Question

Find \((a) r^{\prime \prime}(t)\) and \((b) r^{\prime}(t) \cdot r^{\prime \prime}(t)\). $$ \mathbf{r}(t)=\langle\cos t+t \sin t, \sin t-t \cos t, t\rangle $$

Short answer

The second derivative \(r''(t)\) is \(\langle -t\sin(t), -2\sin(t) - 2t\cos(t), 0 \rangle\) and the dot product \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \) is \(-2t^{2}\cos(t)\sin(t) - 4\cos(t)\sin(t) + 2t\cos^{2}(t)\).

Step-by-step solution

Calculate \( \mathbf{r}'(t) \)

To obtain the first derivative of \( \mathbf{r}(t) \), use the basic rules of differentiation applied to each coordinate of the function. So we have: \n \\( \mathbf{r}'(t) = \langle-\sin(t) + \sin(t) + t\cos(t), \cos(t) + \cos(t) - t\sin(t), 1\rangle \) = \langle t\cos(t), 2\cos(t) - t\sin(t), 1\rangle

Calculate \( \mathbf{r}''(t) \)

To get the second derivative, \( \mathbf{r}''(t) \), take the derivative once again of \( \mathbf{r}'(t) \). We get: \n \ \( \mathbf{r}''(t) = \langle -t\sin(t), 2(-\sin(t) - t\cos(t)), 0\rangle = \langle -t\sin(t), -2\sin(t) - 2t\cos(t), 0 \rangle

Calculate \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \)

The dot product of these two vectors \( \mathbf{r}' \) and \( \mathbf{r}'' \) is calculated as follows: \n \ \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) = t\cos(t)(-t\sin(t)) + [2\cos(t) - t\sin(t)][-2\sin(t) - 2t\cos(t)] + 1 \cdot 0 = -t^{2}\cos(t)\sin(t) - 4\cos(t)\sin(t) + 2t\cos^{2}(t) - 2t^{2}\cos(t)\sin(t) = -2t^{2}\cos(t)\sin(t) - 4\cos(t)\sin(t) + 2t\cos^{2}(t)