### Video Transcript

A particle has a position defined
by the equations π₯ equals π‘ cubed minus five π‘ and π¦ equals three minus two π‘
squared. Find the velocity vector of the
particle at π‘ equals two.

Weβre told that the position of the
particle is defined by a pair of parametric equations. In real terms, this means that,
given a value of time π‘, we obtain a coordinate pair π₯π¦ for the position of the
particle at that time. We could say that, in vector terms,
the position of the particle at time π‘ is π of π‘ equals π‘ cubed minus five π‘ π
plus three minus two π‘ squared π. This isnβt though quite what weβre
looking for. We want to find the velocity vector
of our particle when π‘ is equal to two. So we recall that, given a function
for displacement which of course is essentially the difference in the objectβs
position from one time to another, we find a function for velocity by
differentiating with respect to π‘.

This means that we can achieve a
velocity vector for our particle at time π‘ by differentiating each component for
the position vector with respect to π‘. Thatβs the derivative of π‘ cubed
minus five π‘. And then we differentiate the
vertical component as well, the derivative of three minus two π‘ squared. We know that to differentiate a
polynomial term, we multiply the entire term by the experiment and then reduce the
exposure by one. So π‘ cubed differentiates to three
π‘ squared and negative five π‘ differentiates to negative five.

Similarly, the derivative of three
minus two π‘ squared is negative four π‘. Remember, the derivative of any
constant is simply zero. Our vector function for velocity is
three π‘ squared minus five π plus negative four π‘ π. Remember, we want a velocity vector
at π‘ is equal to two. So weβre going to substitute π‘
equals two into our vector. Thatβs three times two squared
minus five π plus negative four times two π, which simplifies to seven π minus
eight π.