It is convenient to write 2(o) in the form z = ² sin² p, use the fact that 2'(x) = 2'(o)/x'(o) and express the integrand of the functional in terms of o: But 2' Т T-S" do = da do 1+2' (0)²/x' (0)² 2gz (6) 2c² sin o cos o and, since x = x2 +2¹2 = 4c² sin² and = T: = To dov 0 ² (20 - sin 2) + d, a' = 2c² sin² p, so that 2c 2cob T= √29 Jo do= √2g But, from the analysis preceding the exercise, c= √A/sin, and so x¹ (0)² + 2¹(0)² z (0) 2 2А ФЫ g sin o
It is convenient to write 2(o) in the form z = ² sin² p, use the fact that 2'(x) = 2'(o)/x'(o) and express the integrand of the functional in terms of o: But 2' Т T-S" do = da do 1+2' (0)²/x' (0)² 2gz (6) 2c² sin o cos o and, since x = x2 +2¹2 = 4c² sin² and = T: = To dov 0 ² (20 - sin 2) + d, a' = 2c² sin² p, so that 2c 2cob T= √29 Jo do= √2g But, from the analysis preceding the exercise, c= √A/sin, and so x¹ (0)² + 2¹(0)² z (0) 2 2А ФЫ g sin o
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Can you work out more details, I don't understand.
Transcribed Image Text:Solution to Exercise 5.7
It is convenient to write z(p) in the form z = c² sin²o, use the fact that z'(x) =
z'(o)/x'(o) and express the integrand of the functional in terms of o:
T
= 5.²
0
=
do
dx
do
1+2'(o)²/x' (0)²
2gz (6)
=
T =
32 10th
T=
But z' = 2c² sino cos and, since x = =²(26 – sin 26) + d, a' = 2c² sin² 6, so that
x¹² +2¹² = 4c4 sin² and
2c
2cob
√29 16th
/2g
√2g
But, from the analysis preceding the exercise, c = √Ã/ sino, and so
do
=
do
2A b
x'(0)² + 2¹(0)²
z (0)
g sin ob
![Now substitute z = c² sin² to give
x = 2c²
[dósin² 6 = c²
I do
² (26-sin 20)+d and
stationary path the time of passage is
=
do (1
- cos 26)
z =
T[z] =
-
² (1 — cos 26),
2A ob
g
sin ob
to show that on the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59d44c96-efb1-4f3c-83b3-5a6a84cf94cb%2Fb9e86fa0-a771-405a-8ee9-7a616342e896%2Fizw5pu_processed.png&w=3840&q=75)
Transcribed Image Text:Now substitute z = c² sin² to give
x = 2c²
[dósin² 6 = c²
I do
² (26-sin 20)+d and
stationary path the time of passage is
=
do (1
- cos 26)
z =
T[z] =
-
² (1 — cos 26),
2A ob
g
sin ob
to show that on the
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps with 3 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
