My Math Forum Length curve

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 February 23rd, 2009, 12:54 PM #1 Newbie   Joined: Feb 2009 Posts: 2 Thanks: 0 Length curve Hello I have a function, and need to find the distance of that polynomial function. Not sure how to do this any help. Notes: I think its a degree four, it might be a three, i have five points. Starting fron the origin, (0,0) Point O, Point A (2,-2), Point B (6, 2.15) point c (10,-3) Point E (16,7). Im not really good at this stuff. Put it in my calculator and get this a= 6.555e-03 b=-0.1810119 c=1.44681547 d=-3.2220238 e=5.902e-11 and r^2=1 Below it, it says y=ax^4+bx^3+cx^2+dx+e so i figure i sub a,b,c,d,e into that, and that how i get my function y=6.555e-03x^4 - 0.1810119x^3 + 1.44681547x^2 - 3.2220238x + 5.902e-11 So how do i use this in that arc length formula, dad said to use string lol These may help http://www.mathwords.com/a/arc_length_of_a_curve.htm http://www.maths.abdn.ac.uk/~igc/tch/ma ... ode21.html
 February 23rd, 2009, 02:05 PM #2 Member   Joined: Feb 2009 Posts: 34 Thanks: 0 Re: Length curve hi watty08 using your points you can calculate the parameters of your polynom aX^4+bX^3+cX^2+dX+e If they are points laying on your curve it mean than for example for the first points O(0,0) that mean you plug 0 you have to get 0 so f(0)=0 so e=0 for the other point A(2,-2) mean f(2)=-2 that mean you plug 2 you have to get -2 so 16a+8b+4c+2d=-2 and so on You arrive at a system eq with 4 eq and 4 unknown variable a, b, c, d (e=0) the system is: 16a+8b+4c+2d=-2; 1296a+216b+36c+6d=2.15; 10000a+1000b+100c+10d=-3; 65536a+4096b+256c+16d=7 I get the solution for you a=0.0066; b=-0.1810; c=1.4468; d=-3.222; if you plug those numbers in your system every eq is satisfied Ok i think is still a problem with your number here because everything should be very simple you have to check this...So your function is f(x)=0.0066X^4-0.1810X^3+1.4468X^2-3.222X a little bit nasty because you have to do it's derivative df/dx and after that the length is L=integral from 0 to 16 (your last coord point) from (sqrt(1+[df/dx]^2)) is very complicate to do this if the function is not very simple that you can get off the square root Good luck Check your numbers again
 February 23rd, 2009, 09:08 PM #3 Newbie   Joined: Feb 2009 Posts: 2 Thanks: 0 Re: Length curve How did you get a,b,c,d did you get them by using similtanious equations> Are you also saying that i differentiate the f(x)=0.0066X^4-0.1810X^3+1.4468X^2-3.222X And then sub that into the equation (sqrt(1+[df/dx]^2)) If so wont i still have the X's left over. Thanks heaps for your help
February 24th, 2009, 05:07 AM   #4
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Re: Length curve

Quote:
 Originally Posted by watty08 How did you get a,b,c,d did you get them by using similtanious equations
Yes he used simultaneous equations, and came up with the same results as the Quartic regression feature on your calculator.

Quote:
 Originally Posted by watty08 Are you also saying that i differentiate the f(x)=0.0066X^4-0.1810X^3+1.4468X^2-3.222X And then sub that into the equation (sqrt(1+[df/dx]^2)) If so wont i still have the X's left over.
Yes, dy/dx is the derivative of your function f(x)=0.0066X^4-0.1810X^3+1.4468X^2-3.222X

Arc length for this problem is:
$\int^{16}_{0} \sqrt{1+\left(\frac{dy}{dx}\right)^2 \, dx$

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