Question

A man standing on a wharf is hauling in a rope attached to a boat, at the rate of 4 ft/sec. if his hands are 9 ft. above the point of attachment, what is the rate at which the boat is approaching the wharf when it is 12 ft away?

Answer 1

$\sf{\underline{Answer:}}$ $\sf{ - 5 \: ft/s}$

$\sf{\underline{Given:}}$

$\boxed{\sf{\frac{dr}{dt} = - 4ft/s}}$

$\sf{\underline{Here:}}$

$\sf{h = 9}$

$\sf{d = 12}$

$\sf{\underline{We\:need\:to\:find:}}$ $\boxed{\sf{ \frac{d(d)}{dt}}}$

Pythagoras gives us the value of r at that particular instant:

$\implies$ $\sf{ {r}^{2} = {h}^{2} + {d}^{2}}$

$\implies$ $\sf{{r}^{2} = {9}^{2} + {12}^{2}}$

$\implies$ $\sf{{r}^{2} = 81 + 144}$

$\implies$ $\sf{ {r}^{2} = 225}$

$\implies$ $\sf{r = \sqrt{225}}$

$\implies$ $\sf{r = 15 \: ft}$

$\sf{\underline{Now:}}$ r = 15

$\sf{\underline{We\:know\:that:}}$

$\boxed{\sf{ {r}^{2} =81 + {d}^{2}}}$

Differentiating implicitely with respect to t:

$\boxed{\sf{2r \times \frac{dr}{dt} = 2d \times \frac{d(d)}{dt}}}$

Putting in the numbers,

$\sf{\underline{We\:get:}}$

$\implies$ $\boxed{\sf{2 \times 15 \times ( - 4) = 2 \times 12 \times \frac{d(d)}{dt} }}$

$\sf{\underline{That\:is:}}$

$\implies$ $\sf{30 \times ( - 4) = 24 \times \frac{d(d)}{dt} }$

$\implies$ $\sf{- 120 = 24 \times \frac{d(d)}{t}}$

$\implies$ $\sf{ - \frac{120}{24} = \frac{d(d)}{dt}}$

$\implies$ $\boxed{\sf{ \frac{d(d)}{t} = - 5 \: ft/s}}$