Mathematical Immaturity

14.15 Exercises

• (a) Prove that the radius of curvature of a parabola is smallest at its vertex. (b) Given two fixed unit vectors $A$ and $B$ making an angle $\theta$ with each other, where $0 < \theta < \pi.$ The curve with position vector $\mathbf{r}(t) = tA + t^2B$ is a parabola lying in the plane spanned by $A$ and $B.$ Determine (in terms of $A,$ $B,$ and $\theta$) the position vector of the vertex of this parabola. You may use the property of the parabola stated in part (a).
• (a) Suppose we have a parabola described by the position vector $\mathbf{r}$ $$ \begin{align*} \\ \mathbf{r}(x) &= x\,\mathbf{i} + cx^2\,\mathbf{j} \end{align*} $$ where $c$ is a real-valued scalar constant. Now, we find its velocity, speed, and acceleration with respect to $x$ $$ \begin{align*} \\ \mathbf{v}(x) &= \mathbf{i} + 2cx\,\mathbf{j} \\ \\ v(x) &= \sqrt{1 + 4c^2x^2} \\ \\ \mathbf{a}(x) &= 2c\,\mathbf{j} \end{align*} $$ To find the radius of curvature, we first use Equation (14.22) to calculate curvature $$ \begin{align*} \\ \kappa(x) &= \frac{\|\mathbf{a}(x) \times \mathbf{v}(x)\|}{v^3(x)} \tag{14.22} \\ \\ &= \frac{\|-2c\,\mathbf{k}\|}{\left(1 + 4c^2x^2\right)^{3/2}} \\ \\ &= \frac{|2c|}{\left(1 + 4c^2x^2\right)^{3/2}} \end{align*} $$ If $c \neq 0,$ we can take reciprocal of $\kappa(x)$ to give: $$ \begin{align*} \\ \rho(x) &= \frac{\left(1 + 4c^2x^2\right)^{3/2}}{|2c|} \end{align*} $$ But, since $4c^2x^2$ must be greater than or equal to zero for all real $x$ and $c,$ then $\rho(x)$ is smallest at $x = 0.$ But when $x = 0,$ $y = cx^2 = 0,$ which is the vertex of the parabola described by $\mathbf{r}(x). \quad \blacksquare$ (b) Using the result of (a), we know that the vertex of a given parabola is the point at which the radius of curvature is smallest. Thus, we wish to find $t$ that minimizes $\rho(t).$ First, we find velocity, speed, and acceleration with respect to $t$ $$ \begin{align*} \\ \mathbf{v}(t) &= A + 2tB \\ \\ v(t) &= \|A + 2tB\| \\ &= \left[(A \cdot A) + 4t(A\cdot B\right) + 4t^2(B\cdot B)]^{1/2} \\ &= \left[1 + 4t\cos\theta+ 4t^2\right]^{1/2} \\ \\ \mathbf{a}(t) &= 2B \end{align*} $$ Then, we can use Equation (14.22) to obtain $$ \begin{align*} \\ \kappa(t) &= \frac{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}{v^3(t)} \tag{14.22} \end{align*} $$ Calculating the cross product of acceleration and velocity, noting that $B \times B = O$ $$ \begin{align*} \\ \mathbf{a}(t) \times \mathbf{v}(t) &= 2B \times \left(A + 2tB\right) \\ &= 2B \times A \end{align*} $$ Taking the norm of this product, noting that $A$ and $B$ are unit vectors, we get: $$ \begin{align*} \\ \|\mathbf{a}(t) \times \mathbf{v}(t)\| &= 2\|B \times A\| \\ &= 2\|B\|\|A\|\sin\theta \\ &=2\sin\theta \end{align*} $$ Combining this with our previously calculated speed, we can express the radius of curvature as the reciprocal of $\kappa(t)$ $$ \begin{align*} \\ \rho(t) &= \frac{v^3(t)}{\|\mathbf{a}(t) \times \mathbf{v}(t)\|} \\ \\ &= \frac{\left[1 + 4t\cos\theta + 4t^2\right]^{3/2}}{2\sin\theta} \end{align*} $$ To find an extremum of $\rho,$ we take its derivative with respect to $t$ $$ \begin{align*} \\ \rho'(t) &= \frac{3}{2}\frac{\left[1 + 4t\cos\theta + 4t^2\right]^{1/2}}{2\sin\theta}\left(4\cos\theta+8t\right) \end{align*} $$ Setting $\rho' = 0$ and solving for $t$ we see that the radius of curvature has exactly one extremum at $t = -\frac{1}{2}\cos\theta.$ We can see that $\rho$ is decreasing when $t < -\frac{1}{2}\cos\theta$ and increasing when $t > -\frac{1}{2}\cos\theta$ which must mean that this extremum is a local minimum. And since it is the only extremum, it is also the global minimum. But we know from part (a) that the smallest radius of curvature for a parabola occurs at its vertex. Thus, the vertex of the parabola described by $\mathbf{r}(t)$ occurs at $$ \begin{align*} \\ -\frac{1}{2}\cos\theta A + \frac{1}{4}\cos^2\theta B \quad \blacksquare \end{align*} $$