Question

The number of bacteria in a flask grows according to the differential equation (dy)/(dt)= 0.06 y In this question, time is measured in hours and the number of bacteria, y, is measured in millions. The number of bacteria at time t = 0 is 4 million. Enter a formula for the number of bacteria at time t y = Click here to preview your answer. Incorrect: Your answer is incorrect. What is the value of the growth constant? Growth constant : per hour. How long does it take for the number of bacteria to double? (Enter your answer correct to two decimal places.) Doubling time : hours. How many million bacteria will be present after 9 hours have passed? (Enter your answer correct to one decimal place.) Number present after 9 hours : million.

Answer 1

y = 4e^(0.06t).

Explanation:

dy/dt = 0.06y

Solving:

dy = 0.06y dt

dy/y = 0.06dt

Integrating both sides:

ln y = 0.06t + C

y = e^(0.06t + C)

y = Ae^(0.06t) where A is a constant.

At t = 0 , y = 4 million so

y = 4 = Ae^0 = A

So the formula is

y = 4e^(0.06t).

Answer 2

a) y = 4e^(0.06t)

b) 0.06

c) 11.55 hours

d) 6.9 million

Explanation:

When the growth rate (millions per hour) is proportional to the number (millions), the relationship is exponential. The growth rate is the constant of proportionality.

a) Formula for y(t):

y = 4e^(0.06t)

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b) The growth constant is 0.06, the multiplier of t in the exponential function. It is the constant of proportionality in the given differential equation:

y' = 0.06y.

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c) The doubling time is found from ...

2 = e^(0.06t) . . . the multiplying factor is 2 to double the original number

ln(2) = 0.06t . . . . taking natural logs

ln(2)/0.06 = t ≈ 11.55 h . . . . doubling time

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d) Put t=9 into the formula from part (a). After 9 hours, there will be ...

y(9) = 4e^(0.06·9) ≈ 6.9 . . . . million bacteria present