# Video: Finding the Instantaneous Rate of Change of the Velocity of a Particle at a Particular Value of T

If the function π£(π‘) = sin π‘ + π‘Β² β 6π‘ represents the velocity of a particle at any time π‘, in seconds, for π‘ β₯ 0, what is the instantaneous rate of change of the velocity at π‘ = 3 seconds?

02:07

### Video Transcript

If the function π£ of π‘ is equal to sin π‘ plus π‘ squared minus six π‘ represents the velocity of a particle at any time π‘, in seconds, for π‘ greater than or equal to zero, what is the instantaneous rate of change of the velocity at π‘ is equal to three seconds?

Weβve been given the function π£ of π‘ which is sin π‘ plus π‘ squared minus six π‘. This represents the velocity of a particle where time is greater than or equal to zero and is measured in seconds. And weβre asked for the instantaneous rate of change of the velocity at π‘ is equal to three seconds. Now, the instantaneous rate of change of velocity is actually equal to acceleration at a specified time. And acceleration is the derivative of velocity with respect to time. So, all we need to do is to find dπ£ by dπ‘ at π‘ is equal to three seconds.

Our velocity function π£ has three parts: sin π‘, π‘ squared, and negative six π‘. So, we can split our derivative into three parts. dπ£ by dπ‘ is equal to d by dπ‘ of sin π‘ plus d by dπ‘ of π‘ squared plus d by dπ‘ of negative six π‘. We know that d by dπ‘ of sin π‘ is equal to cos π‘. And we can use the power rule for our second term, which says d by dπ₯ of ππ₯ to the π is equal to πππ₯ to the π minus one if π is real and π is a constant. This gives us the derivative of π‘ squared with respect to π‘ is two π‘. And our final term, the derivative with respect to π‘ of negative six π‘, is negative six, so that the acceleration, which is dπ£ by dπ‘, is cos π‘ plus two π‘ minus six.

We need to find the value of dπ£ by dπ‘ at π‘ is equal to three seconds. So, letβs substitute π‘ equal to three into our derivative. And we have dπ£ by dπ‘ at π‘ is equal to three is equal to cos three plus two times three minus six. Thatβs equal to cos three plus six minus six, which is equal to cos three because six minus six is zero. And so, the instantaneous rate of change of the velocity at π‘ is equal to three seconds is equal to dπ£ by dπ‘ at π‘ is equal to three, which is cos three.