/*! 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 12 Find the orthogonal trajectories... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 12

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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Most popular questions from this chapter

A given family of curves is said to be self-orthogonal if its family of orthogonal trajectories is the same as the given family. Show that the family of parabolas \(y^{2}=2 c x+c^{2}\) is self orthogonal.

Two chemicals \(c_{1}\) and \(c_{2}\) react to form a third chemical \(c_{3} .\) The rate of change of the number of pounds of \(c_{3}\) formed is proportional to the amounts of \(c_{1}\) and \(c_{2}\) present at any instant. The formation of \(c_{3}\) requires \(3 \mathrm{lb}\) of \(c_{2}\) for each pound of \(c_{1} .\) Suppose initially there are \(10 \mathrm{lb}\) of \(c_{1}\) and \(15 \mathrm{lb}\) of \(c_{2}\) present, and that \(5 \mathrm{lb}\) of \(c_{3}\) are formed in 15 minutes. (a) Find the amount of \(c_{3}\) present at any time. (b) How many \(\mathrm{lb}\) of \(c_{3}\) are present after 1 hour? Suggestion Let \(x\) be the number of pounds of \(c_{3}\) formed in time \(t>0 .\) The formation requires three times as many pounds of \(c_{2}\) as it does of \(c_{1}\), so to form \(x\) lb of \(c_{3}, 3 x / 4\) lb of \(c_{2}\) and \(x / 4 \mathrm{lb}\) of \(c_{1}\) are required. So, from the given initial amounts, there are \(10-x / 4 \mathrm{lb}\) of \(c_{1}\) and \(15-3 x / 4 \mathrm{lb}\) of \(c_{2}\) present at time \(t\) when \(x\) lb of \(c_{9}\) are formed. Thus we have the differential equation $$ \frac{d x}{d t}=k\left(10-\frac{x}{4}\right)\left(15-\frac{3 x}{4}\right) $$ where \(k\) is the constant of proportionality. We have the initial condition $$ x(0)=0 $$ and the additional condition $$ x(15)=5. $$

Suppose a certain amount of money is invested and draws interest compounded continuously. (a) If the original amount doubles in two years, then what is the annual interest rate? (b) If the original amount increases \(50 \%\) in six months, then how long will it take the original amount to double?

A stone weighing \(4 \mathrm{lb}\) falls from rest toward the earth from a great height. As it falls it is acted upon by air resistance that is numerically equal to \(\frac{1}{2} v(\mathrm{in}\) pounds), where \(v\) is the velocity (in feet per second). (a) Find the velocity and distance fallen at time \(t\) sec. (b) Find the velocity and distance fallen at the end of 5 sec.

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