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 7 m from the dock?

sqrt50/7

To find the rate at which the boat is approaching the dock, we need to determine the rate at which the distance between the boat and the dock is changing. Let's consider the situation at a particular instant when the boat is 7 m from the dock.

Let's define some variables:
- x: horizontal distance between the boat and the dock (measured perpendicular to the dock)
- y: vertical distance between the boat and the dock (measured along the dock, from the pulley to the bow of the boat)
- t: time in seconds

We are given that the rope is being pulled in at a rate of 1 m/s, which means that the rate of change of the length of the rope with respect to time is 1 m/s. Since the rope is attached to the bow of the boat and passes through a pulley on the dock, the length of the rope is equal to the hypotenuse of a right triangle with sides x and y.

By the Pythagorean theorem, we have: x^2 + y^2 = L^2, where L is the length of the rope.

Differentiating both sides of the equation with respect to time, we get:
2x(dx/dt) + 2y(dy/dt) = 2L(dL/dt)

Since we are interested in determining the rate at which the boat is approaching the dock, we need to find dy/dt when x = 7 m. Since the pulley is 1 m higher than the bow of the boat, we have y = 1 m.

At this point, we need to know the value of x and L in order to solve for dy/dt. Unfortunately, the problem does not provide this information.

Please provide the value of x and/or L so we can continue with the calculation.