Video: APCALC04AB-P2A-Q01-907145370841

The rate at which vehicles are entering a certain road for a period of time can be modelled by the function π‘Ÿ given by π‘Ÿ = 8𝑑 (1 βˆ’ (𝑑/25))Β² (1 βˆ’ (𝑑/60))⁸, 0 ≀ 𝑑 ≀ 25, and π‘Ÿ = 0, 𝑑 > 25, where π‘Ÿ(𝑑) is measured in vehicles per minute and 𝑑 is measured in minutes. It is observed that vehicles exit the road at a constant rate of 4 vehicles per minute. Initially, at 𝑑 = 0, there are 45 vehicles on the road. i) How many vehicles enter the road during the time interval 0 ≀ 𝑑 ≀ 25?

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Video Transcript

The rate at which vehicles are entering a certain road for a period of time can be modelled by the function π‘Ÿ given by π‘Ÿ is equal to eight 𝑑 times one minus 𝑑 over 25 squared times one minus 𝑑 over 60 to the power of eight for 𝑑 is greater than or equal to zero and less than or equal to 25, and zero for 𝑑 is greater than 25, where π‘Ÿ of 𝑑 is measured in vehicles per minute and 𝑑 is measured in minutes. It is observed that vehicles exit the road at a constant rate of four vehicles per minute. Initially, at 𝑑 equals zero, there are 45 vehicles on the road. 1) How many vehicles enter the road during the time interval 𝑑 is greater than or equal to zero and less than or equal to 25?

And there are three further parts to this question that we’ll consider in a moment. Whenever we think about rate of change, we think of derivatives. In this case, that’s the rate at which vehicles enter the road. And that’s equal to the derivative of the number of vehicles. This means we can find a function for the number of vehicles entering the road by performing the reverse process, by integrating π‘Ÿ with respect to 𝑑. We’re only interested in the number of vehicles entering in the closed interval for 𝑑, zero to 25. So the number of vehicles is equal to the integral evaluated between zero and 25 of eight 𝑑 times one minus 𝑑 over 25 squared times one minus 𝑑 over 60 to the eighth power, with respect to 𝑑.

Now, we’re going to use our calculator to evaluate this. And we will type in almost exactly as shown. However, since our calculator is going to integrate with respect to π‘₯, we’re going to change 𝑑 for π‘₯. That gives us 124.0074 and so on. We’ll round this correct to the nearest whole number. And we see that, roughly, 124 vehicles enter the road during the time interval 𝑑 is greater than or equal to zero and less than or equal to 25. We’ll now consider part two for this question.

This says, how many vehicles are on the road at 𝑑 equals 25 minutes?

We’re interested in the rate at which vehicles enter the road and the rate at which they leave them this time. That will tell us how many vehicles are on the road at 𝑑 equals 25. And we already know that there are initially 45 vehicles on the road. We calculated that 124 vehicles entered in the first 25 minutes. So that gives us a total of 169 vehicles. We also know that they exit the road at a constant rate of four vehicles per minute. And there’s two ways that we can work out how many leave over our time interval.

We could integrate four with respect to 𝑑 and evaluate between zero and 25. Alternatively, we can use simple proportion to see that if four leave every minute, four times 25 which is 100 will leave in 25 minutes. The number of vehicles on the road at 𝑑 equals 25 minutes is going to be the difference between the number that we already had and the number that entered verses the number that left. That’s 169 minus 100 which is 69. There are 69 vehicles on the road at 𝑑 equals 25 minutes.

Part three says, for 𝑑 is greater than 25, the vehicles continued to exit the road at a rate of four vehicles per minute. What is the first instance of 𝑑 when there are no vehicles on the road?

We know that, we know that our function π‘Ÿ is equal to zero when 𝑑 is greater than 25. So no new vehicles are entering the road. But there are 69 on the road when 𝑑 is equal to 25 minutes. And those continue to leave at a rate of four per minute. We divide 69 by four to work out how long it will take for all of the vehicles to leave the road; that’s 17.25. Since this began happening at 25 minutes, we know that 𝑑 is equal to 25 plus 17.25, which is equal to 42.25. And the first instance of 𝑑 when there are no vehicles on the road was 42.25 minutes.

Part four says, for 𝑑 is greater than or equal to zero and less than 25, at what time is the number of vehicles on the road at its maximum? And what is the number of vehicles on the road at this time, approximated to the nearest whole number?

Remember, we can find the critical points of our function by setting the derivative equal to zero. In this case, the rate of change is the derivative. So we set the difference between the incoming rate versus the outcoming rate equal to zero. That’s eight 𝑑 times one minus 𝑑 over 25 squared times one minus 𝑑 over 60 to the eighth power minus four equal to zero. We can solve this using our calculators. And we see that 𝑑 is equal to 0.56447 or 12.3527 and so on. We know that the absolute maximum will occur at either of these critical points or at the endpoints of our function. So we’re going to evaluate π‘Ÿ plus 45 at these points. And we’ll look for the highest value.

When 𝑑 is equal to zero, there are 45 vehicles on the road. And we know when 𝑑 is equal to 25, there are 69 vehicles on the road. When 𝑑 is equal to 0.564, we get the number of vehicles to be 43.9 and so on. Correct to the nearest whole number, let’s say 44. When 𝑑 is equal to 12.353, we see that the number of vehicles is 107.1 and so on. Let’s say, correct to the nearest whole number, 107 vehicles. The maximum value of our function occurs when 𝑑 is equal to 12.353. And so the number of vehicles on the road is at its maximum when 𝑑 is equal to 12.353 minutes. And this is 107 vehicles.

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