beanstalk doubling every hour what would speed of inchworms need to be? A beanstalk is continuously growing at a constant rate so that after one hour its length 
is doubled, after two hours it is four times the original length, after three hours it is eight 
times as long, and so one. (The growth is uniform: every part of the stalk grows at the 
same rate.) Two bugs start at a distance of one foot away from each other on the beanstalk 
and begin crawling towards each other at the constant speed x ft/hr (x is some number). 
For what values of x will the bugs meet?
 A: Since both inchworms are travelling at the same speed they will meet at a point that is 0.5 feet between them.  So I only have to consider one inchworm trying to get to a point 0.5 feet away.
The beanstalk grows according to $ l_b = 2^t $.  So the location of our inchworm over time will be $ l_i = 0.5 + 2^t - xt $.  We wish the worm to reach the middle so we can change that to being $ 0 = 0.5 + 2^t - xt $.  Rearranging $ x = \frac{0.5 + 2^t}{t} $.  This tells us how fast our worm must travel in order to meet up at time t.
I then used wolfram alpha to work out that the minimum value for x would be 2.20488.
A: Say one worm starts at position $p_1$ climbing upward. If its position at time $t$ is $w_1(t)$, then we know it is accelerating just from the beanstalk growing. And on top of that, it moves with its own speed $x$. The differential equation this suggests is $$w_1'(t)=\ln(2)w_1(t)+x$$ with initial condition $w_1(0)=p_1$. This is a nonhomogeneous linear ODE. Its solution is $$w_1(t)=\left(p_1+\frac{x}{\ln(2)}\right)2^t-\frac{x}{\ln(2)}$$
The other worm is climbing downward, and similarly has position at time $t$ given by $$w_2(t)=\left(p_2-\frac{x}{\ln(2)}\right)2^t+\frac{x}{\ln(2)}$$ Note that $p_2=p_1+1$. We want $w_1(t)=w_2(t)$ to have a solution in $t$. That is: $$\left(p_1+\frac{x}{\ln(2)}\right)2^t-\frac{x}{\ln(2)}=\left(p_2-\frac{x}{\ln(2)}\right)2^t+\frac{x}{\ln(2)}$$ This simplifies to: $$2^t=\frac{2x}{2x-\ln(2)}$$ As time passes, the left side takes values in $[1,\infty)$. So it is required that $\frac{2x}{2x-\ln(2)}>1$ in order for the worms to meet at some time $t$. The numerator is automatically larger than the denominator, and the numerator is positive. So we only need that the denominator is positive. So we need $x>\ln(2)/2=0.3465\ldots$.
A: Motivated by Glen's comment to my other answer, here is an answer that avoids differential equations. View the scenario as the lower worm starting at ground level and not walking, while the upper worm walks with twice as much speed moving downward. [I am avoiding arguing why this specific scenario is equivalent to the posed problem because I find it hard to explain convincingly and clearly.]
Consider the worm who is moving downward. If its speed is just fast enough, then its height above ground will stay constant. It will be walking downward at just the right speed to counter the upward push from the growing bean stalk. So there is a special speed $v$ where this happens. The upper worm reaches the lower worm [who is staying put on the ground] precisely if it can overcome speed $v$.
The upper worm starts at $1$ foot off the ground. If it did not walk, its instantaneous velocity at that moment [from the beanstalk growth] would be $\left.\frac{d}{dt}\left(1\cdot2^t\right)\right|_{t=0}=\ln(2)$. [Or perhaps you avoid calculus althogether if you are studying exponential growth and there is some other way to calculate this speed as $\ln(2)$.] So the worm has to move downward with speed greater than $\ln(2)$.
Converting to the original problem where both worms move, the speed would need to be greater than $\ln(2)/2$, agreeing with my other posted answer.
A: The problem is clearly from physics classical mechanics area.
Let's apply definition of relative velocity to the problem.  See here. I apply it twice. 
First, see the pic. 1-2 bug A is at rest($V_A=0$) and bug B has speed $V_B=2*x (1)$ relative to Bug A.
Second, see pic. 3-4 Bug B now has speed $V_{BR} = V_B-V (2)$ where $V=2^t (3)$feet/hour is beanstalk speed. Bug B $V_{BR}$ is relative to beanstalk, i.e. beanstalk is not growing anymore :) .
The problem now come to a simple one where Bug B needs to cover distance $S=1$ feet. So, $S= V_{BR}*t (4)$. Add equations (1-3) to (4)
$S = t*(2*x-2^t) (5)$ or
$x(t) = \frac{S}{2*t} + 2^{t-1} (6)$, replace S with 1
$x(t) = \frac{1}{2*t} + 2^{t-1} (7)$ 
Let's find the minimum of the function (7) or differentiate 
$x^{'}(t) = -\frac{1}{2*t^2} + 2^{t-1}*ln{2}= 0(8)$  or
$t^2*2^t=\frac{1}{ln{2}} (9)$
(9) is Lambert W-function 
Wolfram Alpha gives this result for $t=0.884124$ or
$x(0.884124) = \frac{1}{2*0.884124} + 2^{0.884124 - 1} (10)$
$x = 0.565532 + 2^{-0.115876}=0.565532 + 0.922822= 1.488354  (10)$
