Why Optimization?

Posted on November 7, 2023 by Lin McMullin

Unit 5 ends with a return to a realistic context. To optimize something means to find the best way to do it. “Best” or “optimum” may mean the quickest, the cheapest, the most profitable, or the easiest way to do something.

For example, you may be asked to build a box of a given volume with the least, and therefore cheapest, amount of material. Thus, these are really problems where you need to find the maximum or minimum of the function that models the situation. There are applications to engineering, finance, science, medicen, and economics among others.

The most difficult part of these problems is often writing the equation to be optimized; not the calculus involved. Once you have the model, finding the extreme value is easy.

The last part of this unit extends the ideas of this unit to implicit relations, those whose graph may not be a function. These too, increase, decrease, and have extreme values. The same techniques help you to find them.

Course and Exam Description Unit 5 Sections 11 and 12

A note for teachers: You are not behind scheduel. Please remember that I am posing this series ahead, probably well ahead, of where you are. This is so that they will be here when you get here.

Why Existence Theorems?

Posted on October 31, 2023 by Lin McMullin

An existence theorem is a theorem that says, if the hypotheses are met, that something, usually a number, must exist.

For example, the Mean Value Theorem is an existence theorem: If a function f is defined on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in the open interval (a, b) such that $\displaystyle {f}'\left( c \right)\left( {b-a} \right)=f\left( b \right)-f\left( a \right)$ .

The phrase “there exists” also means “there is” and “there is at least one.” In fact, it is a good idea when seeing an existence theorem to reword it using each of these other phrases. “There is at least one” reminds you that there may be more than one number that satisfies the condition. The mathematical shorthand for these phrases is an upper-case E written backwards: $\displaystyle \exists$ .

…then there is a number c in the open interval (a, b) such that…
…then there is at least one number c in the open interval (a, b) such that…
…then $\displaystyle \exists$ c in the open interval (a, b) such that…

Textbooks, after presenting an existence theorem, usually follow-up with some exercises asking you to actually find the value that exists: “Find the value of c guaranteed by the Mean Value Theorem for the function … on the interval ….” These exercises may help you remember the formula involved.

But, the important thing about most existence theorems is that the number exists, not what the number is.

Other important existence theorems in calculus

The Intermediate Value Theorem

If f is continuous on the interval [a, b] and M is any number between f(a) and f(b), then there exists a number c in the open interval (a, b) such that f(c) = M.

Another wording of the IVT: If f is continuous on an interval and f changes sign in the interval, then there must be at least one number c in the interval such that f(c) = 0

Extreme Value Theorem

If f is continuous on the closed interval [a, b], then there exists a number c in [a, b] such that $\displaystyle f\left( c \right)\ge f\left( x \right)$ for all x in the interval.

Another wording: Every function continuous on a closed interval has (i.e. there exists) a maximum value in the interval.

If f is continuous on the closed interval [a, b], then there exists a number c in [a, b] such that $\displaystyle f\left( c \right)\le f\left( x \right)$ for all x in the interval. Or: Every function continuous on a closed interval has (i.e. there exists) a minimum value in the interval.

Critical Points

If f is differentiable on a closed interval and $\displaystyle {f}'\left( x \right)$ changes sign in the interval, then there exists a critical point in the interval.

Rolle’s theorem

If a function f is defined on the closed interval [a, b] and differentiable on the open interval (a, b) and f(a) = f(b), then there must exist a number c in the open interval (a, b) such that $\displaystyle {f}'\left( c \right)=0$ .

Mean Value Theorem – Other forms

If I drive a car continuously for 150 miles in three hours, then there is a time when my speed was exactly 50 mph.

If a function f is defined on the closed interval [a, b] and differentiable on the open interval (a, b), then there is a point on the graph of f where the tangent line is parallel to the segment between the endpoints.

Cogito, ergo sum

And finally, we have Descartes’ famous “theorem:” Cogito, ergo sum (in Latin) or in the original French, Je pense, donc je suis, translated as “I think, therefore I am” proving his own existence.

Why Analytical Applications?

Posted on October 27, 2023 by Lin McMullin

The last unit showed you some ways the derivative may be used to solve problems in the context of realistic situations. This unit looks at analytical applications of the derivative – that is applications apparently unrelated to any kind of real situation. This is a bit misleading since the things you will learn are meant to be extended to practical problems. It’s just that for now we will study the ideas and techniques in general, not in any context.

The unit begins with two important theorems. The Mean Value Theorem that relates the average rate of change of a function to the instantaneous rate of change (the derivative), The MVT, as it is called, helps prove other important ideas especially the Fundamental Theorem of Calculus at the beginning of the integration.

The other theorem is the Extreme Value Theorem. The EVT tells you about the existence of maximum and/or minimum values of a function on a closed interval.

Both are existence theorems, theorems that tell you that something important or useful exists and what conditions are required for it to exist. More on existence theorems in my next post.

As with all theorems, learn the hypothesis and conclusion. The graphical interpretation makes these easy to understand.

You will learn how to determine where a function is increasing and decreasing. This leads to finding the maximum or minimum points – where the function changes from increasing to decreasing or vice versa: You will learn three “tests” – theorems really – to justify the extreme value.

Along with that you will learn some more about the second derivative and concavity.

These ideas and theorems will help you accurately draw the graph of a function and nail down the precise location of the important points and tell what is happening between them. Yes, your graphing calculator can do that, but you’re taking this course to learn why.

You will be asked to determine information about the function from its derivatives – plural. The derivative may be given as a function, a graph, or even a table of values.

You will also be asked to justify your reasoning – tell how you can be sure what you say is correct. You do that by citing the theorem that applies and check its hypotheses, not by Paige’s method:

These concepts are tested on the AP Calculus exams and often produce the lowest scores of the six free-response questions. Yet, if you learn these concepts, that question can be the easiest.

P.S. Some books use the Latin words extremum (singular) or extrema (plural). They mean the extreme value(s). Maybe they have hung around so that the uninitiated will think calculus is difficult and confusing. I don’t know. Use them if you like: impress your (uninitiated) friends.

Course and Exam Description Unit 5 Topics 5.1 through 5.9

Why Radian Measure?

Posted on September 22, 2023 by Lin McMullin

I’m sure you’ve been wondering since you first heard about radian measure before calculus why calculus is always done in radian measure. Once you’ve learned the derivatives of the trigonometric functions, you will appreciate why radians are pretty much the only choice for calculus people.

The idea for this series of posts, The Why Series, a this post I wrote a few years ago called “Why Radians?” The post gets a lot of ‘hits’ every year. While that post was written for teachers, there is no reason you, a student, shouldn’t read it as well; it’s not a secret. Or maybe it is, but now that you’re initiated into calculus, you may read it.

If you’re still wondering why calculus people use radians, follow this link: Why Radians?

AP Calculus Course and Exam Description Unit 2 Section 7.

Why the Derivative?

Posted on September 12, 2023 by Lin McMullin

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

Unit 9 – Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Posted on January 11, 2022 by Lin McMullin

Unit 9 includes all the topics listed in the title. These are BC only topics (CED – 2019 p. 163 – 176). These topics account for about 11 – 12% of questions on the BC exam.

Comments on Prerequisites: In BC Calculus the work with parametric, vector, and polar equations is somewhat limited. I always hoped that students had studied these topics in detail in their precalculus classes and had more precalculus knowledge and experience with them than is required for the BC exam. This will help them in calculus, so see that they are included in your precalculus classes.

Topics 9.1 – 9.3 Parametric Equations

Topic 9.1: Defining and Differentiation Parametric Equations. Finding dy/dx in terms of dy/dt and dx/dt

Topic 9.2: Second Derivatives of Parametric Equations. Finding the second derivative. See Implicit Differentiation of Parametric Equations this discusses the second derivative.

Topic 9.3: Finding Arc Lengths of Curves Given by Parametric Equations.

Topics 9.4 – 9.6 Vector-Valued Functions and Motion in the plane

Topic 9.4 : Defining and Differentiating Vector-Valued Functions. Finding the second derivative. See this A Vector’s Derivatives which includes a note on second derivatives.

Topic 9.5: Integrating Vector-Valued Functions

Topic 9.6: Solving Motion Problems Using Parametric and Vector-Valued Functions. Position, Velocity, acceleration, speed, total distance traveled, and displacement extended to motion in the plane.

Topics 9.7 – 9.9 Polar Equation and Area in Polar Form.

Topic 9.7: Defining Polar Coordinate and Differentiation in Polar Form. The derivatives and their meaning.

Topic 9.8: Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Topic 9.9: Finding the Area of the Region Bounded by Two Polar Curves. Students should know how to find the intersections of polar curves to use for the limits of integration.

Timing

The suggested time for Unit 9 is about 10 – 11 BC classes of 40 – 50-minutes, this includes time for testing etc.

Previous posts on these topics:

Parametric and Vector Equations

Implicit Differentiation of Parametric Equations

A Vector’s Derivatives

Adapting 2012 BC 2 (A parametric equation question)

Polar Curves

Polar Equations for AP Calculus

Extreme Polar Conditions

Visualizing Unit 9 Desmos Demonstrations for Polar, Vector and Parametric Curves

Implicit Differentiation of Parametric Equations

Posted on October 31, 2021 by Lin McMullin

I’ve never liked memorizing formulas. I would rather know where they came from or be able to tie it to something I already know. One of my least favorite formulas to remember and explain was the formula for the second derivative of a curve given in parametric form. No longer.

If $\displaystyle y=y(t)$ and, $\displaystyle x=x(t)$ then the traditional formulas give

$\displaystyle \frac{{dy}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$ , and

$\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$

It is that last part, where you divide by $\displaystyle {\frac{{dx}}{{dt}}}$ , that bothers me. Where did the $\displaystyle {\frac{{dx}}{{dt}}}$ come from?

Then it occurred to me that dividing by $\displaystyle {\frac{{dx}}{{dt}}}$ is the same as multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$

It’s just implicit differentiation!

Since $\displaystyle \frac{{dy}}{{dx}}$ is a function of t you must begin by differentiating the first derivative with respect to t. Then treating this as a typical Chain Rule situation and multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$ gives the second derivative. (There is a technical requirement here that given $\displaystyle x=x(t)$ , then its inverse $\displaystyle t={{x}^{{-1}}}\left( x \right)$ exists.)

In fact, if you look at a proof of the formula for the first derivative, that’s what happens there as well:

$\displaystyle \frac{d}{{dx}}y(t)=\frac{{dy}}{{dt}}\cdot \frac{{dt}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$

The reason you do it this way is that since x is given as a function of t, it may be difficult to solve for t so you can find dt/dx in terms of x. But you don’t have to; just divide by dx/dt which you already know.

Here is an example for both derivatives.

Suppose that $\displaystyle x={{t}^{3}}-3$ and $\displaystyle y=\ln \left( t \right)$

Then $\displaystyle \frac{{dy}}{{dt}}=\frac{1}{t}$ and $\displaystyle \frac{{dx}}{{dt}}=3{{t}^{2}}$ and $\displaystyle \frac{{dt}}{{dx}}=\frac{1}{{3{{t}^{2}}}}$

Then $\displaystyle \frac{{dy}}{{dx}}=\frac{1}{t}\cdot \frac{{dt}}{{dx}}=\frac{1}{t}\cdot \frac{1}{{3{{t}^{2}}}}=\frac{1}{3}{{t}^{{-3}}}$

And $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\left( {\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)} \right)\cdot \frac{{dt}}{{dx}}=\left( {-{{t}^{{-4}}}} \right)\cdot \left( {\frac{1}{{3{{t}^{2}}}}} \right)=-\frac{1}{{3{{t}^{6}}}}$

Yes, it’s the same thing as using the traditional formula, but now I’ll never have to worry about forgetting the formula or being unsure how to explain why you do it this way.

Revised: Correction to last equation 5/18/2014. Revised: 2/8/2016. Originally posted May 5, 2014.

Teaching Calculus

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

Category Archives: Derivative Theory

Why Optimization?

Why Existence Theorems?

Why Analytical Applications?

Why Radian Measure?

Why the Derivative?