Question

Find the curvature \(K\) of the curve. $$ \mathbf{r}(t)=a \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j} $$

Short answer

The curvature of the curve defined by \( \mathbf{r}(t)=a \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j} \) is given by \( K = \frac{\omega}{\sqrt{a^2+b^2}} \).

Step-by-step solution

Compute the first derivative

Calculate the first derivative of \( \mathbf{r}(t) \). The result is \( \mathbf{r}'(t)=-a\omega \sin(\omega t) \mathbf{i}+b\omega \cos(\omega t) \mathbf{j} \).

Find the magnitude of the first derivative

To find the magnitude of the first derivative \( ||\mathbf{r}'(t)|| \), use the formula for the magnitude of a vector, that is, the square root of the sum of the squares of its components. What you get is \( ||\mathbf{r}'(t)||=\omega \sqrt{a^2\sin^2(\omega t)+b^2\cos^2(\omega t)}=\omega\sqrt{a^2+b^2} \). Only the sine and cosine components dependent on t are squared.

Compute the second derivative

Calculate the second derivative of the given vector function \( \mathbf{r}(t) \). This gives \( \mathbf{r}''(t)=-a\omega^2 \cos(\omega t)\mathbf{i}-b\omega^2 \sin(\omega t)\mathbf{j} \).

Find the magnitude of the second derivative

The magnitude of the second derivate \(|| \mathbf{r}''(t)||\) is calculated with the same formula as for the first derivative. \( ||\mathbf{r}''(t)||=\omega^2 \sqrt{a^2\cos^2(\omega t)+b^2\sin^2(\omega t)}=\omega^2\sqrt{a^2+b^2} \)

Calculate the curvature

The curvature \(K\) is calculated using the formula \( K=\frac{| \mathbf{r}' \times \mathbf{r}'' |}{| \mathbf{r}' |^3} \). In this case, since the vectors \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) are orthogonal to each other, the magnitude of theircross product is simply the product of their magnitudes. Plug in the found values from previous steps to get \( K=\frac{(\omega\sqrt{a^2+b^2})(\omega^2\sqrt{a^2+b^2})}{(\omega\sqrt{a^2+b^2})^3} \). This simplifies to \( K= \frac{\omega}{\sqrt{a^2+b^2}} \)