A bacteria culture starts with 260 bacteria and grows at a rate proportional to its size. After 4 hours there will be 1040 bacteria.
(a) Express the population after t hours as a function of t.
this is what i put as my answer but i guess its not right do i replace the four with t??
260e^tln4/4
The population apparently doubles in 2 hours.
P = 260 * 2^(t/2) is a simpler way to write it.
Note that e^(t ln4/4)= [e^(ln4)]^(t/4)
= 4^(t/4)= [2^2]^(t/4) = 2^(t/2)
Your version is apparently correct, but needs to be simplified
To express the population after t hours as a function of t, we can use the formula for exponential growth. Let's denote P(t) as the population after t hours.
The problem states that the growth rate is proportional to the size, so we can use the general form of exponential growth: P(t) = P(0)e^(kt), where P(0) is the initial population and k is the growth rate constant.
Given that P(0) = 260 and P(4) = 1040, we can substitute these values into the equation:
1040 = 260e^(k*4)
To solve for k, we can divide both sides by 260 and take the natural logarithm (ln):
4 = ln(e^(k*4))
4 = 4k ln(e)
Since ln(e) = 1, we have:
4k = 4
Dividing by 4:
k = 1
Now that we know k, we can substitute it back into the equation to find the population as a function of t:
P(t) = 260e^(kt)
P(t) = 260e^(1t)
P(t) = 260e^t
Therefore, the population after t hours is given by the function P(t) = 260e^t.
To express the population after t hours as a function of t, let's first set up the model for exponential growth. We know that the growth rate is proportional to the current population and that after 4 hours there will be 1040 bacteria.
Let P(t) represent the population at time t. According to the given information, we can set up the following differential equation:
dP/dt = kP(t)
Where k is the constant of proportionality.
To solve this differential equation, we need an initial condition. Since initially there are 260 bacteria, we have P(0) = 260.
Now we can solve the differential equation:
dP/P = k dt
∫dP/P = ∫k dt
ln(P) = kt + C1
P = e^(kt + C1)
P = e^(kt) * e^(C1)
P = C * e^(kt)
Using the initial condition (P(0) = 260), we can solve for the constant C:
260 = C * e^(k * 0)
260 = C
So our function becomes:
P(t) = 260 * e^(kt)
Now, to find the value of k, we can use the information given that after 4 hours there will be 1040 bacteria:
1040 = 260 * e^(k * 4)
4 = e^(4k)
ln(4) = 4k
k = ln(4)/4
Finally, substituting this value into our equation:
P(t) = 260 * e^((ln(4)/4) * t)
And that is the expression for the population after t hours.