Reading Speed for Constant Time to Finish

Question

You open a very long new book on your e-reader and read a few pages. It helpfully informs you that based on your reading speed you have 16 hours of reading left until you are done. You read the rest of the first chapter a little more slowly, and it informs you there are 16 hours of reading left until you are done. In the second chapter you slow further, and after finishing the chapter you still have 16 hours to go. Exasperating!

What are the equations which govern your reading speed, if based on your speed so far in this book you always have 16 hours left to go? And what is the solution for your reading speed at time $t\ge0$, and the fraction of the book you have finished at time $t$?

Here's my approach: let the book be 1 unit long, for $t\ge0$ let your reading speed be $$f(t).$$ Then at time $t$ you have read $$F(t):=\int_0^t f(s)\,ds$$ of the book and your average rate is $$r(t):=F(t)/t.$$ We have the constraint, for $t>0$, that $$\frac{1-F(t)}{r(t)}=16.$$ How do we proceed in solving this problem?

Answer 1

Hint: Write $r(t)$ as $\frac{F(t)}{t}$ and then solve the equation $\frac{1-F(t)}{r(t)}=16$ for $F(t)$. Then use $F(t)$ to find $f(t)$ using the fundamental theorem of calculus.

Answer 2

It's easiest to solve if you ignore $f$ at first and solve for $F$. Substituting $$r(t):=F(t)/t$$ into $$\frac{1-F(t)}{r(t)}=16$$ and manipulating line by line gives us \begin{align} \frac{t\big(1-F(t)\big)}{F(t)} & = 16 \\ t\big(1-F(t)\big) & = 16F(t) \\ t & = (16+t)F(t) \\ F(t) & = \frac{t}{16+t}. \end{align}

Since we are given $$F(t):=\int_0^t f(s)\,ds$$ we know $$f(t)=\frac{d}{dt}F(t)=\frac{16}{(16+t)^2}.$$