Calculus for the Practical Man: Chapter 4, Problem 7

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A man standing on a wharf is hauling in a rope attached to a boat at the rate of four feet a second. If his hands are nine feet above the point of attachment, how fast is the boast approaching the wharf when it is twelve feet away?

So, one may interpret "twelve feet away" as "twelve feet away from the wharf" or "twelve feet away from the man."

Choosing "twelve feet away from the man,"

\begin{align} x &= \sqrt{h^2 - 81} \\ dx &= \frac{h}{\sqrt{h^2 - 81}}dh\\ \frac{dx}{dt} &= \frac{h}{\sqrt{h^2 - 81}}\frac{dh}{dt}\\ \frac{dx}{dt} &= \frac{12}{\sqrt{12^2 - 81}}*4 = \frac{48}{\sqrt{63}}\ = 6.04743156815 \frac{ft}{sec}\\ \end{align}

Choosing "twelve feet away from the wharf,"

\begin{align} h &= \sqrt{x^2 + 81} \\ dh &= \frac{x}{\sqrt{x^2 + 81}}dx\\ \frac{dh}{dt} &= \frac{x}{\sqrt{x^2 + 81}}\frac{dx}{dt}\\ \frac{dx}{dt} &= \frac{dh}{dt}\frac{\sqrt{x^2 + 81}}{x}\\ \frac{dx}{dt} &= 4 * \frac{\sqrt{12^2 + 81}}{12} = 5 \frac{ft}{sec}\\ \end{align}

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If you interpret it as "twelve feet away from the wharf" you get the book's answer. We could argue whether it's ambiguous, but we should prefer a more parallel interpretation, in which case

how fast is the boat approaching the wharf when it is twelve feet away?

should probably be treated as

how fast is the boat approaching the wharf when [the boat] is twelve feet away [from the wharf]?