This derivation is a little easier to follow if we introduce some new notation. We already have [math]\vec{a}\left(t\right)=\vec{c}\,\,''\left(t\right)[/math]. Extending this idea we can write [math]\vec{c}\,\,'\left(t\right)=\vec{v}\left(t\right)[/math] (v for "velocity" of course). Now what I'll do is rename the speed [math]v\left(t\right)=\left|\left|\vec{v}\left(t\right)\right|\right|=\left|\left|\vec{c}\,\,'\left(t\right)\right|\right|[/math] - in other words the same letter as the velocity vector but without the vector hat to emphasize that this is a scalar valued function.[br]With our new notation we can write [math]\vec{T}\left(t\right)=\frac{\vec{v}\left(t\right)}{v\left(t\right)}[/math][br]Rewrite this expression for the unit tangent as a product to make differentiating a little less tedious:[br][math]v\left(t\right)\vec{T}\left(t\right)=\vec{v}\left(t\right)[/math][br]Differentiate both sides to achieve an expression for the acceleration vector:[br][math]v'\left(t\right)\vec{T}\left(t\right)+v\left(t\right)\vec{T}\,\,'\left(t\right)=\vec{a}\left(t\right)[/math][br][br]Recall that the unit normal vector is defined as:[br][math]\vec{N}\left(t\right)=\frac{\vec{T}\,\,'\left(t\right)}{\left|\left|\vec{T}\,\,'\left(t\right)\right|\right|}[/math] which we can rewrite as a product: [math]\left|\left|\vec{T}\,\,'\left(t\right)\right|\right|\vec{N}\left(t\right)=\vec{T}\,\,'\left(t\right)[/math][br][br]Subbing into our expression for acceleration yields:[br][math]\vec{a}\left(t\right)=v'\left(t\right)\vec{T}\left(t\right)+v\left(t\right)\left|\left|\vec{T}\,\,'\left(t\right)\right|\right|\vec{N}\left(t\right)[/math] [br][br]This actually accomplishes the goal of showing that the acceleration vector is a linear combination of the unit tangent and unit normal vectors. But I'll actually go one step further. If you recall, one of our formulas for curvature was:[br][math]\kappa\left(t\right)=\frac{\left|\left|\vec{T}\,\,'\left(t\right)\right|\right|}{\left|\left|\vec{c}\,\,'\left(t\right)\right|\right|}=\frac{\left|\left|\vec{T}\,\,'\left(t\right)\right|\right|}{v\left(t\right)}[/math][br]Rewriting as a product: [math]v\left(t\right)\kappa\left(t\right)=\left|\left|\vec{T}\,\,'\left(t\right)\right|\right|[/math][br][br]Now subbing this back into our formula for acceleration:[br][br][math]\vec{a}\left(t\right)=v'\left(t\right)\vec{T}\left(t\right)+v\left(t\right)^2\kappa\left(t\right)\vec{N}\left(t\right)[/math][br][br]The coefficients in this expression are given the names:[br][b][color=#ff0000]tangential component of acceleration[/color][/b]: [math]a_T=v'\left(t\right)[/math][br]and[br][b][color=#ff0000]normal component of acceleration[/color][/b]: [math]a_N=v\left(t\right)^2\kappa\left(t\right)[/math][br][br]