91Ó°ÊÓ

To find the expression for $$\mathbf{v}(\sqrt{t})$$, we need to substitute $$t$$ with $$\sqrt{t}$$ in the definition of $$\mathbf{v}(t)$$: $$\mathbf{v}(\sqrt{t}) = e^{\sqrt{t}} \mathbf{i} + 2 e^{-\sqrt{t}} \mathbf{j} - e^{2\sqrt{t}} \mathbf{k}$$

Now let's compute the derivative of $$\mathbf{v}(\sqrt{t})$$ with respect to $$t$$ using the chain rule. The chain rule states, if $$y = g(u)$$ and $$u = f(t)$$, then $$\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}$$. In our case, we can apply this rule component-wise when deriving the expressions for $$\mathbf{v}(\sqrt{t})$$: $$\frac{d}{dt}\mathbf{v}(\sqrt{t}) = \frac{d}{dt}\left(e^{\sqrt{t}} \mathbf{i} + 2 e^{-\sqrt{t}} \mathbf{j} - e^{2\sqrt{t}} \mathbf{k}\right)$$ $$= \left(\frac{d}{d\sqrt{t}} e^{\sqrt{t}} \cdot \frac{d\sqrt{t}}{dt} \mathbf{i}\right) + \left(\frac{d}{d\sqrt{t}} \left(2 e^{-\sqrt{t}}\right) \cdot \frac{d\sqrt{t}}{dt}\mathbf{j}\right) -\left(\frac{d}{d\sqrt{t}} e^{2\sqrt{t}} \cdot \frac{d\sqrt{t}}{dt} \mathbf{k}\right)$$ Now compute the necessary derivatives: $$\frac{d e^{\sqrt{t}}}{d\sqrt{t}} = e^{\sqrt{t}}$$ $$\frac{d (2 e^{-\sqrt{t}})}{d\sqrt{t}} = -2 e^{-\sqrt{t}}$$ $$\frac{d e^{2\sqrt{t}}}{d\sqrt{t}} = 2 e^{2\sqrt{t}}$$ $$\frac{d\sqrt{t}}{dt} = \frac{1}{2\sqrt{t}}$$ Substitute the noted derivatives and simplify the expression: $$\frac{d}{dt}\mathbf{v}(\sqrt{t}) = \left(\frac{1}{2\sqrt{t}}e^{\sqrt{t}} \mathbf{i}\right)+ \left(-\frac{1}{\sqrt{t}}e^{-\sqrt{t}} \mathbf{j}\right) - \left(\frac{1}{\sqrt{t}}e^{2\sqrt{t}} \mathbf{k}\right)$$