Problem 71 - Calculus Early Transcendentals

Question

Assuming that \(A, k,\) and \(L\) are positive constants, verify that the graph of \(y=L /\left(1+A e^{-k t}\right)\) has an inflection point at \(\left(\frac{1}{k} \ln A, \frac{1}{2} L\right)\)

Step-by-Step Solution

Answer

The graph has an inflection point at \(\left(\frac{1}{k}\ln A, \frac{1}{2}L\right)\).

Step 1: Differentiate the Given Equation

Given the function \(y = \frac{L}{1 + A e^{-k t}}\). The first step is to find the first derivative \(\frac{dy}{dt}\) using the quotient rule:\[ y' = \frac{d}{dt} \left( \frac{L}{1 + A e^{-k t}} \right) = \frac{0 \cdot (1 + Ae^{-kt}) - L \cdot (-Ak e^{-kt})}{(1 + Ae^{-kt})^2} \]Simplify this to:\[ y' = \frac{LAke^{-kt}}{(1 + Ae^{-kt})^2} \]

Step 2: Find the Second Derivative

Next, find the second derivative \(\frac{d^2y}{dt^2}\) by differentiating \(y'\) with respect to \(t\):Apply the quotient rule again:\[ y'' = \frac{(1 + Ae^{-kt})^2 \cdot (-LAk^2 e^{-kt}) - LAke^{-kt} \cdot 2(1 + Ae^{-kt})(-Ak e^{-kt})}{(1 + Ae^{-kt})^4} \]Simplify thoroughly:\[ y'' = \frac{-LAk^2e^{-kt}(1 + Ae^{-kt}) + 2LA^2k^2 e^{-2kt}}{(1 + Ae^{-kt})^3} \]

Step 3: Set Up the Condition for Inflection Points

An inflection point occurs where the second derivative changes sign, which is typically at points where \(y'' = 0\):\[ -LAk^2 e^{-kt} (1 + Ae^{-kt}) + 2LA^2k^2 e^{-2kt} = 0 \]

Step 4: Solve for the Inflection Point's \(t\)-coordinate

Re-arrange and solve the equation from Step 3:\[ -LAk^2 e^{-kt} + 2LA^2k^2 e^{-2kt} = 0 \]Factor out common terms:\[ e^{-kt}(-LAk^2 + 2LA^2k^2 e^{-kt}) = 0 \]This simplifies to:\[ -1 + 2Ae^{-kt} = 0 \]\[ 2Ae^{-kt} = 1 \]\[ e^{-kt} = \frac{1}{2A} \]Taking the natural log on both sides:\[ -kt = \ln\left(\frac{1}{2A}\right) \]\[ t = \frac{1}{k} \ln(2A) \]

Step 5: Verify the \(y\)-coordinate of the Inflection Point

Substitute \(t = \frac{1}{k} \ln(A)\) into the original function to find \(y\):\[ y = \frac{L}{1 + A e^{-k(\frac{1}{k} \ln A)}} \]Simplify the expression:\[ y = \frac{L}{1 + A \cdot \frac{1}{A}} = \frac{L}{1 + 1} = \frac{L}{2} \]

Step 6: Conclusion

The coordinates \(\left(\frac{1}{k}\ln A, \frac{1}{2}L\right)\) satisfy both \(y'' = 0\) and the original equation, confirming that this is an inflection point.

Key Concepts

Quotient Rule

Understanding the quotient rule is essential when dealing with functions that are the ratio of two other functions. It's a technique in calculus used to differentiate these types of functions. If you have two functions, say \(u(x)\) and \(v(x)\), their quotient \(\frac{u(x)}{v(x)}\) can be differentiated using the quotient rule. The formula is:

• \( \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

To apply this, you need the derivatives \(u'(x)\) and \(v'(x)\) which are the derivatives of the numerator and the denominator respectively. The quotient rule helps to compute the rate of change of one quantity with respect to the other in rational functions.
In the given exercise, we used the quotient rule for the function \(y = \frac{L}{1 + Ae^{-kt}}\). Here, the numerator is a constant \(L\), and the denominator is \(1 + Ae^{-kt}\). By substituting these into the rule, we determine the first derivative. Understanding this rule simplifies handling complex rational functions effectively in calculus.

Natural Logarithm

The natural logarithm is a crucial mathematical concept used for calculating the power to which the base 'e' (approximately 2.718) must be raised to yield a given number. Its notation is \(\ln(x)\), where \( x \) is the argument of the logarithm. This function is important in calculus and many mathematical modeling scenarios because it easily assists in solving exponential equations.
The property of the natural logarithm \(\ln(e^a) = a\) proves very useful as shown in the exercise when solving for the \(t\)-coordinate of the inflection point. By manipulating exponential expressions through natural logarithms, complex equations can be simplified.

• For example, if you have \(e^{-kt} = \frac{1}{2A}\), taking the natural logarithm of both sides gives: \(-kt = \ln\left(\frac{1}{2A}\right)\).
• This simplifies and helps solve for \(t\) as in the steps shown.

Natural logarithms are particularly useful when linearizing features of cultural or scientific equations, providing an elegant way to describe growth or decay processes seen in real-world phenomena.

Second Derivative

In calculus, the concept of the second derivative is pivotal, especially when analyzing the concavity and inflection points of a function's graph. The second derivative, denoted as \(y''\) or \(\frac{d^2y}{dt^2}\), is the derivative of the derivative of a function. It essentially measures how the rate of change of the function is itself changing.
This comes into play in the exercise when determining the inflection point of \(y = \frac{L}{1 + Ae^{-kt}}\). At an inflection point, the second derivative equals zero. Inflection points show where the graph of a function changes concavity, i.e., from concave up to concave down or vice versa.

• Setting \(y'' = 0\) and solving for \(t\) provides the coordinate of the inflection point.
• As showcased, \( -1 + 2Ae^{-kt} = 0 \) was simplified to find \( t = \frac{1}{k}\ln(2A) \) indicating a feasible point of inflection.

The ability to compute and understand the second derivative is essential in optimizing functions and analyzing their structural changes. By identifying these special points in a function's behavior, you gain deeper insights into its graphical representation.