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Perpendicular lines are two lines that intersect to form a right angle, essentially a 90-degree angle. A key property of these lines is the relationship between their slopes. If two lines are perpendicular, the product of their slopes is

• -1. This means that if you know the slope of one line, you can easily find the slope of a line perpendicular to it by taking the negative reciprocal.

In the context of differential equations, if a line has a slope given by the expression \(-\frac{2y}{3x^2}\), its perpendicular line will have a slope that satisfies the equation \( \text{slope} \cdot \left(-\frac{2y}{3x^2}\right) = -1\). This is an essential step in many calculus problems related to curves, tangents, and normals, as it helps us find the necessary slopes for perpendicular lines that fit the problem's conditions, such as the task of finding a curve that maintains this perpendicularity at every point.

The slope of the tangent line at a point on a curve tells you how steep the curve is at that very point. It is mathematically expressed as the derivative

• \( \frac{dy}{dx} \)

which represents the rate of change of \(y\) with respect to \(x\). In this exercise, the goal is to find a curve whose tangent line slope is perpendicular to a given line slope of \(-\frac{2y}{3x^2}\). To achieve this, we set up the relation \( \frac{dy}{dx} \times \left(-\frac{2y}{3x^2}\right) = -1\), which gives us the equation \( \frac{dy}{dx} = \frac{3x^2}{2y}\). This expression is a crucial part of solving such problems as it relates the tangent line's slope to the specific conditions described in the problem. Understanding how to manipulate these equations helps in separating variables and integrating to solve differential equations.

Initial conditions are specific values that allow us to solve differential equations uniquely. In a mathematical context, they provide information about a function at a particular point, which helps determine any constants that arise during integration.In this exercise, the problem states that the curve must pass through the point \((1, 1)\). This known point is our initial condition. After integrating our solved differential equation, we get an expression involving a constant

• (in this case, \(y^2 = x^3 + C\)).

We substitute the initial condition into this equation: substituting \(x = 1\) and \(y = 1\), we find \(1 = 1 + C\). This simplifies to \(C = 0\), fully determining the curve's equation as \(y^2 = x^3\). Initial conditions are essential for ensuring that a solution to a differential equation is not just a general solution, but one that fits particular real-world or problem-specific scenarios.