A culture of bacteria is growing at the rate of 20e^0.8t cells per day, where t is the number of days since the culture was started. Suppose that the culture began with 50 cells.

(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.)

(a) To find a formula for the total number of cells in the culture after t days, we can integrate the given rate of growth.

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.

(a) To find a formula for the total number of cells in the culture after t days, we need to integrate the rate of growth function.

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.

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