Modeling with differential equations.
If a bacteria culture doubles every 10 minutes and the original culture has 300 cells determine the rate of growth.
y = c e^kt
dy/dt = k c e^kt = k y
when t = 0 e^kt = e^0 = 1
y(0) = c
so
y = 300 e^kt
when t = 10
y = 600
600 = 300 e^10k
2 = e^10 k
ln 2 = 10 k
k = .0693/minute
Thanks
To determine the rate of growth of the bacteria culture, we can use a differential equation that describes the exponential growth. Let's denote the number of bacteria cells at any time t as N(t).
According to the given information, the bacteria culture doubles every 10 minutes. This means that the rate of growth of the culture is proportional to its size. We can express this relationship mathematically using the following differential equation:
dN/dt = k * N(t),
where dN/dt represents the rate of change of N with respect to time t, and k is the constant of proportionality.
To solve this differential equation, we can use separation of variables. Let's rearrange the equation:
dN/N = k * dt.
Now we can integrate both sides. The left-hand side integrates to ln(N), while the right-hand side integrates to kt + C, where C is the constant of integration.
ln(N) = kt + C.
To determine the constant of integration C, we can use the initial condition. We are given that the original culture has 300 cells (N(0) = 300), so we substitute this into the equation:
ln(300) = k * 0 + C,
ln(300) = C.
Now we have the equation:
ln(N) = kt + ln(300).
To find the rate of growth, we need to find the value of k. However, we don't have enough information to determine the exact value of k in this scenario.