Question

Find the curvature \(K\) of the curve. $$ \mathbf{r}(t)=4 \cos 2 \pi t \mathbf{i}+4 \sin 2 \pi t \mathbf{j} $$

Short answer

The curvature of the given curve \(\mathbf{r}(t)\) is \(\frac{1}{2 \pi}\).

Step-by-step solution

Compute the first derivative of \(\mathbf{r}(t)\)

The derivative of a vector function is computed component-wise. So, the first derivative \(\mathbf{r}'(t)\) is given by: \(\mathbf{r}'(t) = -8 \pi \sin(2 \pi t) \mathbf{i} + 8 \pi \cos(2 \pi t) \mathbf{j}\).

Compute the second derivative of \(\mathbf{r}(t)\)

Following the same procedure as in step 1, the second derivative \(\mathbf{r}''(t)\) is given by: \(\mathbf{r}''(t) = -16 \pi^2 \cos(2 \pi t) \mathbf{i} - 16 \pi^2 \sin(2 \pi t) \mathbf{j}\).

Compute the magnitudes of the first and second derivatives

The magnitude of a vector \(\mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j}\) is given by \(|\mathbf{a}| = \sqrt{a_x^2 + a_y^2}\). Therefore, \(|\mathbf{r}'(t)| = \sqrt{(-8 \pi \sin(2 \pi t))^2 + (8 \pi \cos(2 \pi t))^2} = 8 \pi\). Using the same mathematics, we get that \(|\mathbf{r}''(t)| = \sqrt{(-16 \pi^2 \cos(2 \pi t))^2 + (-16 \pi^2 \sin(2 \pi t))^2} = 16 \pi^2\).

Compute the cross product

The cross product of 2-dimensional vectors \(\mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j}\) and \(\mathbf{b} = b_x \mathbf{i} + b_y \mathbf{j}\) is given by \(|\mathbf{a} \times \mathbf{b}| = |a_x b_y - a_y b_x|\). Therefore, \(\mathbf{r}'(t) \times \mathbf{r}''(t) = |-8 \pi \sin(2 \pi t)\times -16 \pi^2 \sin(2 \pi t) - 8 \pi \cos(2 \pi t) \times -16 \pi^2 \cos(2 \pi t)| = 256 \pi^3\).

Evaluate the curvature

Use the formula for curvature, \(K = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}\). Substituting the magnitudes of \(\mathbf{r}'(t) = 8 \pi\), \(\mathbf{r}''(t) = 16 \pi^2\), and cross product \(256 \pi^3\) gives: \(K = \frac{256 \pi^3}{(8 \pi)^3} = \frac{1}{2 \pi}\).