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# Maths Help watch

1. 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.
2. (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
3. (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?
4. (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 = ??
5. (Original post by dtox)
dt/dv = ??
what?
6. (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.
7. (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 ??
8. (Original post by Improvement)
dV/dt = k/v^2 ??
∫v² dv = ∫k dt

It should be easy from there.
9. Cheers

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