] t,t^2 \cos (\frac{\pi}{t^2}), & \text{if $t$ $\in$ [-1,1]\ {0}} \end{cases}$

I need to show that it is:

1) The path is differentiable

2) The path is not rectifiable

I sketched the function $t^2 \cos (\frac{\pi}{t^2})$ to see what it looks like.

enter image description here

1) As far as I know the cosine function and $t^2$ are differentiable functions. Is there any way I can use this fact to argue that $t^2 \cos (\frac{\pi}{t^2})$ is differentiable?

2) Rectifiable means that it is possible to measure the length of the path $\implies$ finite length. By looking at the graph it seems like there are infinitely many oscillations going when $t \rightarrow 0$ so the path becomes infinitely long. Is my reasoning correct? I have no idea how to show this though.

Any help would be greatly appreciated

  • 1
    $\begingroup$ Regarding differentiability, the only point that is not obvious is $0$. Here you have to use the definition of a derivative, i.e. prove the limit $(f(h)-f(0))/h$ when $h\to0$ exists. However, by computing $f'(x)$ on an interval containing $0$, you will show that it's not bounded, and that the integral that appears in the expression of arc length is not convergent at $0$. $\endgroup$ – Jean-Claude Arbaut Apr 18 '15 at 17:06
  • $\begingroup$ @Jean-ClaudeArbaut Thank you for your answer. I am having difficulties showing that the limit exists at 0. $\lim_{h\to 0} \frac{0^2 \cos (\frac{\pi}{0})-0}{0^2}=\frac{0}{0}$ Is there anyway i can get around this? $\endgroup$ – qmd Apr 18 '15 at 18:56
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    $\begingroup$ @Rzeta: See the last addition to my answer. $\endgroup$ – Rory Daulton Apr 18 '15 at 19:13

1) Your function is clearly differentiable at $t$ if $t\ne 0$, since cosine and $t^2$ are differentiable. To show differentiability at $t=0$ use the limit-of-the-difference-quotient definition of the derivative, and use the fact that $|\gamma(t)|\le t^2$. This will quickly show that the derivative at $t=0$ is defined and is zero.

2) Consider that $\gamma(t)$ oscillates between $t^2$ and $-t^2$ infinitely many times in any open interval containing zero. Find the points where $\gamma(t)$ reaches those values, and consider the sum of the line segments between those consecutive points. If $a<0<b$ then the length of the line segment between $(a,\gamma(a))$ and $(b,\gamma(b))$ is at least $|\gamma(a)|+|\gamma(b)|$. You can show the sums of the lengths of those segments is infinite. That will happen since there are "so many points" where $\gamma(t)$ reaches those high and low points.

Answering a comment to this answer: You want to find the points where the $y$-coordinate of $\gamma(t)$ oscillates the most, i.e. where $\cos\frac{\pi}{t^2}=\pm 1$, which is

$$t=\pm\sqrt{\frac 1k}\quad\text{for $k\in\Bbb Z^+$}$$

You find infinitely many of these points in any open interval containing zero, and each one adds at least the value

$$\left|t^2\cos\frac{\pi}{t^2}\right|=\frac 1k$$

to the length of any arc between zero and this $t$. And you already know that the series

$$\sum_{k=n}^{\infty}\frac 1k$$

diverges. Thus, the curve from any such $t$ to zero is not rectifiable.

Answering a comment to a comment to the main question: The derivative at $t=0$ is not $\lim_{h\to 0} \frac{0^2 \cos (\frac{\pi}{0})-0}{0^2}=\frac{0}{0}$ but rather

$$\lim_{h\to 0}\frac{(0+h)^2\cos\frac{\pi}{(0+h)^2}-0}{h}$$

And you can show that the absolute value of the expression inside the limit is less than


so the limit, hence the derivative, is zero.

  • $\begingroup$ Thank you very much for your help. I don't quite understand what you mean when you say " Find the points where γ(t) reaches those values,...". Which values are you referring to? $\endgroup$ – qmd Apr 18 '15 at 17:34
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    $\begingroup$ @Rzeta: The values $t^2$ and $-t^2$. I added more detail to my answer. I see that both of us were sloppy about whether $\gamma(t)$ is a point or a real number: I usually mean it to be a real number, so the point on the curve is $(t,\gamma(t))$. $\endgroup$ – Rory Daulton Apr 18 '15 at 18:58
  • $\begingroup$ Wow, thank you so much. This is great. I just realized that you can relate this problem back to convergent and divergent series that we studied last semester. I undertand that I want to find where the y-coordinate oscillates the most and that this happens when $\cos(\frac{\pi}{t^2})=1$, but how are you getting the relation $t=\sqrt{...}$ $\endgroup$ – qmd Apr 18 '15 at 19:18
  • $\begingroup$ Addition: I think I understand now. Basically I keep plugging in $\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{3}}....\sqrt{\frac{1}{k}}$ into $\cos(\frac{\pi}{t^2})$ and as $k \rightarrow \infty$ my path oscillates infinitely many times. However, one thing I don't understand is how $\lvert t^2 \cos(\frac{\pi}{t^2}) \rvert=\frac{1}{k}$ wouldn't this be $\frac{1}{k^2}$ making it a convergent series? Thanks again for your patience. $\endgroup$ – qmd Apr 18 '15 at 20:04
  • $\begingroup$ @Rzeta: Remember that $t=\pm\sqrt{\frac 1k}$, so $t^2=\frac 1k$ and $\left|t^2\cos\frac{\pi}{t^2}\right|=\frac 1k$. $\endgroup$ – Rory Daulton Apr 18 '15 at 20:08

Sketch of integral approach: If $\gamma:[a,b]\to \mathbb R^2$ is $C^1,$ then the length of the curve is

$$\int_a^b |\gamma'(t)|\,dt.$$

Now your $\gamma$ is $C^1([a,1])$ for each $a,0<a<1.$ So you're done if you show

$$\int_a^1 |\gamma'(t)|\,dt\to \infty\,\, \text {as } a\to 0^+.$$


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