Question

Find \((a) r^{\prime \prime}(t)\) and \((b) r^{\prime}(t) \cdot r^{\prime \prime}(t)\). $$ \mathbf{r}(t)=t \mathbf{i}+(2 t+3) \mathbf{j}+(3 t-5) \mathbf{k} $$

Short answer

The second derivative of the vector function is a zero vector \( r^{\prime \prime}(t) = 0 \mathbf{i} + 0 \mathbf{j} + 0\mathbf{k} \) and the dot product of the first derivative and the second derivative is 0.

Step-by-step solution

Compute the first derivative

The first derivative \(r'(t)\) of a vector \(\mathbf{r}(t)=t\mathbf{i}+(2t+3)\mathbf{j}+(3t-5)\mathbf{k}\) is simply the derivative of each of its components. Thus, \( r^{\prime}(t) = \mathbf{i} + 2 \mathbf{j} + 3\mathbf{k} \).

Compute the second derivative

The second derivative, \(r''(t)\), is obtained by differentiating \(r'(t)\). Since the first derivative is a constant vector, its derivative is the zero vector. So we have \( r^{\prime \prime}(t) = 0 \mathbf{i} + 0 \mathbf{j} + 0\mathbf{k} \).

Compute the dot product

The dot product of \( r^{\prime}(t) \) and \( r^{\prime \prime}(t) \) is calculated by multiplying corresponding components of the two vectors and then summing up these products. However since \( r^{\prime \prime}(t) \) is the zero vector, the dot product is 0.