4. In the class, you learn a bunch of tools. But, in the homework, I need you to be a little bit creative and use all you have learned to solve a seemingly random problem. Let us consider semi-circular paper of radius 2. The area of the whole paper is 2π ≈ 6.28. We want to cut a chord(red line) from this paper such that the area of the leftover paper(shown in blue) is 4.0. Where should I cut the chord (find the value of a where the red line should be such that the area of the blue region is 4.0? Try to get a decent answer. If you are good with geometry you can actually figure this one out analytically, but I want you to do it numerically using what we have learned so far. return sqrt(r**2 - ***2) ▷ 1 def circ_top(x, r=2.): 2 3 4 5 [3] def circ_bot(x, r=2.): return -sqrt(r**2 6 x=np.linspace(0,2,100) ***2) 7 y_top= [circ_top(xx) for xx in x] 8 y_bot = [circ_bot(xx) for xx in x] 9 ax= plt.figure(figsize=(4,4)) 10 plt.gca().set_aspect('equal') #get current axis and set aspect ratio 11 plt.plot(x, y_top, 'k-') 12 plt.plot(x, y_bot, 'k-') 13 plt.axvline (1.1, color='red') 14 plt.fill_between(x,y_top,y_bot, where=x>1.1, alpha=0.3) 2.0 15 10 0.5 0.0 -0.5 -1.0 -1.5 -2.0 1 00 05 10 15 20 Python Python

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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%matplotlib inline
import numpy as np
from matplotlib import pyplot as plt
from math import sin, cos, exp, pi, sqrt
import math

4. In the class, you learn a bunch of tools. But, in the homework, I need you to be a little bit creative and use all you have learned to solve a seemingly random problem.
Let us consider semi-circular paper of radius 2. The area of the whole paper is 2π ≈ 6.28. We want to cut a chord(red line) from this paper such that the area of the leftover paper(shown in blue) is 4.0. Where should I cut the chord (find the value of a where the red line should be such that
the area of the blue region is 4.0? Try to get a decent answer.
If you are good with geometry you can actually figure this one out analytically, but I want you to do it numerically using what we have learned so far.
return sqrt(r**2 - ***2)
▷
1
def circ_top(x, r=2.):
2
3
4
5
[3]
def circ_bot(x, r=2.):
return -sqrt(r**2
6 x=np.linspace(0,2,100)
***2)
7 y_top= [circ_top(xx) for xx in x]
8 y_bot = [circ_bot(xx) for xx in x]
9
ax= plt.figure(figsize=(4,4))
10
plt.gca().set_aspect('equal') #get current axis and set aspect ratio
11 plt.plot(x, y_top, 'k-')
12 plt.plot(x, y_bot, 'k-')
13 plt.axvline (1.1, color='red')
14 plt.fill_between(x,y_top,y_bot, where=x>1.1, alpha=0.3)
<matplotlib.collections.PolyCollection at 0x1060a7690>
2.0
15
10
0.5
0.0
-0.5
-1.0
-1.5
-2.0
1
00 05 10 15 20
Python
Python
Transcribed Image Text:4. In the class, you learn a bunch of tools. But, in the homework, I need you to be a little bit creative and use all you have learned to solve a seemingly random problem. Let us consider semi-circular paper of radius 2. The area of the whole paper is 2π ≈ 6.28. We want to cut a chord(red line) from this paper such that the area of the leftover paper(shown in blue) is 4.0. Where should I cut the chord (find the value of a where the red line should be such that the area of the blue region is 4.0? Try to get a decent answer. If you are good with geometry you can actually figure this one out analytically, but I want you to do it numerically using what we have learned so far. return sqrt(r**2 - ***2) ▷ 1 def circ_top(x, r=2.): 2 3 4 5 [3] def circ_bot(x, r=2.): return -sqrt(r**2 6 x=np.linspace(0,2,100) ***2) 7 y_top= [circ_top(xx) for xx in x] 8 y_bot = [circ_bot(xx) for xx in x] 9 ax= plt.figure(figsize=(4,4)) 10 plt.gca().set_aspect('equal') #get current axis and set aspect ratio 11 plt.plot(x, y_top, 'k-') 12 plt.plot(x, y_bot, 'k-') 13 plt.axvline (1.1, color='red') 14 plt.fill_between(x,y_top,y_bot, where=x>1.1, alpha=0.3) <matplotlib.collections.PolyCollection at 0x1060a7690> 2.0 15 10 0.5 0.0 -0.5 -1.0 -1.5 -2.0 1 00 05 10 15 20 Python Python
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