# Question Video: Differentiating Exponential Functions Using the Chain Rule Mathematics • Higher Education

A particle has a position defined by the equations π₯ = π‘Β³ β 5π‘ and π¦ = 3 β 2π‘Β². Find the velocity vector of the particle at π‘ = 2.

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### 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.

Weβ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 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 isnβt though quite what weβre 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 objectβs 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‘. Thatβs 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 weβre going to substitute π‘ equals two into our vector. Thatβs three times two squared minus five π plus negative four times two π, which simplifies to seven π minus eight π.