Questions: A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 45 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 90 ft) and begins to walk away at a constant rate of 2 ft/s. How fast is the weight rising when the worker has walked 10 feet? ft / s How fast is the weight rising when the worker has walked 30 feet? ft / s [Note: You should consider the triangle involving the pulley, location of the worker and the original position of the weight.]

Solution Steps

Let \( x \) be the horizontal distance the worker has walked from the starting point, and \( y \) be the vertical distance the weight has risen. The total length of the rope is constant at 90 feet. The distance from the connection point to the ceiling is 45 feet.

The relationship between \( x \) and \( y \) can be described using the Pythagorean theorem in the right triangle formed by the worker, the pulley, and the weight: \[ x^2 + y^2 = 45^2 \]

Differentiate both sides of the equation with respect to time \( t \): \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]

Given that the worker walks at a constant rate of 2 ft/s, \(\frac{dx}{dt} = 2\) ft/s. Substitute this into the differentiated equation: \[ 2x(2) + 2y \frac{dy}{dt} = 0 \] \[ 4x + 2y \frac{dy}{dt} = 0 \] \[ 2y \frac{dy}{dt} = -4x \] \[ \frac{dy}{dt} = -\frac{2x}{y} \]

When the worker has walked 10 feet, \( x = 10 \). Use the Pythagorean theorem to find \( y \): \[ 10^2 + y^2 = 45^2 \] \[ 100 + y^2 = 2025 \] \[ y^2 = 1925 \] \[ y = \sqrt{1925} \approx 43.87 \]

Now, substitute \( x = 10 \) and \( y = 43.87 \) into the equation for \(\frac{dy}{dt}\): \[ \frac{dy}{dt} = -\frac{2(10)}{43.87} \approx -0.456 \text{ ft/s} \]

When the worker has walked 30 feet, \( x = 30 \). Use the Pythagorean theorem to find \( y \): \[ 30^2 + y^2 = 45^2 \] \[ 900 + y^2 = 2025 \] \[ y^2 = 1125 \] \[ y = \sqrt{1125} \approx 33.54 \]

Now, substitute \( x = 30 \) and \( y = 33.54 \) into the equation for \(\frac{dy}{dt}\): \[ \frac{dy}{dt} = -\frac{2(30)}{33.54} \approx -1.79 \text{ ft/s} \]

Final Answer