Question

the position vector of a particle is r(t) find the requested vector the velocity at t=2 for r(t)=(2t^2+5t+6)i-3t^3j+(10-t^2)k

Answer 1

remember that
r'(t)=v(t)
thus:
d/dt [r(t)]=v(t) => d/dt <2t^2+5t+6,-3t^3,10-t^2>
\[<\frac{d}{dt}2t^2+5t+6,\frac{d}{dt}-3t^3,\frac{d}{dt}10-t^2>\]
\[r'(t)=v(t)=<4t+5,-9t^2,-2t>\]
or, in vector hat notation:
\[v(t)=4t+5 i-9t^2j-2tk\]
now you have velocity with respect to time, simply plug and chug for the value.

Answer 2

Thank u