$$$\frac{1}{2 \sqrt{t^{2} + 1}}\cdot \left\langle 2 t, 2\right\rangle$$$

计算 $$$\frac{1}{2 \sqrt{t^{2} + 1}}\cdot \left\langle 2 t, 2\right\rangle$$$。

将向量的每个坐标乘以该标量：

$$${\color{Blue}\left(\frac{1}{2 \sqrt{t^{2} + 1}}\right)}\cdot \left\langle 2 t, 2\right\rangle = \left\langle {\color{Blue}\left(\frac{1}{2 \sqrt{t^{2} + 1}}\right)}\cdot \left(2 t\right), {\color{Blue}\left(\frac{1}{2 \sqrt{t^{2} + 1}}\right)}\cdot \left(2\right)\right\rangle = \left\langle \frac{t}{\sqrt{t^{2} + 1}}, \frac{1}{\sqrt{t^{2} + 1}}\right\rangle$$$

$$$\frac{1}{2 \sqrt{t^{2} + 1}}\cdot \left\langle 2 t, 2\right\rangle = \left\langle \frac{t}{\sqrt{t^{2} + 1}}, \frac{1}{\sqrt{t^{2} + 1}}\right\rangle = \left\langle \frac{t}{\left(t^{2} + 1\right)^{0.5}}, \left(t^{2} + 1\right)^{-0.5}\right\rangle$$$A