Initially, we have a pitcher of 2L of unsweetened lemonade.
It has 0 grams of sugar.
Then we pour in sugar water---concentration: 100 grams per liter---at a rate of 0.05L per second.
As we pour, the pitcher leaks at a rate of 0.01L per second.
How many grams of sugar are in the pitcher after 60 seconds?
Let t be time in seconds since the pouring and leaking started.
Let x(t) be the volume of the liquid in the pitcher, in liters.
Since dx/dt = .05-.01 and x(0)=2, x=2+0.04t.
Let s(t) be the mass of the sugar dissolved in the liquid in the picture, in grams.
The rate at which sugar is poured in is the rate at which sugar water is added (0.05L/second)
multiplied by the sugar concentration of the sugar water (100g/L).
The rate at which sugar leaks out is the rate at which the pitcher is leaking (0.01L/second)
multiplied by the sugar concentration of the pitcher (s/x).
Therefore, ds/st=.05*100-.01*(s/x)=5-.01s/(2+0.04t) and s(0)=0.
We can rearrange this differential equation into the form s'+p's=q where s'=ds/dt, p'(t)=1/(200+4t), and q(t)=5.
Integrating p', we get p(t)=(1/4)ln(200+4t) (plus a constant which we may choose to be 0) for all t > -50.
Therefore the integrating factor is e^p(t)=(200+4t)^(1/4) and, setting u=s*e^p, we have u'=q*e^p.
Integrating q*e^p, we have u(t)=(200+4t)^(5/4)+c. (The constant is important.)
Since u=e^p*s, s=u/(e^p), so s(t)=200+4t+c(200+4t)^(-1/4).
Solving s(0)=0 for c, we have c=-200^(5/4).
So, after simplifying, s(t)= 200(1-(1+0.02t)^(-1/4))+4t.
After 60 seconds (t=60), there are approximately 276 grams of sugar in the pitcher.
If there was no leak, then we would simply have ds/dt=5 and s(0)=0, implying s(t)=5t.
Therefore, there would have been 300 grams of sugar after 60 seconds if the pitcher weren't leaking.
The first graph shows s(t) in the case of leak and no leak.
If we divide s(t) by x(t), we get the concentration of sugar.
After 60 seconds, the concentration is about 63 g/L, as you can see in the first graph.
After sixty seconds, the sugar concentration in the pitcher would be 60 g/L if there was no leak in the pitcher,
because in that case s(t)=5t and x(t)=2+0.05t.
The second graph shows s(t)/x(t) in the case of leak and no leak.
The reason the leaky pitcher has sweeter lemonade is that what is poured in has a high sugar concentration (100g/L),
while what leaks out has a lower sugar concentration.
Another way to explain this is that the leak increases the rate at which the contents of the pitcher becomes
more like the sugar water (and less like unsweetened lemonade) because what is poured in is pure sugar water,
while what is leaked out is a mixture of sugar water and unsweetened lemonade.