Question

Find the curvature \(K\) of the plane curve at the given value of the parameter. $$ \mathbf{r}(t)=5 \cos t \mathbf{i}+4 \sin t \mathbf{j}, \quad t=\frac{\pi}{3} $$

Short answer

The curvature \(K\) of the plane curve at \(t=\frac{\pi}{3}\) is given by \( K = \frac{|\mathbf{r'}(t) \times \mathbf{r''}(t)|}{|\mathbf{r'}(t)|^3} \), where \( |\mathbf{r'}(t)| \) and \( |\mathbf{r''}(t)| \) are the magnitudes of the first and second derivatives, respectively, evaluated at \( t=\pi/3 \).

Step-by-step solution

Compute First Derivative

The first derivative \( \mathbf{r'}(t) \) of the curve \( \mathbf{r}(t) \) is obtained by differentiating each component of \( \mathbf{r}(t) \) with respect to \(t\). The derivatives are: \[ \mathbf{r'}(t) = -5 \sin(t) \mathbf{i} + 4 \cos(t) \mathbf{j} \]

Compute Second Derivative

The second derivative \( \mathbf{r''}(t) \) is found by differentiating the first derivative with respect to \(t\). The derivatives are:\[ \mathbf{r''}(t) = -5 \cos(t) \mathbf{i} - 4 \sin(t) \mathbf{j} \]

Find Magnitudes and Curvature

The magnitudes \( |\mathbf{r'}(t)| \) and \( |\mathbf{r''}(t)| \) of the first and second derivatives are needed in the curvature formula. The magnitude of a vector is given by the square root of the sum of the squares of its components. Therefore:\[ |\mathbf{r'}(t)| = \sqrt{(-5\sin t)^2 + (4\cos t)^2} \]\[ |\mathbf{r''}(t)| = \sqrt{(-5\cos t)^2 + (-4\sin t)^2} \]Using the formula for the curvature of the curve \( K = \frac{|\mathbf{r'}(t) \times \mathbf{r''}(t)|}{|\mathbf{r'}(t)|^3} \), substitute the values of \( |\mathbf{r'}(t)| \) and \( |\mathbf{r''}(t)| \) at \( t=\pi/3 \) to calculate the curvature.