Comparing rectangles by their area and by their diagonal

The other day I was in someone’s office waiting for some paperwork to be completed.
It was a small room, about 12 feet by 12 feet.
The back wall was orange, and its lower part was horizontally covered by a desk.
In the middle of the orange wall there was a picture of a green car. I thought that was an interesting color combination.
Looking at the picture on the wall, I wondered how bigger the visible part of the wall (including the picture) was, in area, in relation to the area of the picture.
The horizontal length of the wall was about two-and-a-half times the length of the picture, while the hight of the visible part of the wall was about three times the height of the picture.
Putting

b = length of the picture
h = height of the picture
a = area of the picture
B = length of the wall
H = height of the visible part of the wall
A = area of the visible part of the wall (including the picture)

we have, approximately

B = 5b/2, and H = 3h

therefore

A = BH = (5b/2)(3h) = 15bh/2 = 15a/2

So the proportion between those two areas is 15/2, 0r 7.5

Then I wondered what the proportion would be between the diagonals of those same two rectangles. As it turns out, when the two rectangles (the wall and the picture) do not have the same shape, the proportion between their diagonals depends not only on how many times the sides of the wall are on relation to the sides of the picture, but on the precise shape of the reference rectangle, as well.
If we use

d = measure of the picture’s diagonal
D = measure of the diagonal of the visible part of the wall

Then we have

D = sqrt(B² + H²)
 = sqrt((5b/2)² + (3h)²)
 = sqrt(25b²/4 + 9h²)

while

d = sqrt(b² + h²)

The coefficients of b² and h² inside the square root in the expression for D are not the same, so we cannot simplify that expression completely just in terms of d.

If we introduce a new constant, k, to represent how many times the height of the picture equals its length, then we can write

b = kh

In that case, the diagonals of the two rectangles can be expressed as

d = sqrt((kh)² + h²)  =
 = sqrt(k²•h² + h²) = 
 = h•sqrt(k² + 1)

D = sqrt(25(kh)²/4+ 9h²) = 
 = sqrt(25k²h²/4 + 9h²) = 
 = h•sqrt(25k²/4 + 9)

This shows the proportion between D and d being dependent on k, the proportion between h and b, because

D/d = sqrt(25k²/4 + 9) / sqrt(k² + 1) 

and this is a function of k that is not constant.

A very cool property of binomial coefficients

Last night I was reading a post in this other blog, by Ben Vitale. His post is about some special types of binomial coefficients.

In case you don’t remember, given two non-negative integers m and r, their binomial coefficient  C(m,r) (also known as “m chose r”) counts the number of subsets of size r out of a set of size m, and it is calculated by the formula:

C(m,r) = m!/(r!•(m-r)!)

In particular, when r=2, we get

C(m,2) = m!/(2!•(m-2)!) = m(m-1)/2

This is an expression for counting the number of all possible unordered pairs formed by two elements of a set of size m.
Somewhere near the end of his post, Ben presents the following formula:

C(kn,2) = k•C(n,2) + (1/2)•(k-1)k•n²

On seeing this formula, I noticed the second term on the right-hand side,

(1/2)•(k-1)k•n²

can be rewritten using the binomial coefficient C(k,2), producing the expression

C(kn,2) = k•C(n,2) + C(k,2)•n²

This formula holds true for any two positive integers n and k, each greater than 1.
In particular, by using the trick of swapping n with k in that expression, we get the following one:

C(nk,2) = n•C(k,2) + C(n,2)•k²

Now, multiplication being commutative, we know kn = nk.

Therefore, C(kn,2) = C(nk,2)

And we can connect their two equivalent expressions, making this formula:

 k•C(n,2) + (n²)•C(k,2) = (k²)•C(n,2) + n•C(k,2)

I don’t know about you but I believe this is just amazing!
Let’s illustrate this property with a couple examples:

a) First, we choose two numbers, let’s say n=5 and k=7

b) Then we calculate their squares: 5² = 25, and 7² = 49

c) Now we calculate their binomial coefficients for 2:

C(5,2) = 5•4/2 = 20/2 = 10
C(7,2) = 7•6/2 = 42/2 = 21

d) Now we plug the numbers into the formula, and verify it:

7•C(5,2) + 25•C(7,2) = 49•C(5,2) + 5•C(7,2)

(7)(10) + (25)(21) = (49)(10) + (5)(21)

70 + 525 = 490 + 105

595 = 595

Let’s check some other numbers, for example n=8 and k=14

Their squares are 8² = 64, and 14² = 196

Their binomial coefficients with 2 are:

C(8,2) = 8•7/2 = 28
C(14,2) = 14•13/2 = 91

and we verify:

14•C(8,2) + 64•C(14,2) = 196•C(8,2) + 8•C(14,2)

(14)(28) + (64)(91) = (196)(28) + (8)(91)

392 + 5824 = 5488 + 728

6216 = 6216

The reason I like this property so much is because both expressions

k•C(n,2) + (n²)•C(k,2)

and

(k²)•C(n,2) + n•C(k,2)

are linear combinations of the same two binomial coefficients. These linear combinations have different coefficients:

k and n² in one case, and k² and n, in the other.

In principle, right off the bat, I would not expect them to turn out equal but they are, every single time! What is more, the only change in the coefficients is that you move the square (the exponent 2) from k to n, and viceversa.
I find all this to be pretty cool. I wanted to share it with you, and many thanks to Ben Vitale for his blog post.

Why is substitution so confusing for so many people?

The short answer is, because it is not a mathematical operation. Not in the sense addition, subtraction, multiplication and division are.

For most people, mathematics is about crunching numbers. When they are doing math homework, or preparing for an exam, they expect to be adding, subtracting, multiplying, dividing,  deriving, or integrating, or calculating square roots, or standard deviations, averages, percentages, or otherwise combining two or more numbers into a final numerical answer.

However, when we are following the step-by-step solution to an algebraic equation, oftentimes substitution is one key step in the sequence, and substitution is an editing operation performed on a line of text. It is a special kind of cut-and-paste, find-and-replace action that has found a proper place as a drop-down menu item inside word processing computer software.

If you use a computer-like device to type letters, notes, or messages, you are probably familiar with that menu item, following the sequence of buttons 

Edit > Find > Find and Replace > Replace All

that you can use to change one word or phrase into another all across the selected part of your document. The computer automatically does that for you, saving you the time, effort, and risk of errors you would take if you did it yourself visually and manually, searching line by line, phrase by phrase, and word by word to make your desired replacement in all places.

That is what substitution is about, replacing one expression by another expression, under the assumption that they represent the same numerical value.

What stumps many people when they stumble upon a substitution step while following the solution to an algebraic equation, is their own expectation that every step had to be “mathematical,” meaning: “number-crunching-y” in one way or another. But substitution is editing, it is cutting and pasting, it is not multiplying, it is finding and replacing, it is not adding.

So, in many people’s minds, substitution is not math. They just don’t see the substitution step because they are not expecting to see it. I mean, they see it but they don’t recognize it as math because it is not math in the same sense addition and multiplication are.

They get so confused by the fact they are seeing something they were not expecting to see, that the surprise does not allow them to see whatever else is going on at that step.

It is as if, in some part of their mind they are going like:

“Math, math, math, not-math, .. wait, what?”

It is O.K. There are some seemingly non-mathematical procedures that are part of math, too, especially logic. You just have to learn to expect them to show up every once in a while, so they won’t take you by surprise.

Confusion type: unexpected answer

When the answer to a problem shows up in an unexpected or unfamiliar format, the student very well may fail to realize that result is the answer to the problem.  

Confusion:
Are we there yet?
Where is the answer?
This cannot possibly be the answer, can it?
This type of confusion can occur at any level.
It most commonly shows up in Algebra, the first time students are asked to find equations of lines, circles and so on. Before that point, chances are they had only been asked to find numerical solutions to equations, or systems of equations, or to simplify algebraic expressions.
So, up to that point, for them an equation is understood as the problem, while the answer needs to be a number, or a simplified expression. Therefore, in their minds, an equation cannot be the answer to any problem.
This can happen again and again, in many forms, even in Differential Equation courses. For example, when the answer to some problem takes the form F(x, y) = c, meaning the variable ‘y’ is interpreted to be an implicitly defined function of ‘x.’ The same kind of confusion as mentioned above can take place if the student is strongly expecting the answer to be an explicitly defined function, in the traditional form y = f(x). At that point the solution given by the equation F(x, y) = c may seem like nothing, like no answer at all, just one more step in the process of finding the “real” answer. However, this impression is of course mistaken because in those particular cases F(x, y) = c is the real answer, since an explicit definition of the form y = f(x) for ‘y’ is nowhere to be found.

Associative and distributive properties of multiplication

Illustrating a not uncommon confusion

Recently, working with a student who was preparing for the Algebra portion of a test, we came across this problem about simplifying the radical expression 3*sqrt(27).

He factored 27 = 9*3, and he wrote

3*sqrt(27) = 3*sqrt(9*3) = 3*(sqrt(9)*sqrt(3)) = 3*(3*sqrt(3))

Then he paused, hesitated, stopped, and he silently looked at me, telling me with his look that he didn’t know what to do next, pointing his pencil to the parenthesis, and the multiplication inside the parenthesis.

I said: “Probably the problem here is not asking for decimals, maybe we can do this multiplication first”, and I wrote

9*sqrt(3)

Then he asked: “But, don’t you have to also multiply the first 3 times the sqrt(3), as well?”

At that point I saw his confusion. It was about the associative property of multiplication, and the distributive property of multiplication over addition.

He was thinking that because

a(b+c) = ab + ac

maybe then

a(bc) would have to equal ab*ac, instead of just (ab)c.

So I did a quick example with numbers. First I wrote:

2(3+5) = 6 + 10 = 16

and then

2*(3*5) = 2*15 = 30, to establish a firm reference.

Then, making explicit how the associative property works in this particular case, I wrote:

(2*3)*5 = 6*5 = 30

Finally, showing the mistake of trying to distribute multiplication over itself, I wrote:

(2*3)*(2*5) = 6*10 = 60 which is different from 30.

Looking at this example the student agreed that the first procedure (using the associative property of multiplication) gives the right answer, while the second (trying to distribute the multiplication symbol outside the parenthesis over the multiplication inside) doesn’t work. After having the formats clearly illustrated with whole numbers, I wrote them with variables, for him to see the difference in the abstract level.

In this particular instance, the confusion arose in the first place because of the square root symbol in the factor sqrt(3).

When a triple product a(bc) involves only whole numbers, students don’t have a problem with that. They just multiply the two numbers inside the parenthesis, and then multiply that product by the number outside the parenthesis. However, for many students, the presence of a radical makes the problem a little bit too abstract. They don’t have a ready answer for 3*sqrt(3), so they think: “I don’t know how to do that multiplication, so I don’t know how to do the problem,” or “That multiplication cannot be done without a calculator, so I have to apply some rule but I don’t know which one.” Once in the abstract realm of properties, formulas, and identities, their chances of getting lost increase dramatically.

This type of confusion is actually quite interesting. Students often fail to make a connection between variables and numbers. They seem to forget that variables stand for numbers. However, here the student was applying the connection in one way, from the abstract rule to the concrete example. The problem was the rule he was applying was made up. It is not a rule at all. It only seemed a rule to him because of its typographical similarity with the distributive property of multiplication over addition.

First, the presence of a radical disconnected him from the concrete, familiar realm of whole numbers. Then he went searching his memory banks for an algebra rule to apply. The first rule he remembered was the distributive property. At that point, an oversimplification drive took over, suggesting the idea that the typographically corresponding expression should be a rule as well. So he was considering adopting such a made-up rule as a real number property, without checking its validity with numerical examples before applying it but just going by the typographical similarity with the distributive property, effectively making him want to distribute multiplication over itself.

Solving systems of equations by substitution

A topic that is hard to explain because it is so simple

The substitution method oftentimes works as a powerful technique for solving systems of equations. This method is widely taught in middle and high schools, as part of the Algebra curriculum, along with the other standard methods for solving systems of linear equations in two variables: the elimination method, the graphing method, and the method of determinants (also known as Kramer's rule). Solving systems of equations by substitution is a very interesting process, especially when we consider that the fundamental basis of its concrete execution is not really any algebraic operation at all but a typographical one. Substitution means textual substitution. It is a typographical "find and replace" operation, whereby we combine two strings of characters into a new one, by means of "copy," "cut," and "paste" manipulations. It is a common experience for math teachers noticing many of their students get confused when learning the substitution method. I believe a big part of such confusion in the student's mind comes from the unexpected, unexplained, fundamental difference in nature between algebraic, arithmetic, numerical operations, on one hand, and such a typographical, textual, character-and-string oriented operation like substitution, on the other hand. Most teachers explain the substitution method by doing some examples on the board, and hoping that students will somehow "get it." Indeed, some students do get it. After watching the teacher doing a few examples, something clicks and, that is it, they now know it. They have gotten it. Moreover, usually they not only get it but they love it when they realize how it works. Unfortunately though, these students I refer to in the last few sentences, typically make up only between ten and twenty percent of the class. They are the intellectual high achievers of the class, many of whom will go on to careers in science, engineering, medicine, or money management. The other eighty to ninety percent of the class typically did not get it. They are confused, they do not know what is going on, they have no clue what the teacher did or is talking about. For them this is no happy experience. Actually, it can be really aggravating if the teacher is particularly enthusiastic about substitution but lacks the ability to infect the whole class with his or her enthusiasm. There are some particular examples of systems of equations that, when solved by substitution, seem to yield a spectacularly elegant and short solution. When students have not yet understood the substitution method, watching one of these spectacular solutions makes them feel like the teacher is practicing some mysterious magic trick in front of them. This only adds to their discomfort, and their distaste for math in general, since it is only natural to fear and/or reject what we do not understand. As a math tutor, I have the luxury of working with one student at a time, so I can focus my attention on delivering the particular information my student needs, in the way he or she wants to approach each problem. In the case of substitution, I make sure they understand how to do it, by doing the first example myself so slowly, so carefully, so explicitly, so spelled out, so mechanically, that I make my students feel for sure they can do it faster than me. When I explain solving by substitution I do not try to look smart. Instead, I become a machine, and I consciously take all the magic away from the process, so my student can clearly see how simple it is. The delivery here needs to be accurately tailored to each individual student. It is much harder to do this in front of a whole class, because teachers have to maintain their authority; and making the explanation so explicit that the last student in the class understands it, would probably lower the teachers' own status in the eyes of several other students. In part, substitution is difficult to teach and understand because it is so simple. Compounding the problem, we have to remember all the accumulated deficiencies students are still struggling with, and dragging behind since their first years in elementary school. When solving a system of equations by substitution, the actual substitution is only one step in the process. Even when it is done correctly, students still need to work their way through all the algebraic and arithmetic operations needed to solve the given problem. They very well may do the substitution correctly, only to mess up the problem two steps down the road because they do not know how to add/subtract negative numbers, or they do not know how to divide fractions, or they are still adding with their fingers.

"Made-up" operations

Sparing some test-takers the abstraction of modern algebra

Here is a specific type of problem that usually confuses many students who are preparing for standardized tests like the GMAT, GRE, and SAT:

Let the operation Δ be defined as aΔb = (a² - b)/(a+b) for all real numbers a, b such that a does not equal -b. If a = 15 and aΔb = 5, what is the value of b?

One source of confusion here is the symbol used to represent the operation (either Δ, or θ, or @, or other similar symbol). To the student, these symbols seem unusual, odd, strange, or weird. The main confusion source is the word “operation” itself, referring to the odd-looking symbol. This causes a particularly strong reaction in students who have been away from school a long time, not taking any math classes in the last several years. When they hear or read the word “operation” in connection with math, they automatically think of the four classic operations they are familiar with since elementary school: addition, subtraction, multiplication, and division. They know that weird-looking symbol is none of them.
When they ask me questions about this type of problem, often the conversation unfolds like this (using the example problem above):
~~~~~
Student: What the heck is that symbol Δ? That is not an operation, is it?
Tutor: No, you are right, it is not an operation. Nobody uses that in math. It is nothing like the quadratic formula, or something. No.
Student: So, why are they saying it is an operation?
Tutor: Oh, do not worry about it, it is nothing, they are just making it up. It is a made-up operation.
Student: But, why? Just to confuse me?
Tutor: You got that right. They want to see if you can plug in whatever values they give you, and go along with whatever expression comes out of that. For example: let’s say a=1 and b=2.
Then we have 1Δ2 = (1² – 2)/(1+2) = -1/3. Now, I bet you can do this other example: if a was 3 and b was 5, how much would 3Δ5 equal?
Student: So, is that it? I just have to plug in the numbers?
Tutor: Yes, that is right, the numbers, or the expressions the problem gives you.
Student: O.K., then: (15²-b)/(15 + b) = 5. Oh, well, now I have an equation, and I can solve for b.
Tutor: Perfect.
Student [after solving the equation]: Pfff! That is easy.
Tutor: Good, excellent!
Student: It was just plugging in the numbers, and solving the equation but they make it seem so complicated at the beginning with that weird symbol.
Tutor: Yes, I know. That is exactly what they do. So, just be prepared for those weird-looking, out-of-the-blue, made-up operations. Do not let them surprise you.
~~~~~
In abstract algebra, a binary operation on a given set is a function taking two input values from that set, and returning an output value in the same set. The set does not even have to be a set of numbers. So, if you want to get technical, the question of whether or not a formula like (a² - b)/(a+b) defines an operation, really has to do with the domain and codomain of the function.
In this particular example (a² - b)/(a+b) is not a binary operation on the set of real numbers, because the restriction that the denominator needs to be other than zero excludes the set {(x, -x)} from the function’s domain. You could call it a partially defined operation. Other formulas, like sqrt(ab), the geometric mean of two numbers, are operations only on the set of positive numbers, because the product ab needs to be positive for the square root to be defined.

However, I do not get into any of these abstract concepts with my students, unless they specifically ask, with curiosity, and with an open mind because, otherwise, it would be Greek to them, and it would be a waste of their time. In most cases regarding this particular type of confusion, test takers only want validation that they are not crazy, and that they did not totally miss a whole classic operation (like addition, subtraction, multiplication, and division) during elementary and middle school. So, I want to address their concern, and make sure they know I understand their question; the source of their surprise and confusion. I want to increase their confidence in themselves, that they can successfully solve the problem on their own. To do it, they do not need to know anything about abstract binary operations in algebraic structures. That is a topic CSET takers need to pay some detailed attention to but not GMAT, GRE, or SAT takers. There is no time for me to go into such topics with them. The typical student only wants to know how to solve the problems. They are quite comfortable with their familiar belief that the word “operation” must mean addition, multiplication, subtraction, or division. They are not paying me to make them go through all the mental gymnastics it would take them to overcome their resistance to expand their concept of “operation.” So I just give them what they are looking for, that is, the fastest way for them to be able to solve the problems, and to feel good about it.

Ubiquitous Numbers

One and Zero are always everywhere

The more you tutor math, the more skilled you become in finding good, clear ways to explain all types of math concepts to students. However, some concepts are more elusive than others. The difficulty of grasping a concept depends not only on the concept itself but also on the student who is assimilating it. One person, for example, can easily understand polynomial multiplication, and struggle with percentages, while someone else can find percentages very easy but have trouble with polynomials. There are also some concepts or topics that seem to be hard for a significant majority of students, like word problems for instance.
Substitution is a very powerful problem-solving technique, and it is widely used in a variety of situations. Students who find substitution easy have a clear advantage over students who have trouble understanding it. Substitution comes up in many different ways, some more complex than others. Some students understand the more basic forms of substitution but have problems applying the same techniques when dealing with more complex expressions. In fact, a very consistent general trend is the host of negative reactions students tend to show in varying degrees when facing bigger, longer, more complicated expressions. The more complex the expression, the more likely that students will get confused, or feel overwhelmed by it.
There is a particular way of using substitution to which most students react with a strong resistance: It involves transforming a given expression into another one that is equivalent in value but looks more complicated. This is done with the ultimate goal of simplifying the expression but it starts out by complicating it a little more. It is like climbing up a hill to find a way down the mountain.
The most common way of using this technique is by introducing a representation of the numbers Zero or One into the given expression.
Zero and One are very special numbers. They implicitly are everywhere in any given algebraic expression, even when we do not see them written out.
Zero is called the additive identity because zero plus any number is that same number ( x + 0 = x ).
One is called the multiplicative identity because one times any number equals that same number ( x∙1 = x ).
These two properties make Zero and One algebraically omnipresent in an implicit way.
Further, we have the following properties:
A number subtracted from itself equals zero ( x – x = 0 ).
A number different from zero, when divided by itself gives us one as the result ( x/x = 1 ).
These last two properties give Zero and One an infinite number of representations (“disguises” if you will) to show up in a formula. So, not only are Zero and One ubiquitous, they can come in a dizzying multiplicity of seemingly different forms.
The above properties of Zero and One, and their consequences, make them extremely useful in solving equations, and in manipulating algebraic expressions in general.
However, as I mentioned before, many students present a strong resistance to the idea of making an expression more complicated to be able to reduce it later. This is partly because they do not see the point of multiplying a number times one, or adding zero to it; partly because doing so seems to increase the problem in size and complexity; partly because they feel we are working backwards into some uncalled-for calculation, and finally because they believe they would not know what particular form of Zero or One they are supposed to introduce into the expression if they were doing the problem on their own.
Sometimes, when working with a student on a given problem, it is very easy for me to see a path to the solution using these types of techniques but I have learned to make sure the student does not feel like I was expecting him or her to be able to solve the problem in the same way. I only use these techniques when the student is completely stuck in the problem, not making any progress at all. When they see one possible solution, it gives them some perspective on the different factors playing a role in the problem. After showing them one possible solution method, if they do not feel comfortable that they would be able to successfully use the same method on their own, we focus on finding an alternate method that works better for them.

The Square of a Sum

An arithmetic and geometric approach to Algebra

A very common mistake test takers make when they have been out of school for a while is that they automatically try to expand the square of a sum as if it was the same as the plain sum of the squares of the individual terms. For example, sometimes some students, when presented with say, (x + 5)², they wrongly expand it as x² + 25, leaving out the middle term 10x.
They forget they have to use the foil method to multiply the given sum times itself. Since typographically the sum of the squares looks like something that could equal the square of the sum, they guess they can expand the square that way, as if it was a real math rule, and then they get the problem wrong.
I try to preemptively address the possibility of this mistake, because both sums of squares and squares of sums show up a lot in the math section of standardized tests. For a long time I have used a drawing, with two little squares and two congruent rectangles filling up a larger square, to show the geometric interpretation (in terms of area) for the algebraic formula (a + b)² = a² + 2ab + b². I emphasize the middle term, 2ab, in this expansion, and I point to the corresponding two rectangles in the drawing.
The above approach is useful but many students forget the formula when it comes to actually applying it. Most do not realize it is a universal pattern where to plug any other expressions in.
So, lately I have been experimenting with a slightly different approach: I present two or three numerical examples first, before drawing the squares and rectangles. I ask the student to choose two numbers, and then I guide them through a sequence of calculations that allows them to actually compute and see the numerical difference between the square of the sum, and the sum of the squares. Then I ask them to multiply the original numbers, then to double this product. By asking all the questions in the right order, I have gotten comments from them like: “Wow!” or “How weird!” or “That is so funny!” or “This is very interesting!” or “Does it always work like that?” or “How can that be?” or “What is going on?” or “I’m sure there is a pattern here!” Then I draw the squares and rectangles with all the numbers in their proper places. So, creating the perception of a “mystery” with the numbers, and then explaining it away with the picture seems to work well. It makes sense for the students. They usually say: “Oh, right! Now I get it.”

Calculus is harder than Algebra

A few sources of confusion to be aware of.

Sometimes I see students who used to get good grades in their Algebra classes but who are now struggling with Calculus.
They tell me they are confused by Calculus; they do not know when to apply the Chain Rule; they get uncomfortable around dx; they have no clue how to start the problems, and so on and so forth.
What most amazes them is they know they had no problems with Algebra, they understood Algebra well; so the question in their mind is: “What is it about Calculus that makes it so difficult?”
One problem is their implicit, non-accurate expectations. Sometimes they think they are doing something wrong because they have not gotten to the solution yet, after filling in one page with equations. However, they may very well be on the right track. They are not doing anything wrong. They just never expected for the solution to take so much work, and time. They were expecting to arrive to the solution after three or four steps, like they used to do in their Algebra work. Instead, with Calculus problems they may need to go through ten or twelve similar steps.
Sometimes they ask:
“Another integral? You mean, this is not the result yet?”
Apparently they cannot believe it. I tell them:
“Look, we have to keep going. We are not there yet but we are getting close. O.K.? Think of this as a little marathon. You were used to run the 400 meters, now we are going for three miles. We just have to keep going.”
Another feature that makes Calculus harder than Algebra is the huge number of options when it comes to selecting routes to the solution, many of which may turn to be dead-ends. Calculus incorporates all the operations of Algebra, including exponents, roots, and logarithms; plus all trigonometric functions; and it makes heavy use of function composition. All these ingredients can be thrown into a problem involving limits, or derivatives, or integrals. So Calculus really pushes you to get your Algebra up to speed. And because the expressions get more complex, there are more forks in the road at almost every step of the way. This is confusing for many students.
Now, in my opinion, the major source of discomfort when doing the transition from Algebra to Calculus, is that students who like Algebra like also order, and neatness. This proves to be a disadvantage when it comes to acquiring an intuition for Calculus.
They tell me:
“I do not know what to do, where to start, when to apply what rule, none of that! In Algebra everything was more clear, more precise.”
I ask them:
“And you liked that, didn’t you? Having a set procedure to follow in an orderly manner, right?”
They say:
“Yes!”
Then I say something like:
“Well, I hate to break the news to you but when it comes to Calculus, you are out there in the wild, and everything is moving, all the time, even if it does not seem that way. You have to get used to it.”
Many times they ask:
“But, why? Why is it that way?”
My explanation goes more or less like this:
“Well, Calculus was invented to deal with problems of a very physical nature: motion, volume, pressure, speed, weight, and the like, but not only that. The specific aim of Calculus was to provide answers as to how those physical magnitudes behave when they are changing, either growing or going down, and being related to each other at the same time. Originally, everything in Calculus happens in time, nothing stays the same, everything is changing, moving, decreasing, growing, building up, fading away, speeding up, slowing down, just like in the real world. So when you think of a variable x or y or v in Calculus, it is not just a number that is there, with some fixed value, but a number that wants to go somewhere else, a number that has already started to change, even if only a little tiny bit”
Somehow this bizarre wording paints a picture that makes sense for them, in that it puts their confusion into the proper perspective, eliminating a big part of it.
One of my students, after hearing this type of explanation, said:
“Oh! So, dx is the sneakiness of change! That’s why I don’t like it! But now I know I have to deal with it. I cannot just pretend it’s not there.”
And I said:
“That is exactly right. That is exactly what it is.”
Then she said:
“And, we always have to apply the Chain Rule, we cannot get rid of it, because the variables are connected to each other.”
And I said:
“Yes, the Chain Rule is always there.”
It is a big help to be able to tell students at least some of the reasons why they are confused, because when they do not even know the source of their confusion, everything gets exponentially more confusing for them.

Solving Equations For a Particular Variable

A very basic principle

An equation has one equal sign.

The equal sign divides the equation into left hand side and right hand side.

The two sides may look totally different from each other as expressions but the equal sign says their numerical value has to be the same.

The fundamental principle of equations says that, when two expressions have the same numerical value, if we apply one operation to both expressions, the resulting expressions after the operation is performed will also be equal in value. They will be equal not to the original expressions, but to each other.

So, if A, B and C are three algebraic expressions, and we have the equation A = B, then all of the following will also be valid equations:

A + C = B + C

A - C = B - C

(A)(C) = (B)(C)

A/C = B/C [provided C is not zero]

A^2 = B^2

Square root of A = Square root of B

This fundamental principle is used over and over to solve equations for specific variables, one step at a time.

For example, in solving for x the equation (3x + 1)/2 = 5y - 4, we can do it like this:

1) Multiply both sides by 2 and we get

3x + 1 = 2(5y - 4)

2) Subtract 1 from both sides and we get

3x = 2(5y - 4) - 1

3) Divide both sides by 3 and we get

x = ( 2(5y - 4) - 1)/3

Now the equation has been solved for x in a series of steps, where each step consists of applying one and the same operation to BOTH sides of the equation.

The fact that the resulting expression for x can be simplified to

x = (10y -9)/3

is not relevant here. I am only illustrating the process we use to isolate x one step at a time by applying the same operation to both sides of the equation.

The following YouTube video from InterAlgebra12 shows several more examples: