### Video Transcript

A dairy farmer has 80 gallons of milk to sell. The rate at which customers buy the milk is modeled by π of π‘ equals five plus 0.7π‘ sin of π‘ squared plus two π‘ over 10 for π‘ is greater than zero and less than 10, where π of π‘ is measured in gallons per hour and π‘ 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 π of π‘ equals one plus 1.4 times the natural log of six π‘ squared plus four π‘ for π‘ is greater than six and less than 10, where π of π‘ is measured in gallons per hour and π‘ 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 weβll consider in a moment. Letβs first look at what weβve been given. Weβre told the rate at which the milk leaves the farm. Thatβs given by the function five plus 0.7π‘ times sin of π‘ squared plus two π‘ over 10. And thatβs for π‘ is greater than zero and less than 10. Weβre also told the rate at which the milk is replenished. Thatβs one plus 1.4 times the natural log of 60 squared plus four π‘. And this time, thatβs when π‘ 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 weβre thinking about the rate of change of something, here thatβs 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, weβre going to do the reverse process. Weβll integrate our function with respect to π‘. Weβll evaluate it between the limits of zero and four. And that of course, since this is a calculator question, we donβt need to try and figure out how to integrate our function. Instead, weβll simply type this into our graphical calculator, almost exactly as weβve written it here. The only thing weβre going to need to remember is that our calculator will use dπ₯. So weβre going to replace π‘ with π₯ and integrate with respect to π₯.

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. Letβs clear some space and look at part 2 of this question.

Part 2 says, find π prime of five. Using correct units, explain the meaning of π prime of five in the context of the problem.

Letβs begin by simply finding π prime of π‘. Weβre going to differentiate our function π with respect to time. The derivative of our constant is zero. But the derivative of 0.7π‘ times sin of π‘ squared plus two π‘ over 10 is a little bit more tricky. Weβve got a couple of things going on. Weβve got the product of two functions, one of which is a function of a function or a composite function. So weβre going to need to use both the product and the chain rule here.

The product rule says that for two differentiable functions π’ and π£, the derivative of their product is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. And the chain rule says that if π¦ is a function in π and π is a function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ times dπ by dπ₯. And you might have seen that written with π’ instead of π. I chose π because we have π’ in our first formula. And I didnβt want to confuse you. Weβll let π’ be equal to 0.7π‘ and π£ be equal to sin of π‘ squared plus two π‘ over 10. dπ’ dπ‘ is equal to 0.7. Then, weβre going to let π be equal to π‘ squared plus two π‘ over 10, which means that π£ is equal to sin of π. And Iβm 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 π£ with respect to π‘ is equal to the derivative of π£ with respect to πβthatβs cos of πβtimes the derivative of π with respect to π‘. And I took out a constant of a 10 to make life easier. And that gives us a 10th times two π‘ plus two. We replace π with π‘ squared plus two π‘ over 10. And we see that dπ£ by dπ‘ is equal to a 10th times two π‘ plus two times cos of π‘ squared plus two π‘ over 10. Using the product rule for π prime of π‘, we get π’ times dπ£ by dπ‘ plus π£ times dπ’ by dπ‘. Factoring 0.7 and simplifying a little and we see that π prime of π‘ is equal to 0.7 times sin of π‘ squared plus two π‘ over 10 plus π‘ times π‘ plus one over five times cos of π‘ squared plus two π‘ over 10.

Remember weβre looking to find π prime of five. So weβre going to substitute five every time we see π‘. 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, thatβs negative 4.179. And letβs look to explain the meaning of this within the context of the question.

π of π‘ tells us the rate at which customers buy the milk. And we know that π prime, the derivative of our function, evaluated at π‘ 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 itβs decreasing by 4.179 gallons per hour. Weβll now look at part 3.

Is the number of gallons increasing or decreasing at time π‘ equals eight hours? Give a reason for your answer.

Notice that when π‘ is equal to eight, weβre interested in two functions. Itβs the rate at which milk is being sold and the rate at which itβs being produced. The overall rate of change in the amount of milk in stock is therefore the difference between these two functions. Thatβs π of π‘ minus π of π‘. And at π‘ is equal to eight, thatβs π of eight minus π of eight. Thatβs 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. Thatβs 9.4429 and so on minus 10.5404. Thatβs negative 1.0975 which is negative 1.097 gallons per hour.

Notice weβve evaluated the rate at which the milk is increasing at eight hours minus the rate at which itβs 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 π‘ equals eight hours.

Part 4 says, βHow many gallons of milk are in the farm at time π‘ equals nine?β

Remember we said that to find the amount of milk, we integrate our functions. Since the cows in the farm started producing milk at π‘ equals six hours, the amount of milk is, therefore, equal to the integral evaluated between six and nine of π of π‘ with respect to π‘ minus the integral evaluated between zero nine of π of π‘ with respect to π‘. And we can evaluate this as shown. Typing the first integral into our calculator and remembering to replace π‘ with π₯, when we do, we get 27.756 and so on. And for our second integral, we get 50.007. Thatβs 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 π‘ equals nine.