Explanation:
Given,
Position vector, [tex]c(t)=6ti+((3t)^2)j+((t)^3)k[/tex]
We need to find the velocity vector of the given path
[tex]v=\dfrac{d(c(t))}{dt}\\\\v=\dfrac{d(6ti+((3t)^2)j+((t)^3)k)}{dt}\\\\v=\dfrac{d(6ti)}{dt}+\dfrac{d((3t)^2j}{dt}+\dfrac{d(t^3)k}{dt}\\\\v=6i+18tj+3t^2k[/tex]
Hence, this is the velocity vector of the given path.
