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^−0.5t) where t is measured in hours. At time t = 0 the population is 10 cells and is increasing at a rate of 4 cells/hour. Find the values of a and b.
I still don't understand how you got those values for a and b. What was the final step?
Ah, yeast! Those little busy beings always up to something. Let's get to the bottom of this equation, shall we?
We know that at time t = 0, the population is 10 cells. This means that f(0) = 10.
So, plug in 0 for t in the equation and we get: 10 = a / (1 + be^0) = a / (1 + b)
Now, we also know that the population is increasing at a rate of 4 cells/hour. This means that f'(t) = 4 when t = 0.
Differentiating the equation with respect to t, we get f'(t) = (-0.5ae^(-0.5t)) / (1 + be^(-0.5t))^2.
Plugging in t = 0 gives us: 4 = (-0.5a) / (1 + b)^2 = -0.5a / (1 + b)^2
Now, we have two equations:
(1) 10 = a / (1 + b)
(2) 4 = -0.5a / (1 + b)^2
To solve for a and b, we need to eliminate one variable. Let's solve equation (1) for a and substitute it into equation (2).
From equation (1), we have: a = 10(1 + b)
Substituting this value of a into equation (2), we get:
4 = -0.5(10(1 + b)) / (1 + b)^2
Simplifying, we get:
4 = -5 / (1 + b)
Cross-multiplying, we get:
4(1 + b) = -5
Expanding and rearranging, we get:
4 + 4b = -5
Finally, solving for b:
4b = -9
b = -9/4
Substituting the value of b into equation (1), we get:
10 = a / (1 + (-9/4))
10 = a / (1/4)
10 = 4a
a = 10/4
a = 5/2
So, the values of a and b are a = 5/2 and b = -9/4.
And there you have it! The yeast population equation has been deciphered, with a dash of clownish calculations.
To find the values of a and b, we need to use the information given in the problem.
Given:
Population at time t = 0 (n(0)) = 10 cells
Rate of increase at time t = 0 (f'(0)) = 4 cells/hour
We can find the values of a and b using these two pieces of information.
Step 1: Finding the value of a
At time t = 0, the population is given by the equation f(t) = a/(1 + be^(-0.5t)).
Substituting t = 0, we get:
f(0) = a/(1 + be^(0)) = a/(1 + b) = 10 cells
Since the population at time t = 0 is 10 cells, we can write the equation:
10 = a/(1 + b) ----(Equation 1)
Step 2: Finding the value of b
The rate of increase at time t = 0 is given by the derivative of the population function with respect to time:
f'(t) = 0.5abe^(-0.5t) = 4 cells/hour
Substituting t = 0, we get:
f'(0) = 0.5ab = 4 cells/hour
Simplifying the equation, we have:
ab = 8 ----(Equation 2)
Step 3: Solving the equations
Using Equation 1 and Equation 2, we have the following system of equations:
10 = a/(1 + b)
ab = 8
We can solve these equations simultaneously to find the values of a and b.
Let's solve Equation 2 for a:
a = 8/b ----(Equation 3)
Substituting Equation 3 into Equation 1, we get:
10 = (8/b)/(1 + b)
Multiply both sides of the equation by (1 + b) to eliminate the denominator:
10(1 + b) = 8/b
Expanding and rearranging the equation:
10b + 10 = 8/b
Multiply both sides of the equation by b to eliminate the denominator:
10b^2 + 10b = 8
Rearrange the equation:
10b^2 + 10b - 8 = 0
This is a quadratic equation in terms of b. We can use the quadratic formula to solve for b:
b = (-10 ± √(10^2 - 4(10)(-8)))/(2(10))
Simplifying the equation:
b = (-10 ± √(100 + 320))/20
b = (-10 ± √420)/20
b = (-10 ± 2√105)/20
Now, we have two possible values for b:
b1 = (-10 + 2√105)/20
b2 = (-10 - 2√105)/20
Simplifying these values, we get:
b1 = (-5 + √105)/10
b2 = (-5 - √105)/10
Step 4: Finding the value of a
Now that we have the values of b, we can substitute them into Equation 3 to find the corresponding values of a.
For b = (-5 + √105)/10:
a = 8/b = 8/((-5 + √105)/10) = 80/(-5 + √105)
For b = (-5 - √105)/10:
a = 8/b = 8/((-5 - √105)/10) = 80/(-5 - √105)
So, the values of a and b are:
For b = (-5 + √105)/10: a = 80/(-5 + √105)
For b = (-5 - √105)/10: a = 80/(-5 - √105)
To find the values of a and b in the given population function, we'll use the information provided in the question:
1. First, we need to find the value of a. The population function n = f(t) = a/(1 + be^(-0.5t)) represents the population at time t. We are told that at time t = 0, the population is 10 cells. So, we can substitute these values into the equation:
10 = a/(1 + be^(-0.5 * 0))
Simplifying the equation gives us:
10 = a/(1 + b)
2. Next, we need to find the value of b. The question states that at time t = 0, the population is increasing at a rate of 4 cells/hour. This means the derivative of the population function with respect to time (t) at t = 0 is equal to 4. Taking the derivative of the population function:
f'(t) = (-0.5ab * e^(-0.5t))/(1 + be^(-0.5t))^2
Substituting t = 0 into the derivative and equating it to 4, we get:
4 = (-0.5ab * e^(-0.5 * 0))/(1 + be^(-0.5 * 0))^2
Simplifying the equation gives us:
4 = (-0.5ab)/(1 + b)^2
Now we have two equations:
10 = a/(1 + b)
4 = (-0.5ab)/(1 + b)^2
We can solve these equations simultaneously to find the values of a and b.
well, just plug in some numbers!
f(0) = 10, so
a/(1+b) = 10
df/dt = a(0.5be^(-0.5t))/(1+be^-0.5t)^2, so
a(b/2)/(1+b)^2 = 4
Solve those two equations, and you get
a = 50
b = 4