Word Problems and Unit Conversions

For some people, working with variables that have different units is very easy. They can conceptualize the relationship between the different variables in a way that seems natural to them, yet remains very difficult to explain to others. There are even more people who cannot do this type of math so automatically, and often struggle to solve problems... myself included. However, there is a trick to solving these questions, and the solution lies, quite simply, in keeping track of the units. I will try to demonstrate.

I will start simply, with multiplication and division. You know that if one number is on the top and on the bottom of a division line, you can cancel them out to reduce things:

You also know that you can do this when you are working with variables, like x and y:

So then, it's only natural that you can do this with units as well!

Now, here is the trick. If you are struggling with the concept of a word or other problem that deals with units, change the way that you are looking at it, to focus on the units involved. They will guide you towards the final solution.

Let's try a basic physics problem: How far will an object travel in 10 seconds if it is moving at a constant speed of 25 meters/second? This one might not be too difficult to conceptualize, but it demonstrates my point. Watch what I do with the units at the start, if you are not familiar with this part.

As you can see, the units cancel out to leave just the 'meters.'

Let's try one that is a bit more complicated: If a force of 10 N pushes a 10 kg block along a flat, frictionless surface for 5 seconds, what is the block's average velocity? This one is undoubtedly more difficult to do in your head, but if you keep track of the units (what you have, and what you need), then you should be able to do it. For those who aren't in physics, it is easier to know that the unit Newton, N, is equal to (kg) x (m) / (s)^2... blogger is still terrible with superscripts and equations. :(

As you can see, we wanted velocity (m/s) in the end, and so I just worked with the units until I got what I wanted. From a physics point of view, I used Force = Mass x Acceleration, and Velocity = Acceleration x Time.

Currency conversions are a common everyday example of where you can use this trick. If I say that the exchange rate is 0.985 $US / $CAN, and I want to know how much $US I can exchange for $232 CAN:

Naturally, it's best to try to understand the concepts of the questions, and the more complex questions won't be quite so easy to muddle through just by tracking the units. However, hopefully you can see the benefit of looking at these types of problems from this angle, and it will help you to solve them. I learned this technique in high school, and I still benefit from it. As always, feel free to drop me a line if you would like me to clear anything up. :)

Polynomials

I have a request to go over polynomials, and to specifically explain what FOIL means.

"Polynomials" refer to mathematical expressions that contain multiple terms (poly = multiple, nomials = terms/numbers (roughly)). "Monomial" refers to a polynomial that has only a single (mono) term. Furthermore, the product of multiplying monomials together also results in a monomial. Basically, think of it along these lines:

- 1 term (bunch of things multiplied/divided/raised to a power) = monomial

- more than 1 term (monomials) that are added or substracted = polynomial

Some examples hopefully clear this up:

1) Monomials:
x,
2x,
x^3,
xy,
5xy,
(10xy^4)/3

2) Polynomials:
x+5,
2x-5,
(5xyz^3)/4 + 7x

Doing calculations with polynomials is fairly straight-forward, but you do have to keep a few things in mind.

For addition or subtraction of polynomials, you must remember that you can only combine 'like terms.' 'Like terms' contain the sample variables and differ by the numeric coefficient in the front. 3x and 5x are like terms, and can be added to get 8x. However, 3x and x^2 are not like terms and CANNOT be added like in the first example.

x + 2x = 3x
4x + 4x = 8x
15x + 100x = 115x
50x - 10x = 40x

You cannot add terms in this way that have different variables:
x^2 + 5x
3x^4 - 2x^3
x + 7

**Technically, some expressions such as these can be SIMPLIFIED by pulling out a common factor, but they still cannot be added up, as with like terms:
eg. x^2 + 5x = x(x + 5)

Multiplying polynomials is a little more complicated, but still straight-forward, provided that you keep track of what you are doing. When multiplying two polynomials together, all that you need to do is add up the products of multiplying each term in the first polynomial by each term in the second polynomial. You have probably heard the expression "FOIL" when talking about multiplying polynomials. FOIL is short for "First, Outside, Inside, Last" and refers to which terms you multiply together and add up when multiplying two polynomials, each composed of two monomials (specifically, this is the product of two BINOMIALS).

Here is an example of this:

(2x + 3)(x + 5) = ?

First: 2x * x
Outside: 2x * 5
Inside: 3 * x
Last: 3 * 5

Adding these up gives:
2x^2 + 10x + 3x + 15

Furthermore, as I showed earlier, about adding polynomials with like terms, you can simplify this expression:
2x^2 + 13x + 15

And that's the final, reduced answer. You can apply the FOIL principle to any two binomials to arrive at their product. When you have more complicated polynomials, such as those composed of 3, 4, 5, or more monomials, you do the same type of thing... what I find easiest is to take the first term of the first polynomial, and multiply it with every term of the second polynomial. Then do the same for the second term in the first one, multiplying with every term in the second one, and so on.

Like this:
(x^2 + x + 5)(x^3 + x^2 + 1)

First group (multiply x^2 with all in the second polynomial)
x^2 * x^3
x^2 * x^2
x^2 * 1

Second group (x)
x * x^3
x * x^2
x * 1

Third group (5)
5 * x^3
5 * x^2
5 * 1

Now, you just add up all these terms, and simplify where you can:
(x^2 * x^3) + (x^2 * x^2) + (x^2 * 1) + (x * x^3) + (x * x^2) + (x * 1) + (5 * x^3) + (5 * x^2) + (5 * 1)

(x^5) + (x^4) + (x^2) + (x^4) + (x^3) + (x) + (5x^3) + (5x^2) + (5)

(x^5) + 2(x^4) + 6(x^3) + 6(x^2) + x + 5

And that's it. A little more complicated, but as long as you keep track of what you're doing and work your way through it, you will arrive at the answer!

Graphing Sine and Cosine Curves

As I usually include in most of my posts, please feel free to ask questions or leave comments on any material that I post, and I will try my best to address them. I recently received a request to go over how to graph sine and cosine curves.

Sine and cosine functions are PERIODIC FUNCTIONS, meaning that the function will evaluate to a repeated number after a defined length (period). You've probably seen these graphs several times before and not realized what they were. Sine and cosine graphs are virtually identical, EXCEPT that they are slightly shifted relative to one another... or they are "out of phase." This is an important distinction to make, as confusing which is which, of course, will get you the wrong answer.

The EASIEST way to remember, or to quickly plot out the graph, is to create a table of values. Let's work on the Sine curve first. The cosine one works exactly the same way. Creating this table of values will be a good review of how these trig functions work. On that note, we will evaluate for values for x that represent the angles of our special triangles, and then continue with more values at the same intervals. (I'm going to represent the square root function by "sq.rt.") If you want, you can continue to make the list longer if you wish, but I think this is sufficient to demonstrate the graph.

y = sin(x)

x |||||y
-----------
0 |||||0
30 |||||0.5
60 |||||(sq.rt.3)/2
90 ||||| 1
120||||| (sq.rt.3)/2
150||||| 0.5
180||||| 0
210||||| -0.5
240||||| -(sq.rt.3)/2
270||||| -1
300||||| -(sq.rt.3)/2
330||||| -0.5
360||||| 0

As you can probably see from looking at the list of values, the numbers seem to have a repeating pattern. If you continue to add values, they will begin to repeat themselves again. Essentially, we have gone in a full circle (360 degrees), and so that is why they repeat.

So, taking these values that we have and plotting them on a graph, we can see what a sine curve looks like:



The cosine curve will look very similar, and you can deduce it exactly the same way. As I said before, they look the same but are out of phase. The important difference to remember is that the sine curve passes through (0,0), and the cosine curve passes through (0,1), but the 'periodicity' is the same... that is, they repeat at the same regular intervals. You can extend the curve into the negative x direction, or further in the positive direction.

I hope that explains what the original poster was asking me to go over. These functions, like any other function, can also be modified by adding coefficients etc, in the same way as other functions that I've reviewed before. I will explain these concepts in a future post. Or for fun, add a coefficient to the front of the sine function (eg. A*sin(x)), or to the x value itself (sin(2x)) and create tables of values and compare the resulting graphs!

Pythagorean Identities

Starting with the fundamental trig identities that I've already covered, a little bit of derivation leads to several new identities that will always hold true, and are especially useful in simplifying more elaborate equations.  Expanding upon my last post, there are a few other identities that can be derived from the Pythagorean Identity, which also go by the same name to describe the group of them, due to how they are deduced by making use of the his famous Theorem.  (Click here to get a great explanation about these Pythagorean Identities.)

If we start with the first Pythagorean identity, which I already explained:

Sin²(ɵ) + Cos²(ɵ) = 1

we can divide each side by the sine term to give something new.

[Sin²(ɵ) + Cos²(ɵ)  ] / Sin²(ɵ)  = 1 / Sin²(ɵ)

The first term (sin / sin) reduces to 1:

1 + Cos²(ɵ)  / Sin²(ɵ)  = 1 / Sin²(ɵ)

The remaining terms can be now be simplified.  We can apply the laws of exponents to group the terms and express them in a slightly different way, which allows us to apply additional trig identity substitutions to express things in terms of inverse functions:

1 + [Cos(ɵ)  / Sin(ɵ)]²  = [1 / Sin(ɵ)]²

1 + Cot²(ɵ)  = Csc²(ɵ)

This is the second of the three Pythagorean identities.  Hopefully my explanation is clear enough to understand, and you can follow along with my derivation of it.  As I always say on this site, if you understand the derivation, then you don't have to worry about making mistakes in memorizing the actual formula.  This is a perfect example of that, because the starting point is very easy to remember, but even I sometimes will confuse myself if I try to write down these additional identities without thinking about how to get there.  Learning the derivations of identities, formulas, and equations will always help you!

As I said, there are three Pythagorean identities.  This third one is derived in a similar manner to what we have already done, using the same sort of logical steps but dividing instead by the cosine term (rather than the sine term).  I am not going to write down all of the steps here, but I highly recommend that you grab a pencil and paper and quickly work through it, following along with the process that I outlined above.  There are only a few steps to work through, and in the end, this is what you should end up with:

1 + Tan²(ɵ)  = Sec²(ɵ)

As it is with many identities and equations in mathematics, it doesn't take very many steps to make something look completely different, as I've shown here.  I hope that you've been able to follow along and that everything makes sense.  If the steps in the derivations confused you at all, then I suggest going back to my posts on the basic trig functions, and working with those until you are comfortable.  The steps I have applied in these transformations were very simple substitutions, and I think that the most confusing part knowing what to look for in the first place.  With practice, you will learn to see patterns and recognize where you can apply various rules to make things simpler.

There are still several other trigonometric identities which I will show you how to derive in coming posts, many of which require the same types of thinking.  Stay tuned for more tutorials that explain things like the sine law and cosine law!

Fundamental Trigonometric Identities

Trigonometric identities are specific equalities that express one trigonometric function in terms of other trig functions. These fundamental trigonometric identities (also sometimes called basic trig identities) are fairly straightforward, but they take some work to derive them. If you are comfortable with simple derivations, you shouldn't have any problems though. Personally, I find it easier to remember the basic set of trig identities, and derive the more complex ones from those, rather than trying to memorize all of them... although some people are more comfortable just to memorize them.

These fundamental trigonometric identities are traditionally visualized as a right angle triangle inscribed within a circle of radius r, with sides formed by length x, height y, and the radius as the hypotenuse, r:

The basic trig definitions and relationships can easily be observed.  I have covered these trigonometric functions in other blog posts, so I won't go into them any further here, other than to just remind you of the concept of SOHCAHTOA.  Refer to these specific posts for more information about sine, cosine, and tangent.

Sin(θ) = y/r..... opposite  /hypotenuse
Cos(θ) = x/r..... adjacent / hypotenuse
Tan(θ) = y/x..... opposite / adjacent

For the first identity that I will show you how to derive, we need to make use of the Theorem of Pythagoras (I would also here like to refer you to my post about the Theorem of Pythagoras, which itself is derived from the cosine law, for more information on this trigonometric math concept.)  If we now apply the Theorem of Pythagoras to this right angle triangle we have inscribed within a circle, we can see:

r² = x² + y²

Dividing everything by r² gives:

1 = (x²)/(r²) + (y²)/(r²)
1 = (x/r)² + (y/r)²

And then, substituting in the basic trig identities listed above, we get:

1 = [Cos(θ)]² + [Sin(θ)]²

And that is the first of the 8 fundamental trigonometric identities. Nothing to it. In fact, this is an important mathematical expression.  As such, it has a special name or designation given to it to indicate that it is important.  It's proper name is the Pythagorean Trigonometric Identity. I'll rewrite it below in proper notation to clean it up a bit.  Notice how you can write the 2 (for the squares) in two different ways.  Either as above, like [Cos(θ)]², where you put the trig function in brackets to square it (note that you don't square the angle!), or more commonly as below, with the 2 placed next to the trig function to indicate that it is being squared.

Similarly, we can derive another basic relationship from a standard trigonometric identity.  For this one, it starts with:

Tan(θ) = y/x

Now to substitute in some more expressions.  This one is a little more complicated than how we arrived at the Pythagorean Identity above, but follow along and you should hopefully be able to see what I do.  The next step then, is to substitute in the basic Sine and Cosine definitions from the list at the top of this page (isolated for x and y, respectively) to give:

Tan(θ) = (r x Sin(θ)) / (r x Cos(θ))
Tan(θ) = Sin(θ) / Cos(θ)

And that's it again. Like the Pythagorean Trigonometric Identity I described above, this one also has a specific name.  This time, this one is called the Ratio Identity:

Those are now two of the simplest basic trig identities from which most of the others can be derived.  As I mentioned at the top, I always find that it is easier to memorize HOW to derive these fundamental trigonometric identities, rather than memorizing them explicitly and then possibly getting confused.  You WILL memorize them eventually (which always happens when you work with them so much!), but when you are first trying to understand them, I believe that it is always better to understand what is going on instead of just memorizing the final solutions.  This approach is a very wise strategy to apply throughout your mathematics training!


*Update 03-April-2012:  I have recently come back to this topic of basic trig identities, and have started a completely new post called the Trig Identities Cheat Sheet.  Please visit that site for a more thorough explanation of some of these derivations and fundamental identities of trigonometric functions.  Also, please click the +1 button below if you found this helpful!  Thanks.