91Ó°ÊÓ

To find the velocity vector, differentiate the position vector \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t^2 \mathbf{k} \) with respect to time \( t \).ewline\( \mathbf{v}(t) = \frac{d}{dt}(t^2 \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t^2 \mathbf{k}) = 2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \).

Differentiate the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k} \) to find the acceleration vector.ewline\( \mathbf{a}(t) = \frac{d}{dt}(2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k}) = 2 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k} \).

The tangential component of acceleration \( a_T \) is found by projecting \( \mathbf{a}(t) \) onto \( \mathbf{v}(t) \).ewlineFirst, find the magnitude of \( \mathbf{v}(t) \): \( \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + (2t)^2 + (4t)^2} = \sqrt{4t^2 + 4t^2 + 16t^2} = \sqrt{24t^2} = 2\sqrt{6}t \).ewlineThe unit tangent vector \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\| \mathbf{v}(t) \|} = \frac{1}{2\sqrt{6}t} (2t \mathbf{i} + 2t \mathbf{j} + 4t \mathbf{k}) = \frac{1}{\sqrt{6}}(\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \).ewlineFinally, project \( \mathbf{a}(t) \) onto \( \mathbf{T}(t) \):ewline\( a_T = \mathbf{a} \cdot \mathbf{T} = (2 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k}) \cdot \frac{1}{\sqrt{6}}(\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \).ewlineCalculating the dot product: \( a_T = \frac{1}{\sqrt{6}}(2 \times 1 + 2 \times 1 + 4 \times 2) = \frac{1}{\sqrt{6}}(2 + 2 + 8) = \frac{12}{\sqrt{6}} = 2\sqrt{6} \).

The normal component of acceleration \( a_N \) can be found using the formula \( a_N = \sqrt{\| \mathbf{a} \|^2 - a_T^2} \).ewlineFirst, compute \( \| \mathbf{a} \| = \sqrt{2^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \).ewline\( a_N = \sqrt{\| \mathbf{a} \|^2 - a_T^2} = \sqrt{(2\sqrt{6})^2 - (2\sqrt{6})^2} = \sqrt{24 - 24} = \sqrt{0} = 0 \).