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Magnitude of $$$\left\langle 4 t^{3}, - \sin{\left(t \right)}, 3\right\rangle$$$
Your Input
Find the magnitude (length) of $$$\mathbf{\vec{u}} = \left\langle 4 t^{3}, - \sin{\left(t \right)}, 3\right\rangle$$$.
Solution
The vector magnitude of a vector is given by the formula $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\sum_{i=1}^{n} \left|{u_{i}}\right|^{2}}$$$.
The sum of squares of the absolute values of the coordinates is $$$\left|{4 t^{3}}\right|^{2} + \left|{- \sin{\left(t \right)}}\right|^{2} + \left|{3}\right|^{2} = 16 t^{6} + \sin^{2}{\left(t \right)} + 9$$$.
Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}$$$.
Answer
The magnitude is $$$\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9} = 4 \left(t^{6} + 0.0625 \sin^{2}{\left(t \right)} + 0.5625\right)^{0.5}$$$A.