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Post by Admin on Aug 18, 2023 22:08:26 GMT
I agree; Author says that as the particle orbits round the elipse this is equal to to this formula so how is he sure about this equality? Mondo I've given you the hints to calculate $v$ and $r$. Vasco |
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Post by Admin on Aug 18, 2023 22:11:32 GMT
Mondo
This is extremely straightforward.
Vasco
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Post by mondo on Aug 18, 2023 22:15:38 GMT
Mondo 1. Can you please write your first integration properly so that I can understand it please? You have no differentials in the rhs of your argument. Vasco The differential equation is $\frac{d^2}{d^2t}z(t) = -z$ hence to find a formula for $z$ I need to integrate this twice. So I do $\frac{d}{dt}z = \int -re^{it} = ire^{it}$ |
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Post by Admin on Aug 18, 2023 22:21:55 GMT
Mondo 1. Can you please write your first integration properly so that I can understand it please? You have no differentials in the rhs of your argument. Vasco The differential equation is $\frac{d^2}{d^2t}z(t) = -z$ hence to find a formula for $z$ I need to integrate this twice. So I do $\frac{d}{dt}z = \int -re^{it} = ire^{it}$ Mondo But this is not correct mathematics. Vasco |
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Post by mondo on Aug 18, 2023 22:30:28 GMT
You mean the notation is wrong or where is the mistake? In principle I need to integrate it twice right?
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Post by Admin on Aug 18, 2023 22:30:30 GMT
Mondo
Do you mean
$\displaystyle\frac{d}{dt}z=\int -zdt$?
We can't do what you have done because $r$ is a function of $t$. $r$ is not a constant. Please don't leave out the $dt$s
Vasco
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Post by mondo on Aug 18, 2023 22:47:10 GMT
Mondo Do you mean $\displaystyle\frac{d}{dt}z=\int -zdt$? yes We can't do what you have done because $r$ is a function of $t$. $r$ is not a constant. Please don't leave out the $dt$s Vasco Right! But how to write $r$ as a function of $t$? |
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Post by Admin on Aug 18, 2023 22:58:25 GMT
Mondo
My hints tell you how. But you would be better off doing the other exercise first: calculate the energy. It's easier.
Vasco
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Post by mondo on Aug 18, 2023 23:36:34 GMT
I am trying to work out the energy formula but results are not good:
$z = a cos(t) + ibsin(t)$ so to get the formula for velocity we need to differentiate this in respect to $t$ (time) -> $\frac{d}{dt}z = -asin(t) + ibcos(t)$. Next $v^2 = a^2sin^2(t) - 2asin(t)ibcos(t) - b^2sin^2(t)$
To get radius $r^2 = a^2cos^2(t) + b^2sin^2(t)$
So now when I add $v^2 + r^2$ I should get $a^2 + b^2$ but I don't get it. I can simplify $a^2cos^2(t) + a^2sin^2(t)$ to $a^2(cos^2(t) + sin^2(t)) = a^2$ but the rest I have no idea what to do with.
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Post by Admin on Aug 19, 2023 7:08:34 GMT
I am trying to work out the energy formula but results are not good: $z = a cos(t) + ibsin(t)$ so to get the formula for velocity we need to differentiate this in respect to $t$ (time) -> $\frac{d}{dt}z = -asin(t) + ibcos(t)$. Next $v^2 = a^2sin^2(t) - 2asin(t)ibcos(t) - b^2sin^2(t)$ To get radius $r^2 = a^2cos^2(t) + b^2sin^2(t)$ So now when I add $v^2 + r^2$ I should get $a^2 + b^2$ but I don't get it. I can simplify $a^2cos^2(t) + a^2sin^2(t)$ to $a^2(cos^2(t) + sin^2(t)) = a^2$ but the rest I have no idea what to do with. Mondo The radius at any point is the modulus of $z$ at that point and so $r^2 =|z|^2= |a\cos(t) + ib\sin(t)|=a^2\cos^2(t) + b^2\sin^2(t)$, which is what you have done. So why do you now do the wrong calculation for the speed? This is the correct calculation: $\dot{z}$ is the velocity and $v$ is the speed and so $v^2=|\dot{z}|^2$ I'll let you do the easy bit, the algebra. Try and think a bit more consistently and clearly. Vasco |
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Post by mondo on Aug 19, 2023 7:34:53 GMT
Ahh I forgot that in the formula for the kinetic energy $v$ stands for the magnitude of velocity - speed not velocity itself. I am sorry, yes now it simplifies to $a^2 + b^2$.
What about the differential equation now, I was stuck how to write the formula for $r$ so I ignored it for now, and wrote $z = e^{it}$ to get $\ddot{z} = - e^{it}$, next $z = \int(\int -e^{it}dt)dt = e^{it} + Ct + D$. Next, using figure [22] I tried to find contstants $C,D$ -> when $t = 0$ $z = a$ hence $1 + D = a \rightarrow D = (a-1)$, when $t=\pi$ then $z = -a$ and hence we can write $-1 + c\pi + a -1 = -a \rightarrow c = \frac{2a+2}{\pi}$ but this is far from the result in the book $e^{\pm it}$
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Post by Admin on Aug 19, 2023 8:26:08 GMT
Mondo
Note that $z=e^{it}$ describes a particle moving round the unit circle, not an ellipse.
Vasco
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Post by mondo on Aug 19, 2023 9:05:44 GMT
I am aware of that. But also note that the solution in the book doesn't seem to have any radius function in it.
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Post by Admin on Aug 19, 2023 9:20:17 GMT
I am aware of that. But also note that the solution in the book doesn't seem to have any radius function in it. Mondo You have calculated the radius function as $r=|z|$, a function of $t$. Vasco |
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Post by Admin on Aug 19, 2023 20:29:18 GMT
Mondo
The differential equation $\ddot{z}=-x$ can be integrated directly but it is quite a lot of algebra and doesn't give a lot of insight into what's going on. The more intuitive approach given in the book on pages 241 to 243 is probably the best way of understanding it all.
Vasco
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