### Video Transcript

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

Here weβve been given a pair of
parametric equations to describe the position of the particle. This means that, given a value of
π‘, we obtain an output which is a coordinate pair for the position of the particle
at that time. We could treat these as completely
separate entities. Now, weβre looking to find the
speed of the particle at π‘ equals two. Well, we know that speed is the
magnitude of the velocity. And we also recall that we can find
a function for velocity by differentiating a function for displacement. We might then choose to say that,
in vector terms, the position of the particle at time π‘ is five π‘ squared plus
four π‘ π plus three π‘ minus two π. We could then differentiate our
function for position and find a function for velocity. And in fact, we can achieve this by
differentiating individually the horizontal and vertical components for our
displacement or our position.

Weβll begin by differentiating the
horizontal component. Thatβs five π‘ squared plus four
π‘. The derivative of five π‘ squared
is 10π‘. And the derivative of four π‘ with
respected π‘ is simply four. Letβs repeat this process for the
vertical component. The derivative of three π‘ is three
and the derivative of negative two is zero. In vector terms then, our velocity
is described by 10π‘ plus four π plus three π. All this means is, at π‘ seconds,
the velocity is 10π‘ plus four in the horizontal direction and three in the
vertical, the π direction. Next, we want to work out the
velocity at π‘ equals two. So weβll substitute π‘ into our
vector equation for velocity that gives us 10 times two plus four π plus three
π. Thatβs 24π plus three π.

Remember that weβre looking to find
the speed, which is the magnitude of the velocity. So we recall that we can find the
magnitude of a vector π given as π₯ π plus π¦ π by finding the square root of π₯
squared plus π¦ squared. In this case, thatβs the square
root of 24 squared plus three squared, which is equal to three root 65. Now, we donβt have any units here,
so weβre done. The speed of the particle at π‘
equals two is three root 65.