91Ó°ÊÓ

The tangent vector is the first derivative of the parameterized curve with respect to \(t\). So, differentiate each component of the given curve with respect to \(t\): $$\mathbf{r'}(t) = \frac{d\langle 2t, 4\sin t, 4\cos t\rangle}{dt}$$

Differentiate each component of the given curve with respect to \(t\): $$\frac{d(2t)}{dt} = 2$$ $$\frac{d(4\sin t)}{dt} = 4\cos t$$ $$\frac{d(4\cos t)}{dt} = -4\sin t$$ So, the tangent vector is given by: $$\mathbf{r'}(t) = \langle 2, 4\cos t, -4\sin t\rangle$$

The magnitude of a vector is given by the square root of the sum of the squares of its components. Hence, the magnitude of the tangent vector is: $$||\mathbf{r'}(t)|| = \sqrt{(2)^2+(4\cos t)^2+(-4\sin t)^2}$$

Simplify the expression for the magnitude of the tangent vector: $$||\mathbf{r'}(t)|| = \sqrt{4+16\cos^2t+16\sin^2t}$$ Since \(\sin^2t+\cos^2t = 1\), so $$||\mathbf{r'}(t)|| = \sqrt{4+16(1)} = \sqrt{20}$$

The unit tangent vector is obtained by dividing the given tangent vector by its magnitude: $$\mathbf{T} = \frac{\mathbf{r'}(t)}{||\mathbf{r'}(t)||}$$ Hence, $$\mathbf{T} = \frac{\langle 2, 4\cos t, -4\sin t\rangle}{\sqrt{20}}$$

Now, we will find the curvature by calculating the derivative of the unit tangent vector, taking its magnitude, and dividing this value by the magnitude of the tangent vector: $$\kappa = \frac{||\mathbf{T'}(t)||}{||\mathbf{r'}(t)||}$$ Differentiate each component of the unit tangent vector \(\mathbf{T}\) with respect to \(t\): $$\frac{d(2/\sqrt{20})}{dt} = 0$$ $$\frac{d(4\cos t/\sqrt{20})}{dt} = -4\sin t/\sqrt{20}$$ $$\frac{d(-4\sin t/\sqrt{20})}{dt} = -4\cos t/\sqrt{20}$$ So, $$\mathbf{T'}(t) = \left\langle 0, -\frac{4\sin t}{\sqrt{20}}, -\frac{4\cos t}{\sqrt{20}}\right\rangle$$ The magnitude of T'(t) is $$||\mathbf{T'}(t)|| = \sqrt{0^2 + \left(-\frac{4\sin t}{\sqrt{20}}\right)^2 + \left(-\frac{4\cos t}{\sqrt{20}}\right)^2}=\sqrt{\frac{16\sin^2t + 16\cos^2t}{20}}$$ Again, since \(\sin^2t+\cos^2t = 1\), so $$||\mathbf{T'}(t)|| = \sqrt{\frac{16(1)}{20}} = \frac{2\sqrt{5}}{\sqrt{20}}$$ Finally, the curvature \(\kappa\) is given by $$\kappa =\frac{||\mathbf{T'}(t)||}{||\mathbf{r'}(t)||} = \frac{\frac{2\sqrt{5}}{\sqrt{20}}}{\sqrt{20}}=\frac{1}{\sqrt{5}}$$