Author: Omrit P

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