Problem # 13
A super bread dough increases in volume at a rate proportional to the volume V present. If V increases by a factor of 10 in 2 hours and V(0)=V0, find V at any time t. How long will it take for V to increase to100V0?
Solution:
Given: V Increases by 10 in 2 Hours V(0)=V0
First we rearrange the variables:
dv/dt = kv to dv/v = kdt
Integrate both Sides:
∫dv/v = ∫kdt
Log(v) = kt + Log(c)
v(t) = ce^kt
Plug in 0 to find c:
t=0
v(0) = V0
V0 = C
Substitute C for V0:
V(t) = V0 e^kt
Input 2 hours and the proportional 10 increase:
T = 2
V(2)=10V0
V0 e^2k = 10V0
e^2k = 10
Now we substitute e^k, since it’s 2k we divide and substitute in for V(t):
V(t)= V0 10^t/2
V(t) = 100V0
So, 100V0 = V0 e(10)^t/2
Now we Solve:
100V0 = V0 e(10)^t/2
10^t/2 = 10^2
t/2 = 2
t = 4
Answer: t = 4
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