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2
Que.1) Using Taylor’s series find expression for sin x in
powers of x.
Consider the derivative of f (x) = sin x.
f’(x) = cos x , f’’(x) = - sin x , f’’’(x) = - cos x , f ’’’’(x) = sin x , …….
From the Taylor series expansion we have :
f (x) = f (0) + f ′(0)
𝑥
1!
+ f ′′(0)
𝑥2
2!
+ f ′′′(0)
𝑥3
3!
+ ⋯
In this example we have f (x) = sin x.
Since ,sin 0 = 0 all even powers of x will equal to 0 in this series expansion.
Thus,
f (x) = sin x = cos 0
𝑥
1!
− cos 0
𝑥2
3!
+ cos 0
𝑥3
5!
− ⋯
Ans:-
3
Now , since cos 0 = 1 the series reduces to:
Sin x =
𝑥
1!
−
𝑥3
3!
+
𝑥5
5!
− ⋯
The series is infinite in odd powers of x with alternating sign and thus can be
written as :
Sin x = 𝒏=𝟎
∞
(−𝟏)𝒏 𝒙𝟐𝒏+𝟏
(𝟐𝒏+𝟏)!
4
Que.2) Visit any Dome shape place measure radius take
picture of that shape. Find volume of that Dome shape
using Triple Integration.
Ans:-
V= 0
𝜋/2
0
2𝜋
0
𝑟
𝑟2. 𝑠𝑖𝑛∅. 𝑑𝑟. 𝑑𝜃. 𝑑∅
V= 0
𝜋/2
0
2𝜋
(
𝑟3
3
)0
𝑟. 𝑠𝑖𝑛∅. 𝑑𝜃. 𝑑∅
Taking r=12m
V= 0
𝜋/2
0
2𝜋
(
123
3
)0
𝑟. 𝑠𝑖𝑛∅. 𝑑𝜃. 𝑑∅
V=
123
3 0
𝜋/2
0
2𝜋
( 𝑠𝑖𝑛∅. 𝑑𝜃. 𝑑∅
5
V=
123
3 0
𝜋/2
(𝜃)0
2𝜋
𝑠𝑖𝑛∅. 𝑑∅
V=
123
3
∗ 2𝜋 0
𝜋/2
𝑠𝑖𝑛∅. 𝑑∅
V=
123
3
∗ 2𝜋 ∗ (−𝑐𝑜𝑠∅)0
𝜋/2
V=
123
3
∗ 2𝜋 ∗ (−
cosπ
2
+ cos 0)
V=2 *3.14* 123
3
V=3617.28
6
Que :-3) Solve.
Find y , z at x=1.4
Ans:-
By R K Method :-
o
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦 𝑦 1 = 1 𝑥0 = 1 , 𝑦0 = 1
h = 0.4 , x = 1.4
By R K Method ,
K =
1
6
[𝐾1 + 2𝐾2 + 2𝐾3 + 𝐾4 ]
K1 = h f (x0 , y0 )
=0.4 (2)
7
K1 = 0.8
K2 = h f (x0 + h/2, y0 + K1/2 )
=0.4 [2.6]
=1.04
K3 = h f (x0 + h/2 , y0 + K2/2 )
=0.4 [2.72]
=1.088
K4 = h f (x0 + h , y0 + K3 )
=0.4 [3.488]
=1.3952
K =
1
6
[0.8 + 2 ∗ 1.04 + 2 ∗ 1.088 + 1.3952 ]
K = 1.0752
y1 = y0 + k
=1 + 1.0752
y1 = 2.0752
o
𝑑𝑦
𝑑𝑥
=
𝑥
𝑧
𝑧 1 = 2 𝑧0 = 2 , 𝑥0 = 1
h = 0.4
By R K Method ,
z1 = z0 + k
K =
1
6
[𝐾1 + 2𝐾2 + 2𝐾3 + 𝐾4 ]
K1 = h f (x0 , z0 )
=0.4 (0.5)
= 0.2
K2 = h f (x0 + h/2, z0 + K1/2 )
=0.4 [0.5714]
=0.2286
K3 = h f (x0 + h/2 , z0 + K2/2 )
=0.4 [0.5676]
=0.2270
K4 = h f (x0 + h , z0 + K3 )
=0.4 [0.6286]
=0.2515
8
K =
1
6
[0.2 + 2 ∗ 0.2286 + 2 ∗ 0.2270 + 0.2515]
K =
1
6
[1.3627]
K = 0.2271
z1 = z0 + k
=2 + 0.2271
z1 = 2.2271
9
Que.4) Take picture of D shape curve you observed in
Sanjivani college, take approximate values find double
integral on that shape if f(x,y)=x + y2
2
10
∫ .∫ (x²+y²)dxdy
Area = ∫ [x²y+ y3/3]dx limits 0 to x²
= ∫ [x4+x6/3]dx
= 0.2 + 0.0476
=0.2476 * 2
Area =0.4952
X=1
X=0
y=x²
y=0
X=0
X=1
X=0
X=1
=
15
5
+
17
21
=
𝑥5
5
+
𝑥7
21]
[ Limits from 0 to 1
11
References:-
1) Higher Engineering Mathematics by Dr. B. S. Grewal(42nd Edition).
2) Numericals Methods by M. K. Jain , S. R. K. Iyengar , R K Jain .
3) Higher Engineering Mathematics by B. V. Ramana .
 All Books In E-Book Available on:
http://tutorial.math.lamar.edu.
12
13

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EM 2 Group - 4 engineering mathematics pptx

  • 1. 1
  • 2. 2 Que.1) Using Taylor’s series find expression for sin x in powers of x. Consider the derivative of f (x) = sin x. f’(x) = cos x , f’’(x) = - sin x , f’’’(x) = - cos x , f ’’’’(x) = sin x , ……. From the Taylor series expansion we have : f (x) = f (0) + f ′(0) 𝑥 1! + f ′′(0) 𝑥2 2! + f ′′′(0) 𝑥3 3! + ⋯ In this example we have f (x) = sin x. Since ,sin 0 = 0 all even powers of x will equal to 0 in this series expansion. Thus, f (x) = sin x = cos 0 𝑥 1! − cos 0 𝑥2 3! + cos 0 𝑥3 5! − ⋯ Ans:-
  • 3. 3 Now , since cos 0 = 1 the series reduces to: Sin x = 𝑥 1! − 𝑥3 3! + 𝑥5 5! − ⋯ The series is infinite in odd powers of x with alternating sign and thus can be written as : Sin x = 𝒏=𝟎 ∞ (−𝟏)𝒏 𝒙𝟐𝒏+𝟏 (𝟐𝒏+𝟏)!
  • 4. 4 Que.2) Visit any Dome shape place measure radius take picture of that shape. Find volume of that Dome shape using Triple Integration. Ans:- V= 0 𝜋/2 0 2𝜋 0 𝑟 𝑟2. 𝑠𝑖𝑛∅. 𝑑𝑟. 𝑑𝜃. 𝑑∅ V= 0 𝜋/2 0 2𝜋 ( 𝑟3 3 )0 𝑟. 𝑠𝑖𝑛∅. 𝑑𝜃. 𝑑∅ Taking r=12m V= 0 𝜋/2 0 2𝜋 ( 123 3 )0 𝑟. 𝑠𝑖𝑛∅. 𝑑𝜃. 𝑑∅ V= 123 3 0 𝜋/2 0 2𝜋 ( 𝑠𝑖𝑛∅. 𝑑𝜃. 𝑑∅
  • 5. 5 V= 123 3 0 𝜋/2 (𝜃)0 2𝜋 𝑠𝑖𝑛∅. 𝑑∅ V= 123 3 ∗ 2𝜋 0 𝜋/2 𝑠𝑖𝑛∅. 𝑑∅ V= 123 3 ∗ 2𝜋 ∗ (−𝑐𝑜𝑠∅)0 𝜋/2 V= 123 3 ∗ 2𝜋 ∗ (− cosπ 2 + cos 0) V=2 *3.14* 123 3 V=3617.28
  • 6. 6 Que :-3) Solve. Find y , z at x=1.4 Ans:- By R K Method :- o 𝑑𝑦 𝑑𝑥 = 𝑥 + 𝑦 𝑦 1 = 1 𝑥0 = 1 , 𝑦0 = 1 h = 0.4 , x = 1.4 By R K Method , K = 1 6 [𝐾1 + 2𝐾2 + 2𝐾3 + 𝐾4 ] K1 = h f (x0 , y0 ) =0.4 (2)
  • 7. 7 K1 = 0.8 K2 = h f (x0 + h/2, y0 + K1/2 ) =0.4 [2.6] =1.04 K3 = h f (x0 + h/2 , y0 + K2/2 ) =0.4 [2.72] =1.088 K4 = h f (x0 + h , y0 + K3 ) =0.4 [3.488] =1.3952 K = 1 6 [0.8 + 2 ∗ 1.04 + 2 ∗ 1.088 + 1.3952 ] K = 1.0752 y1 = y0 + k =1 + 1.0752 y1 = 2.0752 o 𝑑𝑦 𝑑𝑥 = 𝑥 𝑧 𝑧 1 = 2 𝑧0 = 2 , 𝑥0 = 1 h = 0.4 By R K Method , z1 = z0 + k K = 1 6 [𝐾1 + 2𝐾2 + 2𝐾3 + 𝐾4 ] K1 = h f (x0 , z0 ) =0.4 (0.5) = 0.2 K2 = h f (x0 + h/2, z0 + K1/2 ) =0.4 [0.5714] =0.2286 K3 = h f (x0 + h/2 , z0 + K2/2 ) =0.4 [0.5676] =0.2270 K4 = h f (x0 + h , z0 + K3 ) =0.4 [0.6286] =0.2515
  • 8. 8 K = 1 6 [0.2 + 2 ∗ 0.2286 + 2 ∗ 0.2270 + 0.2515] K = 1 6 [1.3627] K = 0.2271 z1 = z0 + k =2 + 0.2271 z1 = 2.2271
  • 9. 9 Que.4) Take picture of D shape curve you observed in Sanjivani college, take approximate values find double integral on that shape if f(x,y)=x + y2 2
  • 10. 10 ∫ .∫ (x²+y²)dxdy Area = ∫ [x²y+ y3/3]dx limits 0 to x² = ∫ [x4+x6/3]dx = 0.2 + 0.0476 =0.2476 * 2 Area =0.4952 X=1 X=0 y=x² y=0 X=0 X=1 X=0 X=1 = 15 5 + 17 21 = 𝑥5 5 + 𝑥7 21] [ Limits from 0 to 1
  • 11. 11 References:- 1) Higher Engineering Mathematics by Dr. B. S. Grewal(42nd Edition). 2) Numericals Methods by M. K. Jain , S. R. K. Iyengar , R K Jain . 3) Higher Engineering Mathematics by B. V. Ramana .  All Books In E-Book Available on: http://tutorial.math.lamar.edu.
  • 12. 12
  • 13. 13