Ch. 9 - First-Order Differential Equations

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 9 - First-Order Differential Equations

Problem 9.4.21

Chapter 9, Problem 9.4.21

Write the formula for a logistic function that has values between y = 0 and y = 1, crosses the line y = 1/2 at x = 0, and has slope 5 at this point.

Verified step by step guidance

Recall the general form of a logistic function: \[y = \frac{L}{1 + e^{-k(x - x_0)}}\] where \(L\) is the maximum value (carrying capacity), \(k\) is the growth rate, and \(x_0\) is the x-value of the sigmoid's midpoint.

Since the function has values between 0 and 1, set \(L = 1\). The function simplifies to: \[y = \frac{1}{1 + e^{-k(x - x_0)}}\]

The function crosses \(y = \frac{1}{2}\) at \(x = 0\), which means the midpoint \(x_0 = 0\). So the function becomes: \[y = \frac{1}{1 + e^{-kx}}\]

To find the slope at \(x = 0\), compute the derivative of \(y\) with respect to \(x\): \[y' = \frac{d}{dx} \left( \frac{1}{1 + e^{-kx}} \right)\] Use the chain rule to differentiate.

Evaluate the derivative at \(x = 0\) and set it equal to 5 (the given slope). Solve for \(k\) to find the growth rate parameter that satisfies the slope condition.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Logistic Function Formula

A logistic function is an S-shaped curve commonly written as y = L / (1 + e^{-k(x - x_0)}), where L is the maximum value, k controls the steepness, and x_0 is the midpoint. For values between 0 and 1, L = 1, simplifying the formula to y = 1 / (1 + e^{-k(x - x_0)}).

Midpoint and Symmetry of Logistic Function

The logistic function crosses its midpoint y = L/2 at x = x_0. This point is the inflection point where the curve changes concavity. Setting x_0 = 0 ensures the function passes through y = 1/2 at x = 0, aligning with the problem's requirement.

Slope of Logistic Function at Midpoint

The slope of the logistic function at the midpoint x = x_0 is given by (kL)/4. Since L = 1 and the slope is specified as 5 at x = 0, we solve 5 = k/4 to find k = 20. This parameter controls how steeply the function rises at the midpoint.

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Related Practice

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