Wikipedia:Reference desk/Archives/Mathematics/2012 February 8

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February 8

Problem solving : work backward

How do I solve the problem? In June 2005, a sixth grade class planted a tree in the schoolyard. The tree grew about 3 inched a year. If the tree was 38 inches high in June 2010, about how high was the tree when it was planted? — Preceding unsigned comment added by 67.8.185.89 (talk) 04:16, 8 February 2012 (UTC)

Start with:

PRESENT HEIGHT = INITIAL HEIGHT + GROWTH

Now subtract GROWTH from both sides:

PRESENT HEIGHT - GROWTH = INITIAL HEIGHT + GROWTH - GROWTH

PRESENT HEIGHT - GROWTH = INITIAL HEIGHT

Now expand GROWTH:

PRESENT HEIGHT - (YEARS OF GROWTH)×(GROWTH PER YEAR) = INITIAL HEIGHT

Now just plug in your values to get the answer. StuRat (talk) 04:50, 8 February 2012 (UTC)

Time lag in second-order positive feedback

In a second-order positive feedback loop that would produce hyperbolic growth with no time lag, what's the shape of the curve if there's a constant time lag? And what about a time lag that's inversely proportional to the quantity -- does the latter still produce a singularity? NeonMerlin 04:46, 8 February 2012 (UTC)

I don't have an exact solution, but the double exponential function $e^{2^{t}}$ comes close to solving ${\frac {dx}{dt}}={\bigl (}x(t-1){\bigr )}^{2}$. As for the shrinking delay, $x={\frac {-k}{t}}$ (where $k\approx 2.1479$ solves $(k+1)^{2}=k^{3}$) solves ${\frac {dx}{dt}}=\left[x\left(t-{\frac {1}{x(t)}}\right)\right]^{2}$. --Tardis (talk) 06:32, 16 February 2012 (UTC)

Value of infinite limit

I'm having trouble with this, the equation of motion from the free fall article (position as a function of time) for a body falling under gravity but subject to air resistance proportional to the square of velocity:

$y=y_{0}-{\frac {v_{\infty }^{2}}{g}}\ln \cosh \left({\frac {gt}{v_{\infty }}}\right)$

As t becomes infinite, the equation should become

$y=y_{0}-v_{\infty }{t}$

but I get a -ln2 term appearing. What's wrong?86.174.199.35 (talk) 17:19, 8 February 2012 (UTC)

It doesn't start going at rerminal velocity immediately so you need y0 adjusted by some constant amount. Dmcq (talk) 17:45, 8 February 2012 (UTC)

Of course. 86.174.199.35 (talk) 08:28, 9 February 2012 (UTC)

Actually thinking about it one can have an amount that grew to infinity if it did it slowly enough but that's not what's happening in this case thankfully :) Dmcq (talk) 13:06, 9 February 2012 (UTC)