Good Question 6: 2000 AB 4

2000 AB 4 Water tankAnother of my favorite questions from past AP exams is from 2000 question AB 4. If memory serves it is the first of what became known as an “In-out” question. An “In-out” question has two rates that are working in opposite ways, one filling a tank and the other draining it.

In subsequent years we saw a question with people entering and leaving an amusement park (2002 AB2/BC2), sand moving on and off a beach (2005 AB 2), another tank (2007 AB2), an oil leak being cleaned up (2008 AB 3), snow falling and being plowed (2010 AB 1), gravel being processed (2013 AB1/BC1), and most recently water again flowing in and out of a pipe (2015 AB1/BC1). The in-between years saw rates in one direction only but featured many of the same concepts.

The questions give rates and ask about how the quantity is changing. As such, they may be approached as differential equation initial value problems, but there is an easier way. This easier way is that a differential equation that gives the derivative as a function of a single variable, t, with an initial point \left( {{t}_{0}},y\left( {{t}_{0}} \right) \right) always has a solution of the form

y\left( t \right)=y\left( {{t}_{0}} \right)+\int_{{{t}_{0}}}^{t}{{y}'\left( x \right)dx}.

This is sometimes called the “accumulation equation.” The integral of a rate of change {y}'\left( t \right) gives the net amount of change over the interval of integration [{{t}_{0}},t]. When this is added to the initial amount the result is an expression that gives the amount at any time t.

In a motion context, this same idea is that the position at any time t, is the initial position plus the displacement:

\displaystyle s\left( t \right)=s\left( {{t}_{0}} \right)+\int_{{{t}_{0}}}^{t}{v\left( x \right)dx} where v\left( t \right)={s}'\left( t \right)

The scoring standard gave both forms of the solution. The ease of the accumulation form over the differential equation solution was evident and subsequent standards only showed this one.

2000 AB 4

The question concerned a tank that initially contained 30 gallons of water. We are told that water is being pumped into the tank at a constant rate of 8 gallons per minute and the water is leaking out at the rate of \sqrt{t+1} gallons per minute.

Part a asked students to compute the amount of water that leaked out in the first three minutes. There were two solutions given. The second solves the problem as an initial value differential equation:

Let L(t) be the amount that leaks out in t minutes then

\displaystyle \frac{dL}{dt}=\sqrt{t+1}

L\left( t \right)=\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+C

L\left( 0 \right)=\frac{2}{3}{{\left( 0+1 \right)}^{3/2}}+C=0 since nothing has leaked out yet, so C = -2/3

L\left( t \right)=\frac{2}{3}{{\left( t+1 \right)}^{3/2}}-\frac{2}{3}

L\left( 3 \right)=\frac{14}{3}

The first method, using the accumulation idea takes a single line:

\displaystyle L\left( 3 \right)=\int_{0}^{3}{\sqrt{t+1}dt}=\left. \frac{2}{3}{{\left( t+1 \right)}^{3/2}} \right|_{0}^{3}=\frac{2}{3}{{\left( 4 \right)}^{3/2}}-\frac{2}{3}{{\left( 1 \right)}^{3/2}}=\frac{14}{3}

I think you’ll agree this is easier and more direct.

Part b asked how much water was in the tank at t = 3 minutes.  We have 30 gallons to start plus 8(3) gallons pumped in and 14/3 gallons leaked out gives 30 + 24 – 14/3 = 148/3 gallons.

This part, worth only 1 point, was a sort of hint for the next part of the question.

Part c asked students to write an expression for the total number of gallons in the tank at time t.

Following part b the accumulation approach gives either

\displaystyle A\left( t \right)=30+8t-\int_{0}^{t}{\sqrt{x+1}dx}  or

\displaystyle A\left( t \right)=30+\int_{0}^{t}{\left( 8-\sqrt{x+1} \right)dx}.

The first form is not a simplification of the second, but rather the second form is treating the difference of the two rates, in minus out, as the rate to be integrated.

The differential equation approach is much longer and looks like this:

\displaystyle \frac{dA}{dt}=8-\sqrt{t+1}

A\left( t \right)=8t-\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+C

A\left( 0 \right)=30=8(0)-\frac{2}{3}{{\left( 0+1 \right)}^{3/2}}+C, so C=\frac{92}{3}

A\left( t \right)=8t-\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+\frac{92}{3}

Again, this is much longer. In recent years when asking student to write an expression such as this, the directions included a phrase such as “write an equation involving one or more integrals that gives ….” This pretty much leads students away from the longer differential equation initial value problem approach.

Part d required students to find the time when in the interval 0\le t\le 120 minutes the amount of water in the tank was a maximum and to justify their answer. The usual method is to find the derivative of the amount, A(t), set it equal to zero, and then solve for the time.

{A}'\left( t \right)=8-\sqrt{t+1}

Notice that this is the same regardless of which of the three forms of the expression for A(t) you start with. Thus, an excellent example of the Fundamental Theorem of Calculus used to find the derivative of a function defined by an integral. Or you could just start here without reference to the forms above: the overall rate in the rate in minus the rate out.

{A}'\left( t \right)=0 when t = 63

This is a maximum by the First Derivative Test since for 0 < t < 63 the derivative of A is positive and for 63 < t <120 the derivative of A is negative.

There is an additional idea on this part of the question in the Teaching Suggestions below.

I like this question because it is a nice real (as real as you can hope for on an exam) situation and for the way the students are led through the problem. I also like the way it can be used to compare the two methods of solution.  Then the way they both lead to the same derivative in part d is nice as well. I use this one a lot when working with teachers in workshops and summer institutes for these very reasons.

Teaching Suggestions

  • Certainly, have your students work through the problem using both methods. They need to learn how to solve an initial value problem (IVP) and this is good practice. Additionally, it may help them see how and when to use one method or the other.
  • Be sure the students understand why the three forms of A(t) in part c give the same derivative in part d. This makes an important connection with the Fundamental theorem of Calculus.
  • Like many good AP questions part d can be answered without reference to the other parts. The question starts with more water being pumped in than leaking out. This will continue until the rate at which the water leaks out overtakes the rate at which it is being pumped in. At that instant the rate “in” equals the rate “out” so you could start with 8=\sqrt{t+1}. After finding that t = 63, the answer may be justified by stating that before this time more water is being pumped in than is leaking out and after this time the rate at which water leaks out is greater than the rate at which it is pumped in, so the maximum must occur at t = 63.
  • And as always, consider the graph of the rates.

2000 AB 4

I used this question as the basis of a lesson in the current AP Calculus Curriculum Module entitled Integration, Problem Solving and Multiple Representations © 2013 by the College Board. The lesson gives a Socratic type approach to this question with a number of questions for each part intended to help the teacher not only work through this problem but to bring out related ideas and concepts that are not in the basic question. The module is currently available at AP sponsored workshops and AP Summer Institutes. Eventually, it will be posted at AP Central on the AB and BC Calculus Home Pages.

Local Linearity I

Certain graphs, specifically those that are differentiable, have a property called local linearity. This means that if you zoom in (using the same zoom factor in both directions) on a point on the graph, the graph eventually appears to be a straight line whose slope if the same as the slope (derivative) of the tangent line at that point.

Now we are a little ahead of ourselves here since we haven’t mentioned tangent lines and derivatives yet. But local linearity is the graphical manifestation of differentiability. Functions that are differentiable at a point are locally linear there and functions that are locally linear are differentiable. In the next post we will see how to use the local linear idea to introduce the derivative. For now, we will look at some graphs that may or may not be locally linear. Are the graphs “smooth” everywhere?

(1)\quad f\left( x \right)=1+\sqrt{{{x}^{2}}+0.001}

\displaystyle (2)\quad g\left( x \right)={{x}^{3}}+\frac{\sqrt[3]{{{\left( x-1 \right)}^{2}}}}{7}

The first function is locally linear at (0, 1) but doesn’t look it. Zoom-in several times and you will see that it is smooth and locally linear there eventually the graph looks like a horizontal line near (0, 1). This only looks like an absolute value graph because 1+\sqrt{{{x}^{2}}+0.001}\approx 1+\sqrt{{{x}^{2}}}=1+\left| x \right|.

The second function appears locally linear at (1, 1), but is not. Zoom in a few times at you will see very strange things going on. (Hint: Use a graphing program or a calculator and enter x^3+((x – 1)^2)^(1/3)/7 as simplifying to a power of 2/3 may confuse the calculator.)

The moral is that you can never be sure just looking at a graph whether it is locally linear or not; you’re never sure if you have zoomed in enough.

Nevertheless, the local linearity concept is helpful in introducing the derivative and, when you can be sure the function is not locally linear, knowing the derivative does not exist.

Looking ahead: Take a look at the AP exam from 2005 AB 5. Here you were given a velocity graph “modeled by the piecewise- linear function defined by the graph” copied below.

Student were asked to find v ‘(4) or explain why it does not exist. It does not exist because the graph is not locally linear there.

Later they were asked if the Mean Value Theorem guaranteed a value of c in the interval [8, 20] such that v ‘(c) is equal to the average rate of change over the interval. The answer is “no”, the value does not exist because the function is not differentiable on the interval because it is not locally linear everywhere in the interval.


When considering functions made up of the sums, differences, products or quotients of different sorts of functions (polynomials, exponentials and logarithms), or different powers of the same sort of function we  say that one function dominates the other. This means that as x approaches infinity or negative infinity, the graph will eventually look like the dominating function.

  • Exponentials dominate polynomials,
  • Polynomials dominate logarithms,
  • Among exponentials, larger bases dominate smaller,
  • Among polynomials, higher powers dominate lower,

For example, consider the function x{{e}^{x}}. The exponential function dominates the polynomial. As x\to \infty , the graph looks like an exponential approaching infinity; that is, \displaystyle \underset{x\to \infty }{\mathop{\lim }}\,x{{e}^{x}}=\infty . As  x\to -\infty the graph looks like an exponential with very small but still positive values; that is \displaystyle \underset{x\to -\,\infty }{\mathop{\lim }}\,x{{e}^{x}}=0.

Another example, consider a rational function (the quotient of two polynomials). If the numerator is of higher degree than the denominator, as x\to \pm \infty the numerator dominates, and the limit is infinite. If the denominator is of higher degree, the denominator dominates, and the limit is zero. (And if they are of the same degree, then the limit is the ratio of the leading coefficients. Dominance does not apply.)

Dominance works in other ways as well. Consider the graphs of y=3{{x}^{2}} and y={{2}^{x}}. In a standard graphing window, the graphs appear to intersect twice. But on the right side the exponential function is lower than the polynomial. Look farther out and farther up, the exponential dominates and will eventually lie above the polynomial   (after x = 7.334).

Here’s an example that pretty much has to be done using the dominance approach.

\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}}=0

The polynomial function in the denominator, even with the very small exponent, will dominate the logarithm function. The denominator will eventually get larger than the numerator and drive the quotient towards zero. We will return to this function when we know about finding maximums and points of inflection and find where it starts decreasing. For more on this see my post Far Out!

Why Limits?

There are four important things before calculus and in beginning calculus for which we need the concept of limit.

    1. The first is continuity. Most of the time in pre-calculus mathematics and in the calculus we deal with nice continuous functions or functions that are not continuous at just a few points. Limits give us the vocabulary and the mathematics necessary to describe and deal with discontinuities of functions. Historically, the modern (delta-epsilon) definition of limit comes out of Weierstrass’ definition of continuity.
    2. Asymptotes: A vertical asymptote is the graphical feature of function at a point where its limit equals positive or negative infinity. A horizontal asymptote is the (finite) limit of a function as x approaches positive or negative infinity.

Ideally, one would hope that students have seen these phenomena and have used the terms limit and continuity informally before they study calculus. This is where the study of calculus starts. The next two items are studied in calculus and are based heavily on limit.

3. The tangent line problem. The definition of the derivative as the limit of the slope of a secant line to a graph is the first of the two basic ideas of the calculus. This single idea is the basis for all the concepts and applications of differential calculus.

4. The area problem. Using limits it is possible to find the area of a region with a curved side, even if the curve is not something simple like a semi-circle. The definite integral is defined as the limit of a Riemann sum and gives the area regions with a curved side. This then can be extended to huge number of very practical applications many having nothing to do with area.

So these are the main ways that limits are used in beginning calculus. Students need a good visual  understanding, what the graph looks like,  of the first two situations listed above and how limits describe and define them. This is also necessary later when third and fourth come up.

The First & Second Day of School

Some suggestions for the first days of school:

There has to be a first day, so make the most of it. Take roll; make sure everyone is in the right place. Give out the textbooks. Explain about calculators and so on.

A word about reviewing at the beginning of the year: Don’t!

If you textbook’s first chapter is  a review of pre-calculus, then assign the class to read this chapter. Tell the class that the next day your will answer any questions they have and, after that, for the rest of the year they should refer back to this chapter when they need more information on these topics.

Start day 2 with your first lesson on limits.

Plan to review material from Kindergarten thru pre-calculus when the topics come up during the year – and they will come up. In some cases, plan for them to come up. For instance, the year usually begins with the study of limits. In connection with limits you will be looking at lots of graphs: this is a perfect time to review the graphs of the parent functions, a lot of the terminology related to graphs, discontinuities, asymptotes, and even the values of the trigonometric functions of the special angles.

Months from now you will teach about the derivatives of inverse functions.  That is when you review inverses. Review inverses now and you will have to do it later anyway.