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# Integration question watch

1. Interestingly this now gives the result we want, i.e. v ~ r^-4, so if it a mistake its a pretty good one
Ok now I get

Do you get the same?
Ah sorry ignore that comment I made, if you do this you are differentiating w.r.t r not t, we want it w.r.t t.
3. Ah sorry ignore that comment I made, if you do this you are differentiating w.r.t r not t, we want it w.r.t t.
No you're completely right - use implicit differentiation and product rule, i.e.

and then put in our expression for dr/dt.
4. (Original post by jmz34)
Sorry, in my attempt to keep the question short I've forgotten to mention that. V is a function of r and r only.
If V(r) is a solution, so is V(r)+C for any constant C.

So I don't see how you can deduce that V(r) varies as 1/r^3.
No you're completely right - use implicit differentiation and product rule, i.e.

and then put in our expression for dr/dt.
Ah yeah I see it now. So eventually subbing everything back in and integrating I get:

V=k/r^4 + m/r^2 where k and m are constants. Just to clarify can you then just say V is proportional to r^-4 even though there is an r^-2 term?
6. Ah yeah I see it now. So eventually subbing everything back in and integrating I get:

V=k/r^4 + m/r^2 where k and m are constants. Just to clarify can you then just say V is proportional to r^-4 even though there is an r^-2 term?
We have this:

Sub this into original expression, using that

implies

Notice that the first and last term cancel out, leaving:

i.e.

so we have that v ~ r^-4.

Bit of an effort to get there though hey?!
7. Again, many thanks for your help
8. Again, many thanks for your help
No probs, sorry about all the mistakes - made a bit of a meal of it tbh!

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