Question

A teen chewing bubble gum blows a huge bubble, the volume of which satisfies the differential equation:dV/dt=2(V)^(0.5)Where t is the time in seconds after the teen start to blow the bubble. If the bubble pops as soon as it reaches 529 cubic centimeters in volume, how many seconds does it take for the bubble to pop? Assume that the bubble had no volume when the teen first started blowing.

Answer 1

The ODE is separable:

$\dfrac{\mathrm dV}{\mathrm dt}=2\sqrt V\implies\dfrac{\mathrm dV}{2\sqrt V}=\mathrm dt$

Integrate both sides to get

$\sqrt V=t+C$

Assuming the bubble starts with zero volume, so that $V(0)=0$, we find

$\sqrt0=0+C\implies C=0$

Then the volume of the bubble at time $t$ is

$V(t)=t^2$

and so the time it take for the bubble to reach a volume of 629 cubic cm is

$t^2=529\implies t=23$

or 23 seconds after the teen first starts blowing the bubble.