### Video Transcript

A particle moves along the ๐ฅ-axis. Its velocity ๐ฃ in meters per second is given by ๐ฃ is equal to nine times ๐ก minus four minus 12 times ๐ก minus four all cubed, where ๐ก is the time in seconds. Find the interval over which the particle is accelerating in the positive ๐ฅ-direction.

Weโre told that a particle is moving along the ๐ฅ-axis. This means itโs moving in a straight line. Weโre told the velocity of this particle in meters per second is given by ๐ฃ is equal to nine times ๐ก minus four minus 12 multiplied by ๐ก minus four all cubed. And again, ๐ก here is measured in seconds. We need to find the interval over which this particle is accelerating in the positive ๐ฅ-direction. First, remember, to find the acceleration of a particle moving in a straight line, we want to find the rate of change of velocity with respect to time. In our case, thatโs d๐ฃ by d๐ก. This will give us a function for the acceleration of our particle at time ๐ก.

But remember, weโre looking for where our particle is accelerating in the positive ๐ฅ-direction. And of course, for our particle to be accelerating in the positive ๐ฅ-direction, our rate of change of velocity with respect to time must be positive. So weโre looking for ๐ of ๐ก to be greater than zero. So letโs find an expression for the acceleration of our particle. Remember, thatโs the derivative of the velocity function with respect to time. This gives us the derivative of nine times ๐ก minus four minus 12 multiplied by ๐ก minus four all cubed with respect to ๐ก.

Weโll evaluate this derivative term by term. First, by expanding our first set of parentheses, we get 9๐ก minus 36. And if we differentiate this, we see the derivative of 9๐ก with respect to ๐ก is nine and the derivative of the constant negative 36 is equal to zero. So the derivative of our first term is just equal to nine. Now, to evaluate our second term, we could use the chain rule. However, weโre going to use the general power rule. We recall the general power rule tells us, for any differentiable function ๐ of ๐ฅ and any constant ๐, the derivative of ๐ of ๐ฅ raised to the ๐th power with respect to ๐ฅ is equal to ๐ times ๐ prime of ๐ฅ multiplied by f of ๐ฅ raised to the power of ๐ minus one.

In our case, weโre differentiating with respect to ๐ก. Our inner function ๐ of ๐ก will be ๐ก minus four. And weโre raising this to the power of three. So we have ๐ is equal to three and we see the derivative ๐ prime of ๐ก is just equal to one. So by applying the general power rule, the derivative of our second term is equal to negative 12 times three multiplied by one times ๐ก minus four all squared. And we can simplify the leading coefficient to be negative 36. So weโve shown the acceleration of our particle at times ๐ก is equal to nine minus 36 times ๐ก minus four squared.

Remember, we want to find the values of ๐ก where ๐ of ๐ก is greater than zero. So we need to solve this is greater than zero. Thereโs a few different ways of doing this. Weโll start by adding 36 times ๐ก minus four squared to both sides of the equation. This means we now need to solve the inequality nine is greater than 36 times ๐ก minus four squared. Next, weโll divide both sides of the inequality through by 36. And we can simplify nine divided by 36 to give us one-quarter. So we need to find the values of ๐ก where one-quarter is greater than ๐ก minus four all squared.

We could do this algebraically. However, weโre going to do this with a graphical solution. To find our values of ๐ก, letโs start by sketching a graph of the curve ๐ฆ is equal to ๐ฅ minus four all squared. To sketch this graph, remember that subtracting four from our values of ๐ฅ is the same as translating four units to the right. So our sketch would just be a parabola translated four units to the right.

Now, letโs also sketch the line ๐ฆ is equal to one-quarter. Remember, weโre looking for the values where one-quarter is greater than ๐ก minus four all squared. And we can find these values from our sketch. We can see this will be when our line ๐ฆ is equal to one-quarter is above the curve ๐ฅ minus four all squared. And this is between the two points of intersection between our lines. So letโs find the points of intersection. We want to solve ๐ฅ minus four squared is equal to one-quarter. We want to take the square roots of both sides of this equation. The square root of one-quarter is equal to a half. But remember, weโll get a positive and a negative square root.

So we get ๐ฅ minus four is equal to positive or negative one-half. And to solve for ๐ฅ, weโll add four to both sides of the equation. So the ๐ฅ-coordinates of our points of intersection are four plus a half and four minus a half. Weโll write these as seven over two and nine over two. So from our sketch, one-quarter is greater than ๐ฅ minus four all squared when ๐ฅ is between nine over two and seven over two. This tells our particle will be accelerating in the positive ๐ฅ-direction when ๐ก is between nine over two and seven over two.

And remember, the question wants us to give this as an interval. So weโll write this as the open interval from seven over two to nine over two. Therefore, we were able to show if a particle is moving along the ๐ฅ-axis with a velocity function ๐ฃ is equal to nine times ๐ก minus four minus 12 multiplied by ๐ก minus four all cubed. Then the particle will be accelerating in the positive ๐ฅ-direction when ๐ก is in the open interval from seven over two to nine over two.