Show that the path $\lambda(t)=(t,t^2 \cos (\frac{\pi}{t^2}))$ is differentiable but not rectifiable 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:
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.

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
 A: 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
$$\left|\frac{h^2}{h}\right|=|h|$$
so the limit, hence the derivative, is zero.
A: 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
$$\int_a^1 |\gamma'(t)|\,dt\to \infty\,\, \text {as } a\to 0^+.$$
