Questions: A rope is attached to a boat at water level and a woman on a dock is pulling on the rope at the rate of 50 ft / min. If the woman's hands are 16 ft above the water level, how fast is the boat approaching the dock when the amount of rope out is 20 ft?

Solution Steps

We need to determine how fast the boat is approaching the dock when the woman is pulling the rope at a rate of 50 ft/min and the rope length is 20 ft. The woman's hands are 16 ft above the water level.

We can use the Pythagorean theorem to relate the length of the rope, the height of the dock, and the horizontal distance of the boat from the dock. Let:

• \( y \) be the height of the dock (16 ft),
• \( x \) be the horizontal distance of the boat from the dock,
• \( z \) be the length of the rope (20 ft).

The relationship is given by: \[ x^2 + y^2 = z^2 \]

We need to find the rate at which the boat is approaching the dock, \( \frac{dx}{dt} \). Differentiate both sides of the equation with respect to time \( t \): \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \]

Since the height \( y \) is constant (16 ft), \( \frac{dy}{dt} = 0 \). This simplifies to: \[ 2x \frac{dx}{dt} = 2z \frac{dz}{dt} \]

We know:

• \( \frac{dz}{dt} = -50 \) ft/min (negative because the rope length is decreasing),
• \( z = 20 \) ft,
• \( y = 16 \) ft.

First, solve for \( x \) when \( z = 20 \) ft: \[ x^2 + 16^2 = 20^2 \] \[ x^2 + 256 = 400 \] \[ x^2 = 144 \] \[ x = 12 \text{ ft} \]

Now, substitute \( x \), \( z \), and \( \frac{dz}{dt} \) into the differentiated equation: \[ 2(12) \frac{dx}{dt} = 2(20)(-50) \] \[ 24 \frac{dx}{dt} = -2000 \] \[ \frac{dx}{dt} = \frac{-2000}{24} \] \[ \frac{dx}{dt} = -83.3333 \text{ ft/min} \]

Final Answer