A man in a wharf 6m above the water pulls in a rope to which a boat is tied at the rate of 1.2 m/s. at what rate is the boat approaching the wharf when there is 7.5 m of the rope out ?

Let r = length of rope out x = distance between the boat and the wharf

By Pythagorean theorem: $x = \sqrt{r^2 - 6^2}$

$\dfrac{dx}{dt} = \dfrac{d}{dt} \left( \sqrt{r^2 - 6^2} \right)_{r = 7.5} \cdot \dfrac{dr}{dt}$

$\dfrac{dx}{dt} = 15(1.2) = 18 \text{m/sec}$

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Let

r = length of rope out

x = distance between the boat and the wharf

By Pythagorean theorem:

$x = \sqrt{r^2 - 6^2}$

$\dfrac{dx}{dt} = \dfrac{d}{dt} \left( \sqrt{r^2 - 6^2} \right)_{r = 7.5} \cdot \dfrac{dr}{dt}$

$\dfrac{dx}{dt} = 15(1.2) = 18 \text{m/sec}$

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