This is the logo for my university's math center:
The question is what is the area of the triangle as the circles move closer or further apart, given just the rate at which they move ($d(t)$)? What is the rate of change of the area of the triangle if the function $d(t) = t$? $d(t)= e^t$? This isn't a class question or anything, I just wanted to see if I could figure it out and if my math was valid.
On
There are a couple of issues that arise early on in your discussion. It is reasonable to keep the radii of the two circles, which are equal to one another, constant: it maintains the symmetry of the geometry (the triangle starts out equilateral and remains isosceles). It will be rather more difficult to answer your question if we allowed one of the circles to change size.
I think you may have confused matters by introducing a variable $ \ w \ $ , which is really just the base $ \ b \ $ of the triangle. Also, it is not perhaps the best choice of a variable name to use $ \ d \ $ when working with differential calculus; I will use $ \ \Delta(t) \ $ in what follows.
Your choice for coordinates is fine and we will continue to use them here. If the $ \ x-$ axis passes through both circles' centers and the left end of the left circle's diameter is placed at the origin, then the center of the left circle remains at $ \ x = r \ $ and the right end of its diameter is at $ \ x = 2r \ . $ The original position of the right circle is shown in light orange, with its center at $ \ x = 2r \ $ and the left end of its diameter at $ \ x = r \ ; $ this corresponds to $ \ \Delta(0) = 0 \ . $ When the right circle has been shifted, say, to the left, by an amount $ \ \Delta \ , $ represented by the circle in red, its center is then at $ \ x = 2r - \Delta \ $ and the left end is at $ \ x = r - \Delta \ . $ The base of the triangle (marked in violet) is thus $$ b \ = \ 2r \ - \ (r - \Delta) \ = \ r + \Delta \ \ . $$ [What tells us something has gone wrong in your procedure is that if $ \ r \ , \ h \ , \ \text{and} \ w \ $ are positive measurements, $ \ r^2 = h^2 + (r + \frac{1}{2}w)^2 \ $ cannot be an equation.]
Since the triangle is violet is isosceles, we can work with either half of it to set up the right-triangle relation $$ \ r^2 \ = \ h^2 \ + \ \left[ \frac12 (r + \Delta) \right]^2 \ \ \Rightarrow \ \ h^2 \ = \ r^2 \ - \ \left[ \frac12 (r + \Delta) \right]^2 \ \ . $$
[We see that this checks for the initial arrangement of circles: at $ \ t = 0 \ , \ \Delta = 0 \ , \ b = r \ $ (making the triangle equilateral), and $ \ h^2 = r^2 - \frac{1}{4} r^2 = \frac34 r^2 \ \Rightarrow \ h = \frac{\sqrt{3}}{2} r \ , $ the correct altitude of the equilateral triangle.]
For a problem of the kind you propose, we generally like to avoid dealing with radicals (as much as possible) when differentiating. We will instead start with $ \ A^2 = \frac14 b^2 h^2 \ $ and differentiate this relation implicitly with respect to time: $$ \frac{d}{dt} \ [ A^2 ] \ = \ \frac{d}{dt} \left[ \ \frac14 b^2 h^2 \ \right] \ \ \Rightarrow \ \ 2A · \frac{dA}{dt} \ = \ \frac{1}{4} · \left[ \ 2 b · \frac{db}{dt} · h^2 \ + \ b^2 · 2h · \frac{dh}{dt} \ \right] $$
$$ \Rightarrow \ \ \frac{dA}{dt} \ = \ \frac{1}{4bh} · \left[ \ 2 bh^2 · \frac{db}{d\Delta} · \frac{d\Delta}{dt} \ + \ 2b^2h · \frac{dh}{d\Delta} · \frac{d\Delta}{dt} \ \right] $$ $$ = \ \frac{1}{2} · \left[ \ h · \frac{db}{d\Delta} \ + \ b · \frac{dh}{d\Delta} \ \right] · \frac{d\Delta}{dt} \ \ . $$
We could now insert the expressions
$$ \ b(t) = r + \Delta(t) \ , \ \frac{db}{d\Delta} = 1 \ , \ h(t) = \sqrt{\frac34 r^2 - \frac12 r \Delta(t) - \frac14 [\Delta(t)]^2} \ , \ \frac{dh}{d\Delta} = -[r+\Delta(t)] $$ to obtain
$$ \frac{dA}{dt} \ = \ \frac{1}{2} · \left[ \ \sqrt{\frac34 r^2 - \frac12 r \Delta(t) - \frac14 [\Delta(t)]^2} \ - \ 1 \ \right] · \frac{d\Delta}{dt} \ \ . $$
For your "circle-shifting" functions (which I have made a bit more general), $ \ \Delta(t) \ = \ vt \ \Rightarrow \ \frac{d\Delta}{dt} = v \ $ moves the right circle to the left at constant speed for $ \ v > 0 \ $ and to the right for $ \ v < 0 \ . $ In order to have $ \ \Delta(0) = 0 \ $ for your exponential function, you would want $ \ \Delta(t) = e^{kt} - 1 \ \Rightarrow \ \frac{d\Delta}{dt} = k · e^{kt} \ . $
So suppose for both circles the radius is constant. The area of the triangle is $A = \frac{1}{2}bh$, meaning it's dependent on the y coordinate of the intersection of the circles and the distance between their two edges.
Fixing circle one at the origin, circle two's center is at a distance $d(t)$ from the origin at time $t$. Thus the distance between the two edges of the circle, $w(t)$ can be defined in terms of $r$ and $d$: $$d = r - w + r$$ $$w = 2r - d.$$
Thus $w(t) = 2r-d(t)$, and we have $w$ in terms of a differentiable function.
The next part of this problem is finding $h(t)$ in terms of $r$ and $d(t)$. We can think of a triangle where the base is $r + \frac{1}{2}w$ and the height is $h$. The hypotenuse of this triangle, because the intersection must be on the triangle, is equal to the radius $r$. Thus we can construct the equation $$r^2 = h^2 + (r + \frac{1}{2}w)^2.$$ Substituting our found $w$ in terms of given values and equations, $$r^2 = h^2 + (r + \frac{1}{2}(2r - d))$$ $$h = \sqrt{r^2 - 2r + \frac{1}{2}d}.$$
Thus the area of our triangle can be put in terms of $r$ and $d$: $$A = \frac{1}{2}bh=\frac{1}{2}(2r-d) \Big(\sqrt{r^2 - 2r + \frac{1}{2}d} \Big).$$
For simplicity, suppose $r=1$ and $d(t) = t$. $$A(t) = \frac{1}{2}(2-(d(t)) \Big(\sqrt{-1 + \frac{1}{2}d(t)} \Big) $$ Then the area at any time, $A(t)$, is found by just plugging in $t$.
Now for the rate of change of the area, $A'(t)$, differentiate with respect to $t$. $$A'(t) = -\frac{1}{2} \Bigg(d'(t)\sqrt{-1+\frac{1}{2}d(t)} + \frac{1}{4}\frac{d'(t)}{\sqrt{-1+ \frac{1}{2}d(t)}} \Bigg)$$
For $d(t) = t$:
$$A'(t) = -\frac{1}{2} \Bigg(\sqrt{-1+\frac{1}{2}t} + \frac{1}{4}\frac{1}{\sqrt{-1+ \frac{1}{2}t}} \Bigg)$$
For $d(t) = e^{t}$: $$A'(t) = -\frac{1}{2} \Bigg(e^t\sqrt{-1+\frac{1}{2}e^t} + \frac{1}{4}\frac{e^t}{\sqrt{-1+ \frac{1}{2}e^t}} \Bigg)$$