Problem 165 - Calculus Volume 1

Question

For the following exercises, find the length of the functions over the given interval. $$y=5 x \text { from } x=0 \text { to } x=2$$

Step-by-Step Solution

Answer

The length of the function is \(2\sqrt{26}\).

Step 1: Understand the Arc Length Formula for a Function

The arc length of a function defined by \( y = f(x) \) from \( x = a \) to \( x = b \) is calculated using the formula: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2}\, dx \] In this problem, \( y = 5x \), \( a = 0 \), and \( b = 2 \).

Step 2: Find the Derivative of the Function

We need to find \( \frac{dy}{dx} \) for \( y = 5x \). The derivative is: \( \frac{dy}{dx} = 5 \).

Step 3: Substitute into the Arc Length Formula

Substitute the derivative into the arc length formula: \[ L = \int_0^2 \sqrt{1 + (5)^2} \, dx \] Simplify inside the square root: \[ L = \int_0^2 \sqrt{26} \, dx \].

Step 4: Integrate to Find the Arc Length

Since \( \sqrt{26} \) is a constant, the integral simplifies to: \[ L = \sqrt{26} \int_0^2 \, dx \] Integrate: \[ L = \sqrt{26} [x]_0^2 = \sqrt{26} (2 - 0) = 2\sqrt{26} \].

Key Concepts

Understanding Calculus

Calculus is a branch of mathematics dedicated to the study of continuous change. You can think of it as a way to analyze anything that moves or changes, like motion in physics or growth patterns in biology. The central ideas of calculus focus on concepts such as limits, functions, derivatives, and integrals. Calculus allows us to solve complex problems in a wide array of fields, including engineering, economics, and even the natural sciences. It provides the tools needed to understand and predict how systems evolve over time.

When it comes to arc length, calculus enables us to calculate the length of curves, even when they are not straight lines. This involves using derivatives and integrals, which are fundamental operations in calculus. By breaking down a curve into infinitesimally small pieces, we can calculate the total length by summing these pieces, a process facilitated by integral calculus.

Integral Calculus and Arc Length

Integral calculus is a component of calculus that focuses on the accumulation of quantities. This can involve finding areas under curves, volumes, and other quantities that require addition over an interval. When we calculate arc length, we are essentially summing up small segments of the curve to find the total length.

The formula for arc length involves an integral that sums up these tiny distances over the specified interval. For the exercise with the function \( y = 5x \), defining the interval from \( x = 0 \) to \( x = 2 \), the arc length formula is set up as:

• The integral \( \int_0^2 \sqrt{1 + (\frac{dy}{dx})^2} \, dx \)
• You first find the derivative \( \frac{dy}{dx} \) which in this case is a constant 5.
• Substituting this into the formula gives: \( \int_0^2 \sqrt{1 + 25} \, dx = \int_0^2 \sqrt{26} \, dx \).

Given that \( \sqrt{26} \) is a constant, integrating simply means multiplying it by the width of the interval, resulting in \( 2\sqrt{26} \). Such operations are typical in integral calculus, where the emphasis is on summing continuous quantities.

Differentiation in Arc Length

Differentiation, a major part of calculus, is crucial when dealing with arc length. It involves finding how a function changes, which is expressed through derivatives. A derivative tells us the slope of a function at any point, or essentially, how steep the line is.

In the problem given, where \( y = 5x \), differentiation helps us find \( \frac{dy}{dx} \), the rate of change of \( y \) with respect to \( x \). For linear functions like this one, the derivative is constant, which simplifies calculations greatly. Here, the derivative is simply 5.

• Calculating the derivative is the first step in the arc length process. It provides the value necessary for determining the function's rate of change.
• Substituting this value into the arc length formula allows us to then move on to integrating over the specified interval.
• The whole process ties together differentiation and integration, highlighting the interconnected nature of calculus techniques.

Understanding how differentiation supplies the slope and integrating processes these slopes over an interval is key to mastering calculations involving arc lengths of functions.