Question

When mixing a certain wingspantail a barperson starts with a 10 litre solution
of 20% alcohol. Into this solution the barperson mixes a 40% alcohol
mixture at the rate of 1 litre/minute and pours into glasses at the rate of
2 litres/minute.
(a) Find a formula for the volume V (t) of the mixture at time t minutes.
(b) Find a differential equation that tells the rate of change of the amount of alcohol A(t) at time t minutes.
(c) Solve the differential equation and use the initial condition to find a formula for A(t)

Answer 1

is he adding solution and pouring glasses at same time ?

Answer 2

if so then dV/dt = -1
V(t) = 10-t

Answer 3

i got the same as you for (a), but have no idea about (b) and (c)

Answer 4

hey i think i got it, sorry it took so long, i left but forgot to log off and just woke up a little bit ago
anyway, the rate the alcohol is changing depends on rate its coming in minus rate it is leaving. it also depends on the concentration of the alcohol
.In --> rate is 1 and concentration is a constant 0.4
Out --> rate is 2 and concentration is Alcohol / Volume
\[\frac{dA}{dt} = 0.4 -\frac{2A}{10-t}\]
then to solve differential equation for A(t)
\[\frac{dA}{dt} +(\frac{2}{10-t})A = 0.4\]
find integrating factor
\[e^{2\int\limits_{}^{}(10-t)^{-1}} = e^{-2\ln(10-t)} = (10-t)^{-2}\]
\[\rightarrow (10-t)^{-2}\frac{dA}{dt} + 2(10-t)^{-3}A = 0.4(10-t)^{-2}\]
\[\rightarrow [(10-t)^{-2}A]' = 0.4(10-t)^{-2}\]
integrate both sides
\[(10-t)^{-2}A = \frac{0.4}{10-t}+c\]
\[A = 0.4(10-t)+c(10-t)^{2}\]
solve for constant using initial condition that A= 2 when t=0
2 = 4+100c --> c = -1/50
final solution for A(t)
\[A(t) = 0.4(10-t)-\frac{1}{50}(10-t)^{2}\]