Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

A certain bacterial culture is growing so that it has a mass of \(\frac{1}{2} t^{2}+1\) grams after \(t\) hours. C (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 \(t=2 ?\)

Step-by-Step Solution

Answer

(a) The bacteria grew 0.0201 grams. (b) The average growth rate was 2.01 grams per hour. (c) The instantaneous growth rate was 2 grams per hour.

Step 1: Understand the Growth Formula

The bacterial culture's mass as a function of time, \( t \), is given by the equation \( M(t) = \frac{1}{2}t^2 + 1 \). This formula will help us calculate the mass at various points in time.

Step 2a: Calculate Mass at Each Time

Calculate the mass of the bacteria at \( t = 2 \): \[ M(2) = \frac{1}{2}(2)^2 + 1 = 2 + 1 = 3 \text{ grams} \]Calculate the mass at \( t = 2.01 \):\[ M(2.01) = \frac{1}{2}(2.01)^2 + 1 \approx 2.0201 + 1 = 3.0201 \text{ grams} \]

Step 2b: Determine Growth in Interval

The growth over the interval \( 2 \leq t \leq 2.01 \) is:\[ \Delta M = M(2.01) - M(2) = 3.0201 - 3 = 0.0201 \text{ grams} \]

Step 3: Calculate Average Growth Rate

To find the average growth rate over the interval \( 2 \leq t \leq 2.01 \), use the formula:\[ \text{Average Rate} = \frac{\Delta M}{\Delta t} = \frac{0.0201}{0.01} = 2.01 \text{ grams per hour} \]

Step 4: Calculate Instantaneous Growth Rate

The instantaneous growth rate at \( t = 2 \) is given by the derivative of \( M(t) \). First, find \( M'(t) \):\[ M'(t) = \frac{d}{dt}\left(\frac{1}{2}t^2 + 1\right) = t \]Then, evaluate \( M'(t) \) at \( t = 2 \):\[ M'(2) = 2 \text{ grams per hour} \]

Key Concepts

mass function in time

When studying bacterial growth, the mass function in time is a fundamental concept. It describes how the mass of a bacterial culture changes over time. In this particular exercise, the mass function is given by the equation:\[ M(t) = \frac{1}{2}t^2 + 1 \]This formula tells us the mass of the culture at any time \( t \). It's a mathematical way to see how the bacterial mass increases as time progresses. By plugging different values of \( t \) into the equation, we can determine the mass at those specific times. For example, at \( t = 2 \) hours, the mass is:\[ M(2) = \frac{1}{2}(2)^2 + 1 = 3 \text{ grams} \] Understanding this relationship between time and mass is crucial when analyzing the growth patterns of bacteria.

average growth rate

The average growth rate gives us a sense of how quickly a bacterial culture grows over a specific period. It's calculated by taking the difference in mass at two time points and dividing it by the period during which the change occurred.For the interval between \( t = 2 \) and \( t = 2.01 \), we've already determined that the change in mass, \( \Delta M \), is:\[ \Delta M = 0.0201 \text{ grams} \]And the time interval, \( \Delta t \), is 0.01 hours. Thus, the average growth rate is:\[ \text{Average Growth Rate} = \frac{0.0201}{0.01} = 2.01 \text{ grams per hour} \]This rate helps us understand the overall trend in bacterial growth during that short time, providing a general snapshot of how fast the mass is changing.

instantaneous growth rate

The instantaneous growth rate differs from the average growth rate in that it tells us how fast the mass is growing at an exact moment in time, rather than over an interval. To find this, we need to differentiate the mass function concerning time to get the derivative.The derivative, usually denoted as \( M'(t) \), gives us the rate of change at any specific time \( t \). For this exercise, the derivative of the mass function is:\[ M'(t) = t \]So, at \( t = 2 \), the instantaneous growth rate of the bacteria is:\[ M'(2) = 2 \text{ grams per hour} \]This value means that right at 2 hours, the bacterial mass is increasing at a speed of 2 grams per hour. It's like taking a snapshot of how fast the growth is happening at that exact point.

derivative

The derivative is a core mathematical concept. It's used to determine the rate of change of a function. When dealing with bacterial growth, the derivative of the mass function helps us find how fast the mass is increasing over time.Given the mass function \( M(t) = \frac{1}{2}t^2 + 1 \), to find the derivative, we use basic differentiation rules. The result is:\[ M'(t) = \frac{d}{dt}\left(\frac{1}{2}t^2 + 1\right) = t \]This derivative indicates that the rate of growth of the bacterial mass is directly proportional to time. It means the bacteria grow faster as time increases. Understanding derivatives is essential for quantitatively analyzing changes in growth within precise time frames.