- #1
braindead101
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A tank contains 300 gallons of water and 100 gallons of pollutants. Fresh water is pumped into the tank at the rate of 2 gal/min, and the well-stirred mixture leaves the tank at the same rate. How long does it take for the concentration of pollutants in the tank to decrease to 1/10 of its original value
This problem is really confusing me as it is water that is being pumped in instead of the usual pollutant/waste.
this is my attempt (i just reversed what i usually do, but this is wrong..)
w(t) : amount of water in tank at time t
w'(t) : rate of change of water in tank at time t
water enters at : 2
water leaves at : 2 x w(t)/100 (i think this is wrong... i actually worked this problem out with /100 and /300, and both are wrong as i don't get a positive answer.)
w'(t) = 2 - w(t)/50
w'(t) + 1/50 w(t) = 2
a(t) = 1/50, b(t) = 2
u(t) = exp (integ(1/50)dt)
u(t) = e^(1/50)t
d/dt (e^(1/50)t w(t) ) = e^(1/50)t x 2
e^(1/50)t w(t) = integ(2e^(1/50)t)
e^(1/50)t w(t) = 2(1/50) e^(1/50)t + C
w(t) = [1/25 e^(1/50)t + C] / e^(1/50)t
sub w(0) = 300
w(0) = [1/25 e^(1/50)0 + C] / e^(1/50)0
300 = [1/25 (1) + C] / 1
C = 7499/25
so w(t) = 1/25 + (7499/25)e^(-1/50)t
w(t) = 1/25 [ 1+ 7499e^(-1/50)t]
c(t) = w(t)/100 (again, is this correct?)
c(t) = 1/25000 [1+7499e^(-1/50)t]
1/10 of original value of pollutants is 10.
10 = 1/25000 [1+7499e^(-1/50)t]
250000 = 1+7499e^(-1/50)t
249999/7499 = e^(-1/50)t
ln (249999/7499) = -1/50t
t= -50 ln (249999/7499)
what did i do wrong. i think i did this whole problem wrong actually, any help would be great, thanks.
This problem is really confusing me as it is water that is being pumped in instead of the usual pollutant/waste.
this is my attempt (i just reversed what i usually do, but this is wrong..)
w(t) : amount of water in tank at time t
w'(t) : rate of change of water in tank at time t
water enters at : 2
water leaves at : 2 x w(t)/100 (i think this is wrong... i actually worked this problem out with /100 and /300, and both are wrong as i don't get a positive answer.)
w'(t) = 2 - w(t)/50
w'(t) + 1/50 w(t) = 2
a(t) = 1/50, b(t) = 2
u(t) = exp (integ(1/50)dt)
u(t) = e^(1/50)t
d/dt (e^(1/50)t w(t) ) = e^(1/50)t x 2
e^(1/50)t w(t) = integ(2e^(1/50)t)
e^(1/50)t w(t) = 2(1/50) e^(1/50)t + C
w(t) = [1/25 e^(1/50)t + C] / e^(1/50)t
sub w(0) = 300
w(0) = [1/25 e^(1/50)0 + C] / e^(1/50)0
300 = [1/25 (1) + C] / 1
C = 7499/25
so w(t) = 1/25 + (7499/25)e^(-1/50)t
w(t) = 1/25 [ 1+ 7499e^(-1/50)t]
c(t) = w(t)/100 (again, is this correct?)
c(t) = 1/25000 [1+7499e^(-1/50)t]
1/10 of original value of pollutants is 10.
10 = 1/25000 [1+7499e^(-1/50)t]
250000 = 1+7499e^(-1/50)t
249999/7499 = e^(-1/50)t
ln (249999/7499) = -1/50t
t= -50 ln (249999/7499)
what did i do wrong. i think i did this whole problem wrong actually, any help would be great, thanks.