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Chapter 2: Problem 165

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

Short Answer

Expert verified
The length of the function over the interval is \(2\sqrt{26}\).

Step by step solution

01

Understand the Formula

To find the length of the curve of a function from \(x = a\) to \(x = b\), we use the formula \( L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \). Here, our function is \(y = 5x\) and the interval is from \(x=0\) to \(x=2\).
02

Compute the Derivative

First, calculate the derivative of \(y = 5x\). The derivative of \(5x\) with respect to \(x\) is \(\frac{dy}{dx} = 5\).
03

Substitute Derivative in Formula

Substitute \(\frac{dy}{dx} = 5\) into the formula for arc length. We have:\[ L = \int_{0}^{2} \sqrt{1 + (5)^2} \, dx = \int_{0}^{2} \sqrt{1 + 25} \, dx = \int_{0}^{2} \sqrt{26} \, dx \]
04

Evaluate the Integral

Calculate the integral \(\int_{0}^{2} \sqrt{26} \, dx \). Since \(\sqrt{26}\) is a constant, this is \(\sqrt{26} \times (b - a)\), where \(a = 0\) and \(b = 2\).Thus:\[ L = \sqrt{26} \times (2 - 0) = 2\sqrt{26} \]
05

Substitute and Simplify

After evaluating the integral, the length of the function from \(x = 0\) to \(x = 2\) is \(2\sqrt{26}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of a Function
The derivative of a function is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. For a linear function like \( y = 5x \), finding the derivative is straightforward. In this case, the slope of the line鈥攁nd thus the derivative鈥攊s constant because it's simply the coefficient of \( x \), which is 5.

Understanding derivatives helps us determine how steep a curve is at any given point. For linear functions, this does not vary, but in more complex curves, the derivative can give insights into the entire shape of the curve. A positive derivative means the function is increasing, while a negative one indicates a decrease.
Integral Calculus
Integral calculus deals with the accumulation of quantities, such as areas under curves. It's a significant area of calculus that is often used to find things like voltages, distances, or in this case, the arc length of a curve.

The process of integration essentially "sums up" an infinite number of infinitesimally small quantities to find a total area, length, or volume. In this exercise, integral calculus helps to compute the total length of the straight line segment defined by the function \( y = 5x \) from \( x = 0 \) to \( x = 2 \).
Evaluating Definite Integrals
Evaluating definite integrals involves computing the integral of a function within given limits, \( a \) to \( b \). We use definite integrals to calculate the total value, such as width or length, within these bounds.

In the arc length formula, \( L = \int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx \), you evaluate from \( x = 0 \) to \( x = 2 \) after substituting the derivative. Simplification of the integral becomes easy when constants like \( \sqrt{26} \) appear, as it involves simple multiplication with the interval length. This way, we find that the arc length is \( 2\sqrt{26} \), illustrating the practical use of definite integrals in calculating lengths.

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Most popular questions from this chapter

For the following exercises, use \(y=y_{0} e^{\Lambda t}\). If a bank offers annual interest of \(7.5 \%\) or continuous interest of \(7.25 \%,\) which has a better annual vield?

For the following exercises, use \(y=y_{0} e^{\Lambda t}\). What continuous interest rate has the same yield as an annual rate of \(9 \% ?\)

For the following exercises, find the antiderivatives for the functions. $$ \int \frac{x d x}{\sqrt{x^{2}+1}} $$

Compute \(d y / d x\) by differentiating \(\ln y\). $$ y=\sqrt{x^{2}+1} $$

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