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
Best Answer
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.