Finding Length of Curve Represented by Parametric Equations

Miike012
Messages
1,009
Reaction score
0
I am having difficulty finding the length of the curve represented by parametric equations. The difficulty comes from not knowing how to determine if the curve transverses once, twice,... in a given interval.

The only solution I can think of is (say x = g(t) and y = f(t) and y = F(x) on interval [a,b] )
Look at graph x = g(t) and determine on the interval [a,b] if there are any values of x that repeat
If so then the graph y = F(x) on [a,b] may transverse more than once,
next look at the graph y = f(x) and determine if and y values repeat.

Then from these values one should be able to determine if the graph y =F(x) overlaps its self... but I know there must be a better method...
Help please.
 
Physics news on Phys.org
If the curve is given by x= g(t) and y= f(t) then dx/dt= f'(t) and dy/dt= g'(t) so dx= f'(t)dt and dy= g'(t)dt. The integral for curve length is
\int \sqrt{(dx)^2+ (dy)^2}= \int \sqrt{(f'(t)dt)^2+ (g'(t)dt)^2}= \int\sqrt{(f'(t))^2+ (g'(t))^2}dt

If the graph "overlaps itself" you only want to integrate over values of t that go over the path once. Choose some convenient point, (x_0, y_0), and find succesive solutions to f(t)= x_0, g(t)= y_0.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top