Help please differentiating this equation from a mechanics textbook

Intesar
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Homework Statement
I came across these equations in a mechanics textbook and wish to differentiate the first equation w.r.t. time to obtain that second equation. Any help is much appreciated!


$$c(f_1+f_2)=a^2(\frac{1}{2}\dot{a}^2 -ca+h)+2af'_2$$

$$c(1-\frac{\dot{a}}{c})(f'_1+f'_2)=ca(-2\dot{a}^2 (1-\frac{1}{2}\frac{\dot{a}}{c})-a\ddot{a}(1-\frac{\dot{a}}{c})+2\frac{\dot{a}}{c}h+\frac{a}{c}\dot{h})+2a(1+\frac{\dot{a}}{c}){f''_2}$$

##f_1(x)=f_1(t-\frac{r}{c})##, ##f_2(x)=f_2(t+\frac{r}{c})##
The book didn't explicitly state the functions of time, but from the second equation we can see that ##a## and ##h## are functions of time.

We also know that $$\frac{f'_1+f'_2}{a}+\frac{1}{2}\dot{a}^2+h=0$$ (if that's of any help)
Relevant Equations
Please see above
I tried doing it a few times and this is all I get:

c(˙f1+˙f2)=a˙a2+a2˙a−3ca2+˙ha2+2ha+2˙af′2+2a˙f′2c(f1˙+f2˙)=aa˙2+a2a˙−3ca2+h˙a2+2ha+2a˙f2′+2af2′˙​

Please let me know where I'm going wrong. Thanks
 
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I do not catch the difference of ##f'## and ##\dot{f}## in your explanation. If they are the same, how about try completing differentiation by t ? For example the first term in RHS of the last equation be ##a\dot{a}^3##.
 
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It appears to me that ##f_1## and ##f_2## might be functions of other variables: ##f_1(x_1)## and ##f_2(x_2)##. The primes on the functions denote differentiation with respect to ##x_1## or ##x_2##.

By the chain rule, ##\dot {f_1} = f_1'(x_1) \dot x_1## and similarly for ##f_2##. In order to get their result, it looks like ##x_1## and ##x_2## must depend on time such that ##\dot x_1 = 1- \large \frac{\dot a}{c}## and ##\dot x_2 = 1+ \large \frac{\dot a}{c}##
 
TSny said:
It appears to me that ##f_1## and ##f_2## might be functions of other variables: ##f_1(x_1)## and ##f_2(x_2)##. The primes on the functions denote differentiation with respect to ##x_1## or ##x_2##.

By the chain rule, ##\dot {f_1} = f_1'(x_1) \dot x_1## and similarly for ##f_2##. In order to get their result, it looks like ##x_1## and ##x_2## must depend on time such that ##\dot x_1 = 1- \large \frac{\dot a}{c}## and ##\dot x_2 = 1+ \large \frac{\dot a}{c}##

Yes
TSny said:
It appears to me that ##f_1## and ##f_2## might be functions of other variables: ##f_1(x_1)## and ##f_2(x_2)##. The primes on the functions denote differentiation with respect to ##x_1## or ##x_2##.

By the chain rule, ##\dot {f_1} = f_1'(x_1) \dot x_1## and similarly for ##f_2##. In order to get their result, it looks like ##x_1## and ##x_2## must depend on time such that ##\dot x_1 = 1- \large \frac{\dot a}{c}## and ##\dot x_2 = 1+ \large \frac{\dot a}{c}##

I've attached the page that I am trying to understand. You are right, ##f_1## and ##f_2## are indeed functions of ##x_1## and ##x_2## respectively (see equation 3.71a in the page attached).
I am still however having a hard time to obtain their equation (3.72).
 

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anuttarasammyak said:
I do not catch the difference of ##f'## and ##\dot{f}## in your explanation. If they are the same, how about try completing differentiation by t ? For example the first term in RHS of the last equation be ##a\dot{a}^3##.

Ya my mistake, ##\dot{f_1}=f'_1(x_1)\dot{x_1}##. But I'm still unable to obtain their equation.
 
TSny said:
It appears to me that ##f_1## and ##f_2## might be functions of other variables: ##f_1(x_1)## and ##f_2(x_2)##. The primes on the functions denote differentiation with respect to ##x_1## or ##x_2##.

By the chain rule, ##\dot {f_1} = f_1'(x_1) \dot x_1## and similarly for ##f_2##. In order to get their result, it looks like ##x_1## and ##x_2## must depend on time such that ##\dot x_1 = 1- \large \frac{\dot a}{c}## and ##\dot x_2 = 1+ \large \frac{\dot a}{c}##
[
anuttarasammyak said:
I do not catch the difference of ##f'## and ##\dot{f}## in your explanation. If they are the same, how about try completing differentiation by t ? For example the first term in RHS of the last equation be ##a\dot{a}^3##.

##\dot{f_1}=f'_1(x_1)\dot{x_1}##. I'm still unable to obtain their equation (equation 2). The last equation is what I attempted on doing.
 
Intesar said:
I am still however having a hard time to obtain their equation (3.72).

Consider the left side of the equation $$c(f_1+f_2)=a^2(\frac{1}{2}\dot{a}^2 -ca+h)+2af'_2$$

What do you get for the time derivative of ##c(f_1+f_2)##? Express the result in terms of ##f'_1, f'_2, \dot a##, and ##c##.
 
TSny said:
It appears to me that ##f_1## and ##f_2## might be functions of other variables: ##f_1(x_1)## and ##f_2(x_2)##. The primes on the functions denote differentiation with respect to ##x_1## or ##x_2##.

By the chain rule, ##\dot {f_1} = f_1'(x_1) \dot x_1## and similarly for ##f_2##. In order to get their result, it looks like ##x_1## and ##x_2## must depend on time such that ##\dot x_1 = 1- \large \frac{\dot a}{c}## and ##\dot x_2 = 1+ \large \frac{\dot a}{c}##
[
TSny said:
Consider the left side of the equation $$c(f_1+f_2)=a^2(\frac{1}{2}\dot{a}^2 -ca+h)+2af'_2$$

What do you get for the time derivative of ##c(f_1+f_2)##? Express the result in terms of ##f'_1, f'_2, \dot a##, and ##c##.

##\dot{f_1}=f'_1(1-\frac{\dot{a}}{c})## and ##\dot{f_2}=f'_2(1+\frac{\dot{a}}{c})##
 
Good. I think you should be able to complete it. On the right hand side there is the term ##2af'_2##. What's the time derivative of this term?
 
  • #10
TSny said:
It appears to me that ##f_1## and ##f_2## might be functions of other variables: ##f_1(x_1)## and ##f_2(x_2)##. The primes on the functions denote differentiation with respect to ##x_1## or ##x_2##.

By the chain rule, ##\dot {f_1} = f_1'(x_1) \dot x_1## and similarly for ##f_2##. In order to get their result, it looks like ##x_1## and ##x_2## must depend on time such that ##\dot x_1 = 1- \large \frac{\dot a}{c}## and ##\dot x_2 = 1+ \large \frac{\dot a}{c}##
[
TSny said:
Good. I think you should be able to complete it. On the right hand side there is the term ##2af'_2##. What's the time derivative of this term?

The derivative of ##2af'_2## wrt time is ##2\dot{a}f'_2+2a\dot{f'_2}##, but I'm not sure how to express ##\dot{f'_2}## in terms of ##\dot{a}##, ##c## and ##f'##.
 
  • #11
Intesar said:
The derivative of ##2af'_2## wrt time is ##2\dot{a}f'_2+2a\dot{f'_2}##, but I'm not sure how to express ##\dot{f'_2}## in terms of ##\dot{a}##, ##c## and ##f'##.
##f'_2## is just another function of ##x_2##. So, you can use the chain rule to get ##\dot {f'_2}## just like you did when finding ##\dot {f_2}##.
 
  • #12
anuttarasammyak said:
I do not catch the difference of ##f'## and ##\dot{f}## in your explanation. If they are the same, how about try completing differentiation by t ? For example the first term in RHS of the last equation be ##a\dot{a}^3##.

How did you get ##a\dot{a}^3## for the first term on the RHS?
 
  • #13
TSny said:
Good. I think you should be able to complete it. On the right hand side there is the term ##2af'_2##. What's the time derivative of this term?

So I think I'm almost there, but now I'm having a problem with differentiating the the first term on the RHS (##a^2(\frac{1}{2}\dot{a}^2-ca+h)##) with respect to time. This is what I got after differentiating this term:

##a\dot{a}^3+a^2\dot{a}\ddot{a}-3c{a}^2\dot{a}+\dot{h}a^2+2ha\dot{a}##

It seems to be different from what they got, and I am not sure where I went wrong.
 
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  • #14
Intesar said:
How did you get a˙a3aa˙3a\dot{a}^3 for the first term on the RHS?

\dot{a^2}=2a\dot{a}
 
  • #15
anuttarasammyak said:
\dot{a^2}=2a\dot{a}

Thanks! Now so I tried differentiating the the first term on the RHS (##a^2(\frac{1}{2}\dot{a}^2-ca+h)##) with respect to time. This is what I got after differentiating this term:

##a\dot{a}^3+a^2\dot{a}\ddot{a}-3c{a}^2\dot{a}+\dot{h}a^2+2ha\dot{a}##

It seems to be different from what they got, and I am not sure where I went wrong.
 
  • #16
Please show us the relation between a and r in OP. Then try calculating time derivative of the rest of RHS,
\dot{(2af_2')}
, as well as LHS.
 
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  • #17
anuttarasammyak said:
Please show us the relation between a and r in OP. Then try calculating time derivative of the rest of RHS,
\dot{(2af_2')}
, as well as LHS.

##r=a(t)##
 
  • #18
Intesar said:
##r=a(t)##
Intesar said:
##r=a(t)##

So I tried calculating the derivative now and I'm getting most of the terms (LHS of the equation and the 3 right most terms of the RHS). This is what I have now:

##c(1-\frac{\dot{a}}{c})(f'_1+f'_2)=ca(\frac{\dot{a}^3}{c}+\frac{a\dot{a}\ddot{a}}{c}-3a\dot{a}+\frac{2\dot{a}h}{c}+a\frac{\dot{h}}{c})+2af''_2(1+\frac{\dot{a}}{c})##

I am still trying to figure out the 3 left most terms on the RHS
 
  • #19
TSny said:
##f'_2## is just another function of ##x_2##. So, you can use the chain rule to get ##\dot {f'_2}## just like you did when finding ##\dot {f_2}##.

Ya I just did that. It seems like I'm almost arriving at their expression but I can't seem to get the 2 first terms on the RHS.

##\frac{d}{dt}(a^2(\frac{1}{2}\dot{a}^2-ca)=ca(-2\dot{a}^2(1-\frac{1}{2}\frac{\dot{a}}{c})-a\ddot{a}(1-\frac{\dot{a}}{c}))##

I am not sure how they arrived at that. This is what I keep getting:

##\frac{d}{dt}(a^2(\frac{1}{2}\dot{a}^2-ca)=a\dot{a}^3+a^2\dot{a}\ddot{a}-3ca^2\dot{a}##
 
  • #20
Intesar said:
Thanks! Now so I tried differentiating the the first term on the RHS (##a^2(\frac{1}{2}\dot{a}^2-ca+h)##) with respect to time. This is what I got after differentiating this term:

##a\dot{a}^3+a^2\dot{a}\ddot{a}-3c{a}^2\dot{a}+\dot{h}a^2+2ha\dot{a}##

It seems to be different from what they got, and I am not sure where I went wrong.

It seems like I'm almost arriving at their expression but I can't seem to get the first few terms on the RHS.

##\frac{d}{dt}(a^2(\frac{1}{2}\dot{a}^2-ca)=ca(-2\dot{a}^2(1-\frac{1}{2}\frac{\dot{a}}{c})-a\ddot{a}(1-\frac{\dot{a}}{c}))##

I am not sure how they arrived at that. This is what I keep getting:

##\frac{d}{dt}(a^2(\frac{1}{2}\dot{a}^2-ca)=a\dot{a}^3+a^2\dot{a}\ddot{a}-3ca^2\dot{a}##
 
  • #21
I should have checked more carefully.

Note that in the equation
1588022464558.png

the first and second terms inside the parentheses on the RHS do not match dimensionally. I suspect that there is a typographical error and the middle term should actually be ##-c\dot a## instead of ##-ca##. So the equation should be

1588022601149.png


Then, I believe that taking the time derivative of this will lead to the result as given in the text.
 
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  • #22
TSny said:
##f'_2## is just another function of ##x_2##. So, you can use the chain rule to get ##\dot {f'_2}## just like you did when finding ##\dot {f_2}##.

Ya I just did that. It seems like I'm almost arriving at their expression but I can't seem to get the 2 first terms on the RHS.

##\frac{d}{dt}(a^2(\frac{1}{2}\dot{a}^2-ca)=ca(-2\dot{a}^2(1-\frac{1}{2}\frac{\dot{a}}{c})-a\ddot{a}(1-\frac{\dot{a}}{c}))##

I am not sure how they arrived at that. This is what I keep getting:

##\frac{d}{dt}(a^2(\frac{1}{2}\dot{a}^2-ca)=ca(-2\dot{a}^2(1-\frac{1}{2}\frac{\dot{a}}{c})-a\ddot{a}(1-\frac{\dot{a}}{c}))##
TSny said:
I should have checked more carefully.

Note that in the equation
View attachment 261593
the first and second terms inside the parentheses on the RHS do not match dimensionally. I suspect that there is a typographical error and the middle term should actually be ##-c\dot a## instead of ##-ca##. So the equation should be

View attachment 261594

Then, I believe that taking the time derivative of this will lead to the result as given in the text.

Thank you so so much! I really appreciate all your help. It finally worked :)
 
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