Magnitudo $$$\left\langle 2, \frac{\sqrt{2} \left(1 - 2 t\right) e^{- t}}{4 t^{\frac{3}{2}}}, \frac{\sqrt{2} \left(2 t + 1\right) e^{t}}{4 t^{\frac{3}{2}}}\right\rangle$$$

Temukan besar (panjang) $$$\mathbf{\vec{u}} = \left\langle 2, \frac{\sqrt{2} \left(1 - 2 t\right) e^{- t}}{4 t^{\frac{3}{2}}}, \frac{\sqrt{2} \left(2 t + 1\right) e^{t}}{4 t^{\frac{3}{2}}}\right\rangle.$$$

Besarnya suatu vektor dinyatakan dengan rumus $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\sum_{i=1}^{n} \left|{u_{i}}\right|^{2}}$$$.

Jumlah kuadrat dari nilai mutlak koordinat adalah $$$\left|{2}\right|^{2} + \left|{\frac{\sqrt{2} \left(1 - 2 t\right) e^{- t}}{4 t^{\frac{3}{2}}}}\right|^{2} + \left|{\frac{\sqrt{2} \left(2 t + 1\right) e^{t}}{4 t^{\frac{3}{2}}}}\right|^{2} = \frac{\left(2 t - 1\right)^{2} e^{- 2 t}}{8 \left|{t^{\frac{3}{2}}}\right|^{2}} + \frac{\left(2 t + 1\right)^{2} e^{2 t}}{8 \left|{t^{\frac{3}{2}}}\right|^{2}} + 4.$$$

Oleh karena itu, besar vektor adalah $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\frac{\left(2 t - 1\right)^{2} e^{- 2 t}}{8 \left|{t^{\frac{3}{2}}}\right|^{2}} + \frac{\left(2 t + 1\right)^{2} e^{2 t}}{8 \left|{t^{\frac{3}{2}}}\right|^{2}} + 4} = \frac{\sqrt{64 t^{4} e^{2 t} + 2 \left(2 t - 1\right)^{2} \left|{t}\right| + 2 \left(2 t + 1\right)^{2} e^{4 t} \left|{t}\right|} e^{- t}}{4 t^{2}}.$$$

Besarnya adalah $$$\frac{\sqrt{64 t^{4} e^{2 t} + 2 \left(2 t - 1\right)^{2} \left|{t}\right| + 2 \left(2 t + 1\right)^{2} e^{4 t} \left|{t}\right|} e^{- t}}{4 t^{2}} = \frac{0.25 \left(64 t^{4} e^{2 t} + 2 \left(2 t - 1\right)^{2} \left|{t}\right| + 2 \left(2 t + 1\right)^{2} e^{4 t} \left|{t}\right|\right)^{0.5} e^{- t}}{t^{2}}.$$$A