The Old Pump

Posted on November 5, 2021 by Lin McMullin

A tank is being filled with water using a pump that is old and slows down as it runs. The table below gives the rate at which the pump pumps at ten-minute intervals. If the tank initially has 570 gallons of water in it, approximately how much water is in the tank after 90 minutes?

Elapsed time (minutes)	0	10	20	30	40	50	60	70	80	90
Rate (gallons / minute)	42	40	38	35	35	32	28	20	19	10

And so, integration begins.

Ask your students to do this problem alone. When they are ready (after a few minutes) collect their opinions. They will not all be the same (we hope, because there is more than one reasonable way to approximate the amount). Ask exactly how they got their answers and what assumptions they made. Be sure they always include units (gallons). Here are some points to make in your discussion – points that we hope the kids will make and you can just “underline.”

1. Answers between 3140 and 3460 gallons are reasonable. Other answers in that range are acceptable. They will not use terms like “left-sum”, “right sum” and “trapezoidal rule” because they do not know them yet, but their explanations should amount to the same thing. An answer of 3300 gallons may be popular; it is the average of the other two, but students may not have gotten it by averaging 3140 and 3460.
2. Ask if they think their estimate is too large or too small and why they think that.
3. Ask what they need to know to give a better approximation – more and shorter time intervals.
4. Assumptions: If they added 570 + 42(10) + 40(10) + … +19(10) they are assuming that the pump ran at each rate for the full ten minutes and then suddenly dropped to the next. Others will assume the rate dropped immediately and ran at the slower rate for the 10 minutes. Some students will assume the rate dropped evenly over each 10-minute interval and use the average of the rates at the ends of each interval (570 + 41(10) + 39(10) + … 14.5(10) = 3300).
5. What is the 570 gallons in the problem for? Well, of course to foreshadow the idea of an initial condition. Hopefully, someone will forget to include it and you can point it out.
6. With luck someone will begin by graphing the data. If no one does, you should suggest it; (as always) to help them see what they are doing graphically. They are figuring the “areas” of rectangles whose height is the rate in gallons/minute and whose width is the time in minutes. Thus the “area” is not really an area but a volume (gal/min)(min) = gallons). In addition to unit analysis, graphing is important since you will soon be finding the area between the graph of a function and the x-axis in just this same manner.

Follow up: Flying to Integrationland

Be sure to check the “Thoughts on ‘The Old Pump'” in the comments section below.

Revised from a post of November 30, 2012.

Motion Problems: Same Thing, Different Context

Posted on October 8, 2021 by Lin McMullin

Calculus is about things that are changing. Certainly, things that move are changing, changing their position, velocity, and acceleration. Most calculus textbooks deal with things being dropped or thrown up into the air. This is called uniformly accelerated motion since the acceleration is due to gravity and is constant. While this is a good place to start, the problems are by their nature somewhat limited. Students often know all about uniformly accelerated motion from their physics class.

The Advanced Placement exams take motion problems to a new level. AB students often encounter particles moving along the x-axis or the y-axis (i.e. on a number line) according to a function that gives the particle’s position, velocity, or acceleration. BC students often encounter particles moving around the plane with their coordinates given by parametric equations or their velocity given by a vector. Other times the information is given as a graph or even in a table of the position or velocity. The “particle” may become a car, or a rocket or even chief readers riding bicycles.

While these situations may not be all that “real”, they provide excellent ways to ask both differentiation and integration questions. but be aware that they are not covered that much in some textbooks; supplementing the text may be necessary.

The main derivative ideas are that velocity is the first derivative of the position function, acceleration is the second derivative of the position function and the first derivative of the velocity. Speed is the absolute value of velocity. (There will be more about speed in the next post.) The same techniques used to find the features of a graph can be applied to motion problems to determine things about the moving particle.

So, the ideas are not new, but the vocabulary is. The table below gives the terms used with graph analysis and the corresponding terms used in motion problem.

Vocabulary: Working with motion equations (position, velocity, acceleration) you really do all the same things as with regular functions and their derivatives. Help students see that while the vocabulary is different, the concepts are the same.

Function Linear Motion
Value of a function at x position at time t
First derivative velocity
Second derivative acceleration
Increasing moving to the right or up
Decreasing moving to the left or down
Absolute Maximum farthest right
Absolute Minimum farthest left
yʹ = 0 “at rest”
yʹ changes sign object changes direction
Increasing & cc up speed is increasing
Increasing & cc down speed is decreasing
Decreasing & cc up speed is decreasing
Decreasing & cc down speed is increasing
Speed absolute value of velocity

Here is a short quiz on this idea.

Revised and updated from a post originally published on November 16, 2012

November 2015

Posted on October 31, 2015 by Lin McMullin

Here is the list of “November Topics,” that is what AP classes usually consider from mid-November into December. There has been a lot of discussion about inverses this month at the AP Calculus Community. While not the most read on this blog, the series on inverses may be helpful in considering all the ins and outs of inverses.

The four featured posts on the first page are the most popular from this month. Speed with 3563 hits this year and 6157 hits since it first appear is one of the most popular overall. “Open or Closed?” is another poplar post.

Thinking ahead into December, the first posts on integration are here and will continue into December. (As I’ve mentioned I try to post a few weeks ahead of where most people are now, so you have some time to read and plan.)

October 13, 2014 Extremes without Calculus

November 2, 2012 Open or Closed?

November 5, 2012 Inverses

November 7, 2012 Writing Inverses

November 9, 2012 The Range of the Inverse

November 12, 2012 The Calculus of Inverses

November 14, 2012 Inverses Graphically and Numerically

November 16, 2012 Motion Problems: Same Thing, Different Context.

November 19, 2012 Speed

April 17, 2013 The Ubiquitous Particle Motion Question

September 16, 2014 Matching Motion

November 21, 2012 Derivatives of Exponential Functions

November 26, 2012 Integration Itinerary

November 18, 2012 Antidifferentiation

November 30, 2012 The Old Pump

The Old Pump

Posted on November 30, 2012 by Lin McMullin

Elapsed time (minutes)	0	10	20	30	40	50	60	70	80	90
Rate (gallons / minute)	42	40	38	35	35	32	28	20	19	10

And so begins integration.

1. Answers between 3140 and 3460 gallons are reasonable. Other answers in that range are acceptable. They will not use terms like “left-sum”, “right sum” and “trapezoidal rule” because they do not know them yet, but their explanations should amount to the same thing. An answer of 3300 gallons may be popular; it is the average of the other two, but students may not have gotten it by averaging 3140 and 3460.
2. Ask if they think their estimate is too large or too small and why they think that.
3. Ask what they need to know to give a better approximation – more and shorter time intervals.
4. Assumptions: If they added 570 + 42(10) + 40(10) + … +19(10) they are assuming that the pump ran at each rate for the full ten minutes and then suddenly dropped to the next. Others will assume the rate dropped immediately and ran at the slower rate for the 10 minutes. Some students will assume the rate dropped evenly over each 10-minute interval and use the average of the rates at the ends of each interval (570 + 41(10) + 39(10) + … 14.5(10) = 3300).
5. What is the 570 gallons in the problem for? Well, of course to foreshadow the idea of an initial condition. Hopefully, someone will forget to include it and you can point it out.
6. With luck, someone will begin by graphing the data. If no one does, you should suggest it (as always) to help them see what they are doing graphically. They are figuring the “areas” of rectangles whose height is the rate in gallons/minute and whose width is the time in minutes. Thus the “area” is not really an area but a volume ((gal/min)(min) = gallons). In addition to unit analysis, graphing is important since you will soon be finding the area between the graph of a function and the x-axis in just this same manner.

Next post: Flying to Integrationland

Speed

Posted on November 19, 2012 by Lin McMullin

Speed is the absolute value of velocity: speed = $\left| v\left( t \right) \right|$ .

This is the definition of speed, but hardly enough to be sure students know about speed and its relationship to velocity and acceleration.

Velocity is a vector quantity; that is, it has both a direction and a magnitude. The magnitude of velocity vector is the speed. Speed is a non-negative number and has no direction associated with it. Velocity has a magnitude and a direction. Speed has the same value and units as velocity; speed is a number.

The question that seems to trouble students the most is to determine whether the speed is increasing or decreasing. The short answer is

Speed is increasing when the velocity and acceleration have the same sign.

Speed is decreasing when the velocity and acceleration have different signs.

You should demonstrate this in some real context, such as driving a car (see below). Also, you can explain it graphically.

The figure below shows the graph of the velocity $v\left( t \right)$ (blue graph) of a particle moving on the interval $0\le t\le f$ . The red graph is $\left| v\left( t \right) \right|$ , the speed. The sections where $v\left( t \right)<0$ are reflected over the x-axis. (The graphs overlap on [b, d].) It is now quite east to see that the speed is increasing on the intervals [0,a], [b, c] and [d,e].

Another way of approaching the concept is this: the speed is the non-directed length of the vertical segment from the velocity’s graph to the t-axis. Picture the segment shown moving across the graph. When it is getting longer (either above or below the t-axis) the speed increases.

Thinking of the speed as the non-directed distance from the velocity to the axis makes answering the two questions below easy:

1. What are the values of t at which the speed obtains its (local) maximum values? Answer: x = a, c, and e.
2. When do the minimum speeds occur? What are they? Answer: the speed is zero at b and d

Students often benefit from a verbal explanation of all this. Picture a car moving along a road going forwards (in the positive direction) its velocity is positive.

If you step on the gas, acceleration pulls you in the direction you are moving and your speed increases. (v > 0, a > 0, speed increases)
Going too fast is not good, so you put on your brakes, you now accelerate in the opposite direction (decelerate?), but you are still moving forward, but slower. (v > 0, a < 0, speed decreases)
Finally, you stop. Then you shift into reverse and start moving backwards (negative velocity) and you push on the gas to accelerate in the negative direction, so your speed increases. (v < 0, a < 0, speed increases)
Then you put on the breaks (accelerate in the positive direction) and your speed decreases again. (v < 0, a > 0, speed decreases)

Here is an activity that will help your students discover this relationship. Give Part 1 to half the class and Part 2 to the other half. Part 3 (on the back of Part 1 and Part 2) is the same for both groups. – Added 12-19-17

Also see: A Note on Speed for the purely analytic approach.

Update: “A Note on Speed” added 4-21-2018

Motion Problems: Same Thing, Different Context

Posted on November 16, 2012 by Lin McMullin

Calculus is about things that are changing. Certainly, things that move are changing, changing their position, velocity and acceleration. Most calculus textbooks deal with things being dropped or thrown up into the air. This is called uniformly accelerated motion since the acceleration is due to gravity and is constant. While this is a good place to start, the problems are by their nature, somewhat limited. Students often know all about uniformly accelerated motion from their physics class.

The Advanced Placement exams take motion problems to a new level. AB students often encounter particles moving along the x-axis or the y-axis (i.e. on a number line) according to some function that gives the particle’s position, velocity or acceleration. BC students often encounter particles moving around the plane with their coordinates given by parametric equations or its velocity given by a vector. Other times the information is given as a graph or even in a table of the position or velocity. The “particle” may become a car, or a rocket or even chief readers riding bicycles.

So the ideas are not new, but the vocabulary is. The table below gives the terms used with graph analysis and the corresponding terms used in motion problem.

Function Linear Motion

Value of a function at x position at time t

First derivative velocity

Second derivative acceleration

Increasing moving to the right or up

Decreasing moving to the left or down

Absolute Maximum farthest right

Absolute Minimum farthest left

yʹ = 0 “at rest”

yʹ changes sign object changes direction

Increasing & cc up speed is increasing

Increasing & cc down speed is decreasing

Decreasing & cc up speed is decreasing

Decreasing & cc down speed is increasing

Speed absolute value of velocity

Open or Closed?

Posted on November 2, 2012 by Lin McMullin

About this time of year you find someone, hopefully one of your students, asking, “If I’m finding where a function is increasing, is the interval open or closed?”

Do you have an answer?

This is a good time to teach some things about definitions and theorems.

The place to start is to ask what it means for a function to be increasing. Here is the definition:

A function is increasing on an interval if, and only if, for all (any, every) pairs of numbers x₁ < x₂ in the interval, f(x₁) < f(x₂).

(For decreasing on an interval, the second inequality changes to f(x₁) > f(x₂). All of what follows applies to decreasing with obvious changes in the wording.)

Notice that functions increase or decrease on intervals, not at individual points. We will come back to this in a minute.
Numerically, this means that for every possible pair of points, the one with the larger x-value always produces a larger function value.
Graphically, this means that as you move to the right along the graph, the graph is going up.
Analytically, this means that we can prove the inequality in the definition.

For an example of this last point consider the function f(x) = x². Let x₂= x₁ + h where h > 0. Then in order for f(x₁) < f(x₂) it must be true that

${{x}_{1}}^{2}<{{\left( {{x}_{1}}+h \right)}^{2}}$
$0<{{\left( {{x}_{1}}+h \right)}^{2}}-{{x}_{1}}^{2}$
$0<{{x}_{1}}^{2}+2h{{x}_{1}}+{{h}^{2}}-{{x}_{1}}^{2}$
$0<h\left( 2{{x}_{1}}+h \right)$

This can only be true if ${{x}_{1}}\ge 0$ , Thus, x² is increasing only if $x\ge 0$ .

Now, of course, we rarely, if ever, go to all that trouble. And it is even more trouble for a function that increases on several intervals. The usual way of finding where a function is increasing is to look at its derivative.

Notice that the expression ${{\left( {{x}_{1}}+h \right)}^{2}}-{{x}_{1}}^{2}$ looks a lot like the numerator of the original limit definition of the derivative of x² at x = x₁, namely $\displaystyle {f}'\left( {{x}_{1}} \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( {{x}_{1}}+h \right)}^{2}}-{{x}_{1}}^{2}}{h}$ . If h > 0, where the function is increasing the numerator is positive and the derivative is positive also. Turning this around we have a theorem that says, If ${f}'\left( x \right)>0$ for all x in an interval, then the function is increasing on the interval. That makes it much easier to find where a function is increasing: we simplify find where its derivative is positive.

There is only a slight problem in that the theorem does not say what happens if the derivative is zero somewhere on the interval. If that is the case, we must go back to the definition of increasing on an interval or use some other method. For example, the function x³ is increasing everywhere, even though its derivative at the origin is zero.

Let’s consider another example. The function sin(x) is increasing on the interval $\left[ -\tfrac{\pi }{2},\tfrac{\pi }{2} \right]$ (among others) and decreasing on $\left[ \tfrac{\pi }{2},\tfrac{3\pi }{2} \right]$ . It bothers some that $\tfrac{\pi }{2}$ is in both intervals and that the derivative of the function is zero at x = $\tfrac{\pi }{2}$ . This is not a problem. Sin( $\tfrac{\pi }{2}$ ) is larger than all the other values is both intervals, so by the definition, and not the theorem, the intervals are correct.

It is generally true that if a function is continuous on the closed interval [a,b] and increasing on the open interval (a,b) then it must be increasing on the closed interval [a,b] as well. (There is a proof by Lou Talman of this fact click here .)

Returning to the first point above: functions increase or decrease on intervals not at points. You do find questions in books and on tests that ask, “Is the function increasing at x = a.” The best answer is to humor them and answer depending on the value of the derivative at that point. Since the derivative is a limit as h approaches zero, the function must be defined on some interval around x = a in which h is approaching zero. So answer according to the value of the derivative on that interval.

You can find more on this here.

Case Closed.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Tag Archives: Nov

The Old Pump

Motion Problems: Same Thing, Different Context

November 2015

The Old Pump

Speed

Motion Problems: Same Thing, Different Context

Open or Closed?

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