Chapter 5: Problem 18

For the curve \(y=\sqrt{x},\) between \(x=0\) and \(x=2,\) find: The arc length.

Short Answer

Expert verified

The arc length is 4.

Step by step solution

Understand the arc length formula

For a curve described by the function y = f(x) the arc length from x = a to x = b is given by the integral: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

Find the derivative of the function

For the curve y = \sqrt{x} the derivative is: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \]

Substitute the derivative into the arc length formula

Substitute \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \) into the arc length formula: \[ L = \int_0^2 \sqrt{1 + \left( \frac{1}{2\sqrt{x}} \right)^2} \, dx \] This simplifies to: \[ L = \int_0^2 \sqrt{1 + \frac{1}{4x}} \, dx \]

Simplify the integrand expression

Combine the terms inside the square root: \[ L = \int_0^2 \sqrt{\frac{4x + 1}{4x}} \, dx \] Which can be simplified to: \[ L = \int_0^2 \sqrt{\frac{4x + 1}{4x}} = \int_0^2 \frac{\sqrt{4x + 1}}{2\sqrt{x}} \, dx \]

Perform a u-substitution

Let \( u = 4x + 1 \). Then \( du = 4 \, dx \), and \( dx = \frac{du}{4} \). Also, when \( x = 0 \), \( u = 1 \) and when \( x = 2 \), \( u = 9 \). Substituting these values into the integral gives: \[ L = \int_1^9 \frac{\sqrt{u}}{2\sqrt{\frac{u-1}{4}}} \cdot \frac{du}{4} = \int_1^9 \frac{\sqrt{u}}{\frac{2\sqrt{u-1}}{2}} \cdot \frac{du}{4} = \int_1^9 \frac{\sqrt{u}}{\sqrt{u-1}} \cdot \frac{du}{4} = \int_1^9 \frac{du}{2} \]

Integrate and evaluate the definite integral

Integrate and evaluate: \[ L = \int_1^9 \frac{du}{2} = \frac{1}{2} \, (u)\bigg|_1^9 = \frac{1}{2}(9-1) = 4 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

arc length formula

To find the length of a curve, we use the arc length formula. The arc length \(L\) of a function \(y = f(x)\) from \(x = a\) to \(x = b\) is given by:
\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] This formula provides the total length of the curve by integrating the square root of the sum of 1 and the square of the derivative of the function.
It's a way to measure the distance along the curve rather than a straight line. First, you need to compute the derivative \(\frac{dy}{dx}\) and then square it to substitute back into the formula.

Arc length calculates the actual 'distance' along the curve.
Derivatives play a crucial role in this formula.
Understanding how to manipulate and simplify the integrand is essential.

u-substitution

U-substitution is a method for simplifying integrals by substituting a part of the integrand with a new variable, usually denoted as u. This technique is very handy in integrals that involve complicated expressions.
Let's take an example from the arc length problem:

We had an integral \int_0^2 \frac{\sqrt{4x + 1}}{2\sqrt{x}} \, dx\. To simplify, we set \( u = 4x + 1 \). Then, our differential \( dx \) changes as well: \( du = 4 \, dx \) or \( dx = \frac{du}{4} \).
When \( x = 0 \), \( u = 1\), and when \( x = 2, u = 9\). This transforms the integral into:

The variable limits change according to the new substitution.
The integral becomes simpler and often more manageable to solve.
Finally, involve back-substitution to finish the integration.

definite integral evaluation

Definite integrals are used to find the total value of the function over a specific interval. These integrals have upper and lower limits and are noted as follows: \int_{a}^{b} f(x) \, dx\. They give the net area between the function graph and the x-axis.
Using our example problem, after performing u-substitution, we got the integral: \[ L = \int_1^9 \frac{du}{2} \] This is much simpler to evaluate. The antiderivative of \(\frac{1}{2} u \) is straightforward:
\[ \frac{1}{2} \, (u)\bigg|_{1}^{9} \] Finally, substitute the limits of integration into the antiderivative:
\[ \frac{1}{2}(9) - \frac{1}{2}(1) = 4 \] The definite integral evaluates to 4, which gives us the arc length.

Evaluation turns the integral function value into tangible results.
Consistent simplification and back-substitution are key.