Question

Find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=4 \sqrt{t} \mathbf{i}+t^{2} \sqrt{t} \mathbf{j}+\ln t^{2} \mathbf{k} $$

Short answer

Hence, the derivative of given vector function \(\mathbf{r}(t)\) is \(\mathbf{r}'(t) = 2/\sqrt{t}\mathbf{i} + (2t\sqrt{t} + t^{2} /(2 \sqrt{t}))\mathbf{j} + 2/t\mathbf{k}\)

Step-by-step solution

Identify Functions for Differentiation

The first step is to identify the functions to be differentiated in each of the vector components. \(\mathbf{r}(t) = 4\sqrt{t}\mathbf{i} + t^{2}\sqrt{t}\mathbf{j} + \ln t^{2}\mathbf{k}\) is given to us. We identify that \(\mathbf{i}\) component has '4\sqrt{t}', \(\mathbf{j}\) component has 't²\sqrt{t}' and \(\mathbf{k}\) component has '\ln t²'.

Differentiate the functions

Differentiating the functions of each component using rules of differentiation, we get the following results. For \(\mathbf{i}\) component: The derivative of \(4\sqrt{t}\) will be \(2/\sqrt{t}\). For \(\mathbf{j}\) component: The derivative of \(t^{2}\sqrt{t}\) can be obtained using the product rule, which gives us \((2t\sqrt{t} + t^{2} /(2 \sqrt{t}))\). For \(\mathbf{k}\) component: The derivative of \(\ln t^{2}\) is \(2/t\).

Construct the Derivative Vector

Now, the derivative vector can be assembled by putting these component derivatives to form \(\mathbf{r}^{\prime}(t)\). \(\mathbf{r}^{\prime}(t) = 2/\sqrt{t} \mathbf{i} + (2t\sqrt{t} + t^{2} /(2 \sqrt{t}))\mathbf{j} + 2/t\mathbf{k}\)