91Ó°ÊÓ

Firstly, let's compute the derivative of the vector function \(\mathbf{v}(t)\). To do that, we need to find the derivative of its components with respect to \(t\). Recall that: $$\mathbf{v}(t) = \left\langle t^{2}, -2t, 1\right\rangle$$ Let's find the derivatives of each component: 1. \(\frac{d}{dt}(t^2) = 2t\) 2. \(\frac{d}{dt}(-2t) = -2\) 3. \(\frac{d}{dt}(1) = 0\) Now that we've found the derivatives of each component, we can find \(\mathbf{v}'(t)\) as follows: $$\mathbf{v}'(t) = \left\langle 2t, -2, 0\right\rangle$$

Now, let's find the derivative of the scalar function \(g(t)\). Recall that: $$g(t) = 2\sqrt{t}$$ To find its derivative, we need to find the derivative of the function with respect to \(t\): $$g'(t) = \frac{d}{dt}(2\sqrt{t}) = 2 \cdot \frac{d}{dt}(t^{\frac{1}{2}}) = 2 \cdot \frac{1}{2}t^{-\frac{1}{2}} = t^{-\frac{1}{2}}$$

We have found the derivatives of both \(\mathbf{v}(t)\) and \(g(t)\). Now, we can apply the chain rule to find the derivative of the composite function \(\mathbf{v}(g(t))\). The chain rule states: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \mathbf{v}'(g(t)) \cdot g'(t)$$ Substituting the values of \(\mathbf{v}'(t)\) and \(g'(t)\) calculated above, we get: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 2g(t), -2, 0\right\rangle \cdot t^{-\frac{1}{2}}$$ Now, substitute the expression of \(g(t)\) back into the formula: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 2(2\sqrt{t}), -2, 0\right\rangle \cdot t^{-\frac{1}{2}}$$ Finally, we can simplify the equation by multiplying the components of the vector by \(t^{-\frac{1}{2}}\): $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 4\sqrt{t}t^{-\frac{1}{2}}, -2t^{-\frac{1}{2}}, 0\right\rangle = \left\langle 4, -2t^{-\frac{1}{2}}, 0\right\rangle$$ Thus, the derivative of the given function \(\mathbf{v}(g(t))\) is: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 4, -2t^{-\frac{1}{2}}, 0\right\rangle$$