### Video Transcript

A particle is moving under the action of the force πΉ equals eight π plus five π. Its position vector at time π‘ is given by the relation π as a function of π‘ equals three π‘ squared minus π‘ minus four π plus five halves π‘ squared plus 10π‘ plus five π. Find the rate of the work done by the force at π‘ equals three.

Weβre given the force as well as its components πΉ and the position vector of a particle as a function of time π‘ and we want to solve for the rate at which work is done by the force at a particular time. This rate is a rate with respect to time. So we can call the variable weβre searching for ππ€ ππ‘.

If we recall that work is equal to the dot product of force and displacement, then in terms of the information weβve been given we can rewrite our expression as π ππ‘ the time derivative of πΉ dot π of π‘. And since πΉ is constant with respect to time, we can rewrite this further as πΉ dot π ππ‘ of π of π‘.

Since πΉ and π of π‘ are given in the problem statement, we can insert those values now and make this calculation. Plugging in eight π plus five π for πΉ and the π and π components of π, weβre ready to take the time derivative of π of π‘. When we do, we find that derivative with respect to π‘ gives us six π‘ minus one π plus five π‘ plus 10π.

Weβre told we want to evaluate this rate when time is equal to three. Plugging in that value for π‘ where it appears in our time derivative of π of π‘, our time derivative of the position vector reduces to 17π plus 25π.

Applying the dot product operation between these two vectors, we find a result of eight times 17 plus five times 25 or 261. Thatβs the rate of work done or power exerted by the force at time π‘ equals three.