91Ó°ÊÓ

A curve equation is essentially a mathematical expression that describes a curve on a graph. It represents the relationship between the variables involved, commonly using the coordinates \((x, y)\). In the problem at hand, we have a curve passing through a specific point \((1, \frac{\pi}{6})\). This point helps us identify and verify the particular solution among possible general solutions of the curve's equation.
In our exercise, the equation is eventually found to be \(\sin\left(\frac{y}{x}\right) = \log(x) + \frac{1}{2}\). This shows how the curve's behavior is influenced by both trigonometric and logarithmic functions. The goal is to establish this equation from given mathematical conditions and options.

The slope of a curve at any given point is a measure of how steep the curve is at that point. It reflects the rate of change of the function with respect to one of the variables. In calculus, it is represented by the derivative, denoted by \(\frac{dy}{dx}\). For this exercise, the slope of the curve is described by the function \(\frac{y}{x} + \sec \left(\frac{y}{x}\right)\).
The slope provides crucial information that helps us form a differential equation, which is the first step towards finding the curve's equation. By expressing the slope as a combination of \(x\), \(y\), and a trigonometric function (secant), we explore how varied factors might affect the curve's steepness at any point.

Integration is a fundamental process in calculus used to find the curve given a differential equation. Once we have the slope equation \(\frac{dy}{dx} = \frac{y}{x} + \sec\left(\frac{y}{x}\right)\), we leverage integration to unravel the original curve equation from the differential form.
Using the substitution \(u = \frac{y}{x}\) simplifies the problem, transforming it into \(x \frac{du}{dx} = \sec(u)\). Integrating both sides, \(\int \cos(u) \, du = \int x^{-1} \, dx\), results in the equation \(\sin(u) = \log(x) + C\). This process helps convert the rate of change of the curve back into an equation that represents the curve itself.

Initial conditions in differential equations provide specific values at a particular point on the curve, allowing us to find unique solutions. They are vital to moving from a general to a specific solution. In this exercise, the curve passes through the point \((1, \frac{\pi}{6})\).
Applying this to our integrated equation, \(\sin(u) = \log(x) + C\), where \(u = \frac{\pi}{6}\) and \(x = 1\), helps determine \(C\). Given that \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\log(1) = 0\), we find \(C = \frac{1}{2}\). With \(C\) known, the equation \(\sin\left(\frac{y}{x}\right) = \log(x) + \frac{1}{2}\), providing the specific curve solution for the problem.