Testwiki:Reference desk/Archives/Mathematics/2015 April 16

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April 16

Gross tonnage inverse solution

I always thought this would not even be a recognizable special function. However I derived a solution in terms of the Lambert W function, as follows:

Make the substitution $x=\log V+10,V=10^{x-10}$. Then:

$GT=\frac{10^x(.02x)}{10^{10}}$

$\frac{10^{12}\ln 10}{2}GT\ln 10=e^{x\ln 10}x\ln 10$

$\frac{W(\frac{10^{12}\ln 10}{2}GT\ln 10)}{\ln 10}=x$

from which

$V=10^{\frac{W(\frac{10^{12}\ln 2}{2}GT\ln 10)}{\ln 10}}/10^{10}$.

Wolfram Alpha gives ([1]) the simpler $V=\frac{115.129GT}{W(1.15129*10^{12}GT)}$. Thus I want to ask: firstly, is my solution equivalent to WA's, and if so, how? If not, what did I do wrong? My answer works after substituting it into the formula for GT, but how does WA's?--Jasper Deng (talk) 04:20, 16 April 2015 (UTC)

Nevermind. I answered both questions on my own. For the equivalence to WA's solution, multiply the answer by $\frac{W(\frac{10^{12}\ln 2}{2}GT\ln 10)}{W(\frac{10^{12}\ln 2}{2}GT\ln 10)}$ and use the fact that $W(z)e^{W(z)}=z$. That last identity also aids in checking the solution by substitution.--Jasper Deng (talk) 06:52, 16 April 2015 (UTC)

(ec) The Gross Tonnage (GT) is related to the volume (V) by the equation

GT = 0.02 V ln(V)/ln(10) + 0.2 V

So

y = (V+k) ln(V)

where y = 50 ln(10) GT and k=10 ln(10). Take it from here. Bo Jacoby (talk) 07:13, 16 April 2015 (UTC).