Question

Find the orthogonal trajectories of each of the following families of curves. In each case sketch several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from \(d y / d x\) for the original curves; this constant takes different values for different curves of the original family and you want an expression for \(d y / d x\) which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. . \((y-1)^{2}=x^{2}+k\).

Short answer

The orthogonal trajectories are given by \( y-1 = \frac{k}{x} \).

Step-by-step solution

- Implicit differentiation

Given the family of curves \( (y-1)^{2}=x^{2}+k \), implicitly differentiate both sides with respect to x. This gives: \[ 2(y-1) \frac{dy}{dx} = 2x \].

- Solve for \(\frac{dy}{dx}\)

Isolate \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{x}{y-1} \].

- Find the negative reciprocal

The slopes of the orthogonal trajectories are the negative reciprocals of the given slopes. Therefore: \[ \frac{dy}{dx} = -\frac{y-1}{x} \].

- Set up the differential equation

Set up the differential equation for the orthogonal trajectories: \[ \frac{dy}{dx} = -\frac{y-1}{x} \].

- Separate variables

Separate the variables: \[ \frac{dy}{y-1} = -\frac{dx}{x} \].

- Integrate both sides

Integrate both sides: \[ \int \frac{dy}{y-1} = \int -\frac{dx}{x} \]. This yields: \[ \ln|y-1| = -\ln|x| + C \].

- Solve for the orthogonal trajectories

Exponentiate both sides to solve for the orthogonal trajectories: \[ |y-1| = e^{-\ln|x| + C} = \frac{e^{C}}{|x|} = \frac{k}{|x|} \]. Hence, \[ y-1 = \frac{k}{x} \], where \(k\) is a new constant.

- Sketch curves and trajectories

Sketch several curves of the original family \( (y-1)^{2}=x^{2}+k \) and several of their orthogonal trajectories \[ y-1 = \frac{k}{x} \]. Original curves are hyperbolas centered at \((0,1)\), and their orthogonal trajectories are rectangular hyperbolas.

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly given in the form \( y = f(x) \). Instead, both x and y are mixed together. In our exercise, we start with the given family of curves \( (y-1)^{2} = x^{2} + k \). This equation doesn't explicitly solve for y. To differentiate it, we treat y as a function of x. By using implicit differentiation on both sides, we obtain: \[ 2(y-1) \frac{dy}{dx} = 2x \]. This step is essential because it allows us to find the relationship between x and y through their derivatives.

Differential Equations

Differential equations involve derivatives of functions and are used to describe various phenomena. In this exercise, the goal is to find an orthogonal trajectory, which means finding a family of curves perpendicular to the given one. After finding the derivative \( \frac{dy}{dx} \) for the given curves, we recognize that orthogonal trajectories have slopes which are negative reciprocals. This changes our relationship to: \[ \frac{dy}{dx} = -\frac{y-1}{x} \]. This new equation is our differential equation for the orthogonal trajectories.

Separation of Variables

Separation of variables is a powerful method often used to solve differential equations. Our differential equation was \[ \frac{dy}{dx} = -\frac{y-1}{x} \]. To separate the variables, we rearrange the terms so that all y terms are on one side and all x terms are on the other. It becomes: \[ \frac{dy}{y-1} = -\frac{dx}{x} \]. This manipulation lets us solve each side individually by integrating.

Integration

Integration is the process of finding the original function from its derivative. After separating the variables and rearranging our differential equation, we integrate both sides: \[ \text{Left Side:} \ \int \frac{dy}{y-1} = \ln|y-1| \ \text{Right Side:} \ \int -\frac{dx}{x} = -\ln|x| \]. Combining both results gives us: \[ \ln|y-1| = -\the |x| +C \]. To simplify, exponentiate both sides to remove the logarithms: \[ |y-1| = \frac{e^C}{|x|} = \frac{k}{|x|} \]. Here, k is a constant term representing \ e^C . Now, we have the relationship in the simpler form: \ y-1 = \frac{k}{x} \.

Reciprocal Slopes

When dealing with orthogonal trajectories, the slopes of the two families of curves must be negative reciprocals. From Step 2, we obtained the slope of our original curve: \[ \frac{dy}{dx} = \frac{x}{y-1} \]. For the orthogonal trajectory, we take the negative reciprocal of the given slope. Thus, the slope becomes: \[ \frac{dy}{dx} = -\frac{y-1}{x} \]. This negative reciprocal relationship ensures that the curves intersect at right angles, which is the defining property of orthogonal trajectories. Understanding this concept is crucial for solving problems involving perpendicular curves.