Sunday, September 20, 2020

Division by 0

[Once again, I have written something for my class that I think will be valuable for others.]

Big question: What are the values of LaTeX: \frac{3}{0}, LaTeX: \frac{0}{3}, and LaTeX: \frac{0}{0}?

 

We want to be able to look at each of these fractions, know what it equals, and understand why. This becomes vital in calculus. [Note: Many students have trouble with this. It may be because elementary teachers are often uncomfortable with division, and teach it by memorization, instead of as something deep to understand. Or it may be that this is deep, and our brains need more time to really make sense of it.]

 

To help ourselves understand this, we tie it to something simpler that we understand better. Division is the inverse of multiplication (ie they undo each other). So it will help to explore how the two operations are connected.

We start with a very concrete and simple problem: LaTeX: \frac{6}{3}=2

[Note: One notational problem with division is that it's written in different ways that place the numbers in opposite orders. LaTeX: \frac{6}{3}=6\div3, but these are also equal to. When I was young, I had trouble keeping track of which was which, so I would write down an easy problem, like this one, to help me remember.]

Now we consider the multiplication problem that goes with this division problem: LaTeX: \frac{6}{3}=2\Longleftrightarrow3\cdot2=6, and we can say that 6 divided by 3 is 2 because LaTeX: 3\cdot2=6.

 

Let's use T for top, B for bottom, and A for answer, and rewrite this equivalence of a division problem and its associated multiplication problem, in a way that will always be true: LaTeX: \frac{T}{B}=A\Longleftrightarrow B\cdot A=T

In the fraction (or division), we have top over bottom gives answer, and that gives us a multiplication problem where the original bottom times the answer from the division gives us the original top.  [Note: I am purposely avoiding the proper terms: numerator or dividend, denominator or divisor, and quotient (for the answer). For anyone who gets those terms mixed up, it's easier just to focus on position for the moment.]

Now we are ready to consider each of the three original questions, using this correspondence.

1. Let's think about the multiplication associated with LaTeX: \frac{3}{0}:

LaTeX: \frac{3}{0}=A\Longleftrightarrow0\cdot A=3

So what do we multiply 0 by to get 3? Hmm. It seems that nothing works. There is no number that can multiply with 0 and give us 3. So the division problem (or fraction) has no solution, and we say that LaTeX: \frac{3}{0} is undefined.  This is why we say "division by 0 is undefined".

 

2. LaTeX: \frac{0}{3}=A\Longleftrightarrow3\cdot A=0. Ahh, this one is easier. LaTeX: 3\cdot0=0 so the answer is 0.

 

3. LaTeX: \frac{0}{0}=A\Longleftrightarrow0\cdot A=0. Hmm, this time A could be any number, and the multiplication would be correct. This is still division by 0, so it is still undefined, but it is very different from the first case. We call it indeterminate. We can see why by looking at a rational function example.

Example: LaTeX: y=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}

When x= -2 or 2, this function will be undefined (because we have division by 0). But the function's behavior for x values very close to -2 is very different from its behavior for x values very close to 2.

LaTeX: x=-2 is a vertical asymptote for the graph. This means that as x approaches -2, the y values approach LaTeX: \pm\infty. (This can be written "as LaTeX: x\longrightarrow-2,\:y\longrightarrow\pm\infty".) You can verify this by trying these x values: -2.1, -1.9, -2.01, -1.99,... (You can also use desmos to view the function.)

What happens near LaTeX: x=2? We see that the y value does not depend on the factor LaTeX: \left(x-2\right), because it cancels. So, as long as LaTeX: x\ne2, LaTeX: y=\frac{\left(x-1\right)}{\left(x+2\right)}. At LaTeX: x=2, this would equal 1/4. The function is not defined here, but now we can see that as LaTeX: x\longrightarrow2,\:y\longrightarrow\frac{1}{4}.

So why was LaTeX: \frac{0}{0} called indeterminate? Because the value associated with it in a particular function is determined by other parts of the function. Although LaTeX: \frac{0}{0} is undefined, we saw that, in this particular function the value of the function got close to 1/4 as the x value got close to 2, which is the number that would give us LaTeX: \frac{0}{0}. This concept goes with the concept of limits, one of the 3 major topics in calculus.

 

 

Wednesday, September 16, 2020

Friday, September 11, 2020

Solving Application Problems (in Trigonometry)

I started this blog in 2009, was active for about 6 years, and then not so much for the past 5 years. I wrote two posts in the spring, both related to online teaching. We were all trying to learn how to teach well as we scrambled to do it while learning. I was happy to keep seeing my students online, and Zoom was our class. I used Canvas a little but not much.

Over the summer I learned a lot about effective online teaching. (I'm still not sure it can ever be nearly as effective as in-person, but...) I developed my Canvas shells for each course, and I started the semester readier than I had expected to be. My Canvas shells are not done. I created a "module" that orients students to online learning and my course. And I created a module for our first unit. The rest is still in progress.

Today I added a page for my trig students, on solving application problems. I want to share it here. (And I may share lots of my Canvas "pages" here, sometimes with modifications.)

Years ago, I modified George Polya's wonderful outline of problem solving steps. We start with that. It's a good idea to print it out, and turn to it whenever you're stuck. 


tree with shadow, pretty vs helpful

Draw a Diagram.

Always start by drawing a diagram. This step is vital, and is a major part of "Understanding the Problem".

Your diagram does not need to be artistically good. It does need to show relationships well. An artist might show my shadow going off at an angle. But for a math diagram, it is better to show the right angle involved, as a right angle.

In the diagrams on the right, the top drawing is prettier, and the shadow is more evocative, but the bottom drawing shows the right angle between a vertical object and its horizontal shadow, which is what will help you do your mathematical analysis.

Example (#22 in 2.4, page 93): If the angle of elevation of the sun is 63.4° when a building casts a shadow of 37.5 feet, what is the height of the building?

Draw your diagram now, labeling it with everything given and a variable for the value requested. (My drawing is below.)

 

 

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building with shadow, labeled

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I labeled the height of the building h.

 

 

 

 

 

 

 

 

 

 

Write a Trig Equation.

In a simple problem, with only a few pieces of information this is all you need for the "Devising a Plan" step. We are given the value of the side adjacent (next to) the given angle, and we want to find the value of the side opposite the angle. (The hypotenuse is neither given nor asked for.) Which trig function uses adjacent and opposite? (Two of them do, but the one we use most of the time is...)

 

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... LaTeX: \tan\theta=\frac{opp}{adj}, and this gives us  LaTeX: \tan63.4=\frac{h}{37.5}


 

Do a bit of algebra.

This is the "Carry out the Plan" step. To solve for h, we multiply both sides of the equation by 37.5:

LaTeX: 37.5\cdot\tan63.4=37.5\cdot\frac{h}{37.5}\:\:\Longrightarrow\:\:h=37.5\cdot\tan63.4=74.8857...

I pulled out my calculator for that last step (making sure it was in degree mode). Since our given length was given to tenths of a foot, I round, and give my final answer as 74.9 feet.

 

Check your Solution.

This is the "looking back" step on the handout. If we look at our diagram, does a height of about 75 feet seem reasonable? Well, the height seems bigger than the shadow, and maybe about twice as big, so yes, it seems reasonable.

 

 

Practice.

If you get stuck on application problems, a good way to practice is to re-do problems that you've watched someone else do (perhaps on youtube). Try not to look at your notes. If you need to, go ahead and look. Do as much of the problem on your own as you can. If you looked at your notes at all, do it again the next day.

 
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