The population, p in thousands of bacteria colony can be modelled by the function p(t)=200+20t-t^2, where t is the time, in hours, t is greater than or equal to zero.
a) Determine the growth rate of the bacteria population at each of the following times.
i>3h
ii>8h
Ans: find the 1st derivative which is 20-2t, substitute the times given and it yield a negative number and that in thousands is the answer.
b)What are the implications of the growth rates in part a)
Ans: since the answer are negative they suggest that the population of the bacteria is decreasing.
c)When does the bacteria population stop growing? What is the population at this time.
Ans: set the original equation to zero it will give a time in hours and substitute that time into the first derivative.
It seems to me that your question contains the answers, or at least the means of calculating them. Why just you don't just follow those directions?
i just wanted to know if it is the correct way of going about with them, i forgot to put the (check) in the title.
a) To determine the growth rate of the bacteria population at each of the given times, we need to find the derivative of the population function p(t) with respect to time t.
Given p(t) = 200 + 20t - t^2, we can differentiate it using the power rule and the constant multiple rule:
p'(t) = d/dt (200 + 20t - t^2)
= 20 - 2t
i) To find the growth rate at 3 hours (t = 3), substitute t = 3 into the derivative:
p'(3) = 20 - 2(3)
= 20 - 6
= 14
Therefore, the growth rate of the bacteria population at 3 hours is 14 thousand bacteria per hour.
ii) To find the growth rate at 8 hours (t = 8), substitute t = 8 into the derivative:
p'(8) = 20 - 2(8)
= 20 - 16
= 4
Therefore, the growth rate of the bacteria population at 8 hours is 4 thousand bacteria per hour.
b) The negative growth rates indicate that the bacteria population is decreasing at those times. This implies that the bacteria colony is either dying off or experiencing a decline in its growth rate.
c) To find when the bacteria population stops growing, we need to find the time when the growth rate is zero. Set the first derivative equal to zero and solve for t:
20 - 2t = 0
-2t = -20
t = 10
The population stops growing at t = 10 hours. To find the population at this time, substitute t = 10 into the original equation:
p(10) = 200 + 20(10) - (10)^2
= 200 + 200 - 100
= 400
Therefore, at the time when the bacteria population stops growing (t = 10 hours), the population is 400,000 bacteria.
To determine the growth rate of the bacteria population at each of the given times, we first need to calculate the derivative of the population function. The derivative gives us the rate at which the population is changing with respect to time.
a) i) To find the growth rate at 3 hours (t = 3), we differentiate the function p(t)=200+20t-t^2 with respect to t.
So, p'(t) = 20 - 2t.
Substituting t = 3 into the derivative equation, we get:
p'(3) = 20 - 2(3)
= 20 - 6
= 14.
Therefore, at 3 hours, the growth rate of the bacteria population is 14 thousands per hour.
a) ii) Similarly, to find the growth rate at 8 hours (t = 8), we substitute t = 8 into the derivative equation:
p'(8) = 20 - 2(8)
= 20 - 16
= 4.
Therefore, at 8 hours, the growth rate of the bacteria population is 4 thousands per hour.
b) The implications of the negative growth rates we obtained in part a) indicate that the population of the bacteria is decreasing at those specific times. This means that the bacteria population is declining over time.
c) To find when the bacteria population stops growing, we need to set the derivative equal to zero and solve for t:
p'(t) = 20 - 2t = 0.
Solving the equation:
20 - 2t = 0
-2t = -20
t = 10.
So, when t = 10, the population stops growing. To find the population at this time, substitute t = 10 into the original population function:
p(10) = 200 + 20(10) - (10)^2
= 200 + 200 - 100
= 300.
Therefore, at t = 10 hours, the bacteria population stops growing and the population size is 300 thousand bacteria.