Tractrix & limits I'm in calculus 1 and I'm having quite a bit of trouble with this problem. Any advice as to how to obtain a solution or anything would be much appreciated.
Thanks!

To determine the points on the curve, we need to calculate the intersection point of adjacent threads (the red lines), then determine the limit of that intersection point as the end points of the threads approach zero.
Tractrix 1
This is the situation shown in the example above: a given thread runs from the point with coördinates $(r,0)$ to the point with coördinates $(0,1-r)$, $0 ≤ r ≤ 1$.  In other words, the sum of the $x$ coördinate on the $x$-axis at the lower end of the thread and the $y$-coördinate on the $y$-axis at the upper end of the thread is one.
To determine a point on the tangent curve, consider two adjacent threads: one runs from $(r,0)$ to $(0,1-r)$; the other runs from $(r+h,0)$ to $(0,1-(r+h))$.  These lines intersect at a point $(x_{r+h},y_{r+h})$. The point on the curve is $(x_r,y_r)=\lim\limits_{h\to 0}  (x_{r+h},y_{r+h})$.


*

*Write the equation for the line through the points $(r,0)$ and $(0,1-r)$.

*Write the equation for the line through the points $(r+h,0)$ and $(0,1-(r+h))$.

*Calculate the point $(x_{r+h},y_{r+h})$ where these two lines intersect.

*Calculate $\lim\limits_{h\to 0}  (x_{r+h},y_{r+h})$.

*Use the point just determined to write the equation for the curve that passes through all such points.  (Note: the equation should be symmetric in $x$ and $y$.)

 A: The line joining $(r,0)$ to $(0,1-r)$ is given by
$$ \frac{x}{r}+\frac{y}{1-r} = 1 $$
(This is easy to check: just stick the two points in. To get this, take the general equation of a line $ax+by=1$, put both points in and solve simultaneously.) It's easier to work with if we write it as
$$ (1-r)x+ry-r+r^2 = 0. $$
Now, suppose we have the lines
$$ (1-r)x+ry-r+r^2 = 0 \\
 (1-r-h)x+(r+h)y-(r+h)+(r+h)^2 = 0. $$
Subtracting,
$$ -hx + hy - h + 2hr + h^2 = 0, $$
and dividing by $h$ and taking the limit as $h \to 0$ gives
$$ -x+y -1+2r = 0, $$
so we can solve for the parameter $r$:
$$ r = \frac{1}{2}(1+x-y) \implies 1-r = \frac{1}{2}(1-x+y), $$
We can use this to substitute for $r$ in the equation of the line, which gives
$$ 2(1-x+y)x+2(1+x-y)y = (1-x+y)(1+x-y) $$
after clearing denominators. Expanding gives
$$ 2x-2x^2 +4xy +2y-2y^2 = 1-x^2-y^2+2xy \\
0 = x^2 - 2xy + y^2 - 2x-2y+1 $$
This is symmetric in $x$ and $y$, but can be simplified further: completing the square on $x$ gives
$$ (x-y-1)^2-4y = 0, $$
so $x=1-2\sqrt{y}+y$ after choosing the appropriate root. But the right-hand side is just $(1-\sqrt{y})^2$, so, again choosing the appropriate root, this gives
$$ \sqrt{x}+\sqrt{y} = 1 $$
as the equation.
We can verify that this is the right answer by drawing a picture:

The solution curve is faintly visible in blue, and various straight lines joining $(0,r)$ to $(1-r,0)$ are shown in orange. Note that the blue curve is not a circle; it's actually a segment of a parabola; changing coordinates to $u=x - y-1$, $w=y$ gives the classic equation of a parabola, $u^2=4w$.
