Mathematical Immaturity

14.15 Exercises

• (a) For a plane curve described by the equation $\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j},$ show that the curvature is given by the formula $$ \begin{align*} \\ \kappa(t) &= \frac{|x'(t)y''(t) - y'(t)x''(t)|}{\{[x'(t)]^2 + [y'(t)]^2\}^{3/2}} \\ \\ \end{align*} $$ (b) If a plane curve has the Cartesian equation $y = f(x),$ show that the curvature at the point $(x, f(x))$ is $$ \begin{align*} \\ \kappa &= \frac{|f''(x)|}{\{1 + [f'(x)]^2\}^{3/2}} \end{align*} $$
• These are both specific applications of equation (14.22) of Theorem 14.14: $$ \begin{align*} \\ \kappa(t) &= \frac{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}{v^3(t)} \tag{14.22} \end{align*} $$
• (a) For the plane curve described by the position vector $\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j},$ velocity, speed, and acceleration are: $$ \begin{align*} \\ \mathbf{v}(t) &= x'(t)\,\mathbf{i} + y'(t)\,\mathbf{j} \\ \\ v(t) &= \sqrt{\left[x'(t)\right]^2 + \left[y'(t)\right]^2} \\ \\ \mathbf{a}(t) &= x''(t)\,\mathbf{i} + y''(t)\,\mathbf{j} \end{align*} $$ Using equation (14.22) of Theorem 14.14, curvature is $$ \begin{align*} \\ \kappa(t) &= \frac{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}{v^3(t)} \end{align*} $$ To find the norm $\|\mathbf{a}(t) \times \mathbf{v}(t)\|$ we first calculate the cross product $\mathbf{a}(t) \times \mathbf{v}(t)$ $$ \begin{align*} \\ \mathbf{a}(t) \times \mathbf{v}(t) &= \left[x''(t)y'(t) - y''(t)x'(t)\right]\mathbf{k} \end{align*} $$ Its norm is then: $$ \begin{align*} \\ \|\mathbf{a}(t) \times \mathbf{v}(t)\| &= \sqrt{\left[x''(t)y'(t) - y''(t)x'(t)\right]^2} \\ \\ &= \left|x''(t)y'(t) - y''(t)x'(t)\right| \\ \\ &= \left|-\left[x''(t)y'(t) - y''(t)x'(t)\right]\right| \\ \\ &= \left|x'(t)y''(t) - y'(t)x''(t)\right| \end{align*} $$ Using this and our previously calculated speed value, we see that the curvature at time $t$ is $$ \begin{align*} \\ \kappa(t) &= \frac{\left|x'(t)y''(t) - y'(t)x''(t)\right|}{\left\{\left[x'(t)\right]^2 + \left[y'(t)\right]^2\right\}^{3/2}} \quad \blacksquare \end{align*} $$ (b) If the plane curve is described by the Cartesian equation $y = f(x)$ then its position vector parameterized by $x$ is $$ \begin{align*} \\ \mathbf{r}(x) &= x\,\mathbf{i} + f(x)\,\mathbf{j} \end{align*} $$ Velocity, speed, and acceleration are then defined by: $$ \begin{align*} \\ \mathbf{v}(x) &= \mathbf{i} + f'(x)\,\mathbf{j} \\ \\ v(x) &= \sqrt{1+ \left[f'(x)\right]^2} \\ \\ \mathbf{a}(x) &= f''(x)\,\mathbf{j} \end{align*} $$ Referring back to equation (14.22), curvature can be written as $$ \begin{align*} \\ \kappa(x) &= \frac{\|\mathbf{a}(x) \times \mathbf{v}(x)\|}{v^3(x)} \end{align*} $$ To find the norm $\|\mathbf{a}(x) \times \mathbf{v}(x)\|$ we first calculate the cross product $\mathbf{a}(x) \times \mathbf{v}(x)$ $$ \begin{align*} \\ \mathbf{a}(x) \times \mathbf{v}(x) &= -f''(x)\,\mathbf{k} \end{align*} $$ Then we can see that its norm is $|f''(x)|.$ Using this value along with our previously calculated speed we find that the curvature at $x$ is $$ \begin{align*} \\ \kappa(x) &= \frac{|f''(x)|}{\left\{1 + \left[f'(x)\right]^2\right\}^{3/2}} \end{align*} \quad \blacksquare $$