Question

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

Short answer

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

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: \((x-r)^2 + y^2 = r^2\).

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.

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.

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: \(\frac{dy}{dx}=-\frac{2x- (x^2+y^2)\left(\frac{1-(y^2/x^2)}{x^2}\right)}{2y}\) \[\frac{dy}{dx}=-\frac{x-y^2/x}{y}\]

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.