Mathematical Immaturity

14.9 Exercises

• Prove that the normal component of the acceleration vector is $\|\mathbf{v} \times \mathbf{a}\|/\|\mathbf{v}\|.$
• Use Lagrange's identity for the norm of a cross product $$ \|A \times B\|^2 = \|A\|^2\|B\|^2 - \left(A \cdot B\right)^2 $$
• Recall that by Lagrange's identity for the norm of a cross product, $$ \|\mathbf{v} \times \mathbf{a}\|^2 = \|\mathbf{v}\|^2 \|\mathbf{a}\|^2 - \left(\mathbf{v} \cdot \mathbf{a}\right)^2 $$ Dividing both sides by $\|\mathbf{v}\|^2$ and taking the square root, we get: $$ \begin{align*} \frac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|} &= \left(\|\mathbf{a}\|^2 - \left(\mathbf{a} \cdot T\right)^2\right)^{1/2} \\ \end{align*} $$ Where $T$ is the unit tangent $\mathbf{v}/\|\mathbf{v}\|$ Thus, we wish to show that the normal component of the acceleration vector is equal to $$ \left(\mathbf{a} \cdot \mathbf{a} - \left(\mathbf{a} \cdot T\right)^2\right)^{1/2} $$ First, we recall that component-wise, acceleration is defined as $$ \mathbf{a}(t) = v'(t)T(t) + v(t)\|T'(t)\|N(t) $$ where $N(t) = T'(t)/\|T'(t)\|$ is the principal normal vector. The normal component of the acceleration vector (ie, the coefficient of the principal normal) is: $$ \begin{align*} v\|T'\| &= v\left\Vert\left(\frac{\mathbf{v}}{v}\right)'\right\Vert\\ &= v\left\Vert\frac{\mathbf{v}'}{v} - \frac{v'}{v^2}\mathbf{v}\right\Vert\\ &= \|\mathbf{a} - v'T\| \end{align*} $$ But we know by definition that $T$ and $N$ are perpendicular. If we take the dot product $\mathbf{a} \cdot T,$ noting that $T$ has a length of 1, we get: $$ \begin{align*} \mathbf{a} \cdot T &= v'(T \cdot T) + v(N \cdot T)\\ &= v' \end{align*} $$ This means that we can rewrite $\|\mathbf{a} - v'T\|$ as: $$ \begin{align*} \|\mathbf{a} - v'T\| = \|\mathbf{a} - (\mathbf{a} \cdot T) T\| \end{align*} $$ Now, we expand the right-hand side of the above equation, recalling that $T \cdot T = 1,$ to give us: $$ \begin{align*} \|\mathbf{a} - (\mathbf{a} \cdot T) T\| &= \left(\left(\mathbf{a} - \left(\mathbf{a} \cdot T\right)T\right) \cdot \left(\mathbf{a} - \left(\mathbf{a} \cdot T\right)T\right)\right)^{1/2} \\ &= \left(\mathbf{a} \cdot \mathbf{a} - 2\left(\mathbf{a} \cdot T\right)^2 + \left(\mathbf{a} \cdot T\right)^2 (T \cdot T)\right)^{1/2} \\ &= \left(\mathbf{a} \cdot \mathbf{a} - \left(\mathbf{a} \cdot T\right)^2\right)^{1/2} \end{align*} $$ Thus, we have proven that $$ v\|T'\| = \frac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|} \quad \blacksquare $$