# Curvature of plane curve

This is a motivation for the definition of the curvature of a plane curve:

Suppose then that $\gamma$ is a unit-speed curve in $\mathbb{R}^2$. As the parameter $t$ of $\gamma$ changes to $t+\Delta t$, the curve moves away from its tangent line at $\gamma(t)$ by a distance $(\gamma(t+\Delta t)-\gamma(t)) \cdot \vec{n}$, where $\vec{n}$ is a unit vector perpendicular to the tangent vector $\gamma'(t)$ of $\gamma$ at the point $\gamma(t)$. By Taylor's theorem, $\gamma(t+\Delta t)=\gamma(t)+\gamma'(t)\Delta t+\frac{1}{2}\gamma''(t)(\Delta t)^2+\mathrm{remainder}$, where $(\mathrm{remainder})/(\Delta t)^2$ tends to zero as $\Delta t$ tends to zero. Since $\gamma' \cdot \vec{n}=0$, the deviation of $\gamma$ from its tangent line at $\gamma(t)$ is $\frac{1}{2} \gamma''(t) \cdot \vec{n}(\Delta t)^2+\mathrm{remainder}$...

"the curve moves away by a distance" Why is $(\gamma(t+\Delta t)-\gamma(t)) \cdot \vec{n}$ the distance? If $\gamma(s)=(\gamma_1(s),\gamma_2(s))$ then the distance between $\gamma(t+\Delta t)$ and $\gamma(t)$, as points in the real plane, is $\sqrt{(\gamma_1(t+\Delta t)-\gamma_1(t))^2+(\gamma_2(t+\Delta t)-\gamma_2(t))^2}$, no? But this is off by a factor of $\cos{\theta}$ ($\theta$ being the angle between $\gamma(t+\Delta t)-\gamma(t)$ and $\vec{n}$.

I'm ok with the rest, but where and why did we divide through by $(\Delta t)^2$? I'm sure it does tend to zero, but how is that relevant to reaching the conclusion that $(\gamma(t+\Delta t)-\gamma(t)) \cdot \vec{n}=\frac{1}{2} \gamma''(t) \cdot \vec{n}(\Delta t)^2+\mathrm{remainder}$?

The length $\sqrt{(\gamma_1(t+\Delta t)-\gamma_1(t))^2+(\gamma_2(t+\Delta t)-\gamma_2(t))^2}$ is the distance between the point $\gamma(t + \Delta t)$ and $\gamma(t)$. This is not the distance they're talking about. The author is talking about the distance between $\gamma(t + \Delta t)$ and the tangent line to the curve at the point $\gamma(t)$.
Draw a plane curve like the one under consideration. Mark the points $\gamma(t)$ and $\gamma(t + \Delta t)$. Now draw the tangent line to the curve at $\gamma(t)$, and the unit normal vector $\vec{n}$. This vector should be drawn with its initial point at $\gamma(t)$. The vector $\vec{n}$ has length one and is orthogonal to the tangent line. Now think about the distance between the point $\gamma(t + \Delta t)$ and the tangent line. You should be able to see that this distance is the length of the projection of the vector $\gamma(t + \Delta t) - \gamma(t)$ in the direction $\vec{n}$, which is what the author is calculating.