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 1.4 m/s, how fast is the boat approaching the dock when it is 7 m from the dock?

give me answer

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 decreasing.

Let's assume the bow of the boat is at point B and the pulley on the dock is at point P. The distance between B and P is 7m.

We can determine the speed at which the distance between B and P is decreasing by finding the derivative of that distance with respect to time.

Let's denote the distance between B and P as h(t), where t is the time.

Given that the rope is being pulled in at a rate of 1.4 m/s, the rate at which h(t) is changing can be represented as dh/dt = -1.4 m/s (negative because the distance is decreasing).

Now, we can use the Pythagorean theorem to relate the distance h(t) to the position of the boat.

Let's denote the distance between B and the dock as x(t).

According to the Pythagorean theorem, we have:

x(t)^2 + h(t)^2 = 7^2

Take the derivative of both sides of the equation with respect to time t:

2x(t) * dx/dt + 2h(t) * dh/dt = 0

At the moment we're interested in, x(t) = 7m.

We can rearrange the equation to solve for dx/dt:

2x(t) * dx/dt = -2h(t) * dh/dt

dividing by 2x(t):

dx/dt = (-2h(t) * dh/dt) / (2x(t))

Now we need to find the values of h(t) and x(t) when X = 7m:

Using the Pythagorean theorem, when x = 7m, we can solve for h:

7^2 + h(t)^2 = 7^2

h(t)^2 = 7^2 - 7^2 = 0

h(t) = 0

Substitute these values back into the equation for dx/dt:

dx/dt = (-2 * 0 * -1.4) / (2 * 7)

dx/dt = 0 m/s

Therefore, when the boat is 7m from the dock, it is not approaching the dock. Its speed with respect to the dock is 0 m/s.