# X''(t) required for reciprocal of X'(t) to be constant

1. Oct 2, 2015

### NotASmurf

Hey all, was watching my downloads today and saw that my download speed was dropping at such a rate that the eta of the download was almost constant, so I was wondering what negative acceleration would be required for the eta to be constant, if time left is size/speed, speed is first derivative of time and
acceleration is second derivative of time what acceleration is needed for time left to be constant? So a function where 1/integral( f(x) ) = c Any help appreciated.

Last edited: Oct 2, 2015
2. Oct 2, 2015

### andrewkirk

\begin{align*}
\frac{1}{\int_0^t f(u)\,du}&=C\\
\frac{1}{C}&=\int_0^t f(u)\,du\\
\frac{d}{dt}\frac{1}{C}&=\frac{d}{dt}\int_0^t f(u)\,du\\
0&=f(t)
\end{align*}

3. Oct 3, 2015

### NotASmurf

Thanks, 0 is a solution but is there no f'(t) for f(t) where f(t) will approach zero at a rate that 1/integral (f(t)) is constant until it approaches 0? Or perhaps for a certain interval?

4. Oct 3, 2015

### andrewkirk

The differential equation is wrong. If we write the correct DE then there are nonzero solutions. There are two possibilities that occur to me:

Possibility 1: remaining time R is estimated as size of remaining download, divided by current download rate. Assume total download size is S, f(t) is the instantaneous download rate at time t, and R is the estimated remaining time, which is constant. Then the download will never complete, only asymptotically approaching completeness, and we require that $S=\int_0^\infty f(t)\,dt$. So our DE is

$$R = \frac{S-\int_0^\infty f(t)\,dt}{f(t)}$$

This is easily manipulated to a simple, standard DE, whose solution is $f(t)=\frac{S}{R}e^{-\frac{t}{R}}$.

$$R=\frac{S-\int_0^\infty f(t)\,dt}{\left(\int_0^\infty f(t)\,dt\right)\ /\ t}$$

We can re-arrange this and express it as

$$(R+t)F(t)=St$$

where $F(t)\equiv\int_0^t f(u)\,du$.

That has solution $F(t)=C-\frac{SR}{t+R}$, whence $f(t)=SR(t+R)^{-2}$.

5. Oct 3, 2015

### NotASmurf

My many obligations, thank you so much.