91Ó°ÊÓ

To find the first derivative of the curve, we will take the derivative of each component of the vector \(\mathbf{r}(t)\) with respect to \(t\): $$\mathbf{v}(t)=\mathbf{r'}(t)=\left\langle\frac{d}{dt}(t), \frac{d}{dt}(2t^2)\right\rangle = \left\langle1, 4t\right\rangle$$

To find the magnitude of the velocity vector, we can use the formula \(\|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2}\). In this case, we have: $$\|\mathbf{v}(t)\| = \sqrt{(1)^2 + (4t)^2} = \sqrt{1 + 16t^2}$$

To find the unit tangent vector, we will divide the velocity vector by its magnitude: $$\mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} = \frac{\left\langle1, 4t\right\rangle}{\sqrt{1 + 16t^2}} = \left\langle\frac{1}{\sqrt{1+16t^2}}, \frac{4t}{\sqrt{1+16t^2}}\right\rangle$$

To find the acceleration vector, we will take the derivative of each component of the velocity vector \(\mathbf{v}(t)\) with respect to \(t\): $$\mathbf{a}(t) = \mathbf{v'}(t) = \left\langle \frac{d}{dt}(1), \frac{d}{dt}(4t)\right\rangle = \left\langle 0, 4 \right\rangle$$

To find the curvature, we will use the formula \(\kappa(t) = \frac{\|\mathbf{v}(t) \times \mathbf{a}(t)\|}{\|\mathbf{v}(t)\|^3}\): $$\mathbf{v}(t) \times \mathbf{a}(t) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4t & 0 \\ 0 & 4 & 0 \end{vmatrix} = -16t\hat{k}$$ $$\|\mathbf{v}(t) \times \mathbf{a}(t)\| = 16|t|$$ Now, we can find the curvature: $$\kappa(t) = \frac{\|\mathbf{v}(t) \times \mathbf{a}(t)\|}{\|\mathbf{v}(t)\|^3} = \frac{16|t|}{(1 + 16t^2)^{3/2}}$$ Thus, the unit tangent vector \(\mathbf{T}(t)\) and the curvature \(\kappa(t)\) for the given parameterized curve are: $$\mathbf{T}(t) = \left\langle\frac{1}{\sqrt{1+16t^2}}, \frac{4t}{\sqrt{1+16t^2}}\right\rangle$$ $$\kappa(t) = \frac{16|t|}{(1 + 16t^2)^{3/2}}$$