Volume of balloon - Derivative Question

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  • #1
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Homework Statement


The volume of a spherical balloon of radius 'r' is Vcm^3, where V =4/3pir^3
The volume of the balloon increases with time 't' seconds according to the formula

dV/dt = 1000/(2t+1)^2, t>0

i) Find an expression in terms of 'r' and 't' for dr/dt
ii) Given that V = 0 and t = 0, solve the differential equation

dV/dt = 1000/(2t+1)^2, to obtain V in terms of t

iii) Find the radius of the balloon at time t =5

iv) Find the rate of increases of the radius of the balloon


Homework Equations





The Attempt at a Solution


i) Well we know dV/dt = (dv/dr) x (dr/dt)
Which means that I found dr/dt to be - (250/ (pir^2(2t+1)^2))

For the rest of them, i have no clue what to do. :S

Thanks in advance
 

Answers and Replies

  • #2
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What methods of solving differential equations are you familiar with?
 
  • #3
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What methods of solving differential equations are you familiar with?
I'm not exactly sure what you mean? :S

Usually I just differentiate lol :S
 
  • #4
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So you do not understand what a "differential equation" is?
 
  • #5
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I do, but I'm not sure what kind of "methods" you are referring to
 
  • #6
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Do you know how to solve diff. eq.'s in any way?
 
  • #7
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Yeah, I know how to solve differential equations. As in product rule etc? And using the 2nd/3rd derivative? :S
 
  • #8
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So why can't you solve the differential equation in ii)?
 
  • #9
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I'm not sure what to do?

Am i just meant to differentiate it?

I thought you would have to integrate it so you would get the original equation?

I got V = 1000t/(2t+1)^2 + C

Is that it? :S
 
  • #10
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You are given ## \frac {dV} {dt} ##. You need to find ## V(t) ##. Your answer to this is wrong, which you can easily check by differentiating your "solution".
 
  • #11
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Oh ok,

So you integrate dv/dt to get V(t) which is -500/(2t+1)?
 
  • #12
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Very well.
 
  • #13
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How do i find the radius at t=5?

I know what dr/dt is but i cant integrate it because it has two variables :S
 
  • #14
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Now that you have integrated the equation, you have volume as a function of time. Volume is directly related to the radius. Eh?
 
  • #15
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By methods, voko meant if you can use integrating factor, Laplace transform, etc, I believe.

As for the dr/dt, have you tried separating the variables so you have the 'r' on one side and 't' on the other?
 
Last edited:
  • #16
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OH, so because we now have V = -500/(2t+1)

We can let V = 4/3pir^3 and let t=5!

Ahaha, thanks for being so patient with me Voko! i truly appreciate it! :D

Ugh, So i got the radius as 2.21 from plugging the above in.

iv)
However when I sub this radius and time into dr/dt = (250/ (pir^2(2t+1)^2))
I'm not getting the right answer
:'(

What am i doing wrong?
 
Last edited:
  • #17
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I hope you enjoyed your "aha!" moment :)
 
  • #18
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Went from happiness to sadness within seconds. This world is too cruel.
 
  • #19
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This is a physical forum, not philosophical. So you have to render your sadness in a way we can deal with - if we are supposed to.
 
  • #20
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Lol, maybe you could help me with the last part? :( I did it but the answer book says im wrong

So i got the radius as 2.21 from plugging V = 4/3pir^3 and t = 5 into V = -500/(2t+1)

iv)
However when I sub this radius and time into dr/dt = (250/ (pir^2(2t+1)^2))
I'm not getting the right answer
:'(

What am i doing wrong?
 
  • #21
6,054
391
iv) does not mention any time, so I suspect you have to find the function. Again, you know the volume as a function of time, so you should be able get the radius as a function of time, which I think you did in iii). If then you need to know its value at some particular time, just sub the time.
 

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