AP Accumulation Questions

Accumulation 2: AP Exam Rate/Accumulation Questions

I assume that a number of my readers are AP Calculus teachers. In following up my last post on accumulation, today I’m going to discuss a very common type of AP Calculus exam question, the rate question which is loaded with accumulation ideas. The three posts following this one will show how accumulation can help with graphing problems.

If you do not have them, these two links will take you to them. Click on the question number for links to the question and scoring standard.

2000 AB 4: In this questions water was being pumped into a tank at the constant rate of 8 gallons per minute and leaking out at the rate of \sqrt{t+1} gallons per minute. At time t = 0 we are told there are 30 gallons of water in the tank. (Many AP exams problems have two rates acting at the same time, one increasing the amount and the other decreasing it.)

  • The first part of the question asked for the amount of water that leaked out of the tank in the first 3 minutes. This is like the situation discussed in the last post: If you need an amount, look around for a rate to integrate. The answer is \displaystyle\int_{0}^{3}{\sqrt{t+1}dt}=\tfrac{14}{3}\text{ gallons}\text{.}
  • The next part asked for the amount of water in the tank after t minutes. So we start with 30 gallons add the amount put in which is 8 gallons per minute for 3 minutes of 24 gallons. Of course we could also get this by integrating \displaystyle \int_{0}^{3}{8dt}=24. Then we subtract the amount that leaked out from the first part. The amount is 30 + 24 – 14/3 gallons. This was to help students with the next part.
  • The third part asked for an expression for A(t), the amount of water at any time t. So following on the second part we have either \displaystyle A\left( t \right)=30+8t-\int_{0}^{x}{\sqrt{t+1}dt} or \displaystyle A\left( t \right)=30+\int_{0}^{x}{8-\sqrt{t+1}\,dt}. Either form, especially the latter, is the form of an accumulation function: the initial amount plus the integral of the rate of change. It was not required to actually do the integration, but if someone did then \displaystyle A\left( t \right)=8t-\tfrac{2}{3}{{\left( t+1 \right)}^{3/2}}+\tfrac{92}{3}
  • The last part asked when the maximum amount of water was in the tank. As in any extreme value problem you can find this by differentiating any of the expressions for the amount found in the third part: {A}'\left( t \right)=8-\sqrt{t+1} by the FTC. (This could also be found by simply subtracting the two rates.) This will change from positive to negative when t = 63; this is when the maximum amount of water is in the tank. Notice that this is when the amount leaking out becomes greater than the amount being pumped into the tank; the total change becomes negative.

2009 AB 3: This question had a different twist or two on the accumulation idea. This proved rather difficult for the majority of the students (The mean score was 1.92 out of a possible 9 points.) The problem said that the Mighty Cable Company sold their cable for $120 per meter. The cost of producing the cable was given as 6\sqrt{x} dollars per meter. (Notice that these are rates as evidenced by their units $/m; the word “rate” was not used. It is important that students recognize when something is a rate.) The stem also defined profit as the difference between the amount of money received for the cable and the cost of producing the cable.

  • The first part of the question asked for the profit on 25 meters of cable. The amount the company receives is 25 meters times $120 dollars per meter (an amount so we could integrate the rate here, but that’s overkill). The amount the 25 meters costs to produce is (remember if you need an amount, integrate the  rate): \displaystyle \int_{0}^{25}{6\sqrt{x}\,dx}, so the \displaystyle \text{Profit }=120\cdot 25-\int_{0}^{25}{6\sqrt{x}\,dx}=\$2,500
  • The third part built on the first part by asking for the profit earned for a cable k meters long: \displaystyle \text{Profit =}120k-\int_{0}^{k}{6\sqrt{x}\,dx} or \displaystyle P(x)=\int_{0}^{k}{120-6\sqrt{x}\,dx}.  There is your accumulation function. The initial value is $0.
  • The second part was the most interesting. In it students were asked to explain the meaning of  \displaystyle \int_{25}^{30}{6\sqrt{x}\,dx} in the context of the problem. One way to see what this represent is to think about the FTC. The integral of the rate in dollars per meter is the cost per meter. If we call the cost C, then \displaystyle \int_{25}^{30}{6\sqrt{x}\,dx}=C\left( 30 \right)-C\left( 25 \right). Now students did not need to do a computation here; they just have to read what the symbols mean. C\left( 30 \right)-C\left( 25 \right) is the difference between the cost of manufacturing a 25 meter cable and a 30 meter table.  When you integrate a rate, you get the net amount.
  • As in the previous question the fourth part asked for the maximum profit. This was found by differentiating the profit expression from the third part by the FTC, {P}'\left( x \right)=120-6\sqrt{x}, and finding when the derivative changed from positive to negative, at x = 400 meters and substituting this into the profit equation.
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