After t minutes of growth, a certain bacterial culture has a mass, in grams, of M(t) =t^2.
a. How much does the bacterial culture grow during the time 3<t<3.01?
b. What is its average rate of growth during the time interval 3<t<3.01?
c. What is its rate of growth when t=3?
****Full solutions to please because I do not know what I'm doing and so confused with calculus
To solve these questions, we need to use calculus concepts such as differentiation and integration. Don't worry if you're not familiar with these concepts; I will explain each step in detail.
a) To determine how much the bacterial culture grows during the time interval 3 < t < 3.01, we need to find the difference between the mass at time t = 3.01 and the mass at time t = 3.
M(t) = t^2
Let's calculate the mass at t = 3 and t = 3.01:
M(3) = (3)^2 = 9 grams
M(3.01) = (3.01)^2 = 9.0601 grams
Therefore, the bacterial culture grows by 9.0601 - 9 = 0.0601 grams during the time interval 3 < t < 3.01.
b) The average rate of growth during the time interval 3 < t < 3.01 can be found by dividing the change in mass by the change in time:
Average rate of growth = (Change in mass) / (Change in time)
The change in mass is 0.0601 grams (from part a), and the change in time is 3.01 - 3 = 0.01 minutes.
Average rate of growth = 0.0601 / 0.01 = 6.01 grams per minute
Therefore, the average rate of growth during the time interval 3 < t < 3.01 is 6.01 grams per minute.
c) To find the rate of growth when t = 3, we need to take the derivative of the mass function M(t) with respect to time t and then evaluate it at t = 3.
M(t) = t^2
To find the derivative of M(t), we can use the power rule of differentiation, which states that the derivative of t^n is n*t^(n-1).
dM(t)/dt = d(t^2)/dt = 2t
Now, let's evaluate the derivative at t = 3:
dM(t)/dt = 2t
dM(t)/dt at t = 3 = 2*3 = 6
Therefore, the rate of growth of the bacterial culture when t = 3 is 6 grams per minute.
I hope this explanation helps! Let me know if you have any further questions.