How to determine the equation and length of this curve consistently formed by the intersection of Circles

Question

Consider a Point $A$ that moves linearly on the positive $x$-axis with the velocity $1$ m/s and another Point $B$ at a distance $L$ from $A$ with position $(L,0)$. With each forward motion of point $A$ the Point $B$ moves in an arc upward (i.e. along positive $y$-axis) consistently maintaining the distance $L$ from point $A$. As soon as Point $A$ reaches $x=L$, the Point $B$ is perpendicular to Point $A$ and its position is $(L,L)$.

How do you then determine the equation and length of the path that $B$ traces during the period $A$ traverses from $x=0$ to $L$?

Answer 1

There are an infinity of solutions if there is no special constraint (such as the $x$ coordinate of $B$ is fixed).

If you fix $x_B=L$, then $y_B=x_A$...

EDIT: another interesting case is if $B$ is on a circle of center $(L,\dfrac{L}{2})$ and of radius $\dfrac{L}{2}$. Then of course you know the equation of the curve, but you might want to compute the coordinates of $B$ as a function of $x_A$.

EDIT2 (in response to the fact that this curve is impossible).

$B$ is on the circle, thus $(x_B-L)^2+(y_B-\dfrac{L}{2})^2=\dfrac{L}{2}^2$

And $AB=L$, thus $(x_B-x_A)^2+y_B^2=L^2$

From the two equation you have the following equation in $x_B$:

$(x_B-L)^2+(\sqrt{L^2-(x_B-x_A)^2}-\dfrac{L}{2})^2=\dfrac{L}{2}^2$

Answer 2

This is not a complete solution yet.

Let $v_A=1m/s,x_A(t)=v_A t,y_A(t)=0$.

$$(y_B(t)-y_A(t))^2+(x_B(t)-x_A(t))^2=L^2\tag{1}$$

$$\left(\frac{d x_B(t)}{dt}\right)^2+\left(\frac{d y_B(t)}{dt}\right)^2=v_B^2=1\tag{2}$$

$$x_B(0)=L,y_B(0)=0\tag{3}$$

$$x_B(T)=L,y_B(T)=L\tag{4}$$

These are the two equations and boundary conditions that the coordinates of point B should satisfy.

EDIT: From (1) we get: $$x_B(t)=t+\sqrt{L^2-y_B^2(t)}\tag{5}$$ Notice that this solution satisfies the initial condition (3).

From (5) we obtain: $$\frac{d x_B(t)}{dt}=1-\frac{y_B(t)}{(L^2-y_B^2(t))^{1/2}}\frac{d y_B(t)}{dt}\tag{6}$$

Substitution of (6) into (2) leads to an ODE for $y_B(t)$:

$$\left(\frac{d y_B(t)}{dt}\right)^2+\left(1-\frac{y_B(t)}{(L^2-y_B^2(t))^{1/2}}\frac{d y_B(t)}{dt}\right)^2=0\tag{7}$$

We can now solve $\frac{d y_B(t)}{dt}$ from (7) and obtain:

$$\frac{d y_B(t)}{dt}=\frac{2y_B(t)(L^2-y_B^2(t))^{1/2}}{L^2}\tag{8}$$

The solution of $y_B(t)$ is given by:

$$y_B(t)=\frac{4L\exp\left(\frac{2t}{L}+L C_1\right)}{1+4\exp\left(\frac{4t}{L}+2L C_1\right)}\tag{9}$$

Setting $y_B(T)=L$, we can determine the integration constant $C_1$ in (9) so (9) becomes $$y_B(t)=\frac{2L\exp\left(\frac{2(t+T)}{L}\right)}{\exp\left(\frac{4t}{L}\right)+\exp\left(\frac{4T}{L}\right)}\tag{10}$$.

But I was not able to set $y_B(0)=L$ in (10) to determine $T$.

Answer 3

Let $u$ and $v$ be the speeds of $A$ and $B$ respectively.

Then, if $A\equiv(ut,0)$ and $B\equiv(x,y)$, you have $$\begin {cases} (x-ut)^2+y^2=L^2 \\ \\ \dot x^2 + \dot y^2=v^2 \end {cases}$$Factor $\dot y^2$ in the first side of the second equation to get $$\dot y=\frac v {\sqrt {\left(\frac{dx}{dy}\right)^2+1}}$$ Differentiate the first equation with respect to $\,t$ and divide by $\,\dot y$.

You obtain $$\sqrt{L^2-y^2}\left(\frac{dx}{dy}-\frac uv \sqrt {\left(\frac{dx}{dy}\right)^2+1}\right)+y=0$$ Note that the result depends only on the ratio $\frac uv$ .

After some manipulations you get a quadratic equation in $\frac{dx}{dy}$ with coefficients in $y\,$.

Let $f(y)$ be the right solution $\,$(squaring can introduce extraneous solutions).

Then solve $$\int_0^L f(y)\;dy=0$$ as an equation in $\frac uv$ .

At the end plot $$x(y)=L+\int_0^y f(z)\;dz$$ on $[0,L]$ with that value of the ratio.