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

= 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? Neon Merlin  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 $$k\approx2.1479$$ solves $$(k+1)^2=k^3$$) solves $$\frac{dx}{dt}=\left[x\left(t-\frac1{x(t)}\right)\right]^2$$. --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)