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A crystal suspended in a chemical solution is increasing in size over time. The rate of increase of volume is inversely proportional to the square of its volume. Initially, the crystal had a volume of 3cm^3 and, one day later, its volume was 4cm^3. How long will it take for its volume to increase to 10cm^3.
(edited 8 years ago)
Reply 1
Original post by Improvement
A crystal suspended in a chemical solution is increasing in size over time. The rate of increase of volume is inversely proportional to the square of its volume. Initially, the crystal had a volume of 3cm^3 and, one day later, its volume was 4cm^3. How long will it take for its volume to increase to 10cm^3.


solve the differential equation
dv/dt = k/v2 subject to t=0 v=3, t=1 v=4
Reply 2
Original post by TeeEm
solve the differential equation
dv/dt = k/v2 subject to t=0 v=3, t=1 v=4


So from this can you work out K? Or do you have to do for example, dv/dt = dv/dr x dr/dt?
Reply 3
Original post by Improvement
So from this can you work out K? Or do you have to do for example, dv/dt = dv/dr x dr/dt?


dt/dv = ??
Reply 4
Original post by dtox
dt/dv = ??


what?
Reply 5
Original post by Improvement
what?


Can you figure out what dt/dv is equal to? There's no need to mess about with variables, just look at the original equation.
Reply 6
Original post by dtox
Can you figure out what dt/dv is equal to? There's no need to mess about with variables, just look at the original equation.


dV/dt = k/v^2 ??
Original post by Improvement
dV/dt = k/v^2 ??


∫v² dv = ∫k dt

It should be easy from there.
Reply 8
Cheers :biggrin:

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