Mathematical Immaturity

14.7 Exercises

• Referring to the helix in Exercise 7, prove that the velocity $v$ and acceleration $a$ are vectors of constant length, and that: $$\frac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|^3} = \frac{a}{a^2 + b^2}$$
• Recall from exercise 7 that $$\mathbf{v}(t) = -a\omega\,\sin \omega t\,\mathbf{i} + a\omega \cos \omega t\,\mathbf{j} + b \omega\,\mathbf{k}$$ Taking its derivative, we get $$\mathbf{a}(t) = -a\omega^2\,\cos \omega t\,\mathbf{i} - a\omega^2 \sin \omega t\,\mathbf{j}$$ Taking its norm, we see that $\|\mathbf{a}\| = \left[a^2\omega^4\left(\cos^2 \omega t + \sin^2 \omega t\right)\right]^{1/2} = a\omega^2.$ And we know from exercise 7 that $\mathbf{v}$ is of constant length $\omega \sqrt{a^2 + b^2}.$ This proves that $\mathbf{v}$ and $\mathbf{a}$ are of constant length. To prove that $$\frac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|^3} = \frac{a}{a^2 + b^2}$$ we first calculate $\mathbf{v} \times \mathbf{a}:$ $$\mathbf{v} \times \mathbf{a} = \left(ab\omega^3 \sin \omega t\right)\,\mathbf{i} - \left(ab\omega^3 \cos \omega t\right)\,\mathbf{j} + \left(a^2\omega^3 \sin^2 \omega t + a^2\omega^3 \cos^2 \omega t\right)\,\mathbf{k}$$ Taking its norm, we get $$\displaylines{\|\mathbf{v} \times \mathbf{a}\| = \left(a^2b^2\omega^6 + a^4\omega^6\right)^{1/2}\\ = a\omega^3\sqrt{a^2 + b^2}}$$ Combining this result with $\|\mathbf{v}\|^3 = \omega^3\left(a^2 + b^2\right)^{3/2},$ we find that $$\frac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|^3} = \frac{a}{a^2 + b^2}\quad \blacksquare$$