Ant's quest

Let's solve the general case. The piece of rubber has length $L$ , the stretching speed is $V$ , and the ant's crawling speed with respect to the rubber is $S$ .

At time $T$ , the end being stretched is at distance $L(T) = L + VT$ from the wall. Let the ant's distance from the wall be $A(T)$ , and let $F(T)$ = $A(T)/L(T)$ . If $F(T)$ reaches 0, the ant completes its quest.

$F(T + dT)$ = $((A + (A/L)V dt - S dt)/(L + V dt))$

$F(T+ dT)$ = $F(T) - (S/L) dt$

$dF(T)/dT$ = $-(S/L)$

$L(T) = L + VT$ , and $F(0) = 1$ , so:

$F(T) = (1 - (S/V)) ln ((1 + (V/L))T)$

$F(T) = 0$ when $T = (L/V)(e^{(V/S)} - 1)$

Substituting in, $F(T) = 0$ at $T = 10(e^2 - 1)$ sec = 63.9 sec

I solved this problem by finding a mapping (function) from the location of the original particles of the rubber band (at time t = 0) to the location of the particles at t = t, and then used it to help form and solve a differential equation modeling the relationship of the velocity and position of the ant. I let x be a value from 0 cm to 100 cm that represents the starting distance of each particle of the rubber band. This will essentially be my indexing tool for the particles. I do not find it necessary to write the units from this point further because I will not need to do any unit conversions. As the particles of the rubber band begin to move, each particle's new location can be determined based on the fact that the stretching is uniform (multiplicative). This means the distance of a given particle from the wall remains in the same proportion with another given particle's distance from the wall at every point in time. Let A stand for the furthest particle from the wall, and B stand for some other particle that, at time t = 0, is x cm from the wall. The location of A at time t is given by 100 + 10t, since it's moving away from the wall at 10 cm/s. I find the location of B at time t (which I denote by f(x, t)), by setting up a proportion: $\frac{100+10t}{100}=\frac{f(x,t)}{x} \Rightarrow f(x, t)=\frac{x(10+t)}{100}$ . I find the partial derivative of f(x, t) with respect to t in order to find the velocity of each particle, so that substituting a particle's index, x, (original distance at t = 0) will give the function of its velocity: $\frac{\partial f}{\partial t}=\frac{x}{10}$ . Then I let x = g(t) represent the index (original distance) of the particle that the ant is directly above at time t, and h(t) represent the position of the ant (distance from the wall). So, then the ant's velocity and position are given by (resp.): $h'(t)=\frac{g(t)}{10}-5$ and $h(t)=f(g(t),t)=\frac{g(t)(10+t)}{10}$ . Solving for g(t) in the second of the two equations, I obtain $g(t)=\frac{10h(t)}{10+t}$ .Then I substitute the result into the first equation and put the resulting ODE in standard form: $h'(t)-\frac{h(t)}{10+t}=-5$ . From here, I multiply the equation by $e^{\int \frac{-1}{10+t}dt}=\frac{1}{10+t}$ (ignoring the constant of integration), and proceed to solve the ODE: $\frac{h'(t)}{10+t}-\frac{h(t)}{(10+t)^2}=\frac{-5}{10+t} \Rightarrow \frac{d}{dt}[\frac{h(t)}{10+t}]=\frac{-5}{10+t} \Rightarrow \frac{h(t)}{10+t}=-5 \ln (10+t)+C$ . To find C, let t = 0 and note h(0) = 100: $\frac{100}{10}=-5\ln(10)+C \Rightarrow C=10+5\ln(10)$ . Therefore, the ant's distance from the wall is given by $h(t)=(10+t)(-5\ln(10+t)+10+5\ln(10))$ . Setting h(t)=0, and solving for t, yields the final answer $10e^2-10$ seconds.