91Ó°ÊÓ

To find the velocity vector, we'll take the derivative of the position vector with respect to time: $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \frac{d\left\langle t, t^{2}+1\right\rangle}{dt} = \left\langle \frac{d(t)}{dt}, \frac{d(t^{2}+1)}{dt} \right\rangle = \left\langle 1, 2t \right\rangle$$

Now, we'll find the acceleration vector by taking the second derivative of the position vector with respect to time: $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \frac{d\left\langle 1, 2t \right\rangle}{dt} = \left\langle \frac{d(1)}{dt}, \frac{d(2t)}{dt} \right\rangle = \left\langle 0, 2 \right\rangle$$

To find the tangential component of the acceleration, we'll take the dot product of the acceleration vector and the unit velocity vector: $$a_{T} = \mathbf{a}(t) \cdot \hat{\mathbf{v}}(t)$$ First, we need to find the magnitude of the velocity vector: $$|\mathbf{v}(t)| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1 + 4t^2}$$ Now, we can find the unit velocity vector: $$\hat{\mathbf{v}}(t) = \frac{\mathbf{v}(t)}{|\mathbf{v}(t)|} = \frac{\left\langle 1, 2t \right\rangle}{\sqrt{1+4t^2}}$$ Finally, we can compute the dot product: $$a_{T} = \mathbf{a}(t) \cdot \hat{\mathbf{v}}(t) = \left\langle 0, 2 \right\rangle \cdot \frac{\left\langle 1, 2t \right\rangle}{\sqrt{1+4t^2}} = \frac{4t}{\sqrt{1+4t^2}}$$

To find the normal component of the acceleration, we'll use the following formula: $$a_{N} = \sqrt{|\mathbf{a}(t)|^2 - a_{T}^2}$$ First, find the magnitude of the acceleration vector: $$|\mathbf{a}(t)| = \sqrt{(0)^2 + (2)^2} = 2$$ Next, square the tangential component of the acceleration: $$a_{T}^2 = \left(\frac{4t}{\sqrt{1+4t^2}}\right)^2 = \frac{16t^2}{1+4t^2}$$ Finally, plug these values into the formula for the normal component: $$a_{N} = \sqrt{|\mathbf{a}(t)|^2 - a_{T}^2} = \sqrt{2^2 - \frac{16t^2}{1+4t^2}} = \sqrt{\frac{4(1+4t^2)-16t^2}{1+4t^2}} = \frac{4}{\sqrt{1+4t^2}}$$ So, the tangential component of the acceleration is $$a_{T} = \frac{4t}{\sqrt{1+4t^2}}$$ and the normal component of the acceleration is $$a_{N} = \frac{4}{\sqrt{1+4t^2}}$$.