Problem 166 - Calculus Volume 1

Question

For the following exercises, find the length of the functions over the given interval. $$y=-\frac{1}{2} x+25 \text { from } x=1 \text { to } x=4$$

Step-by-Step Solution

Answer

The length of the function from \( x=1 \) to \( x=4 \) is \( \frac{3\sqrt{5}}{2} \).

Step 1: Understand the Formula for Arc Length

The arc length of a function from \( x = a \) to \( x = b \) can be calculated using the formula \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]. Here, \( \frac{dy}{dx} \) is the derivative of the function \( y \) with respect to \( x \).

Step 2: Find the Derivative of the Function

Given the function \( y = -\frac{1}{2} x + 25 \), find its derivative. Since the equation is linear, the derivative is simply the coefficient of \( x \), which gives \( \frac{dy}{dx} = -\frac{1}{2} \).

Step 3: Plug the Derivative into the Formula

Substitute \( \frac{dy}{dx} = -\frac{1}{2} \) into the arc length formula. Calculate \( 1 + \left( -\frac{1}{2} \right)^2 \) which simplifies to \( 1 + \frac{1}{4} = \frac{5}{4} \). Thus, the integral becomes \[ L = \int_{1}^{4} \sqrt{\frac{5}{4}} \, dx \].

Step 4: Simplify the Integral

Since \( \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \) is a constant, the integral simplifies to \( L = \frac{\sqrt{5}}{2} \int_{1}^{4} 1 \, dx \), which equals \( \frac{\sqrt{5}}{2} [x]_{1}^{4} \).

Step 5: Calculate the Definite Integral

Evaluate the definite integral \( [x]_{1}^{4} = 4 - 1 = 3 \). Therefore, the arc length \( L = \frac{\sqrt{5}}{2} \times 3 = \frac{3\sqrt{5}}{2} \).

Key Concepts

Derivative

In mathematics, a derivative represents the rate at which a function is changing at any given point. It is a fundamental concept in calculus and is used to analyze a wide range of functions. For a linear function like the one in the exercise, finding the derivative is straightforward. The function given is:
\( y = -\frac{1}{2} x + 25 \)
To find the derivative, you only need to identify the coefficient of the variable \( x \), which in this case is \(-\frac{1}{2}\). Thus, the derivative \( \frac{dy}{dx} = -\frac{1}{2} \).
The derivative is crucial for determining the arc length of this function, as it provides us with information about how the function behaves between two points. Understanding how to compute and interpret the derivative dramatically enhances your ability to analyze and work with functions of all types.

Integral Calculus

Integral calculus is the branch of calculus that focuses on the accumulation of quantities and the areas under curves. It complements differential calculus, which is about rates of change. Integral calculus allows us to determine the total accumulation of a function across an interval, which in the context of arc length, helps us to calculate the precise length of a curve, rather than just the distance between endpoints.
The formula for arc length we use in this exercise involves integration:

• \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \)

By integrating, we can add up infinitesimally small segments of the curve to find the total length. It's a magnificent tool for solving complex problems involving lengths, areas, and volumes in various fields such as physics, engineering, and economics.

Definite Integral

A definite integral is a specific kind of integration that provides a real number as an answer, representing the net area under a curve between two points. In this exercise, the definite integral calculates the exact arc length of a linear function from one point to another on a plane.
The computation involved in the exercise is:

• First, simplifying the integral’s function after substituting the derivative into the formula.
• This led to \( L = \int_{1}^{4} \sqrt{\frac{5}{4}} \, dx \), which simplifies further.
• As \( \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \) is constant, the integral becomes \( \frac{\sqrt{5}}{2} \) times the definite integral of a constant over the interval \([1, 4]\).
• The solved definite integral yields \([x]_{1}^{4} = 3\).

Thus, the arc length \( L = \frac{3\sqrt{5}}{2} \). The definite integral helps us to precisely determine this length by adding up all the small pieces along the function between the specified limits.