Problem 3 - Calculus Single Variable

Question

An object that weighs \(W_{0}\) at the surface of the earth weighs $$ 15697444 \cdot \frac{W_{0}}{(3962+y)^{2}} $$ when it is \(y\) miles above the surface of the earth. How much work is done lifting an object from the surface of the earth, where it weighs 100 pounds, straight up to a height 100 miles above Earth?

Step-by-Step Solution

Answer

The work done is approximately 9549.34 foot-pounds.

Step 1: Understand the Problem and Equation

The problem involves calculating the work done in lifting an object through a height. The weight of an object at a height \(y\) miles above the Earth's surface is given by the expression \(15697444 \cdot \frac{W_{0}}{(3962+y)^{2}}\). We need to integrate this weight function from \(y = 0\) to \(y = 100\). Given \(W_0 = 100\) pounds at the surface \((y=0)\).

Step 2: Set up the Integral for Work Done

Work done is the integral of the force over the distance. Since the force is the weight of the object and the distance is from the surface \((y=0)\) to 100 miles \((y=100)\), the work done \(W\) is:\[ W = \int_{0}^{100} 15697444 \cdot \frac{100}{(3962+y)^2} \, dy \]

Step 3: Evaluate the Integral

To solve \( \int_{0}^{100} 15697444 \cdot \frac{100}{(3962+y)^2} \, dy \), recognize that this is a simple rational function integration problem:\[ W = 1569744400 \cdot \left[ -\frac{1}{3962+y} \right]_{0}^{100} \]This simplifies to:\[ W = 1569744400 \cdot \left( -\frac{1}{4062} + \frac{1}{3962} \right) \]

Step 4: Calculate the Result

Calculate the differences from the definite integral:\[ W = 1569744400 \cdot \left( \frac{1}{3962} - \frac{1}{4062} \right) \]Compute the difference inside the parentheses and multiply by \(1569744400\):First, compute \( \frac{1}{3962} \approx 0.00025233\) and \( \frac{1}{4062} \approx 0.00024625\). Then compute \(0.00025233 - 0.00024625 = 0.00000608\).Finally, \( 1569744400 \times 0.00000608 \approx 9549.34 \).

Step 5: Conclusion

The work done in lifting the object from the Earth's surface to 100 miles above the surface is approximately \(9549.34\) foot-pounds.

Key Concepts

Integration

Integration is a fundamental concept in calculus, often described as the reverse of differentiation. In the context of the exercise, integration is used to calculate the total work done by summing an infinite number of small contributions over a continuous interval.

When dealing with physical problems, like the one at hand, we explore the continuous change associated with a varying force. Here, the force is the object's weight at a height \( y \) above the earth's surface. We find the integral of this force from the starting point, \( y = 0 \), to the final point, \( y = 100 \), which gives us the total work.

By evaluating the integral, we essentially add up the infinitesimally small pieces of work needed to lift the object a little higher. \(\int_{0}^{100} 15697444 \cdot \frac{100}{(3962+y)^2} \, dy \) shows an integral with a force function dependent on the distance \( y \). The integration tells us how these small effects together create the total work done over the given height.

Work and Energy

Work and energy are two closely related concepts in physics. Work occurs when a force causes an object to move. In practical terms, work is done whenever you push or pull an object over a distance.

In the given problem, the work is done against the force of gravity to lift an object from the surface of the Earth up to a height of 100 miles. The energy transferred through this process is the work's magnitude, which in physics is expressed in units such as foot-pounds in this context.

Key points about work:

• Force must cause movement: The movement in this problem is lifting the object through a certain height.
• Same direction: Work is done only when the movement is in the direction of the force applied.
• Formula for work: \( W = \int_{a}^{b} F(y) \, dy \) where \( F(y) \) is the force as a function of height \( y \).

Understanding these factors allows us to see how the extracted energy resolves into a calculable work total.

Definite Integrals

Definite integrals are used to calculate the exact value of an integral over a specified interval, \( [a, b] \). In this exercise, we used a definite integral to compute the work done in lifting an object from \( y = 0 \) to \( y = 100 \).

The notation \(\int_{0}^{100} 15697444 \cdot \frac{100}{(3962+y)^2} \, dy \) signifies that we integrate the weight function over the interval where \( y \) changes from 0 to 100 miles.

Definite integrals provide the cumulative sum of infinitesimal increments of a quantity over an interval. Here's how to approach definite integrals:

• Identify the function: Recognize the function you're integrating, here it's the weight formula variable over \( y \).
• Determine the limits of integration: These are the starting and ending points, \( y = 0 \) and \( y = 100 \) respectively.
• Evaluate the integral: Solve analytically or numerically to find the total contribution over the interval.

By evaluating, we find how much work is physically exerted over the 100-mile journey. Definite integrals simplify complex quantities into tangible results through integration over specified limits.