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

  1. Oct 2, 2015 #1
    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. jcsd
  3. Oct 2, 2015 #2

    andrewkirk

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    The download rate has to be zero.
    \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*}
     
  4. Oct 3, 2015 #3
    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?
     
  5. Oct 3, 2015 #4

    andrewkirk

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    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}}##.

    Possibility 2: remaining time is estimated as remaining download, divided by average download rate so far. Then the DE is:

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

    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}##.
     
  6. Oct 3, 2015 #5
    My many obligations, thank you so much.
     
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