A stick AB of length L is placed vertically by the wall. At its lower end sits a
bug. The lower end B of the stick starts moving to the right with speed v, and at
the same moment the bug starts crawling along the stick with speed u relative
to the stick. What is the maximal height above the floor that the bug reaches
while it crawls along the stick? Top end A of the stick does not lose contact with
the wall.
At time t, B is vt units to the right of the wall. A is then sqrt(L^2  (vt)^2) and the bug is the fraction ut/L of that value above the floor, i.e., ut*sqrt(L^2  (vt)^2)/L.
So we wish to maximize (ut/L)sqrt(L^2  (vt)^2)
The derivative of the first factor is u/L.
The derivative of the second factor is
2(v^2)t(1/2)/sqrt(L^2  (vt)^2)
= (v^2)t/sqrt(L^2  (vt)^2)
The derivative of the full expression is then
(u/L)sqrt(L^2  (vt)^2)  (ut/L)(v^2)t/sqrt(L^2  (vt)^2)
We need to set this to zero:
u*sqrt(L^2  (vt)^2) u((vt)^2)
 = 
L L*sqrt(L^2  (vt)^2)
uL(L^2  (vt)^2) = uL((vt)^2))
L^2  (vt)^2 = (vt)^2
L^2 = 2(vt)^2
t^2 = L^2/(2v^2)
t = L/(v*sqrt(2))
I note that u doesn't come into this expression, which caused some concern.
But the height of the bug at this point is
(ut/L)sqrt(L^2  (vt)^2)
and substituting the value of t = L/(v*sqrt(2)),
(u/(v(sqrt(2))) * sqrt(L^2  L^2 / 2)
(u/v) L
=  = uL/(2v)
2
and does involve u as well as v, but this does not seem reasonable as it does not approach L as u/v becomes larger and larger.
derivative = u(2 v^2 t^2  L^2) / (L sqrt(L^2  v^2 t^2))
with a zero (root) at L/(v sqrt(2)), the same as my calculation.
The plot at that site, when actual values are shown for L, u, and v, clued me in as to what was happening. With large values of u, the bug shoots up so fast he follows the full line of the stick even beyond the stick's end, going on into the wall, as the equations did not include a hint of a barrier.
If the time it takes for the bug to reach the upper end of the stick (L/u) is less than the time it takes to reach the theoretic maximum on the infinite line (L/(v*sqrt(2))), then we need to plug that time into either the formula for the height of the bug or the formula for the height of the top of the stick (they should be equal).
So, if u>v(sqrt(2)) the answer is sqrt(L^2  (vt)^2), when the bug reaches the top of the ladder and has to stop.
Otherwise the answer is uL/(2v).
As a check if they match up at u=v(sqrt(2)), the latter is L/sqrt(2) and the former is
sqrt(L^2  ((u/sqrt(2))(L/u))^2)
= sqrt(L^2  (L/sqrt(2))^2)
= sqrt(L^2  L^2 / 2)
= L/sqrt(2)
which matches.

Posted by Charlie
on 20190522 20:12:38 