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

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

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 $\left|{1}\right|^{2} + \left|{2 t}\right|^{2} + \left|{3 t^{2}}\right|^{2} = 9 t^{4} + 4 t^{2} + 1$.

Therefore, the magnitude of the vector is $\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{9 t^{4} + 4 t^{2} + 1}$.

The magnitude is $\sqrt{9 t^{4} + 4 t^{2} + 1}\approx 3 \left(t^{4} + 0.444444444444444 t^{2} + 0.111111111111111\right)^{0.5}.$A