Help please differentiating this equation from a mechanics textbook

[Intesar]

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

 

[anuttarasammyak]

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 $a\dot{a}^3$.

 

[TSny]

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 $\dot x_2 = 1+ \large \frac{\dot a}{c}$

 

[Intesar]

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

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).

 

[Intesar]

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.

 

[Intesar]

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 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.

 

[TSny]

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 $c$.

 

[Intesar]

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 $\dot x_2 = 1+ \large \frac{\dot a}{c}$
[

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 $\dot{f_2}=f'_2(1+\frac{\dot{a}}{c})$

 

[TSny]

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?

 

[Intesar]

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 $\dot x_2 = 1+ \large \frac{\dot a}{c}$
[

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 $f'$.

 

[TSny]

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 $\dot {f_2}$.

 

[Intesar]

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?

 

[Intesar]

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.

 

[anuttarasammyak]

How did you get a˙a3aa˙3a\dot{a}^3 for the first term on the RHS?

$$\dot{a^2}=2a\dot{a}$$

 

[Intesar]

$$\dot{a^2}=2a\dot{a}$$

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.

 

[anuttarasammyak]

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.

 

[Intesar]

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.

$r=a(t)$

 

[Intesar]

$r=a(t)$

$r=a(t)$

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

 

[Intesar]

$f'_2$ 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:

$\frac{d}{dt}(a^2(\frac{1}{2}\dot{a}^2-ca)=a\dot{a}^3+a^2\dot{a}\ddot{a}-3ca^2\dot{a}$

 

[Intesar]

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:

$\frac{d}{dt}(a^2(\frac{1}{2}\dot{a}^2-ca)=a\dot{a}^3+a^2\dot{a}\ddot{a}-3ca^2\dot{a}$

 

[TSny]

I should have checked more carefully.

Note that in the equation

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

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

 

[Intesar]

$f'_2$ 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:

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

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 :)