Problem 13 - Calculus: Early Transcendentals

Question

Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it. A heavy rope, 50 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 120 ft high. (a) How much work is done in pulling the rope to the top of the building? (b) How much work is done in pulling half the rope to the top of the building?

Step-by-Step Solution

Answer

(a) 625 ft-lb, (b) 156.25 ft-lb.

Step 1: Define the Work Formula

The work required to lift an object is defined by the formula \( W = \int F(x) \, dx \), where \( F(x) \) is the force at a distance \( x \). Since the weight of the rope is given per foot, the force to lift a length \( dx \) of the rope is \( f(x) = 0.5 \) lb/ft.

Step 2: Determine the Limits and Setup Riemann Sum

For part (a), consider the total rope (50 ft). The work done to lift the rope is \( W = \sum_{i=1}^{n} 0.5 x_i \Delta x_i \). For part (b), only half the rope (25 ft) is considered. \( \Delta x_i \) represents the small length of the rope each subinterval in the sum.

Step 3: Express Work as an Integral for Part (a)

To find the work done in pulling the full rope, express the work as an integral: \[ W = \int_0^{50} 0.5x \, dx \]

Step 4: Evaluate the Integral for Part (a)

The integral becomes \[ W = \int_0^{50} 0.5x \, dx = 0.5 \left[ \frac{x^2}{2} \right]_0^{50} \]Calculating this gives \[ 0.5 \cdot \left( \frac{50^2}{2} - \frac{0^2}{2} \right) = 0.5 \cdot \frac{2500}{2} = 0.5 \cdot 1250 = 625 \text{ ft-lb} \]

Step 5: Express Work as an Integral for Part (b)

To find the work done for half the rope, set up:\[ W = \int_0^{25} 0.5x \, dx \]

Step 6: Evaluate the Integral for Part (b)

The integral becomes \[ W = \int_0^{25} 0.5x \, dx = 0.5 \left[ \frac{x^2}{2} \right]_0^{25} \]Calculating this gives \[ 0.5 \cdot \left( \frac{25^2}{2} - \frac{0^2}{2} \right) = 0.5 \cdot \frac{625}{2} = 0.5 \cdot 312.5 = 156.25 \text{ ft-lb} \]

Key Concepts

Work Done

Work done is a concept that quantifies the amount of energy expended in moving an object over a distance. In this exercise, work is calculated as the energy required to lift a heavy rope. The formula to calculate work done is \( W = \int F(x) \, dx \), where \( F(x) \) represents the force applied to move an object over a small distance \( dx \).
In our case:

• The force is determined by the weight of the rope, given as 0.5 lb per foot.
• The distance is the length over which the rope is lifted.

Understanding work done helps us determine how much energy is needed for specific tasks, such as lifting objects vertically against gravity.

Integral Calculation

In this problem, integral calculation is utilized to find the total work done by summing infinitely small elements of work. The integral \( \int F(x) \, dx \) helps us find the total work by integrating force over the entire length of the rope.
To set up the integral:

• For the whole rope (50 ft), use the integral setup \( \int_0^{50} 0.5x \, dx \).
• For half the rope (25 ft), setup is \( \int_0^{25} 0.5x \, dx \).

These setups help us break down the continual process of lifting the rope into manageable parts that can be easily calculated.

Force and Distance

Force and distance are crucial elements when calculating work, represented in the formula \( W = \int F(x) \, dx \). The force here is the weight of the rope, while the distance is determined by the rope's length.
Here:

• The force for each small section \( dx \) is constant at 0.5 lb/ft, signifying that each foot of rope weighs this amount.
• The total distance for part (a) is 50 ft, whereas for part (b), it's 25 ft.

These two elements combine to help calculate the incremental work done in lifting consecutive small sections of the rope to the top of the building.

Evaluating Integrals

Evaluating integrals lets us turn the expressions \( \int_0^{50} 0.5x \, dx \) and \( \int_0^{25} 0.5x \, dx \) into concrete numerical work values.
The process is as follows:

• Begin by finding the antiderivative: \( 0.5 \times \frac{x^2}{2} \).
• Plug in the limits of integration, \( x = 50 \) and \( x = 25 \) for their respective integrals.
• Calculate the difference of these antiderivatives at the upper and lower bounds.

For the full rope, the work is 625 ft-lb. For half the rope, it's 156.25 ft-lb. Integrals provide a convenient method to add up all small increments of work, resulting in an overall total value needed for practical purposes.