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Post by mondo on Mar 2, 2023 2:57:25 GMT
Hello, on page 294 there is figure [18] to which I have two questions: 1. Author says it represents a situation where w drag an object by a piece of a string along the edge of the table, this is how I see it  Author also says that $Y$ represents the edge of the table but what is $X$? Figure [18a] does not make sense if $X$ is just time as over time we should get closer to $Y$ while this figure shows us the opposite. 2. On the same page there is equation (30), how was $R \frac{dX}{d\sigma} = Re^{-\sigma/R}$? Thank you. |
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Post by Admin on Mar 2, 2023 10:44:26 GMT
Mondo
1. The book says that the $\boldsymbol{Y}$-axis represents the edge of the table, not $Y$ itself.
As you can see from figure 18a, $X$ is the other coordinate of the position of the paperweight. The position of the paperweight is the point $(X,Y)$.
2. The book does not say what you have written, but rather $\displaystyle R\frac{dX}{d\sigma}=-Re^{-\sigma/R}$.
The LHS of (30), $\displaystyle\frac{dX}{d\sigma}=-\frac{X}{R}$, is a differential equation which must be solved for $X$.
That's where the RHS of (30) comes from. I am assuming you have studied elementary calculus.
Vasco
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Post by mondo on Mar 4, 2023 3:41:07 GMT
1. The book says that the $\boldsymbol{Y}$-axis represents the edge of the table, not $Y$ itself. As you can see from figure 18a, $X$ is the other coordinate of the position of the paperweight. The position of the paperweight is the point $(X,Y)$. Ok, so it shows position of this paperweight at random points of time, or rather it shows us it's trajectory during the entire journey. 2. The book does not say what you have written, but rather $\displaystyle R\frac{dX}{d\sigma}=-Re^{-\sigma/R}$. The LHS of (30), $\displaystyle\frac{dX}{d\sigma}=-\frac{X}{R}$, is a differential equation which must be solved for $X$. That's where the RHS of (30) comes from. I am assuming you have studied elementary calculus. Yes but I am getting a different result -> $\frac{dX}{d\sigma} = -\frac{X}{R} => \frac{dX}{X} = -\frac{d\sigma}{R} => log(X) + C_1 = -\frac{\sigma}{R} + C_2 => X = e^{-\frac{\sigma}{R}} + C_3$ so where is the multiplication of $R$ on the RHS coming from? Thank you. |
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Post by Admin on Mar 4, 2023 8:38:05 GMT
Mondo
Your last step above is incorrect.
If $\displaystyle\log(X)+C_1=-\frac{\sigma}{R}+C_2$, it does not follow that $X=e^{-\sigma/R}+C_3$.
If you differentiate your answer you do not get back to the original differential equation:
$\displaystyle\frac{dX}{d\sigma}=-\frac{X}{R}$, and so your answer must be wrong.
If you differentiate the answer in the book you do get back to the original equation, so the answer in the book must be right.
Also if you set $\sigma=0$ in your answer then you find that $X=1+C_3$, but we can see from figure 18 that it should be $X=R$, and we can also see that the answer in the book also gives us $X=R$ when $\sigma=0$.
These are checks that are always worth doing to see if you have made any mistakes.
Try doing that last step again.
Vasco
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Post by mondo on Mar 4, 2023 23:07:53 GMT
Thank you Vasco, that was a really helpful answer.
I see at least one mistake I made - $log(X) = -\frac{\sigma}{R} + C_2 - C_1$ hence $X = e^{\frac{-\sigma}{R} + C_2 - C_1}$ but how do we get another $R$ from these $C_2 - C_1$ on the RHS?
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Post by Admin on Mar 4, 2023 23:26:09 GMT
Thank you Vasco, that was a really helpful answer. I see at least one mistake I made - $log(X) = -\frac{\sigma}{R} + C_2 - C_1$ hence $X = e^{\frac{-\sigma}{R} + C_2 - C_1}$ but how do we get another $R$ from these $C_2 - C_1$ on the RHS? Mondo If you substitute a specific value for $\sigma$ and the corresponding value of $X$ then you should be able to find the value of $C_2-C_1$, as I pointed out in my last post above. Vasco |
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Post by mondo on Mar 5, 2023 1:03:13 GMT
Ahh right, I got it!
Just for the record, $X = e^{-\sigma/R + C_2 - C_1} = e^{C_2 - C_1} e^{-\sigma/R}$ and we know that at $\sigma = 0$ $X = R$ hence $e^{C_2 - C_1} = R$ and finally $X = Re^{-\sigma/R}$
Thank you.
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