### Video Transcript

A particle moving along a path has a velocity π£ and acceleration π. Given that the equation of the displacement is π₯ equals tan π‘, find π.

Weβve been given an equation for π₯ in terms of its time π‘. So letβs recall the link between π₯ displacement, velocity π£, and acceleration π. Velocity is change in position or change in displacement with respect to time. So we can represent this using derivatives, and π£ is dπ₯ by dπ‘. Similarly, acceleration is change in velocity with respect to time. So we represent that using derivatives when π£ is velocity by saying π is dπ£ by dπ‘. Then we can link acceleration with displacement by saying that since velocity is the derivative of π₯ with respect to time, π must be the second derivative of π₯.

So we have an expression for π₯; letβs differentiate it twice. We see that velocity will be the derivative of tan π‘ with respect to π‘. We know that the derivative of tan π‘ is sec squared π‘, so we have an expression for velocity in terms of π‘. Itβs sec squared π‘. Then this means that our expression for acceleration will be the derivative of sec squared π‘ with respect to π‘. If we rewrite this as sec π‘ squared, we see we can use a special version of the chain rule to find its derivative. This is called the general power rule.

We multiply the entire term inside our parentheses by the exponent, then reduce that exponent by one. So thatβs two times sec π‘ to the power of one or two times sec π‘. We then multiply that by the derivative of the function inside our parentheses. The derivative of sec π‘ is sec π‘ tan π‘. And so acceleration is two sec π‘ times sec π‘ tan π‘, which we can simplify as two sec squared π‘ tan π‘.

So weβve found an expression for the acceleration, but weβre not quite finished. Notice we have sec squared π‘ here and we said sec squared π‘ was equal to π£. Similarly, we have tan π‘ and we saw that tan π‘ is equal to π₯. We can therefore replace sec squared π‘ and tan π‘ in our expression, giving us π equals two π£π₯.