If you take a vector field that is a constant vector field where everything just translates then there is no divergence involved because the derivatives will be zero.

如果取的向量场是处处恒定的，所有点都是平移关系，所以没有散度，因为导数为零。

youdao

So, even if the speed is constant, that means, even if a length of the velocity vector stays the same, the velocity vector can still rotate.

因此即使速率不变，也就是说速度矢量的的大小不变，但它的方向可以做旋转。

youdao

And, in some cases, for example, if you know that the vector field is tangent to the curve or if a dot product is constant or things like that then that might actually give you a very easy answer.

例如，在某些情况下，如果已知向量场与曲线相切，或者内积是一个常数等等，那么结果将会很简单。

youdao

This is how you get the rules for adding a vector to another vector, then taking a vector and multiplying it by some constant.

以上这些就是矢量加法的法则,以及数乘矢量的法则

耶鲁公开课 - 基础物理课程节选

I want to consider a particle which has definite acceleration a, a constant one, but a is now a vector.

我想考虑一个加速度为 a 的质点,加速度恒定,但 a 是一个矢量

耶鲁公开课 - 基础物理课程节选

take the derivative of this, get the velocity vector and you notice his magnitude is a constant Whichever way you do it, you can then rewrite this as v square over r.

对这个式子求一次导,就能得到速度矢量,你会发现其模长是常数,不管用什么方法,加速度也可以写成 v^2 / r

耶鲁公开课 - 基础物理课程节选