i have this question
the population ntion of bacteria is observed to be f(t)=C e^2t where C is some constant. if after one hour a scientist abserves the growth rate to be 744 bacteria/hour what was the initial bacteria population.
i know that C is the inital bacteria population, and initally i thought you would just make the equation = 744 and t = 1 but i don't think that's right
so after thinking about it, the derivative of this function is the growth rate right? so if i set the derivativbe equal to 744 and t = 1 then that will tell me what the initial is will it not?
yes, that is right!
Thanks = )
You're on the right track! The derivative of the population function, f(t), represents the growth rate of the bacteria at any given time t. So, setting the derivative equal to 744 and solving for t will not give you the initial population, but it will give you the time at which the growth rate is 744 bacteria/hour.
To find the initial population, you need to substitute the given information into the equation and solve for C. In this case, the growth rate is given to be 744 bacteria/hour after one hour, so t = 1.
f(t) = C * e^(2t)
Substituting t = 1, f(t) = 744:
744 = C * e^(2 * 1)
Simplifying, we have:
744 = C * e^2
Now, divide both sides by e^2 to solve for C:
C = 744 / e^2
Using a calculator, e^2 is approximately 7.389, so:
C = 744 / 7.389
C ≈ 100.5
Therefore, the initial bacteria population, C, is approximately 100.5.