Problem 7 - Calculus Single Variable

Question

A rocket climbs straight up. Its initial total weight, including fuel, is 7000 pounds. Fuel is consumed at the constant rate of 30 pounds per mile. Taking into account the decrease in weight due to fuel consumption but disregarding the decrease in weight due to increasing elevation, approximate the work done in lifting the rocket the first 20 miles into space.

Step-by-Step Solution

Answer

The work done is 134,000 pound-miles.

Step 1: Understand the Problem

We need to calculate the work done to lift a rocket 20 miles into the air, considering the fuel consumption. The rocket starts with 7000 pounds, consuming fuel at 30 pounds per mile.

Step 2: Set Up the Work Integral

Work is calculated as force times distance. The force here is the weight of the rocket, which decreases as fuel is consumed. We can express weight as a function of distance, where the weight at any distance \( x \) miles is \( 7000 - 30x \) pounds.

Step 3: Define the Work Integral

The total work done lifting the rocket is the integral of the weight function over the distance from 0 to 20 miles. This integral is computed as: \[ W = \int_{0}^{20} (7000 - 30x) \, dx \]

Step 4: Calculate the Integral

Integrate the function: \[ W = \int_{0}^{20} 7000 \, dx - \int_{0}^{20} 30x \, dx \]. The first integral is: \[ 7000x \bigg|_0^{20} = 7000 \times 20 \] and the second integral is: \[ 15x^2 \bigg|_0^{20} = 15 \times 20^2 \].

Step 5: Evaluate the Integral

Substitute to find the work: \[ W = (7000 \times 20) - (15 \times 20^2) \].Calculate each term: \( 7000 \times 20 = 140000 \) and \( 15 \times 20^2 = 6000 \).Thus, the work done is \( 140000 - 6000 = 134000 \) pound-miles.

Key Concepts

Work and Energy in Physics

Work is a fundamental concept in physics, often associated with the force required to move an object over a distance. It is calculated using the formula:

• Work = Force × Distance

For our rocket scenario, the work done can be understood as the energy expended to lift the rocket against gravity. Initially, the force acting is the weight of the rocket, which changes because of fuel consumption.
The principle lies in converting fuel energy into kinetic energy, resulting in the upward motion of the rocket. The work done increases with both the force applied and the distance traveled. Notably, in this problem, we focus on the changing weight due to fuel consumption, impacting the force and thereby the total work done over 20 miles.
This approach gives insight into how engineers estimate energy requirements in actual rocket launches, where minimizing unnecessary fuel consumption is crucial.

Definite Integral

A definite integral is a mathematical tool used to find the total accumulation of a quantity over an interval. In this case, it helps us calculate the total work done on the rocket as it climbs 20 miles.
By employing the definite integral, we can sum up infinitely many small pieces of work performed as the rocket ascends, each corresponding to a very short distance with a slightly different force due to the changing weight.

• Here, we use the integral \[ W = \int_{0}^{20} (7000 - 30x) \, dx \]
• This integral gives the total work done, taking into account the decreasing weight as fuel burns off.

Integrating over the distance from 0 to 20 miles provides an exact measurement of work, capturing the nuances of changing forces, which would be difficult to calculate otherwise.
Nowadays, definite integrals are vital across diverse areas, including physics, engineering, and economics, to solve real-world problems involving continuous change.

Function of Distance

When dealing with changing forces, expressing a quantity as a function of distance becomes essential. In our rocket problem, we defined the weight as a function of distance to capture how the fuel being consumed at a constant rate affects the rocket's weight.
This function can be written as:

• Weight function: \( W(x) = 7000 - 30x \)

Here, \( W(x) \) represents the rocket's weight at any given distance \( x \), where the initial weight is 7000 pounds and decreases by 30 pounds per mile.
Understanding this relationship allows us to accurately model how physical changes, such as fuel consumption, impact the system's dynamics over time or distance.Using such functional expressions, individuals can predict outcomes and optimize scenarios efficiently in practical applications like engineering, aviation, or vehicle design.