8.1 Arc Length
Ok, well, today we're going to figure out how to find the the length of an arc. It really is a beautiful thing. Now, remember that we always like to do things in terms of x and y. So, in order to express a constantly curving line, we will simply use an infinite number of hypotenuses (or is it hypoteni?) of right triangles. In order to express the hypotenuse in terms of y's and x's , we use pythagorean theorem:
*IMPORTANT REMINDER*: The legs must always be in terms of dy and dx, or the change in y and the change in x, for the given interval of the hypotenuse. (So you would start at one end of the hypotenuse and go to the other end, measuring how much the y changed and how much the x changed.)
But of course, we need the sum of all these hypotenuses in a given domain. So we use the integral sign and a dx. But....so that we don't change the nature of the function, we also DIVIDE by the dx (dx/dx =1, and thus the value remains unchanged.)
Note the dividing by dx on the bottom.
So now, we want to try and simplify that devilish little formula. So....we put the bottom dx inside the square root. And to do that, we need to square it, so that we don't change the value of the function. (The square root of dx squared is of course dx, what we had in the first place)
Alright. Excellent. Now that we've done that, we can simplify what's inside those parentheses, since dxsquared over dxsquared is 1:Unfortunately, Gianna, there is no +C when we have it definite like this. But that's ok. You still get it, right?
If you notice, in parentheses, we have (dy/dx), WHICH IS THE SAME THING AS THE DERIVATIVE OF THE FUNCTION!!! ISN'T THAT WILD???? So it looks like this:
*IMPORTANT REMINDER*: f(x) MUST be continuous on [a,b] and x must fall between or equal to a and b.
It's very true that we could have skipped all that and cut to the chase by going right to this formula. But I think that Mr. French and I would both agree that we all should value process over product. PLUS IT'S JUST PLAIN COOL TO SEE HOW WE DERIVED THAT FORMULA!!!
Ok. So let's test this theory.
Let's use this math problem:
Find the length of the following curve.
To sort of illustrate, this is what the graph looks like (Fortunately, unlike vons, my images actually came out pretty decently the first time round):
And this is the specific portion that we have to find the length of.
Let's go ahead and take the derivative. Wait! Just kidding. First we should put the function in more recognizable terms (After all, we don't want it to be so scary-looking that a certain someone would just start frantically trying to change it into slope-intercept form.)
Note that I simply made the coefficeints more recognizable. I changed the divide by 6 and instead multiplied x to the 5th by 1/6 (the same thing). Then I took the xcubed out from the denominator of the second term and multiplied (1/10) by x to the -3 (same thing).
Now take the derivative for real this time:
Let's plug it into our equation:
*IMPORTANT REMINDER*: Dont' forget to add 1 inside the integral!!!! (After all, we know that some people in our class seem to have a bit of a problem with adding.......Go Kristin!)
And plug it into our TI-83's and 84's (Or in John Cynn's case TI-89 Platinum Addition. Man, he thinks he's so cool with it, doesn't he?) with the good ol' fnInt(Y1, X, 1, 2,).
Of course I'm sure you could do the complicated way and do it by hand, taking the antiderivative of that and then plugging in 1 and subtracting it from what you get when you plug in two. But I'm pretty sure the only people arrogant enough to do that would be Mark Chodas. (No offense, Mark. You know we love you.)
This is a great site:
There's obviously not much to explain abou this topic. So I found a site that gives a lot of example problems and shows how to solve them!! :)
So as many know, I am a big fan of Desperate Housewives, and I have no shame in telling people that.....
WHO IS YOUR FAVORITE HOUSEWIFE?
a) Gabrielle Soliz
b) Edie Britt
c) Lynette Scavo
d) Susan Meyer
e) Bree VandeKamp
(Feel free to leave a comment about your choice.)
Whoever can guess where this is from first will get a prize.....!!!!!! (And it'll be good.)