Video Transcript
A particle has a position defined
by the equations 𝑥 equals 𝑡 cubed minus five 𝑡 and 𝑦 equals three minus two 𝑡
squared. Find the velocity vector of the
particle at 𝑡 equals two.
We’re 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 𝑡, we obtain a coordinate pair 𝑥𝑦 for the position of the
particle at that time. We could say that, in vector terms,
the position of the particle at time 𝑡 is 𝑠 of 𝑡 equals 𝑡 cubed minus five 𝑡 𝑖
plus three minus two 𝑡 squared 𝑗. This isn’t though quite what we’re
looking for. We want to find the velocity vector
of our particle when 𝑡 is equal to two. So we recall that, given a function
for displacement which of course is essentially the difference in the object’s
position from one time to another, we find a function for velocity by
differentiating with respect to 𝑡.
This means that we can achieve a
velocity vector for our particle at time 𝑡 by differentiating each component for
the position vector with respect to 𝑡. That’s the derivative of 𝑡 cubed
minus five 𝑡. And then we differentiate the
vertical component as well, the derivative of three minus two 𝑡 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 𝑡 cubed differentiates to three
𝑡 squared and negative five 𝑡 differentiates to negative five.
Similarly, the derivative of three
minus two 𝑡 squared is negative four 𝑡. Remember, the derivative of any
constant is simply zero. Our vector function for velocity is
three 𝑡 squared minus five 𝑖 plus negative four 𝑡 𝑗. Remember, we want a velocity vector
at 𝑡 is equal to two. So we’re going to substitute 𝑡
equals two into our vector. That’s three times two squared
minus five 𝑖 plus negative four times two 𝑗, which simplifies to seven 𝑖 minus
eight 𝑗.
