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 (in m/s) is the boat approaching the dock when it is 5 m from the dock

dy/dt = 1.02 m/s

Step-by-step explanation:

First of all, let the distance of the dock above the bow of the boat be represented by x.

Also, let the distance of the boat from the dock be represented by y.

Let z represent the length of the rope.

Now, we can use pythagoras theorem to find a relationship between x, y and z since they form a right angle triangle where the vertical part is the dock, the base represents the water and the hypotenuse represents the rope.

Thus;

x² + y² = z²

Using implicit differentiation, we have;

2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)

We are given that the dock is 1 meter above the bow. Thus, x = 1

Also, the height is not changing and so dx/dt = 0

We are told the dock is 5 feet away from the boat. Thus, y = 5

dy/dt is is the speed we are looking for which is the rate at which the boat is approaching the dock.

We are not given the value of z but we can find it by using pythagoras theorem since we know the dock is 1 meter above the bow and the boat is 5 m away.

Thus;

z = √(1² + 5²)

z = √26

We are told the rope is being pulled in at the rate of 1 m/s. Thus; dz/dt = 1

Plugging all relevant values into the differentiation equation above gives;

2(1)(0) + 2(5)(dy/dt) = 2(√26)(1)

10(dy/dt) = 2√26

Divide both sides by 10 to get:

dy/dt = (√26)/5

dy/dt = 1.02 m/s

your answer is 40

step-by-step explanation:

step-by-step explanation:

i need to answer questions so i dont know but i think its the first one because it goes up but a consent rate