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**A 500 gallon tank originally contains 100 gallons of fresh water. Beginning at time t=0, water containing 50 percent pollutants flows into the tank at the rate of 2 gal/min and the well stirred solution leaves at the rate of 1 gal/min. Find the concentration of pollutants in the tank at the moment it overflows.**

The answer is 48%

The answer is 48%

**my attempt, something is wrong as i did not get that answer or anywhere near it.**

rate in: 2 x 0.50

rate out: 1 x s(t)/(100+t)

s'(t) = 1- s(t)/(100+t)

s'(t) + s(t)/(100+t) = 1

a(t) = 1/(100+t) , b(t) = 1

using u(t) = exp(integ(a(t)dt)):

u(t) = exp(integ(a(t)dt))

u(t) = exp (ln (100+t))

u(t) = 100+t

using d/dt u(t)s(t) = u(t)b(t):

d/dt [(100+t)s(t)]= (100+t)(1)

(100+t)s(t)=integ(100+t)

(100+t)s(t) = 100t + 1/2t^2 + C

s(t) = [100t + 1/2t^2 +C ]/(100+t)

sub s(0) = 0 into s(t) [is this even correct?]

0 = [0+0+c]/[100+0]

c= 0

s(t) = [100t + 1/2t^2]/(100+t)

c(t) = s(t)/100

c(t) = [100t + 1/2t^2]/100(100+t)

then i found when the tank overflow:

500 = 100+t

t= 400

then found c(t)

c(400) = [100(400)+1/2(400)^2]/100(100+400)

c(400) = 2.4

which is wrong.

can anyone help me?

rate in: 2 x 0.50

rate out: 1 x s(t)/(100+t)

s'(t) = 1- s(t)/(100+t)

s'(t) + s(t)/(100+t) = 1

a(t) = 1/(100+t) , b(t) = 1

using u(t) = exp(integ(a(t)dt)):

u(t) = exp(integ(a(t)dt))

u(t) = exp (ln (100+t))

u(t) = 100+t

using d/dt u(t)s(t) = u(t)b(t):

d/dt [(100+t)s(t)]= (100+t)(1)

(100+t)s(t)=integ(100+t)

(100+t)s(t) = 100t + 1/2t^2 + C

s(t) = [100t + 1/2t^2 +C ]/(100+t)

sub s(0) = 0 into s(t) [is this even correct?]

0 = [0+0+c]/[100+0]

c= 0

s(t) = [100t + 1/2t^2]/(100+t)

c(t) = s(t)/100

c(t) = [100t + 1/2t^2]/100(100+t)

then i found when the tank overflow:

500 = 100+t

t= 400

then found c(t)

c(400) = [100(400)+1/2(400)^2]/100(100+400)

c(400) = 2.4

which is wrong.

can anyone help me?