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Let's say we have a curve given by
$$r(t) = f(t)i + g(t)j + h(t)k$$
We know $r'(t) = |r'(t)|T(s)$ (recall we showed that T(s) is another way to call the unit tangent vector), so $$r'(t) = \frac{ds}{dt}T(s)$$
Now, we take the derivative again, and this time, we must use the product rule.
$$r''(t) = \frac{d^2s}{dt^2}T(s) + \frac{ds}{dt}\frac{d}{dt}T(s)$$
$$\frac{d^2s}{dt^2}T(s) + \frac{ds}{dt}\frac{ds}{dt}T'(s)$$
$T'(s) = KN(s)$, and $K = \frac{1}{\rho}$, which is the radius of curvature. $$\boxed{r''(t) = \frac{dv}{dt}T(s) + \frac{v^2}{\rho}N(s)}$$
The second derivative of arc length equals the derivative of speed, because the derivative of arc length **is** speed ($\sqrt{(f'(t)^2 + g'(t)^2 + h'(t)^2})$).
# Tangential Component of Acceleration

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$\dfrac{dv}{dt}$, also referred to as $a_t$, is equal to the scalar of acceleration.
$$\boxed{\dfrac{dv}{dt} = \dfrac{r'(t) \cdot r''(t)}{|r'(t)|}}$$
# Normal Component of Acceleration

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$$\boxed{\dfrac{v^2}{\rho} = \dfrac{|r'(t) \times r''(t)|}{|r'(t)|}}$$
It is important to note that $v = |r'(t)|$, and $K = \dfrac{1}{\rho}$, so
$$\boxed{K = \dfrac{1}{\rho} = \dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}}$$

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$$a = a_tT + a_nN$$
$$|a|^2 = a \cdot a = (a_tT + a_nN) \cdot (a_tT + a_nN) = a_t^2 + a_n^2$$
That's because $T \cdot N = 0$, $T \cdot T = 1$, and $N \cdot N = 1$.
$$\boxed{|a| = \sqrt{a_t^2 + a_n^2}}$$