### Video Transcript

Find the velocity π― of π‘ and acceleration π of π‘ of an object with the given position vector π« of π‘ equals π‘, π‘ minus sin π‘, one minus cos π‘.

We see that the position of the object π« is given as a vector-valued function. Now, we have a number of ways to represent vectors. In printed script, they are usually given as bold letters. In the question, Iβve used a half arrow, but we can also use this hat to represent the unit vectors π’, π£, and π€. And so we see that our position vector π« has an π’-component of π‘, a π£-component of π‘ minus sin π‘, and a π€-component of one minus cos π‘. So, how do we link position, velocity, and acceleration?

Well, the rate of change of position is known as velocity. And when we think about rate of change, weβre really thinking about the derivative. So, velocity is the derivative of our position vector with respect to time. Now, we donβt actually need to worry that weβve been given π« as a vector-valued function; the same relationship holds. But instead, we simply differentiate each component in turn. We also say that the rate of change of velocity is acceleration, so acceleration is given by the derivative of velocity with respect to time.

So, letβs begin by differentiating each component in our vector function π«. Firstly, we look at the π’-component. The derivative of π‘ with respect to π‘ is simply one. Then, we look at our π£-component. We know the derivative of π‘ is one, and we also know that when we differentiate sin π₯ with respect to π₯, we get cos π₯. So, when we differentiate negative sin π‘ with respect to π‘, we must get negative cos π‘. And so, the π£-component of our velocity vector must be one minus cos π‘.

Finally, we look to differentiate one minus cos π‘ with respect to π‘. When we differentiate a constant, we get zero. So, the derivative of one is zero. We also know that when we differentiate cos π₯ with respect to π₯, we get negative sin π₯. So, the derivative of negative cos π‘ with respect to π‘ must be negative negative sin π‘, which is simply sin π‘. And so, we have our vector function that describes velocity. Itβs one, one minus cos π‘, and sin π‘.

Letβs repeat this process, this time, differentiating each component of our velocity function. The derivative of one, we already said, was zero. And then, we saw that the derivative of one minus cos π‘ was sin π‘, so the π£-component of our acceleration is sin π‘. Finally, we differentiate the π€-component of our velocity function. The derivative of sin π‘ with respect to π‘ is cos π‘. And so, we have our two vectors. The velocity vector π― of π‘ is one, one minus cos π‘, sin π‘. And the acceleration vector π of π‘ is zero, sin π‘, cos π‘.