If N is the number of bacteria
dN/dt = 20e^.8t
N = 25 e^.8t + 50
when N = 700
700 = 25 e^.8t + 50
e^.8t = 26
take ln of both sides
.8t = ln 26
t = appr 4.07
check:
if t = 4.07
N = 25 e^.8(4.07) + 50
= 698.6 , not bad
(a) Find a formula for the total number of cells in the culture after t days.
(b) If the culture is to be stopped when the population reaches 700, when will this occur? (Round your answer to one decimal place.)
dN/dt = 20e^.8t
N = 25 e^.8t + 50
when N = 700
700 = 25 e^.8t + 50
e^.8t = 26
take ln of both sides
.8t = ln 26
t = appr 4.07
check:
if t = 4.07
N = 25 e^.8(4.07) + 50
= 698.6 , not bad
The rate of growth is given as 20e^0.8t cells per day. Integrating this with respect to t gives us:
∫(20e^0.8t) dt
To integrate e^0.8t, we use the rule for integrating exponential functions:
∫e^kt dt = (1/k)e^kt + C
Applying this rule to our integral:
∫(20e^0.8t) dt = (1/0.8)e^0.8t + C
Simplifying:
= 25e^0.8t + C
Since we are given that the culture began with 50 cells (when t = 0), we can substitute this into the formula to find C:
50 = 25e^0.8(0) + C
50 = 25 + C
C = 25
Therefore, the formula for the total number of cells in the culture after t days is:
N(t) = 25e^0.8t + 25
(b) We are given that we want to stop the culture when the population reaches 700. To find when this occurs, we can set N(t) equal to 700 and solve for t:
25e^0.8t + 25 = 700
Subtracting 25 from both sides:
25e^0.8t = 675
Dividing both sides by 25:
e^0.8t = 27
To solve for t, we take the natural logarithm of both sides:
ln(e^0.8t) = ln(27)
0.8t = ln(27)
Dividing both sides by 0.8:
t = ln(27)/0.8
Using a calculator to evaluate this:
t ≈ 4.9047
Rounded to one decimal place, the culture will reach 700 cells after approximately 4.9 days.
The rate of growth is given as 20e^(0.8t) cells per day. In order to find the total number of cells, we integrate this with respect to t.
∫(20e^(0.8t)) dt = 20/0.8 * ∫(e^(0.8t)) dt
Using the property of integration, we can integrate e^(0.8t) to get:
= 20/0.8 * (1/0.8) * e^(0.8t) + C
Simplifying this, we get:
= 25/0.8 * e^(0.8t) + C
Now, since the problem states that the culture began with 50 cells, we can substitute t = 0 into this equation:
50 = 25/0.8 * e^(0.8(0)) + C
50 = 31.25 + C
C = 50 - 31.25
C = 18.75
Therefore, the formula for the total number of cells in the culture after t days is:
N(t) = 25/0.8 * e^(0.8t) + 18.75
(b) To find when the population reaches 700 cells, we can set N(t) equal to 700 and solve for t.
700 = 25/0.8 * e^(0.8t) + 18.75
Subtracting 18.75 from both sides:
681.25 = 25/0.8 * e^(0.8t)
Now, divide both sides by 25/0.8:
(681.25 * 0.8) / 25 = e^(0.8t)
27.65 = e^(0.8t)
To solve for t, take the natural logarithm of both sides:
ln(27.65) = ln(e^(0.8t))
Simplifying:
3.314 = 0.8t
Divide both sides by 0.8:
t ≈ 4.142
Therefore, the population will reach 700 cells after approximately 4.1 days.