Mathematical Immaturity

14.15 Exercises

• A point moves in space according to the vector equation $$ \begin{align*} \\ \mathbf{r}(t) &= 4\cos t\,\mathbf{i} + 4\sin t\,\mathbf{j} + 4\cos t\,\mathbf{k} \end{align*} $$ (a) Show that the path is an ellipse and find a Cartesian equation for the plane containing this ellipse. (b) Show that the radius of curvature is $\rho(t) = 2\sqrt{2}(1 + \sin^2 t)^{3/2}.$
• (a) First, find the Cartesian equation for the plane containing the curve. Let $X = (x, y, z)$ be a point on the curve described by $\mathbf{r}(t)$ and let $N = (a, b, c)$ be the vector normal to the plane containing the curve. Find $N$ such that for all $t$ $$ X \cdot N = 0 $$ Then, use the result to show that the curve describes an ellipse inside the given plane. (b) Recall that the radius of curvature $\rho(t)$ is the reciprocal of the curvature $\kappa(t)$ when $\kappa \neq 0:$ $$ \begin{align*} \rho(t) &= \frac{v(t)}{\|T'(t)\|} \end{align*} $$ Theorem 14.14 allows us to express curvature in terms of the cross product of acceleration and velocity: $$ \begin{align*} \kappa(t) &= \frac{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}{v^3(t)} \end{align*} $$ Which in turn means that the radius of curvature is: $$ \begin{align*} \\ \rho(t) &= \frac{v^3(t)}{\|\mathbf{a}(t) \times \mathbf{v}(t)\|} \end{align*} $$
• (a) To find the Cartesian equation for the plane containing this curve, we first find the normal vector $N = (a, b, c)$ such that for any given point $X = (x, y, z)$ in the plane: $$ \begin{align*} \\ X \cdot N = 0 \end{align*} $$ In other words, we wish to find scalars $a, b, c$ (not all zero) such that $$ \begin{align*} \\ 4a \cos t + 4b \sin t + 4c \cos t &= 0 \end{align*} $$ for all $t.$ This relation can be satisfied for all $t$ by the normal vector $N = (1, 0, -1)$ which gives us the following Cartesian equation for the plane containing the curve: $$ \begin{align*} \\ x = z \end{align*} $$ Now, if we perform the substitution $u = x + z$ we can see that $$ \begin{align*} \\ u^2 + y^2 &= x^2 + 2xz + z^2 + y^2 \\ \end{align*} $$ But, we know from our previous result that the curve described by $\mathbf{r}(t)$ lies in the plane $x = z.$ Thus, we can express the three-dimensional curve $\mathbf{r}(t)$ as a two-dimensional curve lying in the plane $x = z$ and rewrite the above equation as: $$ \begin{align*} \\ u^2 + y^2 &= 4x^2 + y^2 \\ &= 64\cos^2 t + 16 \sin^2 t \end{align*} $$ But, if we set $a = 8$ and $b = 4,$ we can see that this curve is an ellipse with Cartesian equation $$ \begin{align*} \\ \frac{u^2}{a^2} + \frac{y^2}{b^2} &= 1 \quad \blacksquare \end{align*} $$ (b) To find the radius of curvature of $\mathbf{r}(t)$ we must first find its velocity: $$ \begin{align*} \\ \mathbf{v}(t) &= -4\sin t\,\mathbf{i} + 4\cos t\,\mathbf{j} + -4\sin t\,\mathbf{k} \end{align*} $$ Its speed $$ \begin{align*} v(t) &= 4\sqrt{1 + \sin^2 t} \end{align*} $$ And its acceleration: $$ \begin{align*} \\ \mathbf{a}(t) &= -4\cos t\,\mathbf{i} -4\sin t\,\mathbf{j} -4\cos t\,\mathbf{k} \end{align*} $$ From Theorem 14.14, we know that curvature $\kappa(t)$ can be written as: $$ \begin{align*} \\ \kappa(t) &= \frac{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}{v^3(t)} \end{align*} $$ And thus the radius of curvature $\rho(t)$ can be written as: $$ \begin{align*} \\ \rho(t) &= \frac{v^3(t)}{\|\mathbf{a}(t) \times \mathbf{v}(t)\|} \end{align*} $$ By Lagrange's identity on the norm of a cross product, we know that $$ \begin{align*} \\ \|\mathbf{a}(t) \times \mathbf{v}(t)\|^2 &= \|\mathbf{a}(t)\|^2\|\mathbf{v}(t)\|^2 - \left(\mathbf{a}(t)\cdot\mathbf{v}(t)\right)^2 \\ &= 256\left(1 + \cos^2 t\right)\left(1 + \sin^2 t\right) - 256\left(\sin^2 t \cos^2 t\right) \\ &= 512 \end{align*} $$ Thus, $\|\mathbf{a}(t) \times \mathbf{v}(t)\| = 16\sqrt{2}$ which makes the radius of curvature: $$ \begin{align*} \\ \rho(t) &= \frac{v^3(t)}{\|\mathbf{a}(t) \times \mathbf{v}(t)\|} \\ &= \frac{64\left(1 + \sin^2 t\right)^{3/2}}{16\sqrt{2}} \\ &= 2\sqrt{2}\left(1 + \sin^2 t\right)^{3/2} \quad \blacksquare \end{align*} $$