Numerical Methods EE3A – G6
Cruz, Eric John N., Jordan, Kenneth Ivan O., Lebios, Prince Cris O., Magada, Elven
John Lester M., Palay, Vincent H.
YouTube Link:
https://www.youtube.com/playlist?list=PLXouz0XUGglIPeBteT4DS73PJZnHVTwQF
TRAPEZOIDAL RULE OF INTEGRATION
Example 1
4
Find ∫0 (𝑥 2 + 2)𝑑𝑥 approximately using trapezoidal rule with n = 8
a)
b)
c)
d)
27.5
28.5
29.5
30.5
Solution:
Example 2
𝜋
Find ∫0 sin 2𝑥 + 4 𝑑𝑥 approximately using trapezoidal rule with n = 4
a)
b)
c)
d)
𝟐𝜋
𝟒𝜋
𝟔𝜋
𝟖𝜋
Solution:
Example 3
3
Use the trapezoidal rule with n = 5 to approximate the integral ∫0
a)
b)
c)
d)
𝟏. 𝟏
𝟏. 𝟐
𝟏. 𝟑
𝟏. 𝟒
Solution:
1
1+𝑥 4
𝑑𝑥
TRAPEZOIDAL RULE OF INTEGRATION
Engineering Problem
Estimate the total amount of oxygen produced in 10 minutes. Use the trapezoidal rule.
Solution:
Minutes
Oxygen (cu.ft/min)
0
0
1
1.3
2
2.7
3
3.2
4
5.3
5
4.7
6
4.5
7
3.8
8
4.2
9
5.1
10
6.3
Parabolic Partial Differential Equation
1. WHICH OF THE FOLLOWING IS A PARABOLIC EQUATION?
𝜕2 𝑢
𝜕2 𝑢
𝜕2 𝑢
a. 𝜕𝑥 2 + 𝜕𝑥𝜕𝑦 + 𝜕𝑦 2 = 0
𝜕2 𝑢
𝜕2 𝑢
𝜕2 𝑢
𝜕2 𝑢
b. 𝜕𝑥 2 = 9 𝜕𝑥𝜕𝑦
𝜕2 𝑢
𝜕𝑢
𝜕𝑢
c. 𝜕𝑥 2 + 2 𝜕𝑥𝜕𝑦 + 𝜕𝑦 2 + 𝜕𝑥 − 6 𝜕𝑦 = 0
𝜕2 𝑢
d. 𝑎2 𝜕𝑥 2 =
𝜕2 𝑢
𝜕𝑡 2
2. Find the value of X to make our equation a parabolic equation.
2𝑋
a. 6
𝜕 2𝑢
𝜕 2𝑢
𝜕 2𝑢
−
12
+
2
=0
𝜕𝑥 2
𝜕𝑥𝜕𝑦
𝜕𝑦 2
b. 4
c. 3
d. 2
Parabolic Partial Differential Equation
𝐴
𝜕 2𝑢
𝜕 2𝑢
𝜕 2𝑢
+
𝐵
+
𝐶
=0
𝜕𝑥 2
𝜕𝑦 2
𝜕𝑥𝜕𝑦
What values can be associated to A, B, and C to make it a parabolic equation?
a. A = 7, B = 3, C = 2
b. A = 37, B = 52, C = 64
c. A = 32, B = 144, C = 162
d. A = 150, B = 22, C = 9
Solution:
Parabolic Partial Differential Equation
Engineering Problem
The partial differential equation of the temperature in a long thin rod is given by
𝜕𝑇
𝜕 2𝑇
=∝ 2
𝜕𝑡
𝜕𝑥
0
1
2
3
4
T=90℃
T=20℃
8cm
If α = 0.8𝑐𝑚2 /s , the initial temperature of rod is 40°C , the rod is divided into four
equal segments ,and when t =0.04 sec (using ∆t = 0.2s). The temperature at node 2 at the
first interval is
SHOOTING METHOD: ORDINARY DIFFERENTIAL EQUATIONS
Example # 1: Use the shooting method to approximate the solution of the boundary
value problem:
𝑑2 𝑦
𝑑𝑥 2
− 2𝑦 = 0 ,
When: y(0) = 1.2 , y(1) = 0.9 , h = 0.25
SHOOTING METHOD
ENGINEERING PROBLEM
Solve for y(t), altitude of rocket given
y” = -g
g = 9.8 𝑚/𝑠 2
y (0) = 0
y (5) = 0
req’d:
y’(0) = ?
Solution:
y” = -9.8
𝑑2 𝑦
𝑑𝑡 2
𝑑
𝑑𝑦
∗ 𝑑𝑡 = -9.8
𝑑𝑡
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑡
𝑑
𝑑𝑡
= -9.8
=∫ −9.8 dt
= -9.8t + c
*y = -9.8t + c
y = ∫(−9.8𝑡 + 𝑐)𝑑𝑡
y=
−9.8𝑡 2
2
+ Ct + D
To find D?
y (0) =
−9.8(0)2
2
+ C (0) + D
D=0
To find C?
y (5) =
40 =
−9.8(5)2
−245
2
2
+ C (5) + D
+ 5C+ 0
5𝐶
=
5
122.5+ 40
5
C = 32.5
To find y (0)
𝑑𝑦
𝑑𝑡
= -9.8t + c
y’ (0) = -9.8t + 32.5
y’ (0) = -9.8(0) + 32.5
y’ (0) = 32.5