137. Find a curve through the origin in the xy-plane whose length from x = 0 to x = 1 is L = ∫ from 0 to 1 of sqrt(1 + (1/4)e^x) dx.

Verified step by step guidance

1. Recognize that the problem asks for a curve \( y = f(x) \) passing through the origin \( (0,0) \) whose arc length from \( x=0 \) to \( x=1 \) is given by \( L = \int_0^1 \sqrt{1 + \frac{1}{4} e^x} \, dx \).
2. Recall the formula for the arc length of a curve \( y = f(x) \) from \( x=a \) to \( x=b \): \[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
3. Compare the given integral's integrand \( \sqrt{1 + \frac{1}{4} e^x} \) with the arc length formula's integrand \( \sqrt{1 + (y')^2} \). This implies: \[ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{1}{4} e^x \] which simplifies to: \[ \left(\frac{dy}{dx}\right)^2 = \frac{1}{4} e^x \]
4. Take the square root of both sides to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \pm \frac{1}{2} e^{x/2} \] Choose the positive root to find an increasing function (or negative if decreasing is desired).
5. Integrate \( \frac{dy}{dx} = \frac{1}{2} e^{x/2} \) with respect to \( x \) to find \( y(x) \): \[ y = \int \frac{1}{2} e^{x/2} \, dx + C \] Use the initial condition \( y(0) = 0 \) to solve for \( C \).

Key Concepts

Arc Length of a Curve

The arc length of a curve y = f(x) from x = a to x = b is given by the integral L = ∫ from a to b of sqrt(1 + (dy/dx)^2) dx. This formula measures the distance along the curve by summing infinitesimal line segments, accounting for both horizontal and vertical changes.

Differential Equation from Arc Length

Given an arc length integral involving sqrt(1 + (dy/dx)^2), we can form a differential equation by equating the integrand to sqrt(1 + (dy/dx)^2). Solving this equation for dy/dx allows us to find the function y(x) that produces the specified arc length.

Initial Condition and Integration

Since the curve passes through the origin, the initial condition y(0) = 0 is used to determine the constant of integration after solving the differential equation. Integrating dy/dx with this condition yields the explicit form of the curve y(x).