A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the bow of the boat at a point 10 ft below the pulley. If the rope is pulled through the pulley at a rate of 20 ft/min, at what rate will the boat be approaching the dock when 125 ft of rope is out?

let x be the horizontal distance to the dock.

x^2=R^2-100 where R is rope length to the pulley.

2x dx/dt=2RdR/dT

solve for dx/dt when R is 125? (x= sqrt(125^2-100). dR/dt= 20ft/min

yum

To find the rate at which the boat is approaching the dock, we can use related rates. Let's call the distance between the boat and the dock "x" and the length of the rope "y".

Given that the rope is being pulled in at a rate of 20 ft/min, we can say that dy/dt = -20 ft/min (the negative sign indicates that the rope is getting shorter as it is being pulled in).

We also know that the length of the rope is decreasing at a constant rate, so dy/dt, the rate at which the length of the rope is changing, is constant.

Since the rope is attached to the bow of the boat at a point 10 ft below the pulley, the boat is being pulled towards the dock as the rope is being shortened.

We can set up a relationship between x and y using the Pythagorean theorem:

x^2 + (y + 10)^2 = y^2

Now, we can use implicit differentiation to find dx/dt, the rate at which the boat is approaching the dock:

2x(dx/dt) + 2(y + 10)(dy/dt) = 2y(dy/dt)

Simplifying the equation, we have:

2x(dx/dt) + 2(y + 10)(-20) = 2y(-20)

Now, we need to find the values of x and y when 125 ft of rope is out.

Using the relationship we found earlier, we can solve for x when y = 125:

x^2 + (125 + 10)^2 = 125^2

x^2 + 135^2 = 125^2

x^2 + 18225 = 15625

x^2 = 15625 - 18225

x^2 = -2600

Since the length cannot be negative, we can discard this solution.

Now, we can find dx/dt when y = 125:

2x(dx/dt) + 2(125 + 10)(-20) = 2(125)(-20)

2x(dx/dt) - 2(135)(20) = -2(125)(20)

2x(dx/dt) - 5400 = -5000

2x(dx/dt) = 5400 - 5000

2x(dx/dt) = 400

dx/dt = 400 / (2x)

Substituting x = √(-2600):

dx/dt = 400 / (2 * √(-2600))

Since √(-2600) is undefined, it means that at the point when 125 ft of rope is out, the boat is not approaching the dock.

Therefore, the rate at which the boat is approaching the dock when 125 ft of rope is out is 0 ft/min.

2.oo6