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    To cut it short, I'm doing a question and have arrived at the following differential equations:

    r"-r(p')^2=-dv/dr (where the ' indications differentiation w.r.t a variable, say 't')

    and p'=c/r^2 where c is a constant.

    I am also given that r=asinp.

    From all this I have derived the following expressions:

    r'=acos(p)p'
    r''=-asin(p)(p')^2+acos(p)p''
    p''=-2cr'/r^3
    r'=cacosp/r^2

    The question asks me to show that V is proportional to r^-4.

    I can see that this involves seperating variables in the first equation, subbing in those derived expressions and integrating. However, I'm not too sure how to do this- since there are terms like p'' which are related to r.

    Thanks in advance.
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    The question asks me to show that V is proportional to r^-4.

    I can see that this involves seperating variables in the first equation, subbing in those derived expressions and integrating. However, I'm not too sure how to do this- since there are terms like p'' which are related to r.

    Thanks in advance.
    do you mean you have this equation to start:

     \frac{d^2r}{dt^2} - r(\frac{dp}{dt})^2 = -\frac{dv}{dr}

    ??

    If so I would suggest the following:

     r = a sin p implies

     dr = a cos p dp

    so

     dp = \frac{dr}{a cos p}

    eliminate p from this using the identity

     cos p = \sqrt{1-sin^2p} = \sqrt{1-\frac{r^2}{a^2}}

    Now sub this expression for dp into the original equation, which eliminates all references to p. I haven't checked but I imagine it can be solved now..?
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    Or do you mean this one

     \frac{d^2r}{dt^2} - r(\frac{c}{r^2})^2 = -\frac{dv}{dr}

    If this, then p isn't in the expression at all..?
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    (Original post by spread_logic_not_hate)
    do you mean you have this equation to start:

     \frac{d^2r}{dt^2} - r(\frac{dp}{dt})^2 = -\frac{dv}{dr}

    ??

    If so I would suggest the following:

     r = a sin p implies

     dr = a cos p dp

    so

     dp = \frac{dr}{a cos p}

    eliminate p from this using the identity

     cos p = \sqrt{1-sin^2p} = \sqrt{1-\frac{r^2}{a^2}}

    Now sub this expression for dp into the original equation, which eliminates all references to p. I haven't checked but I imagine it can be solved now..?
    Thanks, but as far as I can see r is a function of t so it won't really help doing that. This is the main problem I am confused as to how to integrate differentials w.r.t to a variable that isn't the variable which the differential is w.r.t to, if that makes any sense.
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    Thanks, but as far as I can see r is a function of t so it won't really help doing that. This is the main problem I am confused as to how to integrate differentials w.r.t to a variable that isn't the variable which the differential is w.r.t to, if that makes any sense.
    yeah I get you, like what is

     \int \frac{dr}{dt}dx

    for example?

    At a guess perhaps this


     x \frac{dr}{dt}

    but then I get your problem of what if r is a function of x, i.e. it should also be involved in the integration process. Im afraid im not sure what to do there - might be time to involve a teacher/lecturer/textbook!
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    (Original post by spread_logic_not_hate)
    yeah I get you, like what is

     \int \frac{dr}{dt}dx

    for example?

    At a guess perhaps this


     x \frac{dr}{dt}

    but then I get your problem of what if r is a function of x, i.e. it should also be involved in the integration process. Im afraid im not sure what to do there - might be time to involve a teacher/lecturer/textbook!
    Sort of, more like:

     \int \frac{dr}{dt}dr
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    (Original post by jmz34)
    Sort of, more like:

     \int \frac{dr}{dt}dr
    In that case i'd try to find dr/dt explicitly and substitute in the expression I get for it into the integral, but im guessing you tried this already and it isn't possible or something?
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    (Original post by spread_logic_not_hate)
    In that case i'd try to find dr/dt explicitly and substitute in the expression I get for it into the integral, but im guessing you tried this already and it isn't possible or something?
    That integral expression is just a simplified one to show you what I mean, but in the question its the second differential of r w.r.t t, and if you look back I do have an expression for r'' but its in terms of p'' which wouldn't really help since I would need to change dr to acosdp and I'd be in a similar situation.
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    What is v here? (Unless there is some relationship you haven't disclosed, I don't see how v is unique - as you only reference dv/dr, adding a constant to v doesn't affect anything).
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    (Original post by DFranklin)
    What is v here? (Unless there is some relationship you haven't disclosed, I don't see how v is unique - as you only reference dv/dr, adding a constant to v doesn't affect anything).
    Sorry, in my attempt to keep the question short I've forgotten to mention that. V is a function of r and r only.
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    That integral expression is just a simplified one to show you what I mean, but in the question its the second differential of r w.r.t t, and if you look back I do have an expression for r'' but its in terms of p'' which wouldn't really help since I would need to change dr to acosdp and I'd be in a similar situation.
    Ok I get it now.

     r = a sin p

    gives

     \frac{dr}{dp} = a cos p = \sqrt{a^2-r^2}

    and


     \frac{d^2r}{dp^2} = -a sin p = -r

    We also have

     \frac{dp}{dt} = \frac{c}{r^2}

    Therefore

     dp = \frac{c}{r^2} dt

    and

     dp^2 = \frac{c^2}{r^4} dt

    So sub this into the two above expressions to get

     \frac{r^2}{c}\frac{dr}{dt} = \sqrt{a^2-r^2}

    implies

    \frac{dr}{dt} = \frac{c}{r^2}\sqrt{a^2-r^2}

    and

     \frac{d^2r}{dt^2} = \frac{-r^5}{c^4}

    Now the overall expression can be integrated, as you have expressions for dr/dt and d^2r/dt^2 purely in terms of r.

    I think that is correct - there may be a flaw i've yet to spot!
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    EDIT - found a flaw already lol.

    dp^2 = \frac{c^2}{r^4} dt
    Thats not right... Ah well I tried!
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    EDIT EDIT

    Should be this

     dp^2 = \frac{c^4}{r^4}dt^2

    So we're back on again...
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    (Original post by spread_logic_not_hate)
    EDIT EDIT

    Should be this

     dp^2 = \frac{c^4}{r^4}dt^2

    So we're back on again...
    Isn't it c^2/r^4?
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    Isn't it c^2/r^4?
    Yes it is. lol. Fixed the dt^2 mistake, only to make another...
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    (Original post by spread_logic_not_hate)
    Yes it is. lol. Fixed the dt^2 mistake, only to make another...
    Your expression for the second differential of r w.r.t t is incorrect.
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    Your expression for the second differential of r w.r.t t is incorrect.
    yes - should be

     \frac{d^2r}{dt^2} = \frac{-c^2}{r^3}

    Sorry bout that, seem to be all about the mistakes today...
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    (Original post by spread_logic_not_hate)
    yes - should be

     \frac{d^2r}{dt^2} = \frac{-c^2}{r^3}

    Sorry bout that, seem to be all about the mistakes today...
    But surely r'' is zero now?

    DFranklin please help...

    Also 'spread_logic_not_hate' please feel free to continue to try this, thanks for all the help so far dude.
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    Again this is wrong, r'=(c/r^2)SQRT(a^2-r^2), you have to differentiate this using the product rule to get r''.
    Ok now I get

     \frac{d^2r}{dt^2} = \frac{c^2}{r^3} - \frac{2c^2a^2}{r^5}

    Do you get the same?
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