Category Archives: Derivative Applications

Linear Motion (Type 2)

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AP  Questions Type 2: Linear Motion

We continue the discussion of the various type questions on the AP Calculus Exams with linear motion questions.

“A particle (or car, person, or bicycle) moves on a number line ….”

These questions may give the position equation, the velocity equation (most often), or the acceleration equation of something that is moving on the x– or y-axis as a function of time, along with an initial condition. The questions ask for information about motion of the particle: its direction, when it changes direction, its maximum position in one direction (farthest left or right), its speed, etc.

The particle may be a “particle,” a person, car, a rocket, etc.  Particles don’t really move in this way, so the equation or graph should be considered to be a model. The question is a versatile way to test a variety of calculus concepts since the position, velocity, or acceleration may be given as an equation, a graph, or a table; be sure to use examples of all three forms during the review.

Many of the concepts related to motion problems are the same as those related to function and graph analysis (Type 3). Stress the similarities and show students how the same concepts go by different names. For example, finding when a particle is “farthest right” is the same as finding the when a function reaches its “absolute maximum value.” See my post for Motion Problems: Same Thing, Different Context for a list of these corresponding terms. There is usually one free-response question and three or more multiple-choice questions on this topic.

The position, s(t), is a function of time. The relationships are:

  • The velocity is the derivative of the position, {s}'\left( t \right)=v\left( t \right). Velocity is has direction (indicated by its sign) and magnitude. Technically, velocity is a vector; the term “vector” will not appear on the AB exam.
  • Speed is the absolute value of velocity; it is a number, not a vector. See my post for Speed.
  • Acceleration is the derivative of velocity and the second derivative of position, \displaystyle a\left( t \right)={v}'\left( t \right)={{s}''}\left( t \right). It, too, has direction and magnitude and is a vector.
  • Velocity is the antiderivative of the acceleration.
  • Position is the antiderivative of velocity.

What students should be able to do:

  • Understand and use the relationships above.
  • Distinguish between position at some time and the total distance traveled during the time period.
  • The total distance traveled is the definite integral of the speed (absolute value of velocity) \displaystyle \int_{a}^{b}{\left| v\left( t \right) \right|}\,dt.
  •  Be sure your students understand the term displacement; it is the net distance traveled or distance between the initial position and the final position. Displacement, is the definite integral of the velocity (rate of change): \displaystyle \int_{a}^{b}{v\left( t \right)}\,dt.
  • The final position is the initial position plus the displacement (definite integral of the rate of change from x= a to x = t): \displaystyle s\left( t \right)=s\left( a \right)+\int_{a}^{t}{v\left( x \right)}\,dx Notice that this is an accumulation function equation (Type 1).
  • Initial value differential equation problems: given the velocity or acceleration with initial condition(s) find the position or velocity. These are easily handled with the accumulation equation in the bullet above, but may also be handled as an initial value problem.
  • Find the speed at a given time. The speed is the absolute value of the velocity.
  • Find average speed, velocity, or acceleration
  • Determine if the speed is increasing or decreasing.
    • If at some time, the velocity and acceleration have the same sign then the speed is increasing.If they have different signs the speed is decreasing.
    • If the velocity graph is moving away from (towards) the t-axis the speed is increasing (decreasing). See the post on Speed.
    • There is also a worksheet on speed here
    • The analytic approach to speed: A Note on Speed
  • Use a difference quotient to approximate the derivative (velocity or acceleration) from a table. Be sure the work shows a quotient.
  • Riemann sum approximations.
  • Units of measure.
  • Interpret meaning of a derivative or a definite integral in context of the problem

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

This may be an AB or BC question. The BC topic of motion in a plane, (Type 8: parametric equations and vectors) will be discussed in a later post.

The Linear Motion problem may cover topics primarily from primarily from Unit 4, and also from Unit 3, Unit 5, Unit 6, and Unit 8 (for BC) of the 2019 CED. 

Free-response examples:

  • Equation stem 2017 AB 5,
  • Graph stem: 2009 AB1/BC1,
  • Table stem 2019 AB2

Multiple-choice examples from non-secure exams:

  • 2012 AB 6, 16, 28, 79, 83, 89
  • 2012 BC 2, 89

 

 

 

 

Rate & Accumulation (Type 1)

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The Free-response Questions

There are ten general categories of AP Calculus free-response questions.

NOTE: The type number I’ve assigned to each type DO NOT correspond to the 2019 CED Unit numbers. Many AP Exam questions have parts from different Units. The CED Unit numbers will be referenced in each post.

AP  Questions Type 1: Rate and Accumulation

These questions are often in context with a lot of words describing a situation in which some things are changing. There are usually two rates acting in opposite ways (sometimes called an in-out question). Students are asked about the change that the rates produce over some time interval either separately or together.

The rates are often fairly complicated functions. If they are on the calculator allowed section, students should store the functions in the equation editor of their calculator and use their calculator to do any graphing,  integration, or differentiation that may be necessary.

The main idea is that over the time interval [a, b] the integral of a rate of change is the net amount of change

\displaystyle \int_{a}^{b}{{f}'\left( t \right)dt}=f\left( b \right)-f\left( a \right)

If the question asks for an amount, look around for a rate to integrate.

The final (accumulated) amount is the initial amount plus the accumulated change:

\displaystyle f\left( x \right)=f\left( {{x}_{0}} \right)+\int_{{{x}_{0}}}^{x}{{f}'\left( t \right)}\,dt,

where {{x}_{0}} is the initial time, and  f\left( {{x}_{0}} \right) is the initial amount. Since this is one of the main interpretations of the definite integral the concept may come up in a variety of situations.

What students should be able to do:

  • Be ready to read and apply; often these problems contain a lot of writing which needs to be carefully read.
  • Recognize that rate = derivative.
  • Recognize a rate from the units given without the words “rate” or “derivative.”
  • Find the change in an amount by integrating the rate. The integral of a rate of change gives the amount of change (FTC):

\displaystyle \int_{a}^{b}{{f}'\left( t \right)dt}=f\left( b \right)-f\left( a \right).

  • Find the final amount by adding the initial amount to the amount found by integrating the rate. If x={{x}_{0}} is the initial time, and f\left( {{x}_{0}} \right)  is the initial amount, then final accumulated amount is

\displaystyle f\left( x \right)=f\left( {{x}_{0}} \right)+\int_{{{x}_{0}}}^{x}{{f}'\left( t \right)}\,dt,

  • Write an integral expression that gives the amount at a general time. BE CAREFUL, the dt must be included at the correct place. Think of the integral sign and the dt as parentheses around the integrand.
  • Find the average value of a function
  • Understand the question. It is often not necessary to as much computation as it seems at first.
  • Use FTC to differentiate a function defined by an integral.
  • Explain the meaning of a derivative or its value in terms of the context of the problem. The explanation should contain (1) what it represents, (2) its units, and (3) how numerical argument applies in context.
  • Explain the meaning of a definite integral or its value in terms of the context of the problem. The explanation should contain (1) what it represents, (2) its units, and (3) how the limits of integration apply in context.
  • Store functions in their calculator recall them to do computations on their calculator.
  • If the rates are given in a table, be ready to approximate an integral using a Riemann sum or by trapezoids.
  • Do a max/min or increasing/decreasing analysis.

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

The Rate – Accumulation question may cover topics primarily from Unit 4, Unit 5, Unit 6 and Unit 8 of the 2019 CED.

Typical free-response examples:

Typical multiple-choice examples from non-secure exams:

  • 2012 AB 8, 81, 89
  • 2012 BC 8 (same as AB 8)

 

 

 

 

 

Updated January 31, 2019, March 12, 2021

Good Question 17

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A common question in (older?) textbooks is to give students a function or relation and have them graph it without technology (because in the old days technology was not available). Students had to find all the appropriate information without hints or further direction: they were supposed to know what to do and do it.

AP exam questions, for legitimate and understandable reasons, do not ask for as complete an analysis. Rather, they ask students to find specific information about the function or its graph (extreme values, points of inflection, etc.). For the same legitimate and understandable reason many of the features are not considered; the other concepts are tested on different questions.

I think that the full-analysis-from-scratch approach, while not appropriate for an AP exam question, has its merits. The big difference is that students need to be creative in their investigation; not just focus on a few items pointed to by the question.

With that in mind, let’s consider the following question asked three ways.

A very general form

Consider the relation {{x}^{3}}+3{{x}^{2}}+{{y}^{2}}=4. Discuss the graph of the relation giving reasons for your conclusions. Include a brief mention of any unproductive paths you followed and what you learned from them.

A more directed investigation

Consider the relation {{x}^{3}}+3{{x}^{2}}+{{y}^{2}}=4. Find its extreme values and any asymptotes, and vertical tangents (if any). Discuss in detail the appearance of the graph near x = –2 and discuss the slopes in that area. Explain how you arrived at your results.

Closer to an AP style form. The function here is the top half of the function above.

Consider the function y=\sqrt{{4-{{x}^{3}}-3{{x}^{2}}}}.

(a) Find the local maximum and minimum values of the function. Justify your answer.

(b) Where does the function had a vertical asymptote? Justify your answer.

(c) Find \underset{{x\to -2-}}{\mathop{{\lim }}}\,\frac{{dy}}{{dx}} and \underset{{x\to -2+}}{\mathop{{\lim }}}\,\frac{{dy}}{{dx}}. What does this say about the graph?

The real question is can you ask it the first way?

Here is my solution to the first form; this will also give the answers to the other forms. After the solution, I’ll discuss some ideas on how to score such a solution.

Solution

The graph of the relation  {{x}^{3}}+3{{x}^{2}}+{{y}^{2}}=4.  is shown in figure 1.

Figure 1

The domain of the function is x\le 1. Values greater than 1 will make the left side of the equation greater than 4 regardless of the value of y. The range of the relation is all real numbers.

The function is symmetric to the x-axis since substituting (–y) for y will give the same expression. The equation of the top half is  y=\sqrt{{4-{{x}^{3}}-3{{x}^{2}}}} and the lower half is y=-\sqrt{{4-{{x}^{3}}-3{{x}^{2}}}}

The derivative for the top half is \displaystyle {y}'=-\frac{{3{{x}^{2}}+6x}}{{2y}}=-\frac{{3{{x}^{2}}+6x}}{{2\sqrt{{4-{{x}^{3}}-{3{x}^{2}}}}}} by implicit differentiation or by differentiating the equation for the top half.

y'\left( x \right) does not exist at x = 1 specifically, \displaystyle \underset{{x\to 1-}}{\mathop{{\lim }}}\,{y}'\left( x \right)=-\infty . Therefore, the line x = 1 is a vertical tangent. By symmetry for the lower part of the graph \displaystyle \underset{{x\to 1-}}{\mathop{{\lim }}}\,{y}'\left( x \right)=+\infty .  And x = 1 is its vertical asymptote as well. The slope of the original relation changes from positive to negative at x = 1 by going not through zero but from  -\infty to +\infty .

{y}'\left( x \right)=0 when x = 0 and when x = –2. At (0, 2) the derivative changes from positive to negative; this is a local maximum point by the first derivative test. The lower half has a local minimum point at (0, –2) by symmetry.

At (–2, 0) the derivative is an indeterminate form of the type 0/0.

False step: At first, I thought this meant that the two sides were tangent to the line x = –2 making the point (–2, 0) on the top half a cusp. I tried to see if this was true by graphing in a very narrow window. This did not show anything: the graph looked on zooming in, like an absolute value graph. It turned out that an absolute value was involved.

I then made a table of values for the derivative near the point and found that the values appeared to be approaching a number near –1.7 from the left and near +1.7 from the right. I started thinking maybe \displaystyle \sqrt{3}.

Then I changed the constants to parameters and played with the sliders on Desmos (here). With a slight change in the constants the graph appeared to have a nice rounded local minimum near x= –2. Other values showed two separate pieces with vertical tangents near x= –2. This confused me even more.

Next, I did what I should have done earlier: I graphed the top half and its derivative (figure 2):

Figure 2. The top half in blue and its derivative in black.

The derivative has a finite jump discontinuity at x = –2. I remembered seeing this kind of thing before and thought it involved an absolute value of some kind. Still confused, I decided to investigate the derivative further (which I also should have done sooner).

The derivative at x = –2 is an indeterminate form of the 0/0 type. This means that by substituting you get an expression that really doesn’t help; you may still evaluate the limit (remember derivatives are limits) by other methods. Factoring the derivative (using synthetic division on the radicand) gives

\displaystyle {y}'\left( x \right)=-\frac{{3x\left( {x+2} \right)}}{{2\sqrt{{{{{\left( {x+2} \right)}}^{2}}\left( {1-x} \right)}}}}=-\frac{{3x\left( {x+2} \right)}}{{2\sqrt{{{{{\left( {x+2} \right)}}^{2}}}}\sqrt{{1-x}}}}=-\frac{{3x}}{{\sqrt{{1-x}}}}\frac{{x+2}}{{\left| {x+2} \right|}}

And now we see that

\displaystyle \underset{{x\to -2-}}{\mathop{{\lim }}}\,\frac{{-3x}}{{2\sqrt{{1-x}}}}\cdot \frac{{x+2}}{{\left| {x+2} \right|}}=-\sqrt{3}  and  \displaystyle \underset{{x\to -2+}}{\mathop{{\lim }}}\,\frac{{-3x}}{{2\sqrt{{1-x}}}}\cdot \frac{{x+2}}{{\left| {x+2} \right|}}=\sqrt{3}

This agrees with the derivative’s graph. The top graph (and bottom’s by symmetry) comes to an arrow-like point (sometimes called a node) at (–2, 0). The slope on the left approaches -\sqrt{3} and on the right approaches +\sqrt{3}.

______________________

I admit I’ve had a few more years of experience with this sort of thing that a first-year calculus student. I expect this would be difficult for most students who have never seen a question like this, so working up to it with simpler questions is how to start.

In grading this sort of question, you need to consider:

  1. If the student found and justified all the pertinent features of the graph. If they missed something, what they included may still be pretty good. For example, I would not expect a student to the use the word “node,” so omitting it should not be held against them and using it may call for praise.
  2. The things they (and I) may have overlooked or things they considered that were unnecessary or even wrong.
  3. Their overall approach. This last may be the most important part. Including their missteps and unproductive paths is important and should, I think, receive credit. After all, they did not know the path would be unproductive until they followed it a way. Hopefully, they made some missteps and learned something from them. That’s what investigating something in math entails.

I would be happy to hear if you tried this or something similar and so would the other readers of this blog. Please use the “Comment” button below to share your thoughts.

2019 CED Unit 5 Analytical Applications of Differentiation

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Unit 5 covers the application of derivatives to the analysis of functions and graphs. Reasoning and justification of results are also important themes in this unit. (CED – 2019 p. 92 – 107). These topics account for about 15 – 18% of questions on the AB exam and 8 – 11% of the BC questions.

Reasoning and writing justification of results are mentioned and stressed in the introduction to the topic (p. 93) and for most of the individual topics. See Learning Objective FUN-A.4 “Justify conclusions about the behavior of a function based on the behavior of its derivatives,” and likewise in FUN-1.C for the Extreme value theorem, and FUN-4.E for implicitly defined functions. Be sure to include writing justifications as you go through this topic. Use past free-response questions as exercises and also as guide as to what constitutes a good justification. Links in the margins of the CED are also helpful and give hints on writing justifications and what is required to earn credit. See the presentation Writing on the AP Calculus Exams and its handout

Topics 5.1

Topic 5.1 Using the Mean Value Theorem While not specifically named in the CED, Rolle’s Theorem is a lemma for the Mean Value Theorem (MVT). The MVT states that for a function that is continuous on the closed interval and differentiable over the corresponding open interval, there is at least one place in the open interval where the average rate of change equals the instantaneous rate of change (derivative). This is a very important existence theorem that is used to prove other important ideas in calculus. Students often confuse the average rate of change, the mean value, and the average value of a function – See What’s a Mean Old Average Anyway?

Topics 5.2 – 5.9

Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals

Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.

Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function

Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval

Topic 5.6 Determining Concavity of Functions on Their Domains FUN-4.A.4 defines (at least for AP Calculus) When a function is concave up and down based on the behavior of the first derivative. (Some textbooks may use different equivalent definitions.) Points of inflection are also included under this topic.

Topic 5.7 Using the Second Derivative Test to Determine Extrema Using the Second Derivative Test to determine if a critical point is a maximum or minimum point. If a continuous function has only one critical point on an interval, then it is the absolute (global) maximum or minimum for the function on that interval.

Topic 5.8 Sketching Graphs of Functions and Their Derivatives. First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topic 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative. First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topics 5.10 – 5.11

Optimization is important application of derivatives. Optimization problems as presented in most text books, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not be asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.

Topic 5.10 Introduction to Optimization Problems 

Topic 5.11 Solving Optimization Problems

Topics 5.12

Topic 5.12 Exploring Behaviors of Implicit Relations Critical points of implicitly defined relations can be found using the technique of implicit differentiation. This is an AB and BC topic. For BC students the techniques are applied later to parametric and vector functions.

Timing

Topic 5.1 is important and may take more than one day. Topics 5.2 – 5.9 flow together and for graphing they are used together; after presenting topics 5.2 – 5.7 spend the time in topics 5.8 and 5.9 spiraling and connecting the previous topics. Topics 5.10 and 5.11 – see note above and spend minimum time here. Topic 5.12 may take 2 days.

The suggested time for Unit 5 is 15 – 16 classes for AB and 10 – 11 for BC of 40 – 50-minute class periods, this includes time for testing etc.

Finally, were I still teaching, I would teach this unit before Unit 4. The linear motion topic (in Unit 4) are a special case of the graphing ideas in Unit 5, so it seems reasonable to teach this unit first. See Motion Problems: Same thing, Different Context

Previous posts on these topics include:

Then There Is This – Existence Theorems

What’s a Mean Old Average Anyway

Did He, or Didn’t He?   History: how to find extreme values without calculus

Mean Value Theorem

Foreshadowing the MVT

Fermat’s Penultimate Theorem

Rolle’s theorem

The Mean Value Theorem I

The Mean Value Theorem II

Graphing

Concepts Related to Graphs

The Shapes of a Graph

Joining the Pieces of a Graph

Extreme Values

Extremes without Calculus

Concavity

Reading the Derivative’s Graph

Real “Real-life” Graph Reading

Far Out! An exploration

Open or Closed  Should intervals of increasing, decreasing, or concavity be open or closed?

Others

Lin McMullin’s Theorem and More Gold  The Golden Ratio in polynomials

Soda Cans  Optimization video

Optimization – Reflections   

Curves with Extrema?

Good Question 10 – The Cone Problem

Implicit Differentiation of Parametric Equations    BC Topic

Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

2019 CED – Unit 1: Limits and Continuity

2019 CED – Unit 2: Differentiation: Definition and Fundamental Properties.

2019 CED – Unit 3: Differentiation: Composite , Implicit, and Inverse Functions

2019 CED – Unit 4 Contextual Applications of the Derivative  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 5 Analytical Applications of Differentiation  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations  Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration   Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions 

2019 CED Unit 10 Infinite Sequences and Series

2019 CED Unit 4: Contextual Applications of the Derivative

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Unit 4 covers rates of change in motion problems and other contexts, related rate problems, linear approximation and L’Hospital’s Rule. (CED – 2019 p. 82 – 90). These topics account for about 10 – 15% of questions on the AB exam and 6 – 9% of the BC questions.

Topics 4.1 – 4.6

Topic 4.1 Interpreting the Meaning of the Derivative in Context Students learn the meaning of the derivative in situations involving rates of change.

Topic 4.2 Linear Motion The connections between position, velocity, speed, and acceleration. This topic may work  better after the graphing problems in Unit 5, since many of the ideas are the same. See Motion Problems: Same Thing, Different Context

Topic 4.3 Rates of Change in Contexts Other Than Motion Other applications

Topic 4.4 Introduction to Related Rates Using the Chain Rule

Topic 4.5 Solving Related Rate Problems

Topic 4.6 Approximating Values of a Function Using Local Linearity and Linearization The tangent line approximation

Topic 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms. Indeterminate Forms of the type \displaystyle \tfrac{0}{0} and \displaystyle \tfrac{\infty }{\infty }. (Other forms may be included, but only these two are tested on the AP exams.)

Topic 4.1 and 4.3 are included in the other topics, topic 4.2 may take a few days, Topics 4.4 – 4.5 are challenging for many students and may take 4 – 5 classes, 4.6 and 4.7 two classes each. The suggested time is 10 -11 classes for AB and 6 -7 for BC. of 40 – 50-minute class periods, this includes time for testing etc.

Posts on these topics include:

Motion Problems 

Motion Problems: Same Thing, Different Context

Speed

A Note on Speed

Related Rates

Related Rate Problems I

Related Rate Problems II

Good Question 9 – Related rates

Linear Approximation

Local Linearity 1

Local Linearity 2 

L’Hospital’s Rule

Locally Linear L’Hôpital  

L’Hôpital Rules the Graph  

Determining the Indeterminate

Determining the Indeterminate 2

Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

2019 CED – Unit 1: Limits and Continuity

2019 CED – Unit 2: Differentiation: Definition and Fundamental Properties.

2019 CED – Unit 3: Differentiation: Composite , Implicit, and Inverse Functions

2019 CED – Unit 4 Contextual Applications of the Derivative  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 5 Analytical Applications of Differentiation  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations  Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration   Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions 

2019 CED Unit 10 Infinite Sequences and Series

 

 

 

 

 

 

Did He, or Didn’t He?

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Pierre de Fermat (1607 – 1655)

Since it’s soon time to start derivatives, today we look at one way people found maximums and minimums before calculus was invented. Pierre de Fermat (1607 – 1655) was a French lawyer. His hobby, so to speak, was mathematics. He is considered one of the people who built the foundations of calculus. Along with Descartes, he did a lot of the early work on analytic geometry and did much work leading to calculus, which today we would consider calculus. Neither he nor other mathematicians of his time had the tool of limits to aid him. Derivatives were not yet invented. Nevertheless, he came awfully close to both.

Fermat was working on optimization and that required him to find the maximum or minimum values of functions. Here is how he found extreme values of polynomial functions. Even though he didn’t know limits and derivatives, maybe he used them.

Using Fermat’s method, and some modern notation and terminology, we will demonstrate has method by finding the extreme values of the function:

(1)          \displaystyle f\left( x \right)={{x}^{4}}-3{{x}^{2}}+2x               

Near an extreme value Fermat reasoned that a horizontal line would intersect the graph in two points, say where x = a and x = b. (There is a third intersection on this function, since there are three extreme points. The method finds all three.) He thought of this line moving up or down and the two points coming close to the extreme value. Since these two points are on the same horizontal line f(a) = f(b):

(2)          {{a}^{4}}-3{{a}^{2}}+2a={{b}^{4}}-3{{b}^{2}}+2b

After moving everything to the left side and grouping terms by their powers and factoring we have:

(3)          \displaystyle \left( {{{a}^{4}}-{{b}^{4}}} \right)-3\left( {{{a}^{2}}-{{b}^{2}}} \right)+2\left( {a-b} \right)=0

(4)          \displaystyle \left( {a-b} \right)\left( {\left( {a+b} \right)\left( {{{a}^{2}}+{{b}^{2}}} \right)-3\left( {a+b} \right)+2} \right)=0

Near where a = b, (a – b) is not zero, so we may divide by it.

(5)          \displaystyle \left( {a+b} \right)\left( {{{a}^{2}}+{{b}^{2}}} \right)-3\left( {a+b} \right)+2=0

Again, near the extreme value \displaystyle a\approx b, so, substituting  \displaystyle a=b we have

(6)          \displaystyle 2a\left( {2{{a}^{2}}} \right)-3\left( {2a} \right)+2=4{{a}^{3}}-6a+2\approx 0

Now all that remains is to solve this equation (factoring by synthetic division).

(7)          \displaystyle 4{{a}^{3}}-6a+2=2\left( {a-1} \right)\left( {2{{a}^{2}}+2a-1} \right)=0

(8)          \displaystyle a=1,a=\frac{{-1+\sqrt{3}}}{2},a=\frac{{-1-\sqrt{3}}}{2}

These are the x-coordinates of the critical points.

Now, let’s look at all the “calculus” Fermat did not do.

The left side of equations (3) and (4) are

(9)          \displaystyle f\left( a \right)-f\left( b \right)

And after dividing this by (a – b) he had in (5)

(10)         \displaystyle \frac{{f\left( a \right)-f\left( b \right)}}{{a-b}}

Then by letting a be approximately equal to b, which sounds a lot like finding a limit, he had in modern terms

(11)         \displaystyle \underset{{a\to b}}{\mathop{{\lim }}}\,\frac{{f\left( a \right)-f\left( b \right)}}{{a-b}}={f}'\left( a \right)

Expression (6) is \displaystyle {f}'\left( a \right) – Fermat unknowingly is using the derivative! After which he set it equal to 0 and solved to find the x-coordinates of the critical points.

So, did he or didn’t he use calculus? You decide.

Optimization – Reflections

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First, a new resource has been added to the resource page.  An index of all free-response questions from 1971 – 2018 listed by major topics. These were researched by by Kalpana Kanwar a teacher at Wisconsin Heights High School. Thank you Kalpana! They include precalculus topics that were tested on the exams before 1998. These may be good for your precalculus classes. (Remember that the course description underwent major changes in 1998 and some topics were dropped at that time. These include the “A” topics (precalculus), Newton’s Method, work, volume by cylindrical shells among others. Be careful, when assigning old questions; they’re good, but they may no longer be tested.) 

Optimizations problems are situations in which some item is to be made as large or small as possible. Often this is the minimum cost of producing something, or to maximize profit, or to make the largest area or volume with the least material.

While these problems are found in most of the textbooks, they almost never appear on the AP Calculus Exams. The reason for this is that the first step is to write the equation that models the situation. This step does not involve any “calculus.” If a student cannot do this or does it incorrectly, then there is no way to earn the calculus points that follow. On the exams, students are given an expression and asked to find its maximum or minimum value.

Nevertheless, the problems can be interesting and are useful in a practical sense. Reflection is one of my favorites: show that the angle of incidence equals the angle of reflection. In the figure below, light travels from a point A to point D on a reflecting surface CE and then to point B by the shortest total distance. Show that this implies that the angle α between AD and the normal to the surface is equal to the angle β between the normal and DB. The angle α is called the angle of incidence and the angle β is called the angle of reflection.

Using the lengths marked in the drawing,  \overline{{AC}},\overline{{PD}} and \overline{{BE}} are all perpendicular to  \overline{{CE}} the total distance is AD + DB. Therefore:

AD+DB=\sqrt{{{{a}^{2}}+{{x}^{2}}}}+\sqrt{{{{b}^{2}}+{{{\left( {CE-x} \right)}}^{2}}}}

To find the minimum distance find the derivative of AD + DB and set it equal to zero.

\displaystyle \frac{{2x}}{{2\sqrt{{{{a}^{2}}+{{x}^{2}}}}}}+\frac{{-2\left( {CE-x} \right)}}{{2\sqrt{{{{b}^{2}}+{{{\left( {CE-x} \right)}}^{2}}}}}}=0

Then

\displaystyle \frac{x}{{\sqrt{{{{a}^{2}}+{{x}^{2}}}}}}=\frac{{\left( {CE-x} \right)}}{{\sqrt{{{{b}^{2}}+{{{\left( {CE-x} \right)}}^{2}}}}}}

Now, we need to be clever:

\displaystyle \frac{x}{{\sqrt{{{{a}^{2}}+{{x}^{2}}}}}}=\cos \left( {ADC} \right)=\cos \left( \beta  \right) and

\displaystyle \frac{{\left( {CE-x} \right)}}{{\sqrt{{{{b}^{2}}+{{{\left( {CE-x} \right)}}^{2}}}}}}=\cos \left( {BDE} \right)=\cos \left( \alpha  \right)

And therefore, \displaystyle \alpha =\beta  QED.

See the illustration of this in Desmos here and see an easier way to do this problem.

The conic sections all have interesting reflection properties that are quite useful.

The Ellipse: A light ray leaving one focus of an ellipse is reflected by the ellipse through the other focus of the ellipse. The angle of incidence and the angle of reflection are between the segments to the foci and the normal to the ellipse.

The computation is done using a Computer Algebra System (CAS) and is shown below, the line-byline explanation follows:

  • The first line starts with the ellipse \frac{{{{x}^{2}}}}{{{{a}^{2}}}}+\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1 with a > b > 0. Solving for y there are two equations, the second one is for the upper half that we will use below. The other for the lower half.
  • The second line finds the derivative of y and the third line, m2 is the slope of the normal, the opposite of the reciprocal of the derivative.
  • The fourth and fifth lines are the slopes, m1 and m3, are the slopes from the point on the ellipse to the foci. The “such that” bar, |, indicates that what follows it is substituted into the expression.
  • The next two lines compute the inverse tangent of angle rotated counterclockwise between the segments to the foci and the normal. This uses the formula from analytic geometry: {{\tan }^{{-1}}}\left( {\frac{{{{m}_{2}}-{{m}_{1}}}}{{1+{{m}_{1}}{{m}_{2}}}}} \right)
  • The last line shows that the expression from the two lines above it are equal, indicated by the “true” on the right.

An illustration using Desmos is here. Ellipses are used as reflectors in medical and dental equipment so that a relatively dim light source can be concentrated at the place where the doctor or dentist is working without “blinding” everyone in the room. There are also ceilings that reflect sound from one focus to the other without anyone elsewhere in the room hearing. These are only a few of their uses.

The hyperbola: A light ray from one focus of a hyperbola is reflected as though it came from the other focus. This is true whether the reflection is from the side nearer the focus or farther from the focus (reflection from the convex or the concave side.

The computation is like the ellipse computation with only a few sign changes. I will not reproduce it here. If you want to try use \displaystyle \frac{{{{x}^{2}}}}{{{{a}^{2}}}}-\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1 and  c=\sqrt{{{{a}^{2}}+{{b}^{2}}}}.

There is a Desmos illustration here. Use the p-slider to move the point. The left side shows the reflection from the “outside” surface; the right side shows the reflection from the “inside” surface.

Hyperbolas are used in telescopes and magnifying mirrors to enlarge the image.

The Parabola: A light ray from the focus is reflected parallel to the axis of symmetry of the parabola Or you can go the other way: light traveling parallel to the axis is reflected to the focus.

If you try to prove this on use  x=a{{y}^{2}} to avoid working with an undefined slope. The focus is at  \left( {\frac{a}{4},0} \right).

There is a Desmos illustration here

Parabolic reflectors are used in various kinds of spotlight and telescopes and for radar dishes. They are also used for satellite dishes for cable TV; you may have one at home.