The Calculus of Tea
So the other day I was making a cup of tea, and asked my sister whether she would like some as well. Since I didn’t mind leaving the teabag in my own cup while I drank it, I realized that the easiest way to make two cups of tea of approximately the same strength would be to first put the teabag in her cup, and then to move it to my cup for the remainder of the time. This is because the teabag lets flavor out the fastest at the beginning, and gets increasingly slow at it as time progresses. Of course the tea flavor always increases; there’s no point at which the teabag starts reclaiming flavor for itself.
With all these words like “increases” and “increasingly slow” this sounds like a good opportunity for some calculus.

The graph above represents f(t), the derivative of tea flavor with respect to time. Note that df/dt (which will now be referred to as Tea/Time) is always positive, but decreases in value approaching zero as time approaches infinity. The strength of the tea–denoted F(t)–is the definite integral of this graph over some time interval.
Let’s make some assumptions to make this problem easier. Since I occasionally forget about my tea as I let it cool down before drinking, let’s assume that I don’t start drinking my tea until time t=infinity. Thus, the strength of her tea is equal to the integral from zero to whatever time I remove the teabag (called t=a), and the strength of my tea is equal to the tail integral from a to infinity.
We can of course solve for a algebraically:

Of course, this simply verifies our original observation that I want to remove the teabag at the point at which her tea is half as strong as the teabag would be if I waited until time infinity before removing the teabag from a single cup. We are, however, happy to see our integrals come out to a statement we already knew to be true.
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I guess what I’m asking is, “Is it a bad thing that I started forming integrals in my head when my sister said she wanted tea?”