Related Rates Calculus Question:

Question

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 70 feet.

Suppose the worker stands directly next to the weight (i.e., a total rope length of 140 feet) and begins to walk away at a constant rate of 3 ft/s.

How fast is the weight rising when the worker has walked

10 feet?

30 feet?

Answer 1

It will be helpful to draw a diagram of the situation and label the distances as they are described below.

Let

L = 70 ft be half the the total length of the rope,

h(t) be the vertical distance of the weight from the pulley after time t,

x(t) be the horizontal distance of the worker from the initial position after time t,

v = 3 ft/s be the worker's speed.

We are given

h(0) = L,

x(0) = 0,

and we need to calculate h'(t), and then h'(10/v) and h'(30/v).

Let P be the point representing the pulley.

Consider the right angle triangle having the points P, x(0), x(t).

By the Pythagorean theorem the distance between P and x(t) is

√(L² + x(t)²) = √(L² + (vt)²)

Therefore

h(t) = 2L - √(L² + (vt)²)

After differentiation and simplification

h'(t) = -v²t/√(L² + (vt)²)

The worker has walked 30ft after time t = 10s.

Plugging in actual numbers

h'(10) = -3²×10/√(70² + (3×10)²) = -1.2 ft/s

(The speed is negative because it represents the rate of REDUCTION in the vertical distance.)

Similarly after the worker walked 10 ft.