HELP NEEDED: The number of yeast cells in a laboratory culture increases rapidly initially but levels off eventually. The population is modeled by the function n = f(t) = a/(1 + be−0.9^t) where t is measured in hours. At time t = 0 the population is 20 cells and is increasing at a rate of 16 cells/hour. Find the values of a and b.
you need two equations to solve for two unknowns a and b Solve for f(0) and f(1) then find a and b with this system of equations
Hey, I cant seem to find the two equations
When t=0, n=20 and when t=1, n=36.
Plug in values to make 2 equations and solve the system to find a and b.
where is t1, n=36 coming from
Since, the population is increasing at 16 cells per hr.
We were given that n=20 at t=0, so at t=1, n=20+16.
We could also say that when t=3, n=20+16+16 but since we only have 2 constants to fine, we only need 2 equations.
I see the logic but do you get the same answer if you get two equations f(0) = 20 and f'(0) = 16 since rate of change is first derivative
I think that should work too.
Is the function n=a/(1+be^[-0.9t])?
that looks correct
I got a=79.66 and b=2.98
Using the system of equations method.
is it the right answer?
I am working it again with the derivative method to check it.
20 = a/(1+b) 16 = 9ab/(1+b)^2
.9
Yes, those are the equations I got using the derivative method.
Solving those equations simultaneously, I got a=180, b=8. Let me check my work. Just a minute.
Hi guys, how did the second equation ?
20 = a/(1+b) 16 = .9ab/(1+b)^2
f'(0)
can you show me?
Neither answer that I got is a function that starts at 20, increases rapidly, then levels off. I think I may have misunderstood the original equation.
To get the second equation, you use the quotient rule and plug in 0 for t.
@bobobobobb , do you know how to calculate a derivative using the quotient rule?
Yes, are you just diffrentiating a/1+b?
did you use quadratic formula to find b
No, I differentiated the original equation.
when you take the first derivative plug in t =0 you end up with the second equation if you make a the subject in the first equation and plug it in in the second equation, you end with a quadratic equation to solve for b
Join our real-time social learning platform and learn together with your friends!