help me with a calculus problem...

Aug 10, 2001
10,420
2
0
Why does d(?mu) = m[1-(u^2/c^2)]^(-3/2)du

where ? = [1-(u^2/c^2)]^(-1/2)

and m (mass) and c (the speed of light) are constants?
 

dullard

Elite Member
May 21, 2001
26,196
4,869
126
Originally posted by: Random Variable
So I get m[du(1-u^2/c^2)^(-1/2)] + (u^2/c^2)*(1-u^2/c^2)^(-3/2)].

How does that reduce?
You are close, but not quite correct. Instead of doing many steps at once, lets just do one step at a time:

[*]d(ymu) = m*d(yu) = m*y*du+m*u*dy

Notice that you left off a term. How do we simplify? Well we know what dy/du is:

[*]dy/du = u/c^2 * (1-u^2/c^2)^(-3/2)

Calculus teachers hate this next step even though it works: thus,

[*]dy = du * [u/c^2 * (1-u^2/c^2)^(-3/2)]

Substitute dy into the first equation above

[*]d(ymu) = m*y*du + m*u*du * [u/c^2 * (1-u^2/c^2)^(-3/2)]

Simplify:

d(ymu) = m*du*[y+u^2/c^2*(1-u^2/c^2)^(-3/2)]

Plug in the equation for y, and it simplifies quite easilly to your answer in the original post.
 
Aug 10, 2001
10,420
2
0
Originally posted by: dullard
Originally posted by: Random Variable
So I get m[du(1-u^2/c^2)^(-1/2)] + (u^2/c^2)*(1-u^2/c^2)^(-3/2)].

How does that reduce?
You are close, but not quite correct. Instead of doing many steps at once, lets just do one step at a time:

[*]d(ymu) = m*d(yu) = m*y*du+m*u*dy

Notice that you left off a term. How do we simplify? Well we know what dy/du is:

[*]dy/du = u/c^2 * (1-u^2/c^2)^(-3/2)

Calculus teachers hate this next step even though it works: thus,

[*]dy = du * [u/c^2 * (1-u^2/c^2)^(-3/2)]

Substitute dy into the first equation above

[*]d(ymu) = m*y*du + m*u*du * [u/c^2 * (1-u^2/c^2)^(-3/2)]

Simplify:

d(ymu) = m*du*[y+u^2/c^2*(1-u^2/c^2)^(-3/2)]

Plug in the equation for y, and it simplifies quite easilly to your answer in the original post.

Thank you. And don't worry about offending calculus teachers. This is part of a physics problem. :)