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# Acceleration

MathJax TeX Test Page 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

MathJax TeX Test Page $\dfrac{dv}{dt}$, 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

MathJax TeX Test Page $$\boxed{\dfrac{v^2}{\rho} = \dfrac{|r'(t) \times r''(t)|}{|r'(t)|}}$$ It is important to note that $v = |r'(t)|$, and $K = \dfrac{1}{\rho}$, so $$\boxed{K = \dfrac{1}{\rho} = \dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}}$$
MathJax TeX Test Page $$a = a_tT + a_nN$$ $$|a|^2 = a \cdot a = (a_tT + a_nN) \cdot (a_tT + a_nN) = a_t^2 + a_n^2$$ That's because $T \cdot N = 0$, $T \cdot T = 1$, and $N \cdot N = 1$. $$\boxed{|a| = \sqrt{a_t^2 + a_n^2}}$$
David Witten