How do you integrate over a curve in 2D space that is parameterized by a polynomial?

Question

I have a simple curve in 2D space (x, y) that is parameterized by a polynomial as a function of t and I would like to know the length of that curve. How do I accomplish that? I looked into scipy.integrate and numpy.polyint but I didn’t manage to find a solution. It seems that both of them are only able to integrate over a 1D polynomial. Here is an example for a curve:

import numpy as np
from scipy import integrate


x0, y0 = 0.0, 0.0
vx, vy = 0.1, 0.1
ax, ay = -0.0001, 0
coeff = np.array([[ax, ay], [vx, vy], [x0, y0]])

pos = lambda t: np.polyval(coeff, t)

Asked By: mapf

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Source

Answer 1

The arc length is a single-variable polynomial on the curve parameter. You need to define the expression for the differential of the arc length and then you’ll be able to integrate over it, as explained in the link in the comments. As you can see there, it can be simply expressed as the Euclidean norm of the vector (dx/dt, dy/dt). Here is therefore how you can implement it:

import numpy as np
import scipy

x0, y0 = 0.0, 0.0
vx, vy = 0.1, 0.1
ax, ay = -0.0001, 0
coeff = np.array([[ax, ay], [vx, vy], [x0, y0]])

# Position expression is not really necessary
pos = lambda t: np.polyval(coeff, t)

# Derivative of the arc length
def ds(t):
    # Coefficients of polynomial derivative
    coeff_d = coeff[:-1] * np.arange(len(coeff) - 1, 0, -1)[:, np.newaxis]
    # Norm of position derivatives
    return np.linalg.norm(np.polyval(coeff_d, np.expand_dims(t, -1)), axis=-1)

# Integrate across parameter interval
t_start, t_end = 0, 1
arc_length, err = scipy.integrate.quad(ds, t_start, t_end)
print(arc_length)
# 0.1413506691471052

Of course, you could try to work out the analytical expression of the integral of ds and then you wouldn’t need any integration method.

Answered By: jdehesa

How do you integrate over a curve in 2D space that is parameterized by a polynomial?

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