Suppose in January, 𝑡=1 , a lake is covered by 2 square meters of algae. Every month, the area covered by algae doubles. (Assume that the growth occurs continuously.) Find the growth rate as a function of time with units square meters per month. Enter t for 𝑡 in months.
da/dt=2^t*ln(2)
To find the growth rate as a function of time, we need to determine how the area covered by algae changes over time.
We know that the area covered by algae doubles every month. So, the growth rate, dA/dt, is equal to the current area, A, times 2.
dA/dt = 2A
Substituting A = 2 (since it starts with 2 square meters of algae), we can rewrite the equation as:
dA/dt = 2(2) = 4 square meters per month
Therefore, the growth rate, as a function of time, is given by dA/dt = 4 square meters per month.
To find the growth rate as a function of time, we need to determine how the area covered by algae changes over time. We know that every month, the area covered by algae doubles.
Let's denote the area covered by algae at time 𝑡 as 𝐴(𝑡). In January, 𝑡 = 1, and 𝐴(1) = 2 square meters.
Since the area covered by algae doubles every month, we can write the growth rate as follows:
𝐴(𝑡) = 𝑎 × 2^𝑡,
where 𝑎 is a constant representing the initial area covered by algae in January.
Now, to determine the value of 𝑎, we substitute 𝑡 = 1 into the equation:
𝐴(1) = 𝑎 × 2^1 = 2 square meters.
Solving for 𝑎, we find that 𝑎 = 1 square meter.
Therefore, the growth rate of the algae as a function of time is:
𝐴(𝑡) = 1 × 2^𝑡 square meters per month.
I shall assume that the area is 2 m^2 on January 1.
Then at the end of month t, the area will be
a = 2*2^t = 2^(t+1)
I guess if you want the growth rate, that would be
da/dt = ln2 * 2^(t+1) m^2/mo