Question

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

(a) 0.02005 grams; (b) 2.005 grams/hour; (c) 2 grams/hour.

Step-by-step solution

Identify the Function

The mass of the bacterial culture after \( t \) hours is given by the function:\[m(t) = \frac{1}{2}t^{2} + 1\]This function describes how the mass of the bacterial culture changes over time.

Calculate Growth During the Interval (Part a)

To determine how much the bacterial culture grew from \( t = 2 \) to \( t = 2.01 \), find \( m(2.01) - m(2) \):\[\begin{aligned}m(2) &= \frac{1}{2}(2)^2 + 1 = \frac{1}{2}(4) + 1 = 3,\m(2.01) &= \frac{1}{2}(2.01)^2 + 1 = \frac{1}{2}(4.0401) + 1 = 3.02005.\\end{aligned}\]Hence, the growth during the interval is:\[ m(2.01) - m(2) = 3.02005 - 3 = 0.02005 \, \text{grams.} \]

Calculate Average Growth Rate (Part b)

The average growth rate over a time interval \([a, b]\) is given by:\[\text{Average growth rate} = \frac{m(b) - m(a)}{b - a}.\]For \(a = 2\) and \(b = 2.01\), substitute the values from Step 2:\[\begin{aligned}\text{Average growth rate} &= \frac{m(2.01) - m(2)}{2.01 - 2} = \frac{0.02005}{0.01} = 2.005 \, \text{grams per hour.}\end{aligned}\]

Calculate Instantaneous Growth Rate (Part c) using Derivative

The instantaneous growth rate at time \( t \) is the derivative \( m'(t) \). Differentiate the function \( m(t) = \frac{1}{2}t^{2} + 1 \):\[\begin{aligned}m'(t) &= \frac{d}{dt}\left(\frac{1}{2}t^{2}\right) = t.\end{aligned}\]Evaluate the derivative at \( t = 2 \):\[\begin{aligned}m'(2) &= 2.\end{aligned}\]Thus, the instantaneous growth rate at \( t = 2 \) is \(2 \, \text{grams per hour.} \)

Key concepts

Bacterial Growth

Bacterial growth is a fascinating process where bacteria multiply over time. This growth can be modeled mathematically to understand changes happening in a bacterial culture. In our exercise, the mass of the bacterial culture is given by a quadratic function of time:
\[m(t) = \frac{1}{2}t^{2} + 1\]
This equation tells us that as time, \(t\), increases, the mass of the bacterial culture grows at a predictable rate. Understanding this function is crucial because it allows us to calculate how much a culture grows during a specific time interval. Such growth models are vital in fields like microbiology and biotechnology, where predicting bacterial population sizes can help in experiments and production processes.

Average Growth Rate

The average growth rate measures the change in mass of the bacterial culture over a given time interval. This is helpful as it gives us a simple way to describe how fast the bacteria grow over that period. To calculate it, we use:
\[\text{Average growth rate} = \frac{m(b) - m(a)}{b - a}\]
Where \(m(b)\) and \(m(a)\) are the masses after times \(b\) and \(a\). For the interval from \(2\) to \(2.01\) hours, as calculated, the average growth rate is:
\[2.005 \text{ grams per hour}\]
This result explains how much the bacterial mass increases per hour on average over that short interval. This rate is especially useful for shorter periods, as it can provide a clear picture of growth trends before they might change.

Instantaneous Growth Rate

The instantaneous growth rate is a bit different from the average growth rate. It refers to the rate at which the bacterial culture is growing at a precise moment in time. In mathematics, it's the slope of the tangent line to the mass function's curve at a particular point.
To find it, we compute the derivative of the function with respect to time and evaluate it at the time of interest. For our function:
\[m'(t) = \frac{d}{dt}\left(\frac{1}{2}t^{2} + 1\right) = t\]
At \(t = 2\), the instantaneous growth rate is simply:
\[2 \text{ grams per hour}\]
This tells us how fast the bacterial culture is growing exactly when \(t = 2\). Instantaneous rates are crucial for detecting precise changes in growth dynamics, helping in real-time monitoring of cultures.

Derivative

The derivative is a core concept in calculus, used to determine how a function's output changes with respect to changes in its input. When we talk about derivatives in the context of a bacterial growth model, we express how rapidly the mass of bacteria is changing per unit time.
For a simple function like:
\[m(t) = \frac{1}{2}t^{2} + 1\]
The derivative, \(m'(t)\), is calculated as:
\[m'(t) = t\]
This derivative tells us that the rate of mass increase is directly proportional to time. Calculating the derivative helps us find both average and instantaneous growth rates. Overall, derivatives are essential in many scientific fields, enabling the precise modeling of varying quantities such as growth, speed, and decay.