The value of equals the circumference of a circle with radius 1/2. Suppose we ap
ID: 3588767 • Letter: T
Question
The value of equals the circumference of a circle with radius 1/2. Suppose we approximate the circumference by a polygon through n+1n+1 points on the circle. The length of this polygon can be found using the pathlength function from the previous exercise. Compute n+1n+1 points (xi,yi)(xi,yi) along a circle with radius 1/2 according to the formulae
Write a function circle_pts( n ) in Python which accepts a number of sides n and returns a pair of lists (as a tuple) containing the x and ycoordinates of the approximating polygon. You will need to include your working pathlength function from the previous exercise.
***Which I got :
from math import sqrt
def pathlength(x,y):
n = len(x)
lv = [sqrt((x[i]-x[i-1])**2 + (y[i]-y[i-1])**2) for i in range (1,n)]
L = sum(lv)
return L
Write a function approx_pi( n ) in Python for approximating by the path length LL resulting from the pairs of coordinate lists returned by circle_pts( n )
1 i=0, ,n.Explanation / Answer
from math import sqrt, cos, sin, pi
def pathlength(x,y):
n = len(x)
lv = [sqrt((x[i]-x[i-1])**2 + (y[i]-y[i-1])**2) for i in range (1,n)]
L = sum(lv)
return L
def circle_pts( n ):
xPoints = []
yPoints = []
for i in range(0, n+1):
x = 0.5*cos(2*pi*i/n)
y = 0.5*sin(2*pi*i/n)
xPoints.append(x)
yPoints.append(y)
return (xPoints, yPoints)
def approx_pi( n ):
(x, y) = circle_pts(n)
return pathlength(x, y)
print(approx_pi(100))
# code link: https://paste.ee/p/Y4gc8
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