Magnitude of $\left\langle 1, 3, 2 t\right\rangle$

Find the magnitude (length) of $\mathbf{\vec{u}} = \left\langle 1, 3, 2 t\right\rangle$.

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 $\left|{1}\right|^{2} + \left|{3}\right|^{2} + \left|{2 t}\right|^{2} = 4 t^{2} + 10$.

Therefore, the magnitude of the vector is $\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{4 t^{2} + 10}$.

The magnitude is $\sqrt{4 t^{2} + 10}\approx 3.162277660168379 \left(0.4 t^{2} + 1\right)^{0.5}$A.