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A boat is pulled in to a dock by a rope with one end attached to the front of the boat and the other end passing through a ring attached to the dock at a point $\sqrt{275}$ ft higher than the front of the boat. The rope is being pulled through the ring at the rate of $0.2$ ft/sec. How fast is the boat approaching the dock when 18 ft of rope are out?

I have run through this problem, I found the $x$ value was $\sqrt{18^2-275}$ or $7$ and that $z$ was $18$ and the change in $z$ is $.2$ so I plugged it all in and got $(\frac {18}{7})*.2$ but it says that is wrong. I don't know what I am missing I have done it a couple times can anyone help

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If $l$ is the length of the rope and $s$ is the distance from the boat to the dock, then you can draw a right triangle and see that \begin{equation} l^2 = 275 + s^2 \end{equation} Differentiating both sides by $t$, we have \begin{align} \frac{d}{dt} \left[ l^2 \right] &= \frac{d}{dt} \left[275 + s^2 \right] \\ 2l \frac{dl}{dt} &= 2s \frac{ds}{dt} \\ l \frac{dl}{dt} &= s \frac{ds}{dt} \end{align}

We are given that $\frac{dl}{dt} = -0.2 \text{ ft/s}$ (since $l$ is decreasing) and that $l = 18 \text{ ft}$. Using the first equation, we see that $s = 7 \text{ ft}$, and so \begin{align} l \frac{dl}{dt} &= s \frac{ds}{dt} \\ (18 \text{ ft})(-0.2 \text{ ft/s}) &= (7 \text{ ft}) \frac{ds}{dt} \\ \frac{ds}{dt} &= \frac{(18 \text{ ft})(-0.2 \text{ ft/s})}{(7 \text{ ft})} \\ &= -\frac{18}{35} \text{ ft/s} \\ &\approx -0.5142 \text{ ft/s} \end{align}

The boat is approach the dock at about $0.5142$ feet per second.

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    I tried the same thing, it didn't work, I tried both positive and negative and fraction form and decimal. – 2017-02-22
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Let $d$ be the distance to the dock and $r$ be the length of the rope. Then you have $$d^2 + 275 = r^2.$$ At the instance of interest you have $r = 18$ and $r' = -.2$, where ' means time derivative. Using our relationship we have $324 = d^2 + 275$ so $d = 7$. Now take the time derivative on the pythagorean relationship. $$2dd' = 2rr'.$$ Dropping in the stuff we know we have $$2\cdot 7 \cdot d' = 2\cdot 18 \cdot (-.2),$$ so $d' = -3.6/7 = -.514 \rm{ft}/\rm{sec}$. Note that the negative tells us the boat is drawing nearer to the dock. Your answer looks pretty good, but the nuance about the negative is important.

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    I tried the same thing, it didn't work, I tried both positive and negative and fraction form and decimal. I can't figure out what its wrong – 2017-02-22