What Value of b Solves 1/2b^2 = sin(b)?

  • Context: Undergrad 
  • Thread starter Thread starter Coriolis314
  • Start date Start date
Click For Summary
SUMMARY

The equation 1/2b^2 = sin(b) requires finding a positive constant value of b where the integral of x from 0 to b equals the integral of cos(x) from 0 to b. The discussion concludes that this equation lacks an elementary solution, suggesting the use of numerical techniques or the Lambert W function for approximation. Participants emphasize that calculators or approximation algorithms are likely intended for solving this equation, given its presence in a basic mathematics context.

PREREQUISITES
  • Understanding of definite integrals
  • Familiarity with trigonometric functions
  • Knowledge of numerical methods for solving equations
  • Basic comprehension of the Lambert W function
NEXT STEPS
  • Explore numerical methods for solving transcendental equations
  • Study the properties and applications of the Lambert W function
  • Learn about numerical integration techniques
  • Investigate calculator functions for solving equations
USEFUL FOR

Mathematicians, students studying calculus, and anyone interested in solving transcendental equations or exploring numerical approximation methods.

Coriolis314
Messages
10
Reaction score
0
This is not homework. I ran across this question in a book store a few minutes ago and can't seem to finish it...

R= int(x,0,b
S=int(cos(x),0,b
for what value of b (a positive constant) does R=S?

1/2x^2,0,b=1/2b^2
sin(x),0,b=sin(b)

1/2b^2=sin(b)

Its been a while.. but I should be able to do this... thanks
 
Physics news on Phys.org
The integrals are easy enough, but I don't think the equation you ended with has an elementary solution. One approach would be to use numeric techniquest to get an approximate solution. Another approach might be the Lambert W function.
 
Thank you! I assumed it had a smaller function because it was in a very low level book. I think it is mean to be entered into a calculator at the last point or be cycled through an approximation algorithm.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K