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? (Round your answer to two decimal places.)

This is a problem of related rates in calculus. We can solve it using the Pythagorean theorem and differentiation.

Step 1: Set up the problem Let's denote the distance between the boat and the dock as x (in meters), and the length of the rope as y (in meters). According to the Pythagorean theorem, we have:

y^2 = x^2 + 1

Step 2: Differentiate both sides with respect to time (t) Differentiating both sides with respect to time gives us:

2y(dy/dt) = 2x(dx/dt)

Step 3: Solve for dx/dt We are asked to find dx/dt, which is the speed at which the boat is approaching the dock. Solving the above equation for dx/dt gives us:

dx/dt = y(dy/dt) / x

Step 4: Substitute the given values We are given that dy/dt = -1 m/s (the rope is being pulled in, hence the negative sign), y = sqrt(x^2 + 1) (from the Pythagorean theorem), and x = 5 m. Substituting these values into the equation gives us:

dx/dt = sqrt(5^2 + 1)(-1) / 5 = -sqrt(26) / 5 m/s

Step 5: Round to two decimal places Rounding to two decimal places gives us:

dx/dt = -1.02 m/s

So, the boat is approaching the dock at a speed of 1.02 m/s.

This is a problem of related rates in calculus. We can solve it using the Pythagorean theorem and differentiation.

Step 1: Set up the problem Let's denote the distance between the boat and the dock as x (in meters), and the length of the rope as y (in meters). According to the Pythagorean theorem, we have:

y^2 = x^2 + 1

Step 2: Differentiate both sides with respect to time (t) Differentiating both sides with respect to time gives us:

2y(dy/dt) = 2x(dx/dt)

Step 3: Solve for dx/dt We want to find dx/dt, which is the speed at which the boat is approaching the dock. Solving the above equation for dx/dt gives us:

dx/dt = y(dy/dt) / x

Step 4: Substitute the given values We know that dy/dt = -1 m/s (the rope is being pulled in, hence the negative sign), y = sqrt(x^2 + 1) = sqrt(5^2 + 1) = sqrt(26) m, and x = 5 m. Substituting these values into the equation gives us:

dx/dt = sqrt(26)(-1) / 5 = -sqrt(26)/5 m/s

Step 5: Round to two decimal places Rounding to two decimal places, we get:

dx/dt = -0.51 m/s

So, the boat is approaching the dock at a speed of 0.51 m/s. The negative sign indicates that the distance between the boat and the dock is decreasing.