Problem 28 - Calculus Early Transcendentals

Question

Use a graphing utility to generate a curve that passes through the point \((1,1)\) and whose tangent line at \((x, y)\) is perpendicular to the line through \((x, y)\) with slope \(-2 y /\left(3 x^{2}\right) .\)

Step-by-Step Solution

Answer

The curve is described by the equation \(y^2 = x^3\).

Step 1: Understand the Problem

We are looking for a function whose tangent line is perpendicular to the line with slope \(-2y / (3x^2)\). This means the product of the slopes must be \(-1\) because perpendicular lines have slopes that multiply to \(-1\).

Step 2: Calculate the Slope of the Tangent Line

Let the slope of the tangent line at any point \((x, y)\) be \(m_t = \frac{dy}{dx}\). Since it must be perpendicular to \(m = -\frac{2y}{3x^2}\), we set \(m_t \cdot m = -1\), resulting in \(\frac{dy}{dx} \cdot \left(-\frac{2y}{3x^2}\right) = -1\).

Step 3: Solve for \(\frac{dy}{dx}\)

Rearrange the equation: \(\frac{dy}{dx} = \frac{3x^2}{2y}\). This is a separable differential equation.

Step 4: Separate Variables and Integrate

Separate variables: \(2y \, dy = 3x^2 \, dx\). Integrate both sides: \(\int 2y \, dy = \int 3x^2 \, dx\). This gives \(y^2 = x^3 + C\).

Step 5: Apply the Initial Condition

The curve passes through the point \((1,1)\), so plug in these values: \(1^2 = 1^3 + C\). Thus, \(1 = 1 + C\) which implies \(C = 0\).

Step 6: Write the Function

With \(C = 0\), the final equation of the curve is \(y^2 = x^3\).

Key Concepts

Perpendicular Lines

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.

Slope of the Tangent Line

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

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.