/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Find the orthogonal trajectories... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the orthogonal trajectories of the family of circles which are tangent to the \(y\) axis at the origin.

Short Answer

Expert verified
The family of orthogonal trajectories for the circles tangent to the y-axis at the origin can be represented by the equation: \[\frac{dy}{dx}+\frac{y}{x}=1-\frac{y^2}{x^2}\]

Step by step solution

01

Find the equation of the circles that are tangent to the y-axis at the origin.

To find the equation of this family of circles, we need to use the standard equation of a circle, which is given by: \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle, and \(r\) is the radius. Since the circles are tangent to the y-axis at the origin, this means the center is \((r,0)\), since the distance from the tangent point to the center of the circle is equal to the radius. Thus, the equation becomes: \((x-r)^2 + y^2 = r^2\).
02

Re-write the equation to make it easier to differentiate.

We can re-write the equation as \(x^2 - 2xr + y^2 = 0\). Now, we'll be able to differentiate it with respect to x more easily.
03

Take the orthogonal derivative.

Taking the derivative with respect to \(x\) on both sides (using implicit differentiation), we have: \(\frac{d}{dx}\big(x^2-2xr+y^2\big)=\frac{d}{dx}(0)\) \(2x-2r\frac{dr}{dx}+2y\frac{dy}{dx}=0\) Now, for an orthogonal trajectory, we want \(\frac{dy}{dx}=-\frac{dx}{dy}\): \(\frac{dy}{dx}=-\frac{2x-2r\frac{dr}{dx}}{2y}\) Since the original equation represents the propagation of circles in the x-y plane, the orthogonal trajectories are those which are everywhere perpendicular to them. Thus, we need to eliminate the parameter \(r\) from this equation.
04

Eliminate r from the equation.

To eliminate \(r\) from the equation, we will use the equation of the circles itself: \((x-r)^2 + y^2 = r^2\) \(x^2 - 2xr + r^2 + y^2 = r^2\) \[x^2 - 2xr + y^2 = 0\] Now, we may solve \(-2xr = -x^2-y^2\) for \(r\): \[r= \frac{1}{2}(x^2+y^2)/x\] Substituting this into the equation for the orthogonal derivative, we have: \(\frac{dy}{dx}=-\frac{2x- 2\big(\frac{1}{2}(x^2+y^2)/x\big)\frac{d}{dx}\left(\frac{1}{2}(x^2+y^2)/x\right)}{2y}\) After simplifying and cancelling terms, we have: \(\frac{dy}{dx}=-\frac{2x- (x^2+y^2)\left(\frac{1-(y^2/x^2)}{x^2}\right)}{2y}\) \[\frac{dy}{dx}=-\frac{x-y^2/x}{y}\]
05

Write the final equation for orthogonal trajectories.

Now, we can rearrange the equation for the orthogonal trajectories: \[\frac{dy}{dx}+\frac{y}{x}=1-\frac{y^2}{x^2}\] The equation represents the family of orthogonal trajectories for the family of circles which are tangent to the y-axis at the origin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
A differential equation is an equation that relates a function to its derivatives. These equations are fundamental in understanding how dynamic systems change over time. In our specific problem, we are dealing with the differential equation that comes from the family of circles which are tangent to the y-axis at the origin.
This involves finding the derivative of a function that describes these circles, allowing us to derive the relationship between their orthogonal trajectories.
By working with a differential equation, we can find solutions that describe how a quantity changes with respect to some variable like time or space. In our case, we are looking at spatial variables, so the differential equation helps us find curves (trajectories) that cut the original curves at right angles.
Implicit Differentiation
Implicit differentiation is a technique used when it is difficult or impossible to solve equations for one variable in terms of another before differentiating.
This type of differentiation is particularly useful when we handle equations involving circles or other non-linear objects.
For the circles tangent to the y-axis, the equation is \( (x-r)^2 + y^2 = r^2 \). Differentiating such equations implicitly involves treating all terms involving a variable like y as functions of x, allowing us to find the rate at which y changes with x.
Implicit differentiation becomes crucial in this exercise to find the slope of the curves that are orthogonal to the original family of circles.
Circles Tangent to Y-Axis
When we talk about circles tangent to the y-axis at the origin, it means these circles just "touch" the y-axis without crossing it at the origin point.
The standard circle equation is \( (x-h)^2 + (y-k)^2 = r^2 \), indicating a circle with center at \((h, k)\) and radius \(r\).
For a circle tangent to the y-axis at the origin, the center must lie on the x-axis. This is because the y-axis acts as a line of symmetry, and touching means at zero distance horizontally. Therefore, the center is \( (r, 0) \), with "r" being the radius - the distance from the circle's center to its tangent point.
Orthogonal Derivative
The concept of an orthogonal derivative involves determining the slopes of curves that are perpendicular to a given family of curves at every point of intersection.
This is foundational for finding orthogonal trajectories, which are curve sets that intersect at right angles with each member of the original family of curves.
Simplifying, for any function, if the derivative \( \frac{dy}{dx} \) describes the slope, its orthogonal trajectory will have slope \(-\frac{dx}{dy}\). This transformation gives us a new differential equation representing the orthogonal trajectories.
By using implicit differentiation and eliminating parameters as shown in the original steps, we derived the orthogonal slopes leading towards determining these unique intersecting paths.

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