Problem 27 - Calculus Early Transcendentals

Question

Find an equation of a curve with \(x\) -intercept 2 whose tangent line at any point \((x, y)\) has slope \(x e^{-y} .\)

Step-by-Step Solution

Answer

The equation of the curve is \(e^{y} = \frac{x^2}{2} - 1\).

Step 1: Identify Known Values

We know that the equation of the curve is not explicitly given, but we have some important features: the slope of the tangent is given as \(x e^{-y}\) and the x-intercept of the curve is 2. This means that the point \((2, 0)\) lies on the curve.

Step 2: Set Up the Differential Equation

The slope of the tangent line at any point \((x, y)\) is the derivative of the curve, denoted as \(\frac{dy}{dx}\). Therefore, we can express it as a differential equation: \(\frac{dy}{dx} = x e^{-y}\).

Step 3: Rearrange the Differential Equation

Rearrange the differential equation to separate the variables \(x\) and \(y\). This gives us \(e^{y} \, dy = x \, dx\).

Step 4: Integrate Both Sides

Integrate both sides of the equation. On the left side, integrate \(e^{y} \, dy\) which results in \(e^{y}\). On the right side, integrate \(x \, dx\) which results in \(\frac{x^2}{2}\). We get: \(e^{y} = \frac{x^2}{2} + C\), where \(C\) is the constant of integration.

Step 5: Solve for the Constant of Integration

Use the x-intercept condition \((2, 0)\) to find \(C\). Substitute \(x = 2\) and \(y = 0\) into \(e^{y} = \frac{x^2}{2} + C\). We have \(e^{0} = 1\), so the equation becomes \(1 = \frac{2^2}{2} + C\). Therefore, \(1 = 2 + C\), which simplifies to \(C = -1\).

Step 6: Write the Equation of the Curve

Substitute \(C = -1\) back into the equation \(e^{y} = \frac{x^2}{2} + C\). We obtain \(e^{y} = \frac{x^2}{2} - 1\) as the equation of the curve.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point and has the same slope as the curve at that point. It is essential in calculus because it provides an approximation of the curve near that point, giving us insights into the curve's behavior.

• The slope of the tangent line at the point \( (x, y) \) on a curve is represented by the derivative of the function, denoted as \( \frac{dy}{dx} \).
• For the given exercise, the slope of the tangent line is specified as \( x e^{-y} \), indicating how steep the curve is at any chosen \( x, y \) position.

Understanding the concept of the tangent line is fundamental when working with differential equations since the slope function constrains how the curve can move through different points. By knowing the slope at each point, we can predict the direction and rate of the curve's incline, and this acts as a crucial step in setting up differential equations.

Integration

Integration is the mathematical process of finding the integral of a function, which is essentially the reverse operation of differentiation. It is a powerful tool used to solve differential equations and determine functions from given rates of change.

• In the context of the provided exercise, we separated the variables in the differential equation to facilitate integration: \( e^{y} \ dy = x \ dx \).
• After separation, we integrate each side independently. The left-hand side \( \int e^{y} \, dy \) results in \( e^{y} \), while the right-hand side \( \int x \, dx \) computes to \( \frac{x^2}{2} + C \).

Integration helps us discover the original equation of the curve, as it bridges the gap from differentiated forms (rates of change) back to their underlying functions. Understanding this process is central to building a complete picture of the curve's equation.

Initial Condition

The initial condition is a specific known value for the variables of a differential equation that allows us to find the constant of integration, \( C \). This condition acts as a starting point which uniquely defines the specific solution of the differential equation among the infinite possible solutions.

• In this exercise, the x-intercept \( (2, 0) \) serves as the initial condition. It means the curve passes through the point where \( x = 2 \) and \( y = 0 \).
• By substituting these values into the integrated equation \( e^{y} = \frac{x^2}{2} + C \), we solve for \( C \). With \( e^{0} = 1 \), the equation becomes \( 1 = \frac{4}{2} + C \), leading to \( C = -1 \).

The role of the initial condition is critical because it enables us to determine the constant that tailors the general solution of a differential equation to a precise scenario. This ensures that the curve we derive matches the stipulated condition.