Video: APCALC02AB-P1A-Q01-368131518628

The flow rate of water entering a pipe in a factory can be modeled by the function 𝑅(𝑑) = 25 sin (βˆšπ‘‘/2) + 76, where 𝑅(𝑑) is measured in cubic feet per hour, 𝑑 is measured in hours, and 0 ≀ 𝑑 ≀ 10. The rate at which water flows out from the pipe during the same interval can be modeled by the function 𝐢(𝑑) = βˆ’0.1𝑑³ + 1.2𝑑² + 2𝑑 + 60, where 𝐢(𝑑) is measured in cubic feet per hour and 𝑑 is measured in hours for 0 ≀ 𝑑 ≀ 10. Initially, there are 48 cubic feet of water in the pipe and οΏ½

07:46

Video Transcript

The flow rate of water entering a pipe in a factory can be modeled by the function 𝑅 of 𝑑 equals 25 sin of root 𝑑 over two plus 76, where 𝑅 is measured in cubic feet per hour, 𝑑 is measured in hours, and 𝑑 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 𝐢 of 𝑑 equals negative 0.1𝑑 cubed plus 1.2𝑑 squared plus two 𝑑 plus 60, where 𝐢 of 𝑑 is measured in cubic feet per hour and 𝑑 is measured in hours for 𝑑 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 𝑑 equals zero hours. i) How many cubic feet of water enter the pipe during the 10-hour interval 𝑑 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 𝑑 equals eight hours? Give a reason for your answer. We also have a further two more parts to this question. And we’ll 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 we’re going to work out how many cubic feet of water enter the pipe during the 10-hour interval. Remember, we’re looking at the flow rate. And when we consider the rate of something, here that’s the rate of change of the volume of water over time, we’re thinking about the derivative.

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

Now, we’re going to be able to do this on our calculator. Remember though, we’re performing calculus with a trigonometric function. So it’s really important that we ensure our calculator is in radian mode before performing any calculations. Let’s type this into our calculator, remembering that we’re going to need to replace 𝑑 with π‘₯ when we do so cause our calculator is going to replace d𝑑 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 𝑑 is greater than or equal to zero and less than or equal to 10.

Now let’s consider part two. We’re 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 𝑑 hours will be the difference between the flow rate into the pipe and the flow rate out of the pipe. That’s 𝑅 of 𝑑 minus 𝐢 of 𝑑.

We want to evaluate this at 𝑑 equals eight hours. So that’s 𝑅 of eight minus 𝐢 of eight. So that’s 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. 𝑅 of eight is equal to 100.694. And 𝐢 of eight is 101.6. We don’t 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 𝑑 equals eight hours.

We’ll now consider part three and four. Part three says at what time 𝑑 from 𝑑 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? We’ve 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 𝑅 of 𝑑 minus 𝐢 of 𝑑 is equal to zero. So that’s 25 sin of root 𝑑 over two plus 76 minus negative 0.1𝑑 cubed plus 1.2𝑑 squared plus two 𝑑 plus 60 equals zero.

We’ll use our calculator to solve this. And when we do, we see that 𝑑 equals 9.714 or 7.57. Now, depending on the functionality of your graphical calculator, you may need to plot the graph of 𝑦 equals 25 sin of root 𝑑 over two plus 76 minus negative 0.1𝑑 cubed plus 1.2𝑑 squared plus two 𝑑 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. We’re 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 𝑑 equals zero hours is 48 cubic feet. Just as we did in part two, we can find the volume by integrating the difference between 𝑅𝑑 and 𝐢𝑑 with respect to 𝑑. This time, we’ll 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 we’re 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 𝑑 equals 7.75 hours.

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

Part four) For 𝑑 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‘š 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 let’s say given our π‘š, by evaluating 48 plus the integral evaluated between zero and π‘š over 𝑅 of 𝑑 minus 𝐢 of 𝑑 with respect to 𝑑. 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‘š of 𝑅 of 𝑑 minus 𝐢 of 𝑑 with respect to 𝑑. Replacing 𝑅 of 𝑑 and 𝐢 of 𝑑 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‘š at which the amount of water in the pipe reaches 300 cubic feet is as shown.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.