Problem 57 - Calculus - AP Edition

Question

Let \(p\) denote the number of paramecia in a nutrient solution \(t\) days after the start of an experiment, and assume that \(p\) is defined implicitly as a function of \(t\) by the equation $$0=\ln p+0.83-\ln (2.3-0.0046 p)-2.3 t$$ Use implicit differentiation to show that the rate of change of \(p\) with respect to \(t\) satisfies the equation $$\frac{d p}{d t}=0.0046 p(500-p)$$

Step-by-Step Solution

Answer

The rate of change satisfies \(\frac{dp}{dt} = 0.0046p(500-p)\).

Step 1: Differentiate both sides with respect to t

Start by differentiating the given equation implicitly with respect to time \(t\). The given equation is:\[0 = \ln p + 0.83 - \ln(2.3 - 0.0046p) - 2.3t\]Applying differentiation on both sides:- For \(\ln p\), use the chain rule to find \(\frac{d}{dt}(\ln p) = \frac{1}{p} \cdot \frac{dp}{dt}\).- The constant 0.83 has a derivative of 0.- For \(-\ln(2.3 - 0.0046p)\), use the chain rule: \(\frac{d}{dt}(-\ln(2.3 - 0.0046p)) = -\frac{1}{2.3 - 0.0046p} \cdot (-0.0046) \cdot \frac{dp}{dt}\).- For \(-2.3t\), the derivative is \(-2.3\).

Step 2: Rewrite the differentiation equation

Substitute the differentiated components into the equation:\[\frac{1}{p} \cdot \frac{dp}{dt} + 0 - \frac{-0.0046}{2.3 - 0.0046p} \cdot \frac{dp}{dt} - 2.3 = 0\]Simplify to:\[\frac{1}{p} \cdot \frac{dp}{dt} + \frac{0.0046}{2.3 - 0.0046p} \cdot \frac{dp}{dt} = 2.3\]

Step 3: Factor out dp/dt

Factor \(\frac{dp}{dt}\) out from the left side of the equation:\[\frac{dp}{dt} \left( \frac{1}{p} + \frac{0.0046}{2.3 - 0.0046p} \right) = 2.3\]

Step 4: Simplify the expression inside the parentheses

Combine the terms inside the parentheses by finding a common denominator:\[\frac{1}{p} + \frac{0.0046}{2.3 - 0.0046p} = \frac{2.3 - 0.0046p + 0.0046p}{p(2.3 - 0.0046p)} = \frac{2.3}{p(2.3 - 0.0046p)}\]

Step 5: Derive the equation for dp/dt

Substitute back into the previous equation:\[\frac{dp}{dt} \cdot \frac{2.3}{p(2.3 - 0.0046p)} = 2.3\]Divide both sides by \(\frac{2.3}{p(2.3 - 0.0046p)}\) to solve for \(\frac{dp}{dt}\):\[\frac{dp}{dt} = \frac{2.3 \cdot p(2.3 - 0.0046p)}{2.3} = 0.0046p(500-p)\]

Step 6: Conclusion

Through implicit differentiation, we have shown that the rate of change of \(p\) with respect to \(t\) is \(\frac{dp}{dt} = 0.0046p(500-p)\).

Key Concepts

Rate of Change

In mathematics, the rate of change is a crucial concept that describes how a quantity changes over time. In the context of the given exercise, we are looking at the rate of change of the paramecia population, denoted as \( p \), with respect to time \( t \). Essentially, this is represented by \( \frac{dp}{dt} \), which tells us how the number of paramecia changes for each unit of time.
It is essential to understand the rate of change because it provides insights into various processes, such as growth or decay in biological systems. In our exercise, the formula \( \frac{dp}{dt} = 0.0046p(500-p) \) illustrates how the population of paramecia grows initially and then slows down as it approaches a carrying capacity of 500.
This function is a differential equation that models population dynamics and shows how changes in \( p \) are not constant over time but depend on the current value of \( p \). Understanding this rate of change allows us to predict future population sizes and manage resources in ecological experiments and studies.

Chain Rule

The chain rule is a fundamental technique in calculus used for differentiating compositions of functions. It states that if a variable \( y \) is a function of \( u \), and \( u \) is a function of \( x \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). This rule helps us deal with complex equations where functions are nested within one another.
In our original exercise, the chain rule is applied while differentiating both \( \ln p \) and \(-\ln(2.3 - 0.0046p) \). These terms involve functions inside logarithms, and thus, applying the chain rule allows us to express their derivatives with respect to time \( t \).

• For \( \ln p \), using the chain rule, the derivative becomes \( \frac{1}{p} \cdot \frac{dp}{dt} \).
• For \(-\ln(2.3 - 0.0046p) \), it becomes \(-\frac{1}{2.3 - 0.0046p} \cdot (-0.0046) \cdot \frac{dp}{dt} \).

Understanding and applying the chain rule is vital when dealing with implicit differentiation in problems, where \( p \) is defined implicitly by an equation."

Differential Equations

Differential equations are equations that involve derivatives of functions. They are crucial in modeling real-world problems where change is a key aspect, such as population growth, chemical reactions, or financial markets.
In the context of this exercise, the differential equation derived, \( \frac{dp}{dt} = 0.0046p(500-p) \), represents a classic logistic growth model. This equation captures the dynamics of the paramecia population over time, showing initial exponential growth that gradually slows as the population nears the maximum sustainable size in the given environment.
The presence of \( p(500-p) \) in the equation indicates that the growth rate depends on both the current population \( p \) and the remaining capacity \( 500 - p \), a feature of logistic growth equations. These equations are significant in understanding how populations stabilize over time due to resource limitations.
Differential equations like this one are powerful tools in predicting future behavior of systems, allowing researchers and practitioners to make informed decisions based on these predictions.