To find the derivative of the 3-dimensional vector-valued function $\mathbf{r}(t) = t \mathbf{i} + 3 \mathbf{j} + \left(\frac{2}{3} t^{3/2}\right) \mathbf{k}$, we take the derivative of each component with respect to $t$.

Given $\mathbf{r}(t) = t \mathbf{i} + 3 \mathbf{j} + \left(\frac{2}{3} t^{3/2}\right) \mathbf{k}$:

1. $\mathbf{i}$-component: The derivative of $t$ with respect to $t$ is $1$. $\frac{d}{dt}(t) = 1$

2. $\mathbf{j}$-component: The derivative of $3$ (a constant) with respect to $t$ is $0$. $\frac{d}{dt}(3) = 0$

3. $\mathbf{k}$-component: The derivative of $\frac{2}{3} t^{3/2}$ with respect to $t$ is found using the power rule: $\frac{d}{dt}\left(\frac{2}{3} t^{3/2}\right) = \frac{2}{3} \cdot \frac{3}{2} t^{1/2} = t^{1/2}$

Now, combining all the components, the derivative of $\mathbf{r}(t)$ is:

$\mathbf{r}'(t) = 1 \mathbf{i} + 0 \mathbf{j} + t^{1/2} \mathbf{k}$

Simplifying:

$\mathbf{r}'(t) = \mathbf{i} + t^{1/2} \mathbf{k}$

Final Answer:

$\mathbf{r}'(t) = \mathbf{i} + t^{1/2} \mathbf{k}$

Would you like more details on how any of these steps were carried out?

Here are some related questions:

1. What is the second derivative of $\mathbf{r}(t)$?
2. How can we find the magnitude of the velocity vector?
3. How would you compute the curvature of the 3D curve?
4. What is the physical interpretation of the derivative in this context?
5. How would you integrate this vector function?

Tip: When differentiating power functions, remember that the exponent drops in front, and you subtract 1 from it (power rule).