Integral for calculating length of the curve

In summary, the conversation discusses a curve defined by a parametric equation and the need to calculate its length using the arc length formula. The formula includes an integral that cannot be evaluated explicitly for the given equation, requiring the use of special functions such as the incomplete elliptic integral. The conversation also mentions the possibility of using MATLAB or Mathematica to evaluate the integral and suggests the use of Maxima as well.
  • #1
nawidgc
25
0
I have a curve defined by following parametric equation:

\begin{equation}
\gamma(\theta) = 1 + 0.5 \times \cos (N \theta) (\cos(\theta),\sin(\theta)), 0 \leq \theta \leq 2 \pi \
\end{equation}

I need to calculate the length of the curve between say θ = 0 to θ = 1.0
Formula for calculating the arc length of a curve in polar form is well known (see http://mathworld.wolfram.com/ArcLength.html Eq. 17 therein). Using this formula

\begin{equation}
s = \int\limits_{\theta = 0}^{\theta = 1.0} \sqrt{r^2 +\left(\frac{dr}{d\theta}\right)^2}
\end{equation}

where for N = 10, r is given as

\begin{equation}
r = 1 + 0.5 \times \cos(10 \theta)
\end{equation}

and

\begin{equation}
\frac{dr}{d\theta} = -0.5 \times \sin(10\theta)
\end{equation}

The problem is I can't evaluate the integral in Eq. 17 from the Mathworks link for the above equation explicitly. Obviously one can use Matlab ( with trapz command ) to evaluate the integral for given limits but I want an explicit expression for the indefinite integral in terms of generic θ1 and θ2 ( I have hundreds of such integrals to evaluate and difference between θ1 and θ2 is not constant) How do I evaluate the integral? Is there a substitution possible?

Many thanks for help.
 
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  • #2
nawidgc said:
I have a curve defined by following parametric equation:

\begin{equation}
\gamma(\theta) = 1 + 0.5 \times \cos (N \theta) (\cos(\theta),\sin(\theta)), 0 \leq \theta \leq 2 \pi \
\end{equation}

I need to calculate the length of the curve between say θ = 0 to θ = 1.0
Formula for calculating the arc length of a curve in polar form is well known (see http://mathworld.wolfram.com/ArcLength.html Eq. 17 therein). Using this formula

\begin{equation}
s = \int\limits_{\theta = 0}^{\theta = 1.0} \sqrt{r^2 +\left(\frac{dr}{d\theta}\right)^2}
\end{equation}

where for N = 10, r is given as

\begin{equation}
r = 1 + 0.5 \times \cos(10 \theta)
\end{equation}

and

\begin{equation}
\frac{dr}{d\theta} = -0.5 \times \sin(10\theta)
\end{equation}

The problem is I can't evaluate the integral in Eq. 17 from the Mathworks link for the above equation explicitly. Obviously one can use Matlab ( with trapz command ) to evaluate the integral for given limits but I want an explicit expression for the indefinite integral in terms of generic θ1 and θ2 ( I have hundreds of such integrals to evaluate and difference between θ1 and θ2 is not constant) How do I evaluate the integral? Is there a substitution possible?

Many thanks for help.

I missed dθ in Eq. (2) in my post above. The second equation should read

\begin{equation}
s = \int\limits_{\theta = 0}^{\theta = 1.0} \sqrt{r^2 +\left(\frac{dr}{d\theta}\right)^2} d\theta
\end{equation}
 
  • #3
The desired integral can't be expressed in terms of elementary functions; you need special functions, specifically the incomplete elliptical integral of the second kind. See this WolframAlpha link and this Mathworld page. In order to evaluate this special function, you can do this in MATLAB or this in Mathematica.
 
  • #4
lugita15 said:
The desired integral can't be expressed in terms of elementary functions; you need special functions, specifically the incomplete elliptical integral of the second kind. See this WolframAlpha link and this Mathworld page. In order to evaluate this special function, you can do this in MATLAB or this in Mathematica.

There was an error in dr/dθ:redface:

\begin{equation}
\frac{dr}{d\theta} = -5*\sin(10\theta)
\end{equation}

I tried http://www.wolframalpha.com/input/?i=integrate+sqrt%28%281%2B.5*cos%28N*theta%29%29^2%2B%28-N%2F2*sin%28N*theta%29%29^2%29+d%28theta%29 but the page times out. Do I need a pro version? Can this be done in Maxima?
 

What is the integral for calculating the length of a curve?

The integral for calculating the length of a curve is a mathematical concept used to find the length of a curve on a graph. It involves breaking the curve into smaller segments and finding the sum of their lengths using the Pythagorean theorem.

How is the integral used to find the length of a curve?

The integral is used to find the length of a curve by calculating the sum of the lengths of smaller segments of the curve. This is done by using a definite integral, where the limits of integration represent the starting and ending points of the curve.

What is the formula for finding the length of a curve using the integral?

The formula for finding the length of a curve using the integral is given by: L = ∫√(1 + [f'(x)]²) dx where f'(x) represents the derivative of the function.

Can the integral be used to find the length of any curve?

Yes, the integral can be used to find the length of any curve, as long as the curve can be represented by a continuous function. This includes curves in two or three dimensions.

What are some real-life applications of using the integral to find the length of a curve?

The integral for calculating the length of a curve has many real-life applications, such as finding the length of a roller coaster track, calculating the distance traveled by a car on a curved road, or determining the length of a river or coastline. It is also used in engineering, physics, and other fields to calculate the distance traveled by a moving object along a curved path.

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