A spherical raindrops evaporates at rate proportional to surface area?

  • Thread starter kochibacha
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  • #1
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i want to find V(t)
At first i found this problem was very simple but when i try to write differential equations i ended up with these

V' = kA thats for sure

then i confined the problem only to spherical shape and no other shapes of raindrops involved

as i cant express A in term of V alone( surface area of sphere = 4∏r2, volume of sphere is 4/3∏r3 ) then i have to

use chain rule, dV/dt= dVdrdrdt substitute dV/dt from

V'=k4∏r2

i get

4∏r2r'= K4∏r2

r'=k

r = kt+c
r3 = (kt+c)3
4∏r3/3 = (kt+c)34∏/r=V(t)
im i correct? the answer to this problem is V'=kV2/3 im not sure how they transform Area to variable V alone
 
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Answers and Replies

  • #2
pasmith
Homework Helper
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At first i found this problem was very simple but when i try to write differential equations i ended up with these

V' = kA thats for sure

then i confined the problem only to spherical shape and no other shapes of raindrops involved

as i cant express A in term of V alone( surface area of sphere = 4∏r2, volume of sphere is 4/3∏r3 ) then i have to

use chain rule, dV/dt= dVdrdrdt substitute dV/dt from

V'=k4∏r2

i get

4∏r2r'= K4∏r2

r'=k

im i correct? the answer to this problem is v'=kV2/3 im not sure how they transform Area to variable V alone
[tex]
\left(\frac{3V}{4\pi}\right)^{1/3} = r.
[/tex] Now substitute this into [itex]A = 4\pi r^2[/itex].
 
  • #3
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[tex]
\left(\frac{3V}{4\pi}\right)^{1/3} = r.
[/tex] Now substitute this into [itex]A = 4\pi r^2[/itex].
then how can u solve for V(t) ?
 
  • #4
pasmith
Homework Helper
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then how can u solve for V(t) ?
The ODE [tex]
V' = CV^{2/3}
[/tex] for constant [itex]C[/itex] is separable. Divide both sides by [itex]V^{2/3}[/itex] and then integrate with respect to [itex]t[/itex].
 
  • #5
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The ODE [tex]
V' = CV^{2/3}
[/tex] for constant [itex]C[/itex] is separable. Divide both sides by [itex]V^{2/3}[/itex] and then integrate with respect to [itex]t[/itex].
i mean how they derive for v'=cv^2/3
 

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