### Video Transcript

A rocket is launched in the
air. Its height, in meters, as a
function of time is given by β of π‘ equals negative 4.9π‘ squared plus 229π‘ plus
234. Find the maximum height the rocket
attains.

So, weβve been given an equation
for the height β of this rocket in terms of π‘ time. And weβre asked to determine the
maximum height that the rocket attains. We can follow some key steps in
order to do so. First, we need to find an
expression for the first derivative of our function. In this case, thatβs β prime of
π‘. By applying the power rule of
differentiation, we see that β prime of π‘ is equal to negative two multiplied by
4.9π‘ plus 229, which simplifies to negative 9.8π‘ plus 229.

Next, we recall that at the
critical points of the function, the first derivative is equal to zero. So, weβre going to take that
expression weβve found for β prime of π‘, set it equal to zero, and then solve the
resulting equation for π‘. We add 9.8π‘ to both sides and then
divide by 9.8 to give π‘ equals 229 over 9.8, which, as a decimal, is 23.36734, or
23.37 to two decimal places. Now, this is the value of π‘ at
which our function β of π‘ has a critical point. We donβt yet know whether it is a
maximum. And we donβt yet know the height
that the rocket attains at this point.

Our next step, then, is to evaluate
the function β of π‘ when π‘ is equal to 23.37. Substituting into our equation β of
π‘ gives β of 23.37 equals negative 4.9 multiplied by 23.37 squared plus 229
multiplied by 23.37 plus 234. Which evaluates to 2909.56119, or
2909.56 correct to two decimal places. So, we believe that this is the
maximum height that the rocket attains as this occurs at the only critical point of
the function. But we must confirm that it is
indeed a maximum.

To do so, weβll perform the second
derivative test. Weβll evaluate the second
derivative β double prime of π‘ at this critical point. Differentiating our expression for
β prime of π‘, which was negative 9.8π‘ plus 229, we find that β double prime of π‘
is equal to negative 9.8. And in fact, we donβt need to
evaluate the second derivative when π‘ equals 23.37 because it is a constant. The second derivative is the same
for all values of π‘.

We note though, that this value is
negative. And therefore, by the second
derivative test, our critical point is a maximum. Now, we could also have seen this
if we considered that the expression we were given for β in terms of π‘ is a
quadratic expression with a negative leading coefficient. And therefore, the graph of π‘
against β would be an inverted parabola. And we know that it will therefore
have a maximum point rather than a minimum. So, we can conclude that the
maximum height the rocket attains, and we have confirmed that it is indeed a
maximum, is 2909.56 meters, correct to two decimal places.