Question

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 4 m from the dock

Answer 1

Answer:

-1.031 m/s or  $\frac{-\sqrt{17} }{4}$

Step-by-step explanation:

We take the length of the rope from the dock to the bow of the boat as y.

We take x be the horizontal  distance from the dock to the boat.

We know that the rate of change of the rope length is $\frac{dy}{dt}$ = -1 m/s

We need to find the rate of change of the horizontal  distance from the dock to the boat =  $\frac{dx}{dt}$ = ?

for x = 4

Applying Pythagorean Theorem we have

$1^{2} +x^{2} =y^{2}$    .... equ 1

solving, where x = 4, we have

$1^{2} +4^{2} =y^{2}$

$y^{2} = 17$

$y = \sqrt{17}$

Differentiating equ 1 implicitly with respect to t, we have

$2x\frac{dx}{dt} = 2y\frac{dy}{dt}$

substituting values of

x = 4

y = $\sqrt{17}$

$\frac{dy}{dt}$ = -1

into the equation, we get

$2(4)\frac{dx}{dt} = 2(\sqrt{17} )(-1)$

$\frac{dx}{dt} = \frac{-\sqrt{17} }{4}$ = -1.031 m/s