Tag Archives: units

Why Linear Motion?

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Now that you know how to compute derivatives it is time to use them. The next few topics and my next few posts will discuss some of the applications of derivatives, and some of the things you can use them for.

The first is linear motion or motion along a straight line.

Derivatives give the rate of change of something that is changing. Linear motion problems concern the change in the position of something moving in a straight line. It may be someone riding a bike, driving a car, swimming, or walking or just a “particle” moving on a number line.

The function gives the position of whatever is moving as a function of time. This position is the distance from a known point often the origin. The time is the time the object is at the point. The units are distance units like feet, meters, or miles.

The derivative position is velocity, the rate of change of position with respect to time.  Velocity is a vector; it has both magnitude and direction. While the derivative appears to be just a number its sign counts: a positive velocity indicates movement to the right or up, and negative to the left or down. Units are things like miles per hour, meters per second, etc.

The absolute value of velocity is the speed which has the same units as velocity but no direction.

The second derivative of position is acceleration. This is the rate of change of velocity. Acceleration is also a vector whose sign indicates how the velocity is changing (increasing or decreasing). The units are feet per minute per minute or meters per second per second. Units are often given as meters per second squared (m/s2) which is correct but meters per second per second helps you understand that the velocity in meters per second is changing so much per second.   

Of these four, velocity may be the most useful. You will learn how to use the velocity (the first derivative) and its graph to determine how the particle moves over intervals of time: when it is moving left or right, when it stops, when it changes direction, whether it is speeding up or slowing down, how far the object moves, and so on. You can also find the position from the velocity if you also know the starting position.

You will work with equations and with graphs without equations. Reading the graphs of velocity and acceleration is an important skill to learn.

The reasoning used in linear motion problems is the same as in other applications. What you do is the same; what it means depends on the context.

Not to scare anyone, but linear motion problems appear as one of the six free-response questions on the AP Calculus exams almost every year as well as in several the multiple-choice questions.

So, let’s get moving!

Course and Exam Description Unit 4 Sections 4.1 and 4.2

Why the Derivative?

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You’re now ready to learn about the derivative – one of the two big tools of calculus.

If you graph a function on your calculator and Zoom-In several times at a point on it graph, your graph will eventually look like a straight line. Try it now; pick your favorite function and Zoom in where it looks most curvey. Very close up most functions look like lines. (There are exceptions.) The “slope” of this “line” is the derivative.

A little more precisely, the derivative is the slope of the line tangent to the graph at the point you compute it. It is found by considering a line that intersects the graph at two points and then moving the second point to the first using a limiting process. So, derivatives are always limits.

 The derivative is derived from the function itself. (That’s where the name comes from.)

The difference between the slope of a line and the slope of the curve is this: lines have a constant slope; curves have slopes that change as you move along the graph.

Since the derivative changes as you move along the graph, “derivative” also means the function that gives the derivative (slope) at each point. You will learn how to derive this function from the equation of the curve.

You will often need to write the equations of this tangent line. No, biggie: you have the point and the slope. That’s all you need.

(Hint: Forget slope-intercept! When writing the equation of a line is easiest way is to use the point-slope form, \displaystyle y=f\left( a \right)+{f}'\left( a \right)\left( {x-a} \right) where the point is \displaystyle \left( {a,f\left( a \right)} \right) and the slope is the derivative denoted by \displaystyle {f}'\left( a \right). You only need these three numbers – the two coordinates of a point and the derivative at that point). Drop them into the point-slope equation and you’re done.)

The tangent line is not your geometry teacher’s tangent line. Curves are not circles, so the tangent line may cross the curve at another point, or several other points. Sometimes the tangent line will even cross the curve at the point at the point of tangency!

You will begin by learning how to find the derivative using limits (all derivatives are limits). Then you will learn how to find the derivatives by bypassing the limiting process. That’s a good-news-bad-news thing. The good news is the formulas for finding derivatives make your work very easy and straightforward. The bad news is you’ll have to memorize the formulas. Sorry! Just giving you a heads-up.

Units: Like the slope of a line, the derivative is the instantaneous rate of change of the function at the point you calculate it. Since it is a rate of change, it has rate-of-change units: miles per hour, meters per second, furlongs per fortnight, figs per Newton.

Units are important:

Using the derivative, you will be able to find out useful things about a function. You can find exactly where the function is increasing and decreasing, exactly where it has its extreme values (its maximum and minimum), where it has “problems,” and other things. These in turn lead to practical considerations for the solution of problems in engineering, science, economics, finance, and any field that uses numbers.

Summary: Derivative has several meanings:

  • The slope of the tangent line to the curve, a/k/a “the slope of the curve.”
  • The function that gives the slope at any point.
  • The instantaneous rate of change of the of the dependent variable (y) with respect to the independent variable (x). Therefore, its units are the units of y divided by the units of x.

Derivatives are important and useful. So, let’s drive ahead.

Course and Exam Description Unit 2 topics 2.1 to 2.4

The Old Pump

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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. 

Adapting 2021 AB 6

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Six of nine. Continuing the current series of posts, this post looks at the AB Calculus 2021 exam question AB 6. Like most of the AP Exam questions, there is a lot more you can ask based on the stem of this question and a lot of other calculus you can discuss. This series of post offers suggestions as to how to adapt, expand, and use this question to help your students dig deeper and learn more.

2021 AB 6

This is a standard Differential Equation (Type 6) question and contains topics mainly from Unit 7 (Differential equations) and a little from Unit 3 (implicit differentiation) of the current Course and Exam Description. A differential equation with an initial condition is given in a context. The main part is the solution of the initial value question with three short other questions included.

The stem for 2021 AB 6 is:

Part (a): A slope field in the first quadrant with no scale on either axis is given. Students are asked to sketch the solution curve starting at the initial condition, the point (0, 0).  (I prefer this kind of slope field question to those where students are given a few points and asked to sketch the slope field through them. No one draws slope field by hand; slope field drawn by computers are used to study the approximate shape of the solution and determine its interesting properties as is done here and in part (b)). When drawing slope fields, the sketch should extend to from one border to another and contain the initial condition point.

Discussion and ideas for adapting this question:

  • Have student sketch solution through one or more different points. Copy the slope field and add the initial condition point somewhere else.
  • Add an initial point above the horizontal asymptote.
  • Compare and contrast the solutions drawn through several points.
  • Ask what the horizontal segments (at y = 12) tells you in the context of the problem.

Part (b):  Students are given the limit at infinity for the as yet unknow solution and asked to interpret it in the context of the problem including units of measure.

Discussion and ideas for adapting this question:

  • Discuss why this is so.
  • Discuss how to determine the units of the function from the given information.
  • Discuss how to determine the units of the derivative from the given information.
  • Discuss how to determine the units of the derivative from the units of the function.
  • Discuss how to determine the units of the function from the units of the derivative.
  • Discuss whether the interpretation of the limit makes sense in the context of the question.

Part (c): Students are asked to solve the initial value question using the method of separation of variables.

Discussion and ideas for adapting this question:

  • Since separation of variables is the only method for solving a differential equation that students are responsible for knowing, there is not much you can do to adapt or change this question.
  • The initial condition may be substituted immediately after the integration is done the “+ C” is attached, or it may be done later after the expression is solved for y. Show students both method and discuss which is more efficient and which makes more sense to them.
  • Removing the absolute value signs is another place that may confuse students. While some textbooks suggest using a “ ± “ sign and deciding sign which to use later, the better way is to decide as soon as possible. Ask yourself is the expression enclose by the absolute value signs positive or negative near (at) the initial value. If positive, then the absolute value is replaced by the same expression (as in this question); if negative, then the expression is replaced by its opposite. Then complete the question from there.

Part (d): This part needs careful reading. Students are asked, for a slightly different differential equation, if the rate of change in the amount of medicine is increasing or decreasing at a given time. Therefore, students must find the rate of change of the rate of change (the given derivative): the derivative of the derivative (i.e., the second derivative of the function). This requires implicit differentiation of the derivative using the quotient rule.

Discussion and ideas for adapting this question:

  • The second derivative has the first derivative as one of its factors. Students may (automatically) substitute the first derivative before simplifying or evaluating. This correct, but unnecessarily long. Show the students how to find and substitute the value of the first derivative along with the other numbers.
  • Do as little arithmetic as possible. You need only determine if the second derivative is positive or negative.
  • Discuss the meaning of the answer in the context of the problem.

Next week 2021 BC 2

I would be happy to hear your ideas for other ways to use this question. Please use the reply box below to share your ideas.

Good Question 18: 2018 BC 2(b)

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In this post we look at another part of the AP Calculus BC exam. Good Question 15 discussed the unusual units in 2018 BC 2(a). In this post we look at 2018 BC 2(b) where units help us find the correct integral to answer the question.

How do you answer a question of a type you’ve never seen before? I expect that’s what many of the students taking the 2018 AP Calculus exam were asking when they got to BC 5. If you’ve never done a density question how do you handle this one? 

The question concerns density. Density gives you how much of something exists in a certain length, area, or volume.  Density questions have appeared on the exam now and then, most recently 2008 AB 92 (which really isn’t recent, but then there are a lot of questions we never see). I have a blog post about the density here with several examples. In that post the alternate solution to example 3 explained how I used a unit analysis to find the answer; I used a similar approach here.

2018 BC 2 (b)

The stem of the question tells us that at a depth of h meters, 0 < h < 30, the number of plankton in a cubic meter of sea water is modeled by p\left( h \right)=0.2{{h}^{2}}{{e}^{{-0.0025{{h}^{2}}}}} million cells per cubic meter. Part (b) asks for the number of million of plankton in a column of water whose horizontal cross sections have a constant area of 3 square meters.

If the density were constant, then it is just a matter of multiplying the volume of the column times the constant density. Alas, the density is not constant; it varies with the depth. What to do?

Since an amount is asked for, you usually look around for a rate to integrate. Density is a kind of rate: the units are millions of cells per cubic meter. You need to integrate something concerning the density so that you end up with millions of cells; something that will “cancel” the cubic meters.

Consider a horizontal slice thru the column at depth h meters. While I’m not sure plankton is a good topping for pizza, you could picture this as a rather large pizza box whose sides are \sqrt{3} meters long and whose height is  \Delta h meters. This box has a volume of 3 \Delta h cubic meters. For small values of \Delta h the number of million plankton in the box is nearly constant, so at depth hi , there are p(hi) million plankton per cubic meter or {3p\left( {{{h}_{i}}} \right)\Delta h} million plankton in the box.

Notice how the units of the individual quantities combine to assure you the final quantity has the correct units:

\displaystyle (3\text{ square meters)}\cdot \left( {p\left( {{{h}_{i}}} \right)\text{ }\frac{{\text{million plankton}}}{{\text{cubic meters}}}} \right)\left( {\Delta h\text{ meters}} \right)=3p\left( {{{h}_{i}}} \right)\Delta h\text{ million plankton}

Now to find the amount in the column of water we can add up a stack of “pizza boxes.” The sum is \sum\limits_{{i=1}}^{n}{{3p\left( {{{h}_{i}}} \right)\Delta h}}. Now, if we take thinner boxes by letting \Delta h\to 0, we are looking at a Riemann sum. And calculus gives us the answer.

\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{3p\left( {{{h}_{i}}} \right)\Delta h}}=\int_{0}^{{30}}{{3p\left( h \right)dh}}\approx 1,675 million plankton in the column of water (rounded to the nearest million as directed in the question.)

Previous Good Questions can be found under the “Thru the Year” tab on the black navigation bar at the top of the page, or here.

Renumbered 3-14-24 Was Question 16.

Good Question 15: 2018 BC 2(a)

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My choices for the Good Question series are somewhat eclectic. Some are chosen because they are good, some because they are bad, some because I learned something from them, some because they can be extended, and some because they can illustrate some point of mathematics. This question and the next, Good Question 16, are in the latter group. They both concern units. They both are taken from this year’s AP calculus BC exam; both are suitable for AB classes. In this question 2018 BC 2(a) has some unusual units and in the next 2018 BC 2(b) the units help you figure out what to do. Part (c) concerns an improper integral and pard (d) is about parametric equation, neither of these are AB topics. 

2018 BC 2(a)

2018 BC 2 gave an equation that modeled the density p(h) of plankton in a sea in units of millions of cells per cubic meter, as a function of the depth, h, in meters.  Specifically, p\left( h \right)=0.2{{h}^{2}}{{e}^{{-0.0025{{h}^{2}}}}} for 0\le h\le 30. Part (a) asked for the value of {p}'\left( {25} \right) and also asked students “Using correct units, [to] interpret the meaning of {p}'\left( {25} \right) in the context of the problem.”

Plankton

This was a calculator active question, so the computation is easy enough: {p}'\left( {25} \right)=-1.17906

Now units of the derivative are always very easy to determine; this should be automatic. The derivative is the limit of a difference quotient, so its units are the units of the numerator divided by the units of the denominator. In this case that’s millions of cells per cubic meter per meter of depth.

While “millions of cells per meter to the fourth power” is technically correct and will probably earn credit, what is a meter to the fourth power?

It is similar to the better-known situation with velocity and acceleration. I never liked the idea of saying the acceleration is so many meters per square second. What’s a square second? Are there round seconds? Acceleration is the change in velocity in meters per second per second; that is, at a particular time the velocity is changing at the rate of so-many meters per second each second. 

Returning to the question, a cubic meter (volume) and a meter of depth (linear) are not things that you should combine. The notational convenience of writing meters to the fourth power hides the true meaning. So, a better interpretation is “At depth of  25 meters, the number cells is decreasing at the rate of 1.179 million cells per cubic meter per meter of depth.” or “The number of cells changing at the rate of -1.179 million cells per cubic meter per meter of depth.”

Had the model been given using volume units such as millions of cells per liter, then the units of the derivative would be millions of cells per liter per meter. That makes more sense.

But what does it mean?

Let’s look at the graph of the derivative. The window is 0 < h < 30 and –2.5 < p(x) < 2.5

It means, that as we pass thru that thin (thickness \Delta h\to 0) film of water 25 meters down, there are approximately 1.179 million cells per cubic meter less than in the thin film right above it and more than in the thin film right below it.

For reference, p\left( {25} \right)\approx 26.2014 million cells per cubic meter. Of course, that thin (thickness \Delta h\to 0) film of water has very little volume; it is kind of difficult to think of a cubic meter exactly 25 meters below the surface (maybe a cube extending from 24.5 meters to 25.5 meters?). As \Delta h\to 0 does a cubic meter approach a square meter?

The cubic meter above h = 25 has \displaystyle \int_{{24}}^{{25}}{{p\left( h \right)(1)dh}}=26.763 cells and the cubic meter below has 25.586 million cells. This is a decrease of 1.1767 million cells. So, the derivative is reasonable.

(To make the units of \displaystyle \int_{{24}}^{{25}}{{p\left( h \right)(1)dh}} correct, I included a factor of 1 square meter, this multiplied by p(h) million cells per cubic meter and by dh in meters give a result of millions of cells. More on why this is necessary in Good Question 16 on density.)

Previous Good Questions can be found under the “Thru the Year” tab on the black navigation bar at the top of the page, or here.

Units

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I had a question from a reader recently asking about how to determine the units for derivatives and integrals.

Derivatives: The units of the derivative are the units of dy divided by the units of dx, or the units of the dependent variable (f(x) or y) divided by the units of the independent variable (x). The reason for this comes from the definition of the derivative:

\displaystyle {f}'\left( x \right)=\underset{{\Delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+\Delta x} \right)-f\left( x \right)}}{{\Delta x}}

In the quotient the numerator has the units of f and in the denominator the has the same units as x.

Definite Integrals: The units of a definite integral are the units of the integrand f(x) multiplied by the units of dx. This comes directly from the definition of a definite integral:

\displaystyle \int_{a}^{b}{{f\left( x \right)dx}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{f\left( {{{x}_{i}}^{*}} \right)\frac{{b-a}}{n}}}

The factor (b – a) has the same units a x, the independent variable, and the f(x) has whatever units it has. From the Riemann sum we can see that since these factors are multiplied, that product is the units of the definite integral.

The integrand is the derivative of its antiderivative (by the FTC) and so its units are often derivative units (miles per hour, furlongs per fortnight, etc.). When multiplied by (b – a)/n its units “cancel” the units of the denominator of f(x) and the result is the units of the numerator of f(x).  This is not always the case*, therefore, multiplying the units is safest.

*The definite integral \displaystyle \int_{{-2}}^{2}{{\sqrt{{4-{{x}^{2}}}}dx}} gives the area of a semi-circle of radius 2 feet. The units of the radical are feet and represent the vertical distance from the x-axis to a point in the semi-circle; the dx is the horizontal side of the Riemann sum rectangles also in feet. Both are measured in the same linear units and the area is their product: feet times feet or square feet.