### Video Transcript

A moving particle along a curve is defined by the two equations π₯ is equal to π‘ squared minus three π‘ cubed plus four and π¦ is equal to seven π‘ squared minus three. Find the acceleration vector of the particle at π‘ is equal to one.

The question tells us a particle is moving along a curve where its π₯-position at time π‘ is given by π‘ squared minus three π‘ cubed plus four and its π¦-position at time π‘ is given by π¦ is equal to seven π‘ squared minus three. We need to use this information to find the acceleration vector of the particle when π‘ is equal to one.

To start, we recall the acceleration of our particle will be the rate of change in its velocity. And, of course, the velocity is the rate of change of the position of our particle. So, to find the acceleration, we want to differentiate the position function twice. However, we want to find the acceleration vector of our particle. And weβre given two functions for the position of our particle, one to tell us the π₯-coordinate and one to tell us the π¦-coordinate. So, weβll find the acceleration component-wise; weβll find the acceleration in the π₯-direction and the acceleration in the π¦-direction and then represent this as a vector.

First, letβs find the acceleration of our particle in the π₯-direction. Weβll start by differentiating π₯ with respect to π‘ to get to a function for our velocity. And this is equal to the derivative of π‘ squared minus three π‘ cubed plus four with respect to π‘.

We can do this term by term by using the power rule for differentiation. We have two π‘ minus nine π‘ squared. Now, we want to find the acceleration of our particle in the π₯-direction. Weβll do this by differentiating the velocity of our particle in the π₯-direction. Thatβs the derivative of two π‘ minus nine π‘ squared with respect to π‘. Again, we can do this by using the power rule for differentiation. We get two minus 18π‘.

In fact, since we want to find the acceleration vector when π‘ is equal to one, we can substitute π‘ is equal to one into this expression to find the acceleration of our particle in the π₯-direction when π‘ is equal to one. Substituting in π‘ is equal to one, we get two minus 18 times one, which we know is equal to negative 16.

We now want to do the same in our π¦-direction. Weβll first find the velocity in the π¦-direction by finding the derivative of π¦ with respect to π‘. Thatβs the derivative of seven π‘ squared minus three with respect to π‘. We can do this term by term by using the power rule for differentiation. We get 14π‘. Weβll differentiate this one more time to find the acceleration of our particle in the π¦-direction. Thatβs the derivative of 14π‘ with respect to π‘, which we can calculate is equal to 14. So, our particle has a constant acceleration of 14 in the π¦-direction.

Weβre now ready to find the acceleration vector of our particle when π‘ is equal to one. Weβve found the π₯-component of our acceleration vector when π‘ is equal to one is negative 16. So, we start with negative 16π. And we found the particle was accelerating at a constant acceleration of 14. So, when π‘ is equal to one, our acceleration will be 14. This gives us 14π in our acceleration vector. And this gives us our final answer.

We were able to show if a particle is moving on a curve defined by the two equations π₯ is equal to π‘ squared minus three π‘ cubed plus four and π¦ is equal to seven π‘ squared minus three. Then the acceleration vector of this particle when π‘ is equal to one is π is equal to negative 16π plus 14π.