91Ó°ÊÓ

First, we will find the first and second derivatives of \(r(t)\) with respect to the parameter \(t\): $$r(t) = \left\langle t, 2 t^{2}\right\rangle$$ $$r'(t) = \frac{d}{dt}\left\langle t, 2 t^{2}\right\rangle = \left\langle 1, 4t \right\rangle$$ Now, find the second derivative: $$r''(t) = \frac{d^2}{dt^2}\left\langle 1, 4t \right\rangle = \left\langle 0, 4 \right\rangle$$

Now, we will find the unit tangent vector \(\mathbf{T}\) at any point on the curve. This is given by the formula: $$\mathbf{T}(t) = \frac{r'(t)}{\|r'(t)\|}$$ First, find the magnitude of \(r'(t)\): $$\|r'(t)\| = \sqrt{(1)^2 + (4t)^2} = \sqrt{1 + 16t^2}$$ Then, compute the unit tangent vector \(\mathbf{T}(t)\): $$\mathbf{T}(t) = \frac{\left\langle 1, 4t \right\rangle}{\sqrt{1 + 16t^2}} = \left\langle \frac{1}{\sqrt{1 + 16t^2}}, \frac{4t}{\sqrt{1 + 16t^2}} \right\rangle$$

We will now compute the curvature \(\kappa\) using the formula: $$\kappa(t) = \frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|^3}$$ First, find the cross product \(r'(t) \times r''(t)\): $$r'(t) \times r''(t) = \left\langle 1, 4t\right\rangle \times \left\langle 0, 4 \right\rangle = \left\langle -16t, 4-0\right\rangle = \left\langle -16t, 4\right\rangle$$ Calculate the magnitude of the cross product: $$\|r'(t) \times r''(t)\| = \sqrt{(-16t)^2 + (4)^2} = \sqrt{256t^2 + 16}$$ Now, find the cube of the magnitude of \(r'(t)\): $$\|r'(t)\|^3 = (\sqrt{1 + 16t^2})^3 = (1 + 16t^2)^{\frac{3}{2}}$$ Finally, compute the curvature \(\kappa(t)\): $$\kappa(t) = \frac{\sqrt{256t^2 + 16}}{(1 + 16t^2)^{\frac{3}{2}}}$$ Thus, the unit tangent vector \(\mathbf{T}(t)\) and the curvature \(\kappa(t)\) for the parameterized curve \(r(t)=\left\langle t, 2 t^{2}\right\rangle\) are given by: $$\mathbf{T}(t) = \left\langle \frac{1}{\sqrt{1 + 16t^2}}, \frac{4t}{\sqrt{1 + 16t^2}} \right\rangle$$ $$\kappa(t) = \frac{\sqrt{256t^2 + 16}}{(1 + 16t^2)^{\frac{3}{2}}}$$