Question

Find the curvature \(K\) of the plane curve at the given value of the parameter. $$ \mathbf{r}(t)=t \mathbf{i}+\cos t \mathbf{j}, \quad t=0 $$

Short answer

The curve's curvature at the given parameter \( t=0 \) is 1.

Step-by-step solution

Compute the First Derivative

We begin by analyzing the expression \( \mathbf{r}(t)=t \mathbf{i}+\cos t \mathbf{j} \). The derivative of \( t \) with respect to \( t \) is \( 1 \), and the derivative of \( \cos t \) with respect to \( t \) is \( -\sin t \). Thus, the first derivative \( \mathbf{r}'(t) \) is given by \( \mathbf{r}'(t) = \mathbf{i}-\sin t \mathbf{j} \). When \( t = 0 \), this becomes \( \mathbf{r}'(0) = \mathbf{i} \). This derivative indicates the tangent to the curve at the specified \( t \) value.

Compute the Second Derivative

The second derivative is the derivative of \( \mathbf{r}'(t) \). Thus, the derivative with respect to \( t \) of \( -\sin t \) is \( -\cos t \). Therefore, \( \mathbf{r}''(t) = -\cos t \mathbf{j} \). Given \( t = 0 \), we find \( \mathbf{r}''(0) = -1 \mathbf{j} \). This represents the rate of change of the curve's tangent direction at the specified value of \( t \).

Compute the Curvature

Given the formula for the curvature \( K = \frac{||\mathbf{r}' \times \mathbf{r}''||}{||\mathbf{r}'||^3} \). For \( t=0 \), we need to compute the cross-product \( \mathbf{r}' \times \mathbf{r}'' \) equals to \( (\mathbf{i} \times -1\mathbf{j}) = 1 \mathbf{k} \). The magnitude of that cross-product \( ||\mathbf{r}' \times \mathbf{r}''|| = ||\mathbf{k}|| = 1 \). The magnitude of \( \mathbf{r}' \) equals to \( ||\mathbf{i}||=1 \). Therefore, the curvature \( K = \frac{1}{1^3} = 1 \)