Problem 23 - Precalculus

Question

The number of bacteria in a person's body, after they begin a regimen of antibiotics, can be modeled by the function \(b(t)=6-\sqrt{t},\) where \(b\) is measured in tens of thousands and \(t\) is the number of hours after the first dose. Find the limit of the difference quotient for \(b\) to obtain a function \(b(t)\) that represents the instantaneous rate of change of number of bacteria at time \(t.\)

Step-by-Step Solution

Answer

The instantaneous rate of change function is \( \frac{-1}{2\sqrt{t}} \).

Step 1: Understanding the Problem

We need to find the instantaneous rate of change of the function \( b(t) = 6 - \sqrt{t} \). This means we are looking for the derivative of \( b(t) \), which describes the rate of change of the bacteria count at a specific time \( t \).

Step 2: Calculating the Difference Quotient

The difference quotient is defined as \( \frac{b(t + h) - b(t)}{h} \), where \( h \) is a small increment in time. We need to substitute our function \( b(t) \) into this formula.

Step 3: Substituting into the Difference Quotient

For the function \( b(t) = 6 - \sqrt{t} \), we substitute to get:\[ b(t + h) = 6 - \sqrt{t + h} \]Thus, the difference quotient becomes:\[ \frac{(6 - \sqrt{t + h}) - (6 - \sqrt{t})}{h} = \frac{-\sqrt{t + h} + \sqrt{t}}{h} \]

Step 4: Rationalizing the Numerator

To simplify the expression, multiply the numerator and the denominator by the conjugate \( \sqrt{t+h} + \sqrt{t} \):\[ \frac{(-\sqrt{t+h} + \sqrt{t})(\sqrt{t+h} + \sqrt{t})}{h(\sqrt{t+h} + \sqrt{t})} \]This results in:\[ \frac{t - (t + h)}{h(\sqrt{t+h} + \sqrt{t})} = \frac{-h}{h(\sqrt{t+h} + \sqrt{t})} \]

Step 5: Canceling and Simplifying

Cancel \( h \) from the numerator and the denominator:\[ \frac{-1}{\sqrt{t+h} + \sqrt{t}} \]

Step 6: Taking the Limit as h Approaches Zero

Find the limit of the expression as \( h \to 0 \):\[ \lim_{{h \to 0}} \frac{-1}{\sqrt{t+h} + \sqrt{t}} = \frac{-1}{2\sqrt{t}} \]Hence, the instantaneous rate of change of the function \( b(t) \) is \( \frac{-1}{2\sqrt{t}} \).

Key Concepts

Derivative

In mathematics, the derivative is a fundamental concept that represents the rate at which a function changes at any given point. Specifically, in the context of the function \[ b(t) = 6 - \sqrt{t} \]the derivative helps us understand how quickly the number of bacteria is decreasing as time progresses. It's similar to the speed of a car on a road - it tells us how fast or slow the bacteria count is shifting in a very small time interval. By calculating the derivative, we obtain a new function that gives us the instantaneous rate of change. This is the function that can inform us about exactly how the bacteria count is changing, moment to moment, right at time \( t \).

Difference Quotient

The difference quotient is a technique used to approximate the derivative of a function. It's essentially the formula that helps us find how to compute this rate of change. For our function \( b(t) = 6 - \sqrt{t} \), the difference quotient is:\[ \frac{b(t + h) - b(t)}{h} \]where \( h \) is a very small number that approaches zero. This formula calculates the average rate of change over the interval \( h \). The difference quotient sets the stage for finding the derivative, because as the interval \( h \) becomes smaller and smaller, our approximation becomes more precise. This gradually leads us to the instantaneous rate of change when \( h \) is reduced to infinitesimally small.

Limit

Limits are used to mathematically investigate what happens to a function as its inputs approach a certain value. In the context of derivatives, particularly with the difference quotient, a limit allows us to make \( h \) very small so we can compute the precise rate of change. Essentially, by taking the limit of the difference quotient as \( h \) approaches zero, we define the derivative of the function.For instance, with our function, the process involves finding:\[ \lim_{{h \to 0}} \frac{-1}{\sqrt{t+h} + \sqrt{t}} = \frac{-1}{2\sqrt{t}} \]This limit helps us transition from the difference quotient to the derivative, allowing us to see exactly how rapidly the bacterial population decreases at any precise moment \( t \).

Rationalizing the Numerator

Rationalizing the numerator is a strategy we use to simplify expressions, especially when they involve square roots. In the exercise of calculating the derivative via the difference quotient, we had to handle terms like \(-\sqrt{t+h} + \sqrt{t}\). By multiplying by the conjugate, \( \sqrt{t+h} + \sqrt{t} \), we eliminate the square roots in the denominator, which helps in simplifying the expression.This simplification step is crucial because it allows for cancellation of terms like \( h \), which otherwise would make the limit incoherent as \( h \) approaches zero. Once simplified, the expression is suitable for applying the limit and getting the final derivative in an elegant and understandable form. Such algebraic manipulation is fundamental in calculus to make intricate limits manageable and clear.