1. ## Exponential Growth

Suppose that a colony of bacteria is growing exponentially. Let M represent the mass in grams of the colony after t hours. If M satisfies M'(t)=.12M(t) and M(6)=37, find M(t).

2. Originally Posted by Velvet Love
Suppose that a colony of bacteria is growing exponentially. Let M represent the mass in grams of the colony after t hours. If M satisfies M'(t)=.12M(t) and M(6)=37, find M(t).
the derivative models the law of natural exponential growth ...

$\displaystyle y' = ky$

which has the general solution ...

$\displaystyle y = y_0e^{kt}$

you have $\displaystyle k = .12$ and $\displaystyle M(6) = 37$ ... solve for $\displaystyle M_0$

3. Originally Posted by Velvet Love
Suppose that a colony of bacteria is growing exponentially. Let M represent the mass in grams of the colony after t hours. If M satisfies M'(t)=.12M(t) and M(6)=37, find M(t).
If the colony is growing exponentially then the mass at time $\displaystyle t$ satisfies:

$\displaystyle M(t)=M(0)e^{kt}$

where $\displaystyle M(0)$ is the mass at time $\displaystyle t=0$ and $\displaystyle k$ is a constant.

Now use the given information to find $\displaystyle M(0)$ and $\displaystyle k$

CB

4. When you say M sub 0, do you mean y sub 0?

5. Originally Posted by skeeter
the derivative models the law of natural exponential growth ...

$\displaystyle y' = ky$

which has the general solution ...

$\displaystyle y = y_0e^{kt}$

you have $\displaystyle k = .12$ and $\displaystyle M(6) = 37$ ... solve for $\displaystyle M_0$

Alright. I solved for M sub 0, and I got ln(37)/.72

6. Originally Posted by Velvet Love
When you say M sub 0, do you mean y sub 0?
Please quote the post you are refering to, that way there will be no confusion.

CB

7. Originally Posted by Velvet Love
When you say M sub 0, do you mean y sub 0?
I expect that you are able to take the general model (where y indicates the dependent variable), and change the dependent variable to suit your specific problem, like so ...

$\displaystyle M = M_0e^{0.12t}$