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Archive for the ‘Formulas’ Category

How to quickly find the y-intercept (b-value) of a line


Of course there’s a standard way to find the y-intercept of any line, and there’s nothing wrong with using that approach.

High-Octane Boost for Math
High-Octane Boost for Math Ed

But the method I’ll present here is a bit faster and therefore easer. And hey, if we can save time when doing math, it’s worth it … right?

So first let’s recall that the y-intercept of any function is the y-value of the function when the x-value = 0. That’s because the y-intercept is the y-value where the function crosses or touches the old, vertical y-axis, and of course all along the y-axis the x-value is always 0 (zero).

So the standard slope-intercept formula is y = mx + b. In a problem asking for the y-intercept, you’ll be given one point that the line passes through (that point’s coordinates will provide you with an x-value and a y-value), and you will also be told the slope of the line (the line’s m-value).
So then, to get the b-value, which is the value of the y-intercept, you just grab your y = mx + b equation (dust it off if you haven’t used it in a while), and plug in the three value you’ve been given: those for x, y and m. Then you solve the equation for the one variable that’s left: b, the value of the y-intercept.

Let’s look at an example: a line with a slope of 2 passes through the point (3, 10). What is this line’s y-intercept.

Now, according to the problem, the m-value = 2, the x-value = 3, and the y=value = 10. We just take these values and plug them into the equation:
y = mx + b, like this:

10 = (2)(3) + b

After doing these plug-ins, you just solve the equation for b, finding that
b = 4. That means that the y-intercept of the line = 4.

Now let’s see how you can do the same problem, but a little bit faster.
To do so, we first need to play around with the y = mx + b equation by subtracting the mx-term from both sides, like this:

y = mx + b [Standard equation.]
– mx = – mx [Subtracting mx from both sides.]
y – mx = b [Result after subtracting.]
b = y – mx [Result after flipping left & right sides
of the equation above.]

Aha! Look at that final, beautiful equation. This equation has b isolated on the left-hand side. So now if we want to solve for b, all we do is plug in the x, y and m values into the right-hand side of the equation and simplify the value, and the value we get will be the b-value.

For the problem we just solved, with x = 3, y = 10, m = 2, watch how easy it is to solve:

b = y – mx
b = 10 – (2)(3)
b = 10 – 6
b = 4

So notice that this technique, just like the first technique, reveals that the
y-intercept of the line is 4, or (0, 4). The techniques agree, they just get to the same end in slightly different ways.

Notice that with the second, quicker technique, you don’t need to add or subtract any terms. And that’s a key reason that this technique is faster and easier to use than the standard method. So try it out and stick with it if you like it.



Fun Math Problem #2


Here is the second in my series of “Fun Math Problems.”

Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit! I will post the answer to the problems two days later, after people have had time to respond.

To provide your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.
Please show how you worked the problem. Thanks. I will post the names of the first three people who get this right.

The Problem:  Before you go out to lunch, you glance at the clock above your desk. When you come back from lunch, you glance at the clock again, and you notice something strange. The minute and the hour hand have exchanged places from the positions they had just before you went to lunch.

The question is:  how long were you away?

Rusting clock face

Image by The Hidaway (Simon) via Flickr

Answer, Fun Math Problem


Answer to problem about the circular and square pegs and holes.

The “fit” for each situation is the following ratio:
(Area of Inner Figure) ÷ (Area of Outer Figure)

For the square peg in a round hole —
Call the radius of the circle r.
Then the diagonal of square “peg” = 2r
Notice that by slicing the square along its diagonal,
we get a 45-45-90 triangle, with the diagonal being
the hypotenuse and the sides being the two equal legs.
Using the proportions in a 45-45-90 triangle,
side of square peg = r times the square root of 2
Multiplying this side of the square by itself gives
us the area of the square, which comes out as:
2 times the radius squared

This being the case,
Area of square is: 2 times radius squared, and
Area of circle is: Pi times radius squared, and so …

Cancelling the value of the radius squared, we get:
Ratio of (Area of square) to (Area of circle) is:
2÷Pi = 0.6366

For the round peg in a square hole —
Call radius of the circle r.
And since the diameter of the circle is the same length as
the side of the square, the side of the square = 2r
Multiplying the side of the square by itself to get the
area of the square, we find that the area of the square
is given by: 4 times radius squared.

This being the case,
Area of circle is: Pi times radius squared
Area of square is: 4 times radius squared, and so …

Ratio of (Area of circle) to (Area of square) is therefore:
Pi ÷ 4 = 0.7854

Of the two ratios, the ratio of the circular peg in a square hole
is greater than that of the square peg in a circular hole.

Therefore we can say that the circular peg in a square hole
provides a better fit than a square peg in a circular hole.

And that is the answer!

FUN MATH PROBLEM — Circling the Square & Vice-Versa


From time to time I will post interesting math problems.

Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit!

I will post the answer to the problems two days later, after people have had time to respond.

To post your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.

The problem: Which provides the fuller fit? Putting a circular peg in a square hole, or putting a square peg in a circular hole? To get credit, show all work, and justify your answer by expressing each “fit” as a percent.

A few term-clarifications, to help you do this correctly:

a) By “fit,” I mean the ratio of the smaller shape to the larger shape, expressed as a percent. For
example, if a ratio is 4 to 5, that would represent a “fit” of 80 percent.

b) For the circular peg in the square hole, assume that the diameter of the circle equals the side of the
square. For the square peg in a circular hole, assume that the diameter of the circle equals the diagonal of the square.

c) By “fuller fit,” I mean the larger of the two ratios.

Have fun!

Not all variables are created equal


Are all variables the same?

Does every variable serve the same purpose?

When you think about it, you’ll see that the answer is “no.” Variables serve different purposes. When we explain this to students, we help them understand how variables work. Explaining this helps students understand how algebra “works.” You’ll see what I mean in a moment.

Consider the famous slope-intercept equation:  y = mx + b

A student recently asked me:  Are the  x and y variables the same as the m and b variables? What a great opportunity to explain something important!

I explained that the x and y variables serve completely different purposes than  the m and b variables. Here’s how.

The variables m and b are what I call “identifier” variables. By which I mean that they help us identify a specific line. To explain that, I asked the student a set of questions about something everyone understands — home addresses.

What would happen, I asked, if someone wanted to know where I live, and I told him that I live at 942? The student replied that this would not be enough info.

Then I asked, what if I told this person only that I live on Vuelta del Sur (a street name where I live in Santa Fe, NM)? Again the student said that this would not be enough info.

But what if I told this person that I live at 942 Vuelta del Sur. This, the student realized, would be enough information to enable someone to find my house. (All they have to do is Google me, and they’ll have my house AND directions!)

I pointed out that a similar situation applies to lines.

If I have a specific line in mind, and I want someone else to know the line I’m thinking of, is it enough to give this person just the line’s slope? No, for it could be any line with this slope, of which there are infinitely many parallel lines. What if I don’t give the slope but I do give the line’s y-intercept? Still not enough, as there are infinitely many lines that run through this y-intercept. But what if I tell the person both the slope and the y-intercept. Aha! The student could see — through drawings I made of this situation on a coordinate plane — that when you provide both slope and the y-intercept, there is one and only one line that could be indicated.

 

Three lines — the red and blue lines have the ...

Red & blue lines have same slope, so slope alone does not indicate a specific line; Red and green lines have same y-intercept, so y-intercept alone does not identify a specific line.

 

I explained that variables like m and b, which help identify a specific line, are “identifier” variables; their job is to identify a specific line. If your students are more advanced, you can explain that there are other identifier variables in different kinds of equations. For example, in the equation of a parabola:   y – k  = a(x – h)^2, the identifier variables would be the variables a, h, and k.

But what about variables like x and y? What do they do? What is their purpose?

These variables, I explained, have a completely different purpose. I call variables like x and y “ordered-pair generators.”

To explain this, I show students a simple linear equation like  y = 2x, and demonstrate how, using a “T-table,” you can use this equation to generate as many ordered pairs as you’d like, ordered pairs like (0,0), (1,2), (2,4), (3,6), etc. Point out that you can keep going and going. And then explain that the purpose of the x and y variables is to generate the infinitely many points that make up the line.

So the m and b variables tell us where the line is, and the x and y variables allow us to find the infinitely many actual points on the line. The two sets of variables, while different in purpose, work together toward a common goal:  to give us the equation of a line.

There are other purposes that variables serve, of course. And I’ll probably describe some of the other purposes in future posts. But the main point is that it helps students to recognize that variables do serve different purposes. Armed with that understanding, they can make much more sense of algebra’s formulas and equations.

ANSWER TO FRIDAY’S PROBLEM:


ANSWER TO FRIDAY’S PROBLEM:

The problem, once again:

For any polygon, a “diagonal” is defined as a line segment that runs from one vertex  to another, running  through the polygon’s interior. Find a formula that determines the number of diagonals in any convex polygon with n sides. Once you have the formula, use it to figure out the number of diagonals in a convex polygon with 1,000 sides (don’t try this by hand! — that’s why algebra was invented).

The winning answer was provided by Chris Mark. The formula, for a convex polygon with n sides, is this:  Number of Diagonals =  [n(n– 3)]/2. For n = 1000, the number of diagonals = 498,500.

The reasoning behind the formula. A polygon has as many vertices as sides. So a polygon with n sides also has n vertices. Now, consider any vertex of the n-gon. From that vertex the number of diagonals that can be drawn is (n – 3). That is because we cannot draw a diagonal to 3 vertices:  the vertex chosen, and the two adjacent vertices. So the expression (n – 3) = the number of diagonals that can be drawn from any vertex. We multiply (n – 3) by n to obtain the total number of diagonals that can be drawn from all n vertices. But if we simply multiply n by (n – 3), we’d be counting each diagonal twice. To eliminate that problem, we divide the product, [ n(n – 3)], by 2, and that provides the correct formula:

Diagonals  =  [n(n – 3)]/2

Applying this formula to a convex polygon with 1,000 sides, we see that the number of diagonals =  498,500.

In addition to providing the answer, Chris pointed out that the problem need not be restricted to regular polygons, as it was when posted. This formula works for all convex polygons, regular or not.

Thanks, Chris. And thanks to everyone who submitted answers.

Challenge Problem – Polygon Diagonal Formula


Here’s a challenge problem for anyone who’d like to try it.

On Monday I will post the answer and the names of the first five people who got this right. So good luck, everyone.

A regular polygon is a polygon all of whose sides are congruent and all of whose  angles are congruent. For any polygon, a “diagonal” is defined as a line segment that runs from one vertex of the polygon to another, and which runs through the interior of the polygon.

Find a formula that tells how to determine the number of diagonals there are in any regular convex polygon with n sides.

Once you have the formula, use it to figure out the number of diagonals in a regular convex polygon with 1,000 sides (don’t try this by hand! — that’s why algebra was invented).

Good luck!