Best Answer plasmid, 31 July 2014 - 03:17 PM

Spoiler for using differential equations

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Initial mass of salt = S(0) = 10 kg

Initial volume = V(0) = 3000 L

Change in volume over time = dV/dt = 5 L/min (net from water coming in and going out)

Volume at any specific time = V(0) + dV/dt * t = 3000 + 5t

Change in mass of salt over time = dS/dt =

= 0.5 kg/min (that's 0.05 kg/L * 10 L/min coming in) - S(t)/V(t) * 5 L/min (going out)

= 0.5 - 5S(t) / (3000 + 5t)

dS/dt + 5S(t) / (3000 + 5t) = 0.5

This is of the form dy/dx + y*f(x) = g(x), which can in general be solved as

y = (Integral [ g(x) e

S(t) = (Integral [0.5 * e

Use the following substitution

e

S(t) = (Integral [ 0.5 * (3000+5t) dt] + C) / (3000+5t)

S(t) = (Integral [(1500+2.5t) dt] + C) / (3000+5t)

S(t) = (1500t + 1.25t

Since S(0) = 10 kg, C must be 300000

Check that by calculating the derivative

S(t) = (1500t + 1.25t

By the multiplication rule

dS/dt = (2.5t + 1500) / (5t + 3000) - 5 (1500t + 1.25t

= 0.5 - 5 (1500t + 1.25t

= 0.5 - 5 S(t) / (3000 + 5t)

Which is exactly what it should be.

Initial volume = V(0) = 3000 L

Change in volume over time = dV/dt = 5 L/min (net from water coming in and going out)

Volume at any specific time = V(0) + dV/dt * t = 3000 + 5t

Change in mass of salt over time = dS/dt =

= 0.5 kg/min (that's 0.05 kg/L * 10 L/min coming in) - S(t)/V(t) * 5 L/min (going out)

= 0.5 - 5S(t) / (3000 + 5t)

dS/dt + 5S(t) / (3000 + 5t) = 0.5

This is of the form dy/dx + y*f(x) = g(x), which can in general be solved as

y = (Integral [ g(x) e

^{Integral [f(x) dx]}dx] + C) / e^{Integral[f(x) dx]}S(t) = (Integral [0.5 * e

^{Integral [(5/(3000+5t) dt]}dt] + C) / e^{Integral [5/(3000+5t) dt]}Use the following substitution

e

^{Integral [(5/(3000+5t) dt]}= e^{ln(3000+5t)}= 3000+5tS(t) = (Integral [ 0.5 * (3000+5t) dt] + C) / (3000+5t)

S(t) = (Integral [(1500+2.5t) dt] + C) / (3000+5t)

S(t) = (1500t + 1.25t

^{2}+ C) / (3000+5t)Since S(0) = 10 kg, C must be 300000

**S(t) = (1500t + 1.25t**^{2}+ 30000) / (3000+5t), with time in units of minutes and salt in units of kgCheck that by calculating the derivative

S(t) = (1500t + 1.25t

^{2}+ 30000) / (3000+5t)By the multiplication rule

dS/dt = (2.5t + 1500) / (5t + 3000) - 5 (1500t + 1.25t

^{2}+ 30000) / (3000+5t)^{2}= 0.5 - 5 (1500t + 1.25t

^{2}+ 30000) / (3000+5t)^{2}= 0.5 - 5 S(t) / (3000 + 5t)

Which is exactly what it should be.