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Magnitude of $$$\left\langle 2 e^{2 t}, 0\right\rangle$$$
Your Input
Find the magnitude (length) of $$$\mathbf{\vec{u}} = \left\langle 2 e^{2 t}, 0\right\rangle$$$.
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 $$$\left|{2 e^{2 t}}\right|^{2} + \left|{0}\right|^{2} = 4 e^{4 t}$$$.
Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{4 e^{4 t}} = 2 e^{2 t}$$$.
Answer
The magnitude is $$$2 e^{2 t}$$$A.