a tank initially holds 100 gallons of brine solution. at t=0, fresh water is poured into the tank at teh rate of 5 gpm, while the well-stirred mixture leaves the tank at the same rate. after 20 minutes, the tank contains 20/e lbs of salt. find the initial concentration of the brine solution inside the tank. Answer: 0.20 lb/gal
if it helps the formula is \[\huge \frac{dQ}{dt} = r_1c_1 + r_2[\frac{Q}{V_o + (r_2-r_1)t}]\] i think...
hmm...im thinking V = 100 gallons r1 = 5 gpm r2 = 5 gpm at t = 20, Q = 20/e lbs req'd c1 am i right?
t=0 dt=20 min
dt = 20 min? why?
do i evaluate the equation at t = 20?
because dt=change in time
\[\large \frac{dQ}{dt} = (5)c_1 + 5[\frac{\frac{20}{e}}{10 + (0)(20)}]\]
ohh you got dt from t_2 - t_1?
yeah
but t=0 the thing u jus solved has a problem that in there u put t=20 its 0
@hba if t = 0 then what will be Q?
It seems to be an easy question ..--> false wait !! let me think abt this
lol
Hmn i am trying to do that : will these examples work ? http://fredmath.net/DifferentialEq/lecture1/sec1.5/sec-1.5.pdf
if they did i wouldnt have asked now would i :p lol
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