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Magazine Chapter 5: Problem 5
Berechnen Sie die orthogonalen Trajektorien folgender Kurvenscharen a) \(x^{2}+y^{2}=c^{2}\); b) \(y=c x^{2}\) c) \(y=c \ln x\); d) \(y=c \frac{(x-1)^{2}}{x}\).
Short Answer
Expert verified
a) Orthogonal trajectory: \( y = Kx \); b) \( y^2 + \frac{x^2}{2} = C \); c) Complex integration; d) Complex integration.
Step by step solution
01
Differentiate the Given Family
Each family of curves can be represented explicitly by the parameter \( c \). Start by differentiating both sides of the equations with respect to \( x \) to find the slope \( \frac{dy}{dx} \).
02
Equation - a) Finding Slope
For \( x^2 + y^2 = c^2 \), differentiate with respect to \( x \) to get the slope: \( 2x + 2y \frac{dy}{dx} = 0 \). Thus, \( \frac{dy}{dx} = -\frac{x}{y} \).
03
Equation - a) Orthogonal Trajectories
The slope of the orthogonal trajectory is the negative reciprocal: \( \frac{dy}{dx} = \frac{y}{x} \). Integrate to find the orthogonal trajectories: \( \int \frac{dy}{y} = \int \frac{dx}{x} \), leading to \( \ln |y| = \ln |x| + K \) or \( y = Kx \).
04
Equation - b) Finding Slope
For \( y = cx^2 \), differentiate: \( \frac{dy}{dx} = 2cx = \frac{2y}{x} \).
05
Equation - b) Orthogonal Trajectories
The slope of the orthogonal trajectory is \( -\frac{x}{2y} \). Integrate to find: \( \int 2y \, dy = - \int x \, dx \), leading to \( y^2 + \frac{x^2}{2} = C \).
06
Equation - c) Finding Slope
For \( y = c \ln x \), differentiate: \( \frac{dy}{dx} = \frac{c}{x} = \frac{y}{x \ln x} \).
07
Equation - c) Orthogonal Trajectories
The orthogonal slope becomes \( -\frac{x \ln x}{y} \). Integrate to find: \( \int y \, dy = \int - x \ln x \, dx \); resulting in more complex integration forms, yielding non-elementary integration.
08
Equation - d) Finding Slope
For \( y = c \frac{(x-1)^2}{x} \), differentiate: \( \frac{dy}{dx} = \frac{c(2(x-1)x - (x-1)^2)}{x^2} \). Simplify to find the slope \( \frac{2y}{x} + c\frac{x-1}{x} \).
09
Equation - d) Orthogonal Trajectories
The orthogonal slope becomes the negative reciprocal: \(-\left(3x - 2 \right)^{-1}\). Integration of this might follow using trigonometric substitution or specialized tables due to complexity.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation Techniques
Differentiation is a fundamental concept in calculus that allows us to determine the rate of change of a quantity. To find orthogonal trajectories, we begin by differentiating the given equation. This involves finding the derivative of the functions with respect to the variable, typically denoted as \( x \). By calculating the first derivative \( \frac{dy}{dx} \), we achieve the slope of the curve at any point.
- For a relation like \( x^2 + y^2 = c^2 \), implicit differentiation is employed. Here, for both sides, the derivative with respect to \( x \) gives us: \( 2x + 2y \frac{dy}{dx} = 0 \).
- Simplifying and solving the equation provides us with \( \frac{dy}{dx} = -\frac{x}{y} \), which is a crucial step in finding orthogonal trajectories.
Integration Techniques
Integration serves as the counterpart to differentiation in calculus. To find orthogonal trajectories, integration is used to reconstruct a curve from its slope. The integration process often involves calculating indefinite integrals based on the determined slopes of orthogonal trajectories. Let's delve into some key integration techniques:
- For basic slopes, like those obtained from \( \frac{dy}{dx} = \frac{y}{x} \), we set up integrals to solve. This results in \( \int \frac{dy}{y} = \int \frac{dx}{x} \), leading to solutions like \( y = Kx \).
- For more complex equations, such as those derived from \( -\frac{x \ln x}{y} \), integration might require substitution techniques or may result in non-elementary forms. These calculations often involve skills in tackling integrals with logarithmic or trigonometric expressions.
Curve Families
Curve families are sets of curves that are related by a parameter. The main idea is that by changing the parameter, we generate different members of the family. For example, the equation \( y = cx^2 \) represents a family of parabolas, each determined by a different \( c \) value.
- The purpose is to understand the geometrical and analytical properties shared among the curves, like intersections, symmetries, and trajectories.
- Finding orthogonal trajectories involves analyzing these curve families and determining new curves that intersect at right angles. For instance, if we start with circles described by \( x^2 + y^2 = c^2 \), solving for orthogonal trajectories might yield straight lines that are tangent at 90 degrees.
Slope of a Curve
The slope of a curve is a critical concept in understanding the geometry of functions. It represents the steepness or incline of the curve at any given point and is calculated as the derivative \( \frac{dy}{dx} \).
- For orthogonal trajectories, slopes are particularly important because they guide us to determine the lines that intersect these curves at right angles.
- For example, from \( \frac{dy}{dx} = -\frac{x}{y} \), the orthogonal slope found is \( \frac{dy}{dx} = \frac{y}{x} \). This is obtained by taking the negative reciprocal, a common technique in dealing with slopes in calculus.
