Problem 222 - Calculus Volume 1

Question

For the following exercises, find the work done. Compute the work done for a force \(F=12 / x^{2} \mathrm{N}\) from \(x=1\) to \(x=2 \mathrm{m} .\)

Step-by-Step Solution

Answer

The work done is 6 Joules.

Step 1: Understand the Problem

We need to find the work done by a variable force \( F(x) = \frac{12}{x^2} \) as it moves an object from \( x = 1 \) to \( x = 2 \) meters.

Step 2: Set Up the Work Integral

The work done by a variable force is given by the integral \( W = \int_{a}^{b} F(x) \, dx \). For our exercise, we need to calculate \( W = \int_{1}^{2} \frac{12}{x^2} \, dx \).

Step 3: Integrate the Function

To solve the integral \( \int \frac{12}{x^2} \, dx \), rewrite it as \( \int 12x^{-2} \, dx \). The antiderivative of \( x^{-n} \) is \( \frac{x^{-n+1}}{-n+1} \). Thus, we get \( W = 12 \left[-x^{-1}\right] = -\frac{12}{x} + C \).

Step 4: Evaluate the Definite Integral

Evaluate the definite integral \( \int_{1}^{2} \frac{12}{x^2} \, dx = \left(-\frac{12}{x} \right)\Bigg|_1^2 = \left(-\frac{12}{2} - (-\frac{12}{1})\right) = -6 + 12 = 6 \). Therefore, the work done is 6 Joules.

Key Concepts

Variable Force

A variable force is a force that changes in magnitude or direction as an object moves. In our given problem, we have a force defined by the equation \( F(x) = \frac{12}{x^2} \). This formula shows that the force depends on the position \( x \), which means the force isn't constant.

• Imagine pushing a carton up a ramp; the force might decrease as you go higher.
• A variable force requires using calculus to find work instead of just multiplying force by distance.
• This is because the force isn't uniform across the entire distance.

In many physics problems, you'll come across forces that change with position, and understanding how to handle them is crucial for calculating work done.

Definite Integral

The definite integral is a powerful tool in calculus that helps compute the accumulated quantity from a continuous function over an interval, like accumulating small pieces to form a complete picture.

• Think of it as the sum of an infinite number of infinitesimally small quantities.
• In this context, we're summing up the small work contributions from the force at each point between two positions.

To solve the problem of finding the work done, integrate \( F(x) \) from \( x = 1 \) to \( x = 2 \). The integration symbol \( \int \) signifies that we're accumulating the values of \( F(x) \) across this interval. This calculation will give us the total work done by the force as the object moves between these two points.

Antiderivative

An antiderivative, also known as an indefinite integral, is essentially the reverse of differentiation. If differentiating a function gives you a derivative, finding an antiderivative means you're looking for a function that has a given derivative.

• For example, if \( g(x) = rac{12}{x^2} = 12x^{-2} \), the antiderivative is \( G(x) = -\frac{12}{x} + C \).
• Here, \( C \) is called the constant of integration, representing an unknown constant since any constant differentiates to zero.
• Understanding how to find antiderivatives is essential for solving definite integrals and calculating work done by variable forces.

In our exercise, knowing that the antiderivative of \( x^{-n} \) is \( \frac{x^{-n+1}}{-n+1} \) is crucial to solve \( \int 12x^{-2} \, dx \).

Integration Techniques

Integration techniques include various methods for solving integrals, but a basic and essential one is the power rule for integration.

• We rewrite a given function to make use of the power rule easily.
• For instance, we rewrite \( \frac{12}{x^2} \) as \( 12x^{-2} \) to apply the power rule confidently.
• The power rule states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for any real number \( n eq -1 \).

By rewriting the integrand appropriately, it becomes easier to apply this technique. After obtaining the antiderivative, the final step in solving a definite integral involves substituting the upper and lower bounds into the antiderivative function and calculating the difference. This method works beautifully to solve for the total work done over a specified interval.