Distance formula - Part II

Continuing from my last post, where I told you that the distance between two points can be determined by finding the length of the hypotenuse of the right angle triangle formed from the two points.

We already know that to find the length of the hypotenuse, we apply the Theorem of Pythagoras, which says that c^2 = a^2 + b^2. (The square of the hypotenuse is equal to the sum of the squares of the remaining two sides). So from this, then, we see that we need to determine the lengths of the two sides that we have created by extending lines through our points to join at a right angle. And to find these lengths, all we need to know are the coordinates of our ! 2 points!

For a general triangle, then, we have something like this:

To find the horizontal length, it is just the difference between the two x-coordinates (ie. x2-x1). Think of it as taking a stick that is x2 units long, and chopping off a length of that stick that is x1 units long. The stick that you are left with, x2-x1, is the length of our horizontal side.

The same reasoning applies to find the vertical length, which is the difference between the two y-coordinates (ie. y2-y1). One comment I will make here, is that since we are talking ab! out a length of a side, the length has to be the absolute diff! erence b etween the two points (ie. you can't have a negative side length).

So then, for our general triangle, we have our 2 lengths, and let's call the hypotenuse "d" (as in, the distance between the two points).

So then, if we apply the Theorem of Pythagoras to this triangle we have created, we can come up with the distance formula very easily!

c^2 = a^2 + b^2...... which we can change to read:

d^2 = (x2-x1)^2 + (y2-y1)^2

And so we have:

d = sqrt [(x2-x1)^2 + (y2-y1)^2]

Let's quickly try with 2 points. You can draw the triangle out as I have above to follow along more closely. I will just do the quick calculation for you though.

Find the distance between the points (1,2) and (3,5).

d = sqrt [(3-1)^2 + (5-2)^2]

d = sqrt [(2)^2 + (3)^2]

d = sqrt [4 + 9]

d = sqrt [13]

And that's all there is to it. I hope that I've been able to clearly explain how to derive the distance formula. Once you know where a lot of these formulas come from, you'll never have to worry about memorizing them again! :)

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Square Roots and Great Comebacks

From the time I learned about them, I've been fascinated (probably to an unseemly amount) by the square root. I remember reading about a method for calculating square roots by long hand. There's no point, really; we have calculators to do that for us. (If you have some spare time and you enjoy this sort of thing, see if you can figure out the algorithm from the example at left.)

What use are square roots, anyway, aside from solving math problems about ! the diameters of circular lawns? (Have you ever seen any of those? They must encircle those conical swimming pools we dealt with in calculus class.) Here's one use: They can tell you when how big a lead your favorite basketball team needs to be secure in a win.

A few years ago, I derived a rule for determining when a lead was safe in a basketball game--specifically, an NBA game. (It matters, because the shot clock is different between an NBA game and a WNBA game and a NCAA men's game and a NCAA women's game.) You take the square root of the number of seconds left, and add three. For instance, if there's 3:45 left in the game, that's 225 seconds. Square root of 225 is 15, and you add 3, so an 18-point lead is pretty darned safe with 3:45 left. The "add 3" is for a trey at the buzzer. Go ask the Miami Heat about that 'round about now.



Pretty keen, huh? Although--not to put too fine a point on it--well-known sports statistician Bill James also came up with this very same rule. We'll call it independent discovery, at least on my part. I have no idea whether James stole it from me. Give him the benefit of the doubt, though.

But why? Why should this rule work? Why isn't it just the time remaining divided by some rate at which the team that's behind catches up? If a team can make up a 15 points in 225 seconds and then cap that with a trey to make up the 18, why can't it make up 33 points in 7:30? Or 63 points in 15:! 00?

And the sort-of answer to that is, it can. It's just terribly unlikely. Of course, it's already unlikely that a team can make up 15 points in 3:45, but it's still in the realm of possibility. Asking a team to do that twice in a row is just too much. If it was 100 to 1 against doing it once, doing it twice in a row would be 10,000 to 1 against. On the other hand, making up the same 15 points in twice the time is obviously easier. So in twice the time (7:30, natch), you should be able to make up some deficit in between. According to both me and Bill James--and honestly, are you going to go against both of us?--that deficit is 15 times the square root of 2. That's about 21, and if you add the 3 at the end it makes it 24.

Where on earth does this come from? One place is the drunkard's walk, otherwise known as the random walk (but I think "drunkard's walk" is more evocative). In this mathematical scenario, the eponymous drunkard starts off at some placemark--a lamppost, say. Each moment in time, he takes a step, but in a completely random direction. Might be in the same direction as the last step, might be in the opposite direction, might be anything. So after a bunch of steps, he might end up back at the lamppost where he started...or he might be home.

Odds are, though, he'll be at some intermediate distance. How far from the lamppost? Well, the first step is g! oing to take him one step away for sure. We'll represent this by saying that d(1) = 1, where d(t) is the distance of the drunkard from the lamppost at time t. OK, now what about d(2)? Before that second step, he's one step away from the lamppost. His second step might take him two steps away, if he walks in the same direction, or zero steps away, if he walks in the opposite direction (back toward the lamppost). On average, though, he'll walk in some intermediate direction: let's say, perpendicular to his current progress from the lamppost. The Pythagorean theorem says then that

[d(2)]² = [d(1)]² + 1² = 1² + 1² = 1 + 1 = 2

or, in other words! , d(2) = √2. We can go further. We've already got two examples where d(t) = √t and we'd like to get more. To do that, we'll use a process called induction. Suppose that you have a value of t for which d(t) = √t ; we'll now try to show that d(t) = √(t + 1) . Using the same argument as before--that the drunkard walks in some intermediate direction, we get

[d(t +! 1)]² = [! d(t)]² + 1² = [√t]² + 1² = t + 1

and then we directly get d(t + 1) = √(t + 1) . So as long as we can find a t wh! ere d(t) = √t , we're set; it's true for all greater values of t. But we already have such a value: t = 1! (And t = 2, for that matter.) It turns out, then, that the drunkard's walk, after time t, takes him a distance √t away from the lamppost.



Now, a couple of things. First, this isn't anything like a rigorous demonstration of the square root property of the drunkard's walk. You can look that up if you like. But if you work at it a little, it gives you an inkling of the intuition behind it. Secondly, though, and here we're back on track a bit: What has all this got to do with basketball games?

A basketball game is an alternating sequence of possessions. In each possession, the team with the ball is of course trying to score, and the other team is of course trying to prevent it from scoring. When the ball changes hands, the roles are reversed. In each individual possession, the effect on the score is biased: Only the team with the ball can score, usually. But in each pair of possessions, that bias cancels out, since both teams get a chance with the ball. The margin in the game can move in any direction--just like the drunkard's walk.

If the drunkard starts off 50 steps from home, he could conceivably get home in just 50 steps. But it's ridiculously unlikely: Each of those 50 steps would have to be in exactly the right direction. The square root property ! tells us he'll probably be just a bit over 7 steps from the la! mppost; it would take 2500 steps to get him, on average, 50 steps from his starting point. After those 2500 steps, is he guaranteed to be home? Nope. He still has to be walking in the right direction. But it's at least plausible now.

In the same way, a basketball team that's down 18 points could conceivably make that up by scoring six three-pointers in a row while holding their opponents scoreless. If they did that by fouling and their opponents obliged by missing all of their free throws, the whole deficit could be made up in half a minute or so. But that's as unlikely as the drunkard walking 50 steps in exactly the right direction. Instead, a team will make up its deficit in halting fashion, sometimes making up three points, but other times giving up a point, or staying even, in any particular pair of possessions. The drunkard's walk, in other words, and that's why the square root rules great combacks.

I was going to follow this up with a discussion of sociology and mobilizing people, but this post is getting long (see, I do notice it!) and I'll defer that till next time.

square root rules

Solving Equations Using Cramer's Rule

Here is the video on how to solve equations using Cramer's Rule:





The slides for the! above video are as follows:

Slide 1 of 6:



Slide 2 of 6:



Slide 3 of 6:



Slide 4 of 6:



Slide 5 of 6:



Slide 6 of 6:

solving linear equation

Algebra Tutor will help you solve your problem

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Knowledge of Math

Introduction to algebra fraction calculator:

The term algebra states the constants and variables in the equation. Algebra is one of the familiar topics in mathematics. Algebraic fraction is similar to the usual fraction but a difference is that the numerator and the denominator are algebraic expressions. Algebraic fraction simplifying is similar to that of algebraic fractions. Simplifying algebraic expressions is very easy and simple. In online, students can learn about algebraic expressions. In online, algebraic calculator is helpful to solve the given problem. In online calculator, when the denominator and numerator value is entered, calculator automatically simplifies the problem.

Example : Simplifying Algebraic Fractions

Simplify 3b/9b².

Solution:

The given problem is an algebraic expression.,

3b = 3*b

9b² 9*b*b

Like terms gets cancelled,

On simplifying this,

Answer = 1/3b

Need More help with Algebra 1

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Example 7.28: Bubble plots

A bubble plot is a means of displaying 3 variables in a scatterplot. The z dimension is presented in the size of the plot symbol, typically a circle. The area or radius of the circle plotted is proportional to the value of the third variable. This can be a very effective data presentation method. For example, consider Andrew Gelman's recent re-presentation of health expenditure/survival data/annual number of doctor visits per person. On the other hand, Edward Tufte suggests that such representations are ambiguous, in that it is often unclear whether the area, radius, or height reflects the third variable. In addition, he reports that humans tend not to be good judges of relative area.

However, other means of presentin! g three dimensions on a flat screen or piece of paper often rely on visual cues regarding perspective, which some find difficult to judge.

Here we demonstrate SAS and R bubble plots using the HELP data set used in our book. We show a plot of depression by age, with bubble size proportional to the average number of drinks per day. To make the plot a little easier to read, we show this only for female alcohol abusers.

SAS

In SAS, we can use the bubble statement in proc gplot. We demonstrate here the use of the where data set option (section 1.5.1) for subsetting, which allows us to avoid using any data steps. SAS allows the circle area or radius to be proportional to the third variable; we choose the radiu! s for compatibility with R. We alter the size of the circles ! for the same reason. We also demonstrate options for coloring in the filled circles.

libname k "c:\book";

proc gplot data = k.help (where=((female eq 1) 
                    and (substance eq "alcohol")));
bubble cesd*age=i1 / bscale = radius bsize=60 
                           bcolor=blue bfill=solid;
run;

R

In R, we can use the symbols() function for the plot. Here we also demonstrate reading in data previously saved in native R format (section 1.1.1), as well as the subset() function and the with() function (the latter appears in section 1.3.1). The inches option is an arbitrary scale factor. We note that the symbols() function has a great deal of additional capability-- it can substitute squares for circles for plotting the third variable, and add additional dimensions with rectangles or stars. ! Proportions can be displayed with thermometers, and boxplots can also be displayed.

load(url("http://www.math.smith.edu/sasr/datasets/savedfile"))
femalealc = subset(ds, female==1 & substance=="alcohol")
with(femalealc, symbols(age, cesd, circles=i1, 
                                 inches=1/5, bg="blue"))

The results are shown below. It appears that younger women with more depressive symptoms tend to report more drinking.

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