Question

In Exercises \(17-24,\) find \((a) r^{\prime \prime}(t)\) and \((b) r^{\prime}(t) \cdot r^{\prime \prime}(t)\). $$ \mathbf{r}(t)=t^{3} \mathbf{i}+\frac{1}{2} t^{2} \mathbf{j} $$

Short answer

Therefore the second derivative \(r''(t)\) is \(6t \mathbf{i} + \mathbf{j}\), and the dot product \(r'(t) \cdot r''(t)\) is \(18t^{3} + t\).

Step-by-step solution

Compute the first derivative

First differentiate the given vector function component-wise: \(\mathbf{r}'(t) = \frac{d}{dt} [t^{3} \mathbf{i} + \frac{1}{2} t^{2} \mathbf{j}] = 3t^{2} \mathbf{i} + t \mathbf{j}\). Here, the power rule was applied for differentiation.

Compute the second derivative

Next, differentiate the first derivative to get the second derivative: \(\mathbf{r}''(t) = \frac{d}{dt} [\mathbf{r}'(t)]= \frac{d}{dt}[3t^{2} \mathbf{i} + t \mathbf{j}]= 6t \mathbf{i}+ \mathbf{j}\). Here again, the power rule was applied for differentiation.

Find the dot product

According to the definition of dot product, \(\mathbf{r}'(t) \cdot \mathbf{r}''(t) = [3t^{2} \mathbf{i} + t \mathbf{j}] \cdot [6t \mathbf{i} + \mathbf{j}]= 18t^{3} + t.\) This dot product can be computed by performing elementwise multiplication using the coefficients and then summing up the products.