Integration problem?

  • Thread starter kashiark
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  • #1
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Homework Statement


∫ 1/r³ dr/dt


Homework Equations



∫ xa dx = x(a+1)/(a+1)

The Attempt at a Solution


I have no clue how to solve it like this. I don't have the equation for r in terms of t, so I can't just substitute. How would I do it?
 

Answers and Replies

  • #2
lanedance
Homework Helper
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Homework Statement


∫ 1/r³ dr/dt


Homework Equations



∫ xa dx = x(a+1)/(a+1)

The Attempt at a Solution


I have no clue how to solve it like this. I don't have the equation for r in terms of t, so I can't just substitute. How would I do it?

is that how the question is actually written & what are you integrating with respect to?

it doesn't really make sense how you have written it, do you mean:
[tex] \int dt (\frac{1}{r(t)^3}\frac{dr}{dt}) [/tex]

if so have a think about how chain rule differentiation works & how you could try & reverse it using FTC...
 
  • #3
210
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OK if it were like that, wouldn't the dt's just divide out allowing you to integrate with respect to r?
∫ dt(1/r³ dr/dt)
∫ 1/r³ dr
-1/(2r²) + C
 
Last edited:
  • #4
lanedance
Homework Helper
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you can, but its a bit of an abuse of notation, athough it gives the same answer, a better way to think of it is to write:

[tex] \frac{d}{dt} (-\frac{1}{2r(t)^2}) = \frac{1}{r(t)^3}\frac{dr}{dt}[/tex]

then the intergal becomes
[tex] \int dt (\frac{d}{dt} (-\frac{1}{2r(t)^2})) [/tex]

so by FTC, the anti-derivative is
[tex] = (-\frac{1}{2r(t)^2}) +C[/tex]

though the question was to you - how is it written in the actual problem?
 
  • #5
210
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It's not on a worksheet or anything; I was just messing with a particular physical situation, and I came to something in that form that I needed to integrate.
 

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