A dairy farmer has 80 gallons of milk to sell. The rate at which customers buy the milk is modeled by 𝑓 of 𝑡 equals five plus 0.7𝑡 sin of 𝑡 squared plus two 𝑡 over 10 for 𝑡 is greater than zero and less than 10, where 𝑓 of 𝑡 is measured in gallons per hour and 𝑡 is the number of hours after the farmer starts selling the milk. Six hours after he starts selling the milk, the cows in the farm produce milk at a rate modeled by 𝑔 of 𝑡 equals one plus 1.4 times the natural log of six 𝑡 squared plus four 𝑡 for 𝑡 is greater than six and less than 10, where 𝑔 of 𝑡 is measured in gallons per hour and 𝑡 is the number of hours after the dairy farm is open. 1) How many gallons of milk did the customers buy during the first four hours the diary farm was open?
And there are three further parts of this question that we’ll consider in a moment. Let’s first look at what we’ve been given. We’re told the rate at which the milk leaves the farm. That’s given by the function five plus 0.7𝑡 times sin of 𝑡 squared plus two 𝑡 over 10. And that’s for 𝑡 is greater than zero and less than 10. We’re also told the rate at which the milk is replenished. That’s one plus 1.4 times the natural log of 60 squared plus four 𝑡. And this time, that’s when 𝑡 is greater than six and less than 10. For part 1, we are interested in working out purely how much milk the customers bought during the first four hours.
And remember when we’re thinking about the rate of change of something, here that’s the rate of change in the volume of milk, we are actually thinking about the derivative. That means if we want to work out the actual volume of milk sold, we’re going to do the reverse process. We’ll integrate our function with respect to 𝑡. We’ll evaluate it between the limits of zero and four. And that of course, since this is a calculator question, we don’t need to try and figure out how to integrate our function. Instead, we’ll simply type this into our graphical calculator, almost exactly as we’ve written it here. The only thing we’re going to need to remember is that our calculator will use d𝑥. So we’re going to replace 𝑡 with 𝑥 and integrate with respect to 𝑥.
When we do, we see that our integral evaluated between zero and four is 24.3822 and so on. Correct to three decimal places, we can say that the customers bought 24.382 gallons of milk during the first four hours the dairy farm was open. Let’s clear some space and look at part 2 of this question.
Part 2 says, find 𝑓 prime of five. Using correct units, explain the meaning of 𝑓 prime of five in the context of the problem.
Let’s begin by simply finding 𝑓 prime of 𝑡. We’re going to differentiate our function 𝑓 with respect to time. The derivative of our constant is zero. But the derivative of 0.7𝑡 times sin of 𝑡 squared plus two 𝑡 over 10 is a little bit more tricky. We’ve got a couple of things going on. We’ve got the product of two functions, one of which is a function of a function or a composite function. So we’re going to need to use both the product and the chain rule here.
The product rule says that for two differentiable functions 𝑢 and 𝑣, the derivative of their product is 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. And the chain rule says that if 𝑦 is a function in 𝑝 and 𝑝 is a function in 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑝 times d𝑝 by d𝑥. And you might have seen that written with 𝑢 instead of 𝑝. I chose 𝑝 because we have 𝑢 in our first formula. And I didn’t want to confuse you. We’ll let 𝑢 be equal to 0.7𝑡 and 𝑣 be equal to sin of 𝑡 squared plus two 𝑡 over 10. d𝑢 d𝑡 is equal to 0.7. Then, we’re going to let 𝑝 be equal to 𝑡 squared plus two 𝑡 over 10, which means that 𝑣 is equal to sin of 𝑝. And I’m going to clear some space for this. So you might wish to take a moment to write down the chain rule and the product rule.
The derivative of 𝑣 with respect to 𝑡 is equal to the derivative of 𝑣 with respect to 𝑝—that’s cos of 𝑝—times the derivative of 𝑝 with respect to 𝑡. And I took out a constant of a 10 to make life easier. And that gives us a 10th times two 𝑡 plus two. We replace 𝑝 with 𝑡 squared plus two 𝑡 over 10. And we see that d𝑣 by d𝑡 is equal to a 10th times two 𝑡 plus two times cos of 𝑡 squared plus two 𝑡 over 10. Using the product rule for 𝑓 prime of 𝑡, we get 𝑢 times d𝑣 by d𝑡 plus 𝑣 times d𝑢 by d𝑡. Factoring 0.7 and simplifying a little and we see that 𝑓 prime of 𝑡 is equal to 0.7 times sin of 𝑡 squared plus two 𝑡 over 10 plus 𝑡 times 𝑡 plus one over five times cos of 𝑡 squared plus two 𝑡 over 10.
Remember we’re looking to find 𝑓 prime of five. So we’re going to substitute five every time we see 𝑡. That looks a little something like this. And when we put this into our calculator, we get negative 4.1786 and so on. Correct to three decimal places, that’s negative 4.179. And let’s look to explain the meaning of this within the context of the question.
𝑓 of 𝑡 tells us the rate at which customers buy the milk. And we know that 𝑓 prime, the derivative of our function, evaluated at 𝑡 equals five hours is less than zero. So this tells us the rate at which milk is being bought by these customers is decreasing. And it’s decreasing by 4.179 gallons per hour. We’ll now look at part 3.
Is the number of gallons increasing or decreasing at time 𝑡 equals eight hours? Give a reason for your answer.
Notice that when 𝑡 is equal to eight, we’re interested in two functions. It’s the rate at which milk is being sold and the rate at which it’s being produced. The overall rate of change in the amount of milk in stock is therefore the difference between these two functions. That’s 𝑔 of 𝑡 minus 𝑓 of 𝑡. And at 𝑡 is equal to eight, that’s 𝑔 of eight minus 𝑓 of eight. That’s one plus 1.4 times the natural log of six times eight squared plus four times eight minus five plus 0.7 times eight times sin of eight square plus two times eight over 10. That’s 9.4429 and so on minus 10.5404. That’s negative 1.0975 which is negative 1.097 gallons per hour.
Notice we’ve evaluated the rate at which the milk is increasing at eight hours minus the rate at which it’s decreasing at eight hours. And this has given us a negative. Since the overall rate of change of the number of gallons is less than zero, this means the actual number of gallons must be decreasing at time 𝑡 equals eight hours.
Part 4 says, “How many gallons of milk are in the farm at time 𝑡 equals nine?”
Remember we said that to find the amount of milk, we integrate our functions. Since the cows in the farm started producing milk at 𝑡 equals six hours, the amount of milk is, therefore, equal to the integral evaluated between six and nine of 𝑔 of 𝑡 with respect to 𝑡 minus the integral evaluated between zero nine of 𝑓 of 𝑡 with respect to 𝑡. And we can evaluate this as shown. Typing the first integral into our calculator and remembering to replace 𝑡 with 𝑥, when we do, we get 27.756 and so on. And for our second integral, we get 50.007. That’s negative 22.251.
Now, we know that there are initially 80 gallons of milk on the farm. So we subtract 22.251 from 80 to get 57.749. And correct to three decimal places, we can say that there are 57.749 gallons of milk on the farm at time 𝑡 equals nine.