### Video Transcript

The flow rate of water entering a
pipe in a factory can be modeled by the function π
of π‘ equals 25 sin of root π‘
over two plus 76, where π
is measured in cubic feet per hour, π‘ is measured in
hours, and π‘ is greater than or equal to zero and less than or equal to 10. The rate at which water flows out
from the pipe during the same interval can be modeled by the function πΆ of π‘
equals negative 0.1π‘ cubed plus 1.2π‘ squared plus two π‘ plus 60, where πΆ of π‘
is measured in cubic feet per hour and π‘ is measured in hours for π‘ is greater
than or equal to zero and less than or equal to 10. Initially, there are 48 cubic feet
of water in the pipe and π‘ equals zero hours. i) How many cubic feet of water enter
the pipe during the 10-hour interval π‘ is greater than or equal to zero and less
than or equal to 10? ii) Is the amount of water in the pipe increasing or decreasing
at time π‘ equals eight hours? Give a reason for your answer. We also have a further two more
parts to this question. And weβll consider those in a
moment.

So we have the flow rate of water
entering the pipe, the flow rate of the water leaving the pipe. And we know that, initially, there
were 48 cubic feet of water in the pipe. So how weβre going to work out how
many cubic feet of water enter the pipe during the 10-hour interval. Remember, weβre looking at the flow
rate. And when we consider the rate of
something, here thatβs the rate of change of the volume of water over time, weβre
thinking about the derivative.

In this case then, if weβre just
looking to find the volume of water that enters the pipe, weβre going to need to
perform the reverse process. We will integrate our function for
the flow rate into the pipe with respect to π‘. And since weβre looking to work out
the total volume of water that enters the pipe during zero to 10 hours, weβll
evaluate it between the limits zero and 10. So weβre looking to integrate
between the limits of zero and 10, 25 sin of root π‘ over two plus 76 with respect
to π‘.

Now, weβre going to be able to do
this on our calculator. Remember though, weβre performing
calculus with a trigonometric function. So itβs really important that we
ensure our calculator is in radian mode before performing any calculations. Letβs type this into our
calculator, remembering that weβre going to need to replace π‘ with π₯ when we do so
cause our calculator is going to replace dπ‘ with dπ₯. When we do, we get 963.259 and so
on. Correct to two decimal places, we
can say that 963.26 cubic feet of water enter the pipe during the 10-hour interval
π‘ is greater than or equal to zero and less than or equal to 10.

Now letβs consider part two. Weβre looking to work out whether
the amount of water in the pipe is increasing or decreasing. The rate of change of the amount of
water in the pipe at π‘ hours will be the difference between the flow rate into the
pipe and the flow rate out of the pipe. Thatβs π
of π‘ minus πΆ of π‘.

We want to evaluate this at π‘
equals eight hours. So thatβs π
of eight minus πΆ of
eight. So thatβs 25 sin of root eight over
two plus 76 minus negative 0.1 times eight cubed plus 1.2 times eight squared plus
two times eight plus 60. π
of eight is equal to
100.694. And πΆ of eight is 101.6. We donβt even need to work this out
to see that this is less than zero. So the rate of change of our volume
is less than zero. And this means the actual amount of
water in the pipe, the volume of water in the pipe, must be decreasing at π‘ equals
eight hours.

Weβll now consider part three and
four. Part three says at what time π‘
from π‘ is greater than or equal to zero and less than or equal to 10 hours, is the
amount of water in the pipe at its maximum? What is the amount of water in the
pipe at this time? Weβve already said that the rate of
change corresponds to the derivative. And we know that we can find the
critical points of a function by setting its derivative equal to zero. This means we can say that a
critical point on the function that represents the amount of water will occur when
π
of π‘ minus πΆ of π‘ is equal to zero. So thatβs 25 sin of root π‘ over
two plus 76 minus negative 0.1π‘ cubed plus 1.2π‘ squared plus two π‘ plus 60 equals
zero.

Weβll use our calculator to solve
this. And when we do, we see that π‘
equals 9.714 or 7.57. Now, depending on the functionality
of your graphical calculator, you may need to plot the graph of π¦ equals 25 sin of
root π‘ over two plus 76 minus negative 0.1π‘ cubed plus 1.2π‘ squared plus two π‘
plus 60 and look for the roots. Look for the points at which our
curve crosses the π₯-axis. Next, we recall that an absolute
maximum can only occur at the endpoints of our interval or at a relative
maximum. Weβre therefore going to evaluate
the volume of water at zero hours, 7.57 hours, 9.714 hours, and 10 hours.

We were told that the volume of
water in the pipe at π‘ equals zero hours is 48 cubic feet. Just as we did in part two, we can
find the volume by integrating the difference between π
π‘ and πΆπ‘ with respect to
π‘. This time, weβll evaluate it
between the limits of zero and 7.57. And now, we need to take into
account that there were already 48 cubic feet of water in the pipe. So weβre going to add 48 to this
number. And when we evaluate this on our
calculator, we get 163.254. So there are 163.254 cubic feet of
water in the pipe at π‘ equals 7.75 hours.

To find the volume of water in the
pipe at π‘ equals 9.714 hours, weβre going to calculate 48 plus the integral
evaluated between zero and 9.714 of π
of π‘ minus πΆ of π‘ with respect to π‘. That gives us 161.124 cubic
feet. And if we repeat this process for
π‘ equals 10, we see that the volume of water in the pipe at π‘ equals 10 hours is
161.26 cubic feet. And this tells us that, at π‘
equals 7.57 hours, the amount of water is at its maximum level. Weβll now clear some space and
consider the final part of this question.

Part four) For π‘ is greater than
10 hours, the water continues to enter and exit the pipe at the given rates until
the water in the pipe reaches 300 cubic feet. Write an equation without solving
it, involving one or more integrals to find the time π at which the amount of water
in the pipe reaches 300 cubic feet.

In part three of this question, we
found the volume of water in the pipe, at letβs say given our π, by evaluating 48
plus the integral evaluated between zero and π over π
of π‘ minus πΆ of π‘ with
respect to π‘. We can find the time at which it
reaches 300 cubic feet by setting this equal to 300. And we could say that the equation
is therefore 300 equals 48 plus the integral evaluated between zero and π of π
of
π‘ minus πΆ of π‘ with respect to π‘. Replacing π
of π‘ and πΆ of π‘
with the functions given, and we get the equation shown.

We distribute our parentheses. And we see that the equation to
find the time π at which the amount of water in the pipe reaches 300 cubic feet is
as shown.