Let $\gamma : (a,b) \rightarrow \mathbb{R}^2$ be a regular curve. Let $\iota : \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be the map \begin{equation}\iota\left(\begin{pmatrix}x \\y\end{pmatrix}\right) = \begin{pmatrix}x\\y\\0\end{pmatrix}.\end{equation} Then, $(\iota \circ \gamma)$ is also a regular curve. Prove that the curvature of $\gamma$ is the same as that of $(\iota \circ \gamma)$.
It is clear to me that the curvature of these two curves are equal, but I'm struggling to prove this...
So, if we let $\gamma(t) = \begin{pmatrix}x(t)\\y(t)\end{pmatrix}$, then $(\iota \circ \gamma)(t) = \begin{pmatrix}x(t)\\y(t)\\0\end{pmatrix}$. And so, using the formula for the curvature of space curves we get that the curvature of $(\iota \circ \gamma)$ is given by \begin{equation} \kappa_{\iota \circ \gamma} = \frac{\lVert \ddot{\gamma} \times \dot{\gamma}\rVert}{\lVert \dot{\gamma} \rVert^3} = \frac{\ddot{x}\dot{y}-\dot{x}\ddot{y}}{(\dot{x}^2 + \dot{y}^2)^{3/2}}. \end{equation} But then, the curvature of $\gamma$, using the formula for the curvature of a unit-speed plane curve, is given by \begin{equation} \lVert \tilde{\gamma}''(s)\rVert = \sqrt{\frac{d}{dt}\left[\gamma(t(s))\right]} = \sqrt{\gamma^{''}(t(s))t^{'}(s)^2 + \gamma^{'}(t(s))t^{''}(s)} = .... \end{equation}
Where $\tilde{\gamma}(s)$ is an arc-length reparameterisation of $\gamma$. I'm really struggling with showing that these are equal, I'm getting lost in the derivatives.