Question

A boat is pulled toward a dock by means of a rope wound on a drum that is located 6 ft above the bow of the boat. If the rope is being pulled in at the rate of 5 ft/s, how fast (in ft/s) is the boat approaching the dock when it is 34 ft from the dock?

Answer 1

Answer:

Step-by-step explanation:

Given that:

The height of the dock (h) = 6

Let represent d to be the distance between the boat and the dock

Let the length of the rope between the boat and the drum be denoted by (l)

Then, the rate of change for the length of the rope be:

dl/dt = -5 ft/s

Using Pythagoras rule to determine the relationship between these values, we have:

$l^2 = h^2 +d^2$

$l^2 = 6^2 + d^2$

$l^2 = 36 + d^2$

We relate to:  $2l * \dfrac{dl}{dt} = 2d* \dfrac{dd}{dt}$

From the question;

l = 34,

So to find $\dfrac{dd}{dt}$, we get;

$d = \sqrt{l^2 - 36}$

$d = \sqrt{34^2 - 36}$

$d = \sqrt{1156- 36}$

$d = \sqrt{1120}$

d = 33.46

So, we have:

$\dfrac{dd}{dt}= \dfrac{l}{d} \times \dfrac{dl}{dt}$

$\dfrac{dd}{dt}= \dfrac{34}{33.46} \times - 5$

$\dfrac{dd}{dt}= 1.016 \times - 5$

$\dfrac{dd}{dt}=-5.08 \ ft/sec$