Asked by Mischa
The radius of a sphere is increasing at a rate proportional to its radius. If the radius is 4 initially, and the radius is 10 after two seconds, what will the radius be after three seconds?
dR/dt = c R
Integrating this differential equation gives you
ln R/Ro = c t
R/Ro = e^(ct)
At t = 4 seconds, ln 2.5 = 2c
Therefore the constant is
c = 0.4581
D = Ro e^(0.4581 t)
When t = 3, R/Ro = 3.9523
R = 15.81
dR/dt = c R
Integrating this differential equation gives you
ln R/Ro = c t
R/Ro = e^(ct)
At t = 4 seconds, ln 2.5 = 2c
Therefore the constant is
c = 0.4581
D = Ro e^(0.4581 t)
When t = 3, R/Ro = 3.9523
R = 15.81
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