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Chapter 2: Problem 17

A certain bacterial culture is growing so that it has a mass of \(\frac{1}{2} t^{2}+1\) grams after \(t\) hours. (a) How much did it grow during the interval \(2 \leq t \leq 2.01 ?\) (b) What was its average growth rate during the interval \(2 \leq t \leq 2.01 ?\) (c) What was its instantaneous growth rate at \(t=2 ?\)

Short Answer

Expert verified
(a) 0.0201 grams, (b) 2.01 grams/hour, (c) 2 grams/hour.

Step by step solution

01

Determine the Mass at Specific Times

To find out how much the bacterial culture grew during the interval \( 2 \leq t \leq 2.01 \), we first calculate the mass at \( t = 2 \) and \( t = 2.01 \). The formula for mass is \( \frac{1}{2} t^2 + 1 \).Calculate the mass at \( t = 2 \):\[ \text{Mass at } t = 2 = \frac{1}{2} \times (2)^2 + 1 = 2 + 1 = 3 \] grams.Calculate the mass at \( t = 2.01 \):\[ \text{Mass at } t = 2.01 = \frac{1}{2} \times (2.01)^2 + 1 \approx 2.0201 + 1 = 3.0201 \] grams.
02

Calculate the Growth during the Interval

Now we find how much the bacterial culture grew by subtracting the mass at \( t = 2 \) from the mass at \( t = 2.01 \):\[ \text{Growth} = 3.0201 - 3 = 0.0201 \] grams.
03

Calculate the Average Growth Rate

The average growth rate over the interval \( 2 \leq t \leq 2.01 \) can be found using the formula for average rate of change, which is the change in mass divided by the change in time:\[ \text{Average Growth Rate} = \frac{\text{Change in Mass}}{\text{Change in Time}} = \frac{0.0201}{2.01 - 2} = 2.01 \] grams per hour.
04

Find the Instantaneous Growth Rate at \( t = 2 \)

The instantaneous growth rate at \( t=2 \) is found by calculating the derivative of the mass function, which gives the rate of change at any time \( t \). The mass function is given by \( M(t) = \frac{1}{2}t^2 + 1 \).The derivative of \( M(t) \) is:\[ M'(t) = \frac{d}{dt}\left(\frac{1}{2}t^2 + 1\right) = t \]Substitute \( t = 2 \) into the derivative:\[ M'(2) = 2 \]Thus, the instantaneous growth rate at \( t = 2 \) is 2 grams per hour.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Function
In the context of bacterial growth, the **mass function** is a mathematical equation used to describe how the mass of the bacterial culture increases over time. For the given problem, the mass function is expressed as \( M(t) = \frac{1}{2} t^2 + 1 \). This formula allows us to determine the mass of the bacteria at any specific time \( t \). By plugging in different values of \( t \), you can see how the bacterial mass evolves:
  • Plug \( t = 2 \) into the function: You get \( M(2) = \frac{1}{2} \times 2^2 + 1 = 3 \) grams.
  • For \( t = 2.01 \), the mass becomes \( M(2.01) = \frac{1}{2} \times (2.01)^2 + 1 \approx 3.0201 \) grams.
The mass function is essential in understanding how bacteria grow since it captures the overall accumulation of mass over time. This provides a foundation for other calculations such as the average and instantaneous growth rates.
Average Growth Rate
The **average growth rate** gives an overall idea of how fast the bacterial mass is increasing over a specific time interval. In the exercise, we calculate the average growth rate between the times \( t = 2 \) and \( t = 2.01 \).
To find this, you subtract the initial mass from the final mass and then divide by the time difference:
  • Growth during the interval = Final mass at \( t = 2.01 \) - Initial mass at \( t = 2 \) = \( 3.0201 - 3 = 0.0201 \) grams.
  • Time interval = \( 2.01 - 2 = 0.01 \) hours.
Thus, the average growth rate is calculated as:\[\text{Average Growth Rate} = \frac{0.0201}{0.01} = 2.01 \text{ grams per hour}\]This average rate is useful for making predictions about future growth over similar time intervals.
Instantaneous Growth Rate
The **instantaneous growth rate** describes how quickly the bacterial culture's mass is increasing at any exact moment in time, rather than over a time interval. This rate is found by differentiating the mass function. For our problem, we differentiate the function \( M(t) = \frac{1}{2} t^2 + 1 \):
  • The derivative, \( M'(t) \), represents the instantaneous rate: \( M'(t) = t \).
  • At \( t = 2 \), substitute into the derivative: \( M'(2) = 2 \).
This means at exactly \( t = 2 \), the bacteria are growing at a rate of 2 grams per hour, illustrating how swiftly the changes in mass happen exactly at that point in time. This instantaneous rate provides a snapshot of bacterial growth dynamics, essential for understanding more detailed fluctuations in growth patterns.

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