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
1- m(t)= t^2
- m(3)= 3^2= 9
- m(3.01)= 3.01^2 =9.0601
- 9.0601-9=0.0601
your symbols came out strange.
Is 3<t<3.01 supposed to say
3 ≤ t ≤ 3.01 ?
yes
a. How much does the bacterial culture grow during the time 3 < t < 3.01?
To find the growth during a specific time interval, we can subtract the initial mass from the final mass. In this case, the initial mass at t = 3 is M(3) = 3^2 = 9 grams, and the final mass at t = 3.01 is M(3.01) = (3.01)^2 = 9.0601 grams.
Therefore, the growth during the time interval 3 < t < 3.01 is:
Growth = Final mass - Initial mass
Growth = M(3.01) - M(3)
Growth = 9.0601 - 9
Growth = 0.0601 grams
Therefore, the bacterial culture grows approximately 0.0601 grams during the time interval 3 < t < 3.01.
b. What is its average rate of growth during the time interval 3 < t < 3.01?
To find the average rate of growth, we need to divide the growth during the time interval by the length of the time interval. In this case, the length of the time interval is 3.01 - 3 = 0.01 minutes.
Therefore, the average rate of growth during the time interval 3 < t < 3.01 is:
Average Rate of Growth = Growth / Time interval
Average Rate of Growth = 0.0601 grams / 0.01 minutes
Average Rate of Growth = 6.01 grams/minute
Therefore, the average rate of growth of the bacterial culture during the time interval 3 < t < 3.01 is approximately 6.01 grams/minute.
c. What is its rate of growth when t = 3?
To find the rate of growth at a specific point, we can take the derivative of the growth function M(t) with respect to t.
In this case, M(t) = t^2.
Differentiating M(t) with respect to t, we get:
dM/dt = 2t
Therefore, the rate of growth when t = 3 is:
Rate of Growth = dM/dt at t = 3
Rate of Growth = 2 * 3
Rate of Growth = 6 grams/minute
Therefore, the rate of growth of the bacterial culture when t = 3 is approximately 6 grams/minute.
No worries! I'll walk you through the solutions step by step.
a. To find the growth of the bacterial culture during the time 3 < t < 3.01, we need to subtract the initial mass at t = 3.01 from the mass at t = 3. Let's calculate it:
M(3) = 3^2 = 9 grams
M(3.01) = (3.01)^2 ≈ 9.0601 grams
So, the growth during the time interval 3 < t < 3.01 is approximately 9.0601 - 9 = 0.0601 grams.
b. The average rate of growth during the time interval 3 < t < 3.01 is the change in mass divided by the change in time. In this case, it is the growth divided by the time interval.
Average rate of growth = (0.0601 grams) / (0.01 minutes) = 6.01 grams/minute
c. To find the rate of growth when t = 3, we need to take the derivative of the mass function M(t) = t^2 and evaluate it at t = 3.
M'(t) = d/dt(t^2) = 2t
Replacing t with 3:
M'(3) = 2(3) = 6
Therefore, the rate of growth when t = 3 is 6 grams/minute.
In summary:
a. The growth during the time 3 < t < 3.01 is approximately 0.0601 grams.
b. The average rate of growth during the time 3 < t < 3.01 is approximately 6.01 grams/minute.
c. The rate of growth when t = 3 is 6 grams/minute.